## April 12, 2007

### Schur Functors

#### Posted by John Baez

As part of the Tale of Groupoidification, I’ll need to talk about Schur functors. As usually defined, these are simply functors

$F: Vect_{\mathbb{C}} \to Vect_{\mathbb{C}}$

where $Vect_{\mathbb{C}}$ is the category of finite-dimensional complex vector spaces.

An example of a Schur functor is ‘take the antisymmetrized 3rd tensor power’. In the category of Schur functors, $hom(Vect,Vect)$, every object can be expressed as a direct sum of certain ‘irreducible’ objects, which correspond to Young diagrams. The example I just mentioned corresponds to this Young diagram:

Example Young Tableaux in SVG

Given any group representation

$R: G \to Vect_{\mathbb{C}}$

we can compose it with any Schur functor

$F: Vect_{\mathbb{C}} \to Vect_{\mathbb{C}}$

and get a new representation

$F R : G \to Vect_{\mathbb{C}}$

This is a great method of getting new reps from old.

There’s much more to say… but first, Allen Knutson has a question!

For as long as I’ve understood Schur functors, I’ve thought about them as functors $Vect_{\mathbb{C}} \to Vect_{\mathbb{C}}$. But now that we’re going through them in a reading course on Fulton’s Young Tableaux, I discover that the input isn’t really a complex vector space, but an arbitrary module over a commutative ring. (And maybe, just maybe, a bimodule over a noncommutative one, but I doubt it.) In particular, the Schur functor commutes with base change AKA extension of scalars.

What is the right way to describe this object, categorically? (Or should I say, 2- or 3-categorically?)

Here’s a first stab at answering this.

Presumably ‘commuting with base change’ means that our Schur functor is not merely a functor

$F : Mod_R \to Mod_R$

from a single category of $R$-modules to itself, but a family of functors, one for each commutative ring $R$, depending naturally on $R$. So, we have some 2-category $Mod$ of ‘module categories’, and our Schur functor gives some (pseudo)natural transformation

$F : 1_{Mod} \Rightarrow 1_{Mod}$

Thus, for each object $Mod_R$ in $Mod$, we get

$F_R : Mod_R \to Mod_R$

and this is natural in some suitably weak sense.

Working out the details here would be a lot of fun:

1. I’m a bit worried about what happens in nonzero characteristic. Do things like symmetrizing and antisymmetrizing really work the same when you can’t divide by some prime $p$? Or do you get different sorts of ‘Schur functors’ in nonzero characteristic?
2. What really matters is not so much ‘module categories’ but symmetric monoidal abelian categories, perhaps enriched over $\Vect_{\mathbb{Q}}$ or something — whatever is the minimal context where you can do stuff like ‘symmetrize’ or ‘antisymmetrize’ tensor powers of objects.
Posted at April 12, 2007 9:10 PM UTC

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### Re: Schur Functors

I’m a bit worried about what happens in nonzero characteristic. Do things like symmetrizing and antisymmetrizing really work the same when you can’t divide by some prime p? Or do you get different sorts of ‘Schur functors’ in nonzero characteristic?

You never do divide by p. In the most basic examples, Sym2 resp. Alt2, you tensor the tensor square and mod out by {a⊗ b-b⊗ a} resp. by {a⊗a}, for all a,b in the module.

The more general definition, on pp. 104-107, is also as a quotient, of a big tensor product by a submodule generated by a bunch of things with leading coefficient 1. Perhaps they’re a module Grobner basis and so those coefficients 1 are the only ones you need worry about. (I don’t know this to be true.)

The characteristic p issues I understand to arise here are that you can’t argue that the resulting modules have unique “highest” weights any more, with which to prove irreducibility, now that weights wrap around.

Posted by: Allen Knutson on April 12, 2007 10:13 PM | Permalink | Reply to this

### Re: Schur Functors

Allen wrote:

You never do divide by p.

Right, if I’d been more awake I would have remembered that you don’t need to act like a physicist and symmetrize $V^{\otimes 2}$ to define the symmetric tensor square $Sym^2 V$ of a vector space $V$ — you can just mod out by the skew-symmetric tensors.

(Bott used to joke that physicists don’t know about quotient spaces, only subspaces — so they had to invent Hilbert spaces as a way of ignoring the difference.)

Okay, so the division involved in symmetrization is not a problem for defining Schur functors in characteristic $p$. But it seems like some sort of peculiarity of characteristic $p$. Could it be a problem with defining certain ‘Schur natural transformations’? It seems symmetrization comes from some sort of natural splitting of

$0 \to \langle a \otimes b - b \otimes a\rangle \to V^{\otimes 2} \to Sym^2 V \to 0$

Am I saying this natural splitting doesn’t exist in characteristic $2$? That could affect the category of Schur functors

$hom(Vect_k, Vect_k)$

when $char k = 2$, since the morphisms here are natural transformations.

I feel confused about this. I seem to be saying that not every short exact sequence in $hom(Vect_k, Vect_k)$ splits for certain fields $k$. This seems somehow shocking to me.

Chris Brav points out that $Sym$ and $Alt$ interact differently with duals in nonzero characteristic. It would be fun getting a good category-theoretic understanding of this — and maybe a bit tricky, since Schur functors are covariant while dualization is contravariant. Maybe we should study a ‘$\mathbb{Z}_2$-graded’ monoidal category of Schur functors which includes both covariant and contravariant functors from $Vect$ to $Vect$!

Questions, questions… and I should be grading homework.

Posted by: John Baez on April 13, 2007 6:57 PM | Permalink | Reply to this

### Re: Schur Functors

I feel confused about this. I seem to be saying that not every short exact sequence in hom(Vectk,Vectk) splits for certain fields k. This seems somehow shocking to me.

I don’t know much about this area but I think this is indeed the case – this is one reason why in some sense Hochschild cohomology is the wrong thing to use for algebras over non-zero char fields, and you need to go to a different cohomology theory. (Algebraic geometers please correct me if this is rubbish!)

Tangentially I’d like to mention that there seem to be similar subtleties involved in defining Schur functors on normed spaces (that is, if we want to work with isometric isomorphism rather than just topological isomorphism). It seems that one should work with quotients rather than subspaces if I recall correctly if one wants the right universal properties to hold.

Does anyone know if someone’s worked through the relevant details for, say, Schur functors on the contractive category of Banach spaces?

Posted by: Yemon Choi on April 13, 2007 11:00 PM | Permalink | Reply to this

### Re: Schur Functors

To p or not to p?

The correct (independent of characteristic) definition of the symmetric or graded symmetric or skew symmetric algebra is a quotient of the tensor algebra. Only in char 0 can this be identified with a subalgebra.

On the other hand, the symmetric or graded symmetric or skew symmetric COalgebra is a sub COalgebra of the tensor COalgebra, i.e. the invariants under the action of the symmetric groups.

I’m not aware of hochschild having any problem in char p BUT for the cohomology of commutative algebra, Harrison works in char 0 but in general one needs Andre-Quillen cohomology.

Posted by: jim stasheff on April 15, 2007 1:58 AM | Permalink | Reply to this

### Re: Schur Functors

I’m not aware of hochschild having any problem in char p

OK, not a problem as such. I think what I was thinking of is the use of H2 to classify certain extensions (there’s a point in proving that square-zero extensions are the same as cocycles where you seem to need every surjection to be a split surjection, otherwise you have to restrict explicitly to those extensions which do split in Vect). To try and be more coherent: if char k=p, then H2(R, –) would only classify the k-split square-zero extensions and not all of them.

BUT for the cohomology of commutative algebra, Harrison works in char 0 but in general one needs Andre-Quillen cohomology.

I remember reading (a reference to) this a while back. One thing that’s just struck me: is this ‘just’ because Harrison cohomology is defined in terms of certain splitting idempotents on the cochain complex, which in general are only well-defined in char 0; and if so, might it be that the ‘right’ definition of Harrison cohomology is as a quotient
rather than a summand?

(Sorry this has got a bit off-topic!)

Posted by: Yemon Choi on April 15, 2007 2:38 AM | Permalink | Reply to this

### Re: Schur Functors

From a functorial point of view, the Harrison CHAIN complex has as underlying the free Lie COalgebra, the Koszul dual in characteristic 0. Anything Lie requires constraint outside of characteristic 0, cf. restricted Lie algebras — meaning having
a ‘restriction’ operation corresponding to p-th power.

Posted by: jim stasheff on April 15, 2007 1:10 PM | Permalink | Reply to this

### Re: Schur Functors

It is true that there is no problem with constructing
Sym and Alt satisfying the right universal properties. This is done using the quotient construction described above. But one does have to be careful when trying to make certain kinds of identifications that one is used to from characteristic zero.

For example, there are always natural maps

Alti(V*) → Alti(V)*

and

Symi(V*) → Symi(V)*

given by a signed sum over permutations.

The first is always an isomorphism (regardless of characteristic), but the second can fail to be depending on the value of i. To see this, you can choose a basis for V, take the induced bases for everything else, and write down the matrices. In the first case, you get the identity matrix. In the second case, binomial coefficients intervene, some of which will could be zero depending on the characteristic.

In other words, antisymmetrization commutes with dualization, by the same is not always true for symmetrization.

I don’t know how one would guess that this is the case without first examining the maps in question. Does anyone have an illuminating explanation for why symmetrization should not necessarily commute with dualization?

Also, what does this mean for Schur functors?

Posted by: Chris Brav on April 12, 2007 11:11 PM | Permalink | Reply to this

### Re: Schur Functors

I don’t have SGA 1 at hand, but I think one way of packaging all this would be the formalism of fibered categories. The category of all finitely generated free modules over all (commutative) rings is (I think) fibered over the category of rings. So a Schur functor is probably just a morphism of fibered categories.

I think that if you look at locally free modules instead of free ones, then your fibered category will be a stack and your Schur functors will be morphisms of stacks. The first just means that a vector bundle $V$ (= locally free finitely generated module) can be described locally and the second that $F(V)$, where $F$ is your Schur functor, can also be described locally.

(Apologies for the vagueness.)

Posted by: James on April 12, 2007 11:49 PM | Permalink | Reply to this

### Re: Schur Functors

we have some 2-category Mod of ‘module categories’

So somehow ModR becomes a stand-in for R itself?

I was used to thinking of R as a one-object additive category, but that’s naturally a description of associative rings with unit, not commutative ones. So maybe R “is” a one-object, one-morphism additive 2-category.

What’s ModR, then? I guess it’s 2-functors from R to a 2-category with one object, morphisms = abelian groups, but then I don’t know how to compose them.

Anyway, a wimpier idea than yours, but I think capturing the same idea:

Let AddCat be the 2-category of additive categories. It contains the category Ring of 1-object additive categories, which further contains CommRing.
Let Mod be the functor Hom(*,Ab): CommRing -> MonCat, so called because it takes R |-> ModR, a monoidal category.
Then a Schur functor should be a (special) natural transformation Mod -> Mod. So any commutative ring R is handed a functor ModR -> ModR in a natural way.

Posted by: Allen Knutson on April 13, 2007 5:28 AM | Permalink | Reply to this

### Re: Schur Functors

I’m no particular expert on these things, but I would like to point out that there are indeed problems in nonzero characteristic. (This is orthogonal to the issues of change of ground ring which you are discussing.) I find this to be one of the places where the categorical perspective is really helpful. Consider the example of vector spaces over a field of characteristic 2, and consider the degree 2 Schur functors. (And let’s stick to finite dimensional vector spaces, to avoid issues about Choice.)

There are two Schur functors, corresponding to the partitions (2) and (1,1); I’ll denote them by $S^2$ and $S^{1,1}$. The definition (according to Fulton) is that $S^2(V)$ is the quotient of $V^{\otimes 2}$ by the subspace spanned by expressions of the form $u \otimes v - v \otimes u$; the definition of $S^{1,1}(V)$ is the quotient of $V^{\otimes 2}$ by the subspace spanned by elements of the forms $u \otimes u$ and $u \otimes v + v \otimes u$. It is easy enough to see how $S^2$ and $S^{1,1}$ should act on morphisms.

Now, any construction involving quotients should have a dual construction involving subspaces. So let $S_2(V)$ be the kernel of $V^{\otimes 2} \to S^{1,1}(V)$ and let $S_{1,1}(V)$ be the kernel of $V^{\otimes 2} \to S^2(V)$. $S_2$ and $S_{1,1}$ are also functors in a clear way.

I’d seen it pointed out in various contexts that $S^2$ and $S_2$ should be thought of as different in characteristic 2. But I never understood why until I thought about it from the category perspective. For every (finite dimensional) vector space $V$, $S^2(V)$ is isomorphic to $S_2(V)$. But the functors $S^2$ and $S_2$ are not naturally equivalent!

To see this, consider three maps $a_1$, $a_2$ and $a_3$ from the one dimensional vector space spanned by $e$ to the two dimensional vector space spanned by $f_1$ and $f_2$; we map $e$ to $f_1$, to $f_2$ and to $f_1 + f_2$. Then $S^2(a_1)(e \otimes e)$, $S^2(a_2)(e \otimes e)$ and $S^2(a_3)(e \otimes e)$ are linearly dependent, but the analogous quantities for $S_2$ are not! So no isomorphism between $S^2(k^2)$ and $S_2(k^2)$ can commute with the functors.

Of course, one can define similar upper and lower versions for any of the Schur functors.

Two consequences for algebraic geometers:

Given a vector space $V$, one could define the symmetric algebra on $V$ to be either $\oplus S^i(V)$ or $\oplus S_i(V)$. Both have natural ring structures – the former is the polynomial algebra we are used to and the latter is the divided power algebra.

Given a vector bundle $E$ of rank two over a variety in characteristic $2$, $S^2(E)$ and $S_2(E)$ are two different vector bundles of rank $3$. I seem to remember determining that the counter-examples to Kodaira vanishing in characteristic $p$ are ultimately due to this phenomenon, but I don’t remember the details.

PS: I tried to enter this post with iTeX but the subscripts and superscripts didn’t show up correctly in preview and the direct sums turned into boxes. So I’m entering raw TeX. Sorry!

Posted by: David Speyer on April 13, 2007 6:28 AM | Permalink | Reply to this

### Re: Schur Functors

For some reason ‘bigoplus’ doesn’t work. So, I replaced it by ‘oplus’ and reposted your comment using iTeX. This was the only way to fix it, given that you didn’t enter it using iTeX in the first place. (A somewhat annoying feature of this setup is that while I can go in and correct typos in people’s comment, I can’t change the original choice of ‘Text Filter’ without deleting the original comment and reposting it. So, I urge people to always choose a text filter that allows for LaTeX.)

Anyway: I now understand what you’re saying, and it confirms what I was gradually realizing, in a rather shocked way, earlier today. In characteristic two not every short exact sequence in

$hom(Vect_k, Vect_k)$

splits, so the usual Schur functor $S^2$ has an nonisomorphic evil twin $S_2$.

Of course, one can define similar upper and lower versions for any of the Schur functors.

And maybe a necessary condition for them to be non-isomorphic in characteristic $p$ is that $p$ divides $n$!, where $n$ is the number of boxes in the Young diagram?

Has anybody worked out the complete theory of ‘characteristic $p$ Schur functors’?

More precisely, has anyone worked out the complete structure of the abelian category

$hom(Vect_k, Vect_k)$

where $Vect_k$ is the category of finite-dimensional vector spaces over $k = \mathbb{F}_p$? (Or maybe any field of characteristic $p$?)

I can imagine all sorts of functors like

$S^2 \circ S_2$

and

$S_2 \circ S^2$

and so on… do we get a big mess or just some nice manageable generalization of the theory of Young diagrams?

While you say this is orthogonal to the ‘change of base ring’ issue, it seems at least slightly relevant to the issue of finding the Schur-like functors that work in $Mod_R$ regardless of the commutative ring $R$, and are natural with respect to change of $R$. Such things would have to work in every characteristic, for starters.

Posted by: John Baez on April 13, 2007 11:55 PM | Permalink | Reply to this

### Re: Schur Functors

Has anybody worked out the complete theory of “characteristic p Schur functors”?

I recall that the abelian category of not-necessarily-additive endofunctors of the category of vector spaces over F_p is related to the Steenrod algebra A_p. There’s a lot of interest in computing ext groups between things like “the exterior algebra functor”, etc. I wouldn’t be surprised if Schur functors showed up in the literature on this.

Posted by: Changbao on April 14, 2007 3:00 AM | Permalink | Reply to this

### Re: Schur Functors

Cool! That might help me understand the Steenrod algebra better!

Posted by: John Baez on April 14, 2007 9:19 PM | Permalink | Reply to this

### Re: Schur Functors

At least in the parts of math I’m familiar with :-), THE symmetric algebra is exemplified by the polynomial algebra. The divided power algebra is another commutative algebra. Both played a prominent role in the Seminaire Cartan 1954-55. Note that in characteristic 0, the divided poly algebra splits as a tensor product of truncated polynomial algebras.

Posted by: jim stasheff on April 15, 2007 1:14 PM | Permalink | Reply to this

### Re: Schur Functors

Schur functors can be defined even more generally. Let $C$ be an arbitrary linear symmetric monoidal category. Let $X$ be an object of $C$. Consider $X^{\otimes n}.$ There is an action of $S_n$ on this object given by the commutors. Let $\pi_\lambda$ in the group ring of the symmetric group be the projection to an isotypic component. Since $C$ is linear, it makes sense to apply $\pi_\lambda$ to $X^{\otimes n}$. This is defined to be the Schur functor $S_\lambda(X)$.

I learned about all this in Ostrik’s exposition of a Deligne’s paper on the existence of superfiber functors.

Posted by: Noah Snyder on April 13, 2007 9:11 PM | Permalink | Reply to this

### Re: Schur Functors

A few more thoughts on my last comment.

First, one needs to assume that C is Karoubian in order for $\pi$ to have an image. This is a mild assumption since one can always take the Karoubian envelope.

Second, I want to emphasize that by linear I only mean that the Hom spaces are linear (no assumptions on a fiber functor) so this is strictly more general than the FR construction in your post.

Third, one should be able to replace “linear” with “additive.” This takes a little more thought as the definition I gave doesn’t work in certain characteristics (since those projections use denominators). The definition Ostrik uses (though, over $\mathbb{C}$ not $\mathbb{Z}$) is slightly different:

Let $V_\lambda$ be the $\mathbb{Z}[S_n]$-module corresponding to the partition $\lambda$ (I’m not sure if this is still a simple module, but it should still be defined over $\mathbb{Z}$, right?) We think of $V_\lambda$ as an object in $C$ which is a sum of trivial objects (I’m a bit confused about how this is done cannonically). Let $\pi = \sum g$. Ostrik defines

(1)$S_\lambda(X) = \pi(V_\lambda \otimes X^{\otimes n}).$
Posted by: Noah Snyder on April 13, 2007 9:50 PM | Permalink | Reply to this

### Re: Schur Functors

This seems like an opportune time to ask a very naive question. Can anyone tell me, given two Schur functors with their associated Young diagrams, how to find their composition–and its associated Young diagram? Also their tensor product, and its associated Young diagram? There seems to be a lot in the literature about plethysms, which are various decompositions of the composition of two Schur functors into a direct sum of functors, but I can’t find the answer to the simpler question. It may be that it is in one of the basic sources (perhaps Fulton-Harris) and I’m being admittedly lazy by asking here without having headed to the library first. That said, are there any other must-read sources for this subject? Thanks!

Posted by: Stefan on April 13, 2007 10:38 PM | Permalink | Reply to this

### Re: Schur Functors

The short answer to your simpler question is ‘the Littlewood-Richardson rule’. You may find the statement of the rule here or in section 2 here easier to follow.

Posted by: John Baez on April 14, 2007 12:09 AM | Permalink | Reply to this

### Re: Schur Functors

The Littlewood Richardson rule and its many equivalent formulations are for tensor products of Schur functors. Composition of Schur functors is Plethysm and nobody knows a combinatorial rule to do it in general (although some special cases are known).

Posted by: David Speyer on April 15, 2007 4:53 AM | Permalink | Reply to this

### Re: Schur Functors

Whoops. David is right and I was wrong.

(I meant to say this years ago.)

Posted by: John Baez on October 25, 2009 6:07 PM | Permalink | Reply to this
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### Re: Schur Functors

Allen wrote:

You never do divide by $p$. In the most basic examples, $Sym^2$ resp. $Alt^2$, you tensor the tensor square and mod out by $\{a \otimes b - b \otimes a\}$ resp. by $\{a \otimes a\}$, for all $a,b$ in the module.

The more general definition, on pp. 104-107, is also as a quotient, of a big tensor product by a submodule generated by a bunch of things with leading coefficient 1.

Okay. Given this, I think we should be able to define Schur functors on any symmetric monoidal abelian category $C$. ‘Symmetric monoidal’ lets us define the $n$th tensor power $x^{\otimes}$ of any object $x$ in $C$ and get an action of $S_n$ on it. ‘Abelian’ should then let us form the desired quotient of $x^{\otimes n}$ for any Young diagram.

So, given a Young diagram $Y$ and a symmetric monoidal abelian category $C$, I think we get an endofunctor

$Y_C : C \to C$

Moreover, this is ‘pseudonatural’ with respect to $C$. A bit more precisely, suppose we have an exact functor

$F: C \to C^'$

Then we should get a square of functors

$\array{ C & \stackrel{F}{\to} & C^' \\ Y_C \downarrow & & \downarrow Y_{C^'} \\ C & \stackrel{F}{\to} & C^' }$

which commutes up to a natural isomorphism satisfying the usual coherence laws.

So, I’m hoping we can slickly define the category of Schur functors as the category of all endofunctors on symmetric monoidal abelian categories which are pseudonatural with respect to exact functors.

Actually, ‘exact’ should be overkill. At most, I bet we’ll need the kind of ‘half-exactness’ one gets for a functor

$F : Mod_R \to Mod_{R^'}$

that comes from a ‘change of base rings’ of the sort you’re considering. I’m too tired to remember if this is left or right exactness… but I guess it should be related to how Schur functors are defined using a cokernel rather than a kernel.

Posted by: John Baez on April 15, 2007 8:02 AM | Permalink | Reply to this

### Re: Schur Functors

I must be really tired. Schurely functors that preserve cokernels are right exact.

$[symmetric monoidal abelian categories, functors, natural transformations]$

and

$[symmetric monoidal abelian categories, right exact functors, natural transformations]$

The Schur functors

$Y_C : C \to C$

are morphisms in the first 2-category — they’re not usually right exact, since (being ‘nonlinear’) they don’t preserve direct sums. But, they’re usually pseudonatural only with respect to morphisms in the second 2-category — that is, right exact functors.

Somehow I’ve never thought about this much before: transformations that are natural only with respect to morphisms in a subcategory.

Posted by: John Baez on April 15, 2007 8:36 AM | Permalink | Reply to this

### Re: Schur Functors

Somehow I’ve never thought about this much before: transformations that are natural only with respect to morphisms in a subcategory.

Do you not have examples where the subcategory consists of the invertible morphisms? I don’t have an example in mind, but I thought I knew some, like using diffeomorphisms to push forward vector fields, that sort of thing.

Posted by: Allen Knutson on April 16, 2007 6:45 PM | Permalink | Reply to this

### Re: Schur Functors

John wrote:

Somehow I’ve never thought about this much before: transformations that are natural only with respect to morphisms in a subcategory.

A kind of dual notion of this concept is when you think about transformations that might be forced to take values in a (sub)category.

For ordinary categories (i.e. not 2-categories), a nice example of this phenomenon is with representations of groups, as I think can be found in the quantum gravity seminar somewhere.

The algebra of automorphisms of the fiber functor

(1)$i : Rep(G) \rightarrow Vect$

is the group ring,

(2)$Aut (i) = \mathbb{C} [G]$

(I think!), while the algebra of automorphisms of the identity functor

(3)$id : Rep(G) \rightarrow Rep(G)$

is the center of the group ring,

(4)$Aut(id) = Z( \mathbb{C}[G] )$

So you get different answers depending on the freedom your natural transformations have in ‘target space’.

There appears to be a completely analagous result for 2-representations .

Thus, the monoidal category of automorphisms (objects are weak transformations, morphisms are modifications) of the fiber 2-functor

(5)$i : 2Rep(G) \rightarrow 2Vect$

is equal to the `group category’ (the monoidal category spanned by objects of the form $g \in G$),

(6)$Aut (id) = C[G]$

(I think), while one can indeed prove that the braided monoidal category of automorphisms of the identity 2-functor

(7)$id : 2Rep (G) \rightarrow 2Rep (G)$

is equivalent to the monoidal center of $C[G]$, which can also be thought of as $Rep (\Lambda G)$,

(8)$Aut(id) = Z(C[G]) = Rep (\Lambda G)$

This all has a nice interpretation in a TQFT context.

Posted by: Bruce Bartlett on April 21, 2007 11:02 AM | Permalink | Reply to this

### Re: Schur Functors

I get the feeling that you now understand “the” answer to my original question… but haven’t come right out and stated “here’s the complete answer”, or at least, not in one place. Please do!

Posted by: Allen Knutson on April 16, 2007 5:24 PM | Permalink | Reply to this

### Re: Schur Functors

I was afraid that if I wrote it down all in one place you’d run away screaming. But yesterday, walking to work, I realized a way to simplify it.

To enjoy this, you’ll still need to like ‘pseudonatural transformations’ between 2-functors between 2-categories, and modifications between those. But they’re not bad — you can see the definitions on page 534 here, for example.

Okay, here goes the conjecture:

A Schur functor is a pseudonatural transformation

$Y: i \Rightarrow i$

where

$i: C \to D$

is the 2-functor given by the inclusion of the 2-category

$C = [symmetric monoidal abelian categories, right exact functors, natural transformations]$

in the 2-category

$D = [symmetric monoidal abelian categories, functors, natural transformations] .$

More generally and concisely: the category of Schur functors and Schur natural transformations is

$End(i).$

In other words, it’s the category whose objects are pseudonatural transformations from $i$ to $i$, and whose morphisms are modifications between those.

I wish I had time to prove (or disprove) this.

Posted by: John Baez on April 21, 2007 1:15 AM | Permalink | Reply to this

### Re: Schur Functors

Nice! It would be great if this works out.

Posted by: Bruce Bartlett on April 21, 2007 10:24 AM | Permalink | Reply to this

### Re: Schur Functors

I wish I had time to prove (or disprove) this.

Am I right in thinking that you know how to categorify (one level further) each of the objects you mentioned, so you could ask the same question one level up?

Or how about one level down?

Posted by: Allen Knutson on April 23, 2007 4:27 AM | Permalink | Reply to this

### Re: Schur Functors

If we have an object $X$ in a symmetric monoidal abelian category of characteristic 0, and we want to ask what subobjects of $X ^{\otimes n}$ are canonically present thanks to the categorical structure, then I see that the answer to this is given by the Young diagrams. However, it’s a bit more complicated than this, because each Young diagram exists inside its $X ^{\otimes n}$ once for each possible way of numbering it — so isn’t it more accurate to say that the subobjects correspond to the Young tableaux, rather than the Young diagrams?

Of course, I guess it must be possible to show in the group algebra that the subobjects corresponding to different numberings of the same Young diagram are isomorphic. So in this sense it’s valid to say that the subobjects correspond to the Young diagrams.

If this is the case, though, I imagine that there should be many more Schur functors — the ones described above are surely just the isomorphism classes of Schur functors. For example, consider the decomposition of $\mathbb{C}[S_3] \simeq R_{(123)} \oplus R_{(1)(2)(3)} \oplus R_{(12)(3)} \oplus R_{(13)(3)}$ into irreducible subrepresentations corresponding to Young tableaux, where the irrep $R_{(12)(3)}$ is isomorphic to $R_{(13)(2)}$ since they come from the same Young diagram. We can construct a Schur functor for any term in this decomposition of $\mathbb{C}[S_3]$, so that gives us 4.

But shouldn’t there be lots of other embeddings of the irrep $R_{(12)(3)}$ into $\mathbb{C}[S_3]$, apart from the one where it just hits $R_{(12)(3)}$ and the one where it just hits its isomorphic copy $R_{(13)(2)}$? I think there should also be ‘oblique’ ones, which partially hit both, in just the same way as there are more injections of $\mathbb{C}$ into $\mathbb{C}^2$ than just $(1,0)$ and $(0,1)$. Wouldn’t these also give Schur functors, admittedly ones which are isomorphic to ones already obtained? Or can these not actually necessarily be constructed on an arbitrary symmetric monoidal abelian category?

Posted by: Jamie Vicary on October 23, 2009 11:48 AM | Permalink | Reply to this

### Re: Schur Functors

Hi, Jamie — we haven’t heard from you in a long while!

… isn’t it more accurate to say that the subobjects correspond to the Young tableaux, rather than the Young diagrams?

Yes, you’re right.

Of course, I guess it must be possible to show in the group algebra that the subobjects corresponding to different numberings of the same Young diagram are isomorphic.

Sure.

If this is the case, though, I imagine that there should be many more Schur functors — the ones described above are surely just the isomorphism classes of Schur functors.

Sure. When we say vector spaces are classified by their dimension we mean their isomorphism classes are in 1-1 correspondence with dimensions. And when we say Schur functors are classified by Young diagrams we mean their isomorphism classes are in 1-1 correspondence with dimensions.

But what you’re really trying to tell me, I think, is that we shouldn’t just think about Schur functors: we should think about Schur functors and ‘Schur natural transformations’, and understand this category.

I made a highbrow conjecture about this category here. There are, of course, less ridiculously abstract characterizations. But the main thing I’m taking from what you said is that the phrase ‘Young tableaux’ should appear somewhere in this story!

Posted by: John Baez on October 23, 2009 4:14 PM | Permalink | Reply to this

### Re: Schur Functors

Thanks for that reply, John. It seems that my intuition more-or-less matches reality, anyway, thanks for settling that.

we shouldn’t just think about Schur functors: we should think about Schur functors and ‘Schur natural transformations’, and understand this category … I made a highbrow conjecture about this category here. There are, of course, less ridiculously abstract characterizations.

Right! I like your characterization, I think it’s neat.

I haven’t really hung around here for a while, I know. I think that’s probably because the kind of posts I really enjoy — the ones which take some pretty fundamental bit of category theory–related mathematics and turn it inside out, until it’s burned into your brain — have become less frequent as the cafe has got older. Maybe this is just because all the possible topics have been covered already! In fact, I sort of got the impression that this change coincided with the instigation of the $n$Lab, which I don’t have the same affection for as I do for this place … but that could all just be in my head. It would be sad if these sorts of fun technical discussion are happening hidden away in $n$Lab comment boxes instead of being chatted about out here in the open.

Posted by: Jamie Vicary on October 23, 2009 5:26 PM | Permalink | Reply to this

### Re: Schur Functors

There are at least two current topics here at the n-Community that could definitely use some more eyes and minds that I think fall under the description of taking

some pretty fundamental bit of category theory–related mathematics and turn it inside out, until it’s burned into your brain

They are:

Posted by: Eric Forgy on October 23, 2009 9:16 PM | Permalink | Reply to this

### Re: Schur Functors

Jamie wrote:

I think that’s probably because the kind of posts I really enjoy — the ones which take some pretty fundamental bit of category theory–related mathematics and turn it inside out, until it’s burned into your brain — have become less frequent as the cafe has got older. Maybe this is just because all the possible topics have been covered already!

I know it’s not that, because I know dozens more topics that would be great fun to talk about.

For more on this, subject, go here.

Posted by: John Baez on October 23, 2009 6:50 PM | Permalink | Reply to this

### Re: Schur Functors

In particular, Urs has almost entirely moved over there. He seems to feel that information here gets ‘buried’ under new layers of blogging, while information there is easier to find. That’s probably true. But I think I prefer talking here.

I also prefer talking here.

Remember the whole idea:

- here on the $n$Café we talk

- whenever we reach some stable insight we archive that on the $n$Lab, so that it doesn’t get lost.

For instance this very entry here. Why does no-one of you think of it highly enough to produce a coherent version of the information scattered here on the $n$Lab? It wouldn’t even be much effort, just a little copy and pasting.

Posted by: Urs Schreiber on October 24, 2009 10:53 AM | Permalink | Reply to this

### Re: Schur Functors

a little copy and pasting !
and a significant investment in connective tissue to achieve coherence!

Posted by: jim stasheff on October 24, 2009 2:22 PM | Permalink | Reply to this

### Re: Schur Functors

Why does no-one of you think of it highly enough to produce a coherent version of the information scattered here on the nLab? It wouldn’t even be much effort, just a little copy and pasting.

I’m attempting to take you up on that, but I think it’s a little harder than just copying and pasting (as Jim Stasheff surmised it would be). While I was sleeping the first time around when this stuff was discussed, I’m now trying to assemble some of this in an nLab entry, and a number of questions arise.

Here is what I’ve written so far on this; I have some queries in boxes, which I’ll be asking about below as well.

Posted by: Todd Trimble on October 25, 2009 3:49 AM | Permalink | Reply to this

### Re: Schur Functors

Here is what I’ve written so far on this;

Thanks, Todd.

Ah, so that is already very useful. From just the discussion above I had no idea that there are so many questions that even you find are unresolved here.

I have to dash off now. All I did was to add a table of contents to your entry. This way it is easier to see at a glance what is there.

Posted by: Urs Schreiber on October 25, 2009 11:02 AM | Permalink | Reply to this

### Re: Schur Functors

One-way discussions are no fun. Lately I feel that I don’t get any replies here anymore.

Why does no-one of you think of it highly enough to produce a coherent version of the information scattered here on the nLab?

Mmm. If I were to face this question right now, the first answer that comes to my mind, personally, is basically just the selfish reason that I would get no credit for this kind of thing on the nLab. My name wouldn’t appear on it. I couldn’t present it in a personal, witty or stylized way. You might say I could or should present the material in a personalized way on the nLab anyway, but it’s a bit weird to do that in a wiki environment where your name doesn’t appear at the bottom, and in any case someone will and should alter your text later.

Of course, that is a general comment; to answer your specific question about Schur functors: well, Todd has thankfully done it, and I haven’t absorbed this thread about Schur functors yet at all!

Posted by: Bruce Bartlett on October 25, 2009 10:01 PM | Permalink | Reply to this

### Re: Schur Functors

I couldn’t present it in a personal, witty or stylized way. You might say I could or should present the material in a personalized way on the nLab anyway, but it’s a bit weird to do that in a wiki environment

I don’t think I agree. There are plenty of pages on the nlab written in a personal, witty, or stylized way. They do get edited later on, but I think usually the different stylistic approaches get preserved as at least part of the page, unless people think they’re misleading.

Posted by: Mike Shulman on October 25, 2009 10:09 PM | Permalink | PGP Sig | Reply to this

### Re: Schur Functors

Bruce wrote:

If I were to face this question right now, the first answer that comes to my mind, personally, is basically just the selfish reason that I would get no credit for this kind of thing on the $n$Lab.

I think that could be the secret answer to Ben Webster’s question Why do I find MathOverflow fun and the nLab not?

Personally I don’t find MathOverflow fun — at least not yet. But if you answer a question there, your name is attached to the answer — and everyone gets to see how many people like your answer. So, you get instant social status from answering questions there. And let’s face it: most academics aren’t in it for the money, they’re in it for the social status! They want the small group of experts in their chosen field to think they’re cool. (Or if they’re grad students, they’re happy if any mathematicians think they’re cool.) That’s what they live for.

Similarly, when I write This Week’s Finds, I’m not getting publications out of it, but I am getting social status — at least if it’s any good. And to be embarrassingly honest, that’s why I write it.

On the other hand, work on the $n$Lab tends to be anonymized as time passes… except of course work done by Urs, because he’s the ruling genius of the $n$Lab. So perhaps it’s not surprising that some of the people who do the most work there, other than Urs, are not currently professional academics.

And maybe more academics would get involved in the $n$Lab if it were possible to get ones name permanently attached to an article, or portion of an article — or something like that. Something that confers social status.

Again, this might be a conversation for the n-Forum.

Posted by: John Baez on October 26, 2009 5:14 AM | Permalink | Reply to this

### not about Schur Functors any more

So, you get instant social status from answering questions [on MathOverflow]

Perhaps that is precisely why I don’t like it—it puts out into the open this embarrassing aspect of academia. (-:

Again, this might be a conversation for the n-Forum.

Actually, I think there’s value to having it here on the blog. I could be wrong, but I think a lot more people read the blog than the forum. And this isn’t just a technical question about how the lab works, but a more fundamental question about how online collaboration can work and how credit can be assigned. We might want to start a new post about it, though, instead of hijacking this one.

Posted by: Mike Shulman on October 26, 2009 5:43 PM | Permalink | PGP Sig | Reply to this

### Re: Schur Functors

Okay, this blog entry is over two and a half years old, but I’ve been reading it over and trying to put it together for the nLab. Ignorance compels me to ask some questions.

First, to me, these Schur functors were always the Schur functors $S_\lambda$ attached to Young diagrams, and I didn’t know that every endofunctor on $Vect_{fd}$ (finite-dimensional vector spaces over $\mathbb{C}$ – let me stick to the classical case for a minute) was a direct sum of $S_\lambda$’s.

I’d like to understand that a bit better. First: is it literally every endofunctor? At first glance, that’s surprising to me. Is there a nice way of seeing that? Is it that every endofunctor is a finite direct sum of irreducibles?

In this blog discussion, there was some talk about the generality in which one could work: there were some comments on what new subtle features arise in nonzero characteristic over what happens in the classical case. For various reasons (which I’m coming to), I decided to play it safe by restricting attention to symmetric monoidal categories enriched in vector spaces over $\mathbb{Q}$ (with some exactness assumptions, pretty mild ones will do I think). Because of subtleties that can arise in characteristic $p \gt 0$, it felt to me like hoping for a clean general theory which applied to general symmetric monoidal abelian categories might be overreaching or asking for trouble, but I’d like to be better educated about this.

On the other hand, as Noah Snyder pointed out, you can carry the basic constructions quite far just by working with symmetric monoidal additive categories in which idempotents split. This is a very mild exactness assumption (it just means we pass if need be to the “Karoubi envelope” or Cauchy completion in the sense of Lawvere). Notice in particular that we get the Schur functor $S_\lambda$ ($\lambda$ a partition of an $n$-element set) by splitting the idempotent symmetrizer

$\frac1{n!} \sum_{g \in S_n} g: V_\lambda \otimes X^{\otimes} \to V_\lambda \otimes X^{\otimes n}$

that is available if homs are valued in vector spaces over $\mathbb{Q}$.

This may have consequences for the conjecture John made about defining the category of Schur functors, which was:

A Schur functor is a pseudonatural transformation

$Y: i \Rightarrow i$

where

$i: C \to D$

is the 2-functor given by the inclusion of the 2-category

$C=[symmetric monoidal abelian categories, right exact functors, natural transformations]$

in the 2-category

$D = [symmetric monoidal abelian categories, functors, natural transformations].$

More generally and concisely: the category of Schur functors and Schur natural transformations is

$End(i).$

In other words, it’s the category whose objects are pseudonatural transformations from $i$ to $i$, and whose morphisms are modifications between those.

First of all, I wonder whether John meant the 1-cells of $C$ to be symmetric monoidal (right exact) functors. I think of the 1-cells in $C$ as being suitable “change of base” functors which should preserve the construction of the basic gadgets $S_\lambda$, which involves tensor products and tensor powers.

Second: the colimits involved in the construction are absolute coequalizers. Namely, we have that

$S_\lambda(X) = e(V_\lambda \otimes X^{\otimes n})$

where $e = \frac1{n!} \sum_{g \in S_n} g$ is an idempotent symmetrizer projection, but this image can be construed as the coequalizer of the pair

$e, 1: V_\lambda \otimes X^{\otimes n} \overset{\to}{\to} V_\lambda \otimes X^{\otimes n}$

which is a split coequalizer since $e$ is idempotent. Now, split coequalizers are preserved by any functor!

So if what I’m thinking is correct, we really don’t need right exactness at all in the definition of 1-cells of $C$. To define $C$, we can take perhaps

$C = [symmetric monoidal abelian categories, symmetric monoidal functors, natural transformations]$

and $D$ as before.

It seems to me there may be some point to weakening the abelian hypothesis on 0-cells of $C$ to just Cauchy completeness (along the lines of what Noah said in his comment), and perhaps strengthening the hypothesis on the 1-cells to include preservation of direct sums (insofar as the claim is that general Schur functors are direct sums of irreducible ones, perhaps we should demand compatibility with direct sums, i.e., linearity of the functors). So, 1-cells would be additive, but right exactness may be more than what we really need.

Comments on the current nLab entry are welcome and appreciated.

Posted by: Todd Trimble on October 25, 2009 6:11 AM | Permalink | Reply to this

### Re: Schur Functors

Todd wrote:

Ignorance compels me to ask some questions.

Ignorance compels me to guess some answers.

By the way: I’d love to join forces with you and prove this conjecture about Schur functors, or some improved version of it. I think it’s very important, so if we make progress, I’d like us to publish a — short and sweet — paper about it. Interested?

I think it would be cool for there to be a paper with us as coauthors.

First, to me, these Schur functors were always the Schur functors $S_\lambda$ attached to Young diagrams, and I didn’t know that every endofunctor on $Vect_{fd}$ (finite-dimensional vector spaces over $\mathbb{C}$ – let me stick to the classical case for a minute) was a direct sum of $S_\lambda$’s.

I’d like to understand that a bit better. First: is it literally every endofunctor? At first glance, that’s surprising to me.

Yeah, it’s surprising that you can classify them so neatly, but I seem to recall that it’s a well-known fact.

Is there a nice way of seeing that? Is it that every endofunctor is a finite direct sum of irreducibles?

Let’s see. Either I don’t understand this situation (quite possible) or your guess is right: the endofunctor category $[FinVect_{\mathbb{C}}, FinVect_{\mathbb{C}}]$ is an abelian category where every object is a finite direct sum of simple objects.

But here’s one way to get started. Every endofunctor

$FinVect_{\mathbb{C}} \to FinVect_{\mathbb{C}}$

produces, for each $n$, a representation of $GL(n,\mathbb{C})$. Moreover by functoriality these representations fit together in a certain way coming from the standard inclusions $\mathbb{C}^n \hookrightarrow \mathbb{C}^{n+1}$.

That’s why people often talk about Schur functors as giving representations of $GL(\infty,\mathbb{C})$. But right now I just want to point out something about which representations of $GL(n,\mathbb{C})$ we can get. We can’t get them all!

Every (finite-dimensional) representation of $GL(n,\mathbb{C})$ is a finite direct sum of irreducibles. Every irreducible is a tensor product of 1) a representation coming from a Young diagram with $\le n$ boxes and 2) a 1-dimensional representation of the form $g \to det(g)^k$ for some $k \in \mathbb{Z}$. We can see this by combining standard stuff about irreps of $SL(n,\mathbb{C})$ and the way $\SL(n,\mathbb{C})$ sits inside $GL(n,\mathbb{C})$.

Now here’s the cute part: the only representations of $GL(n,\mathbb{C})$ we can get are those that extend to representations of the monoid $End(\mathbb{C}^n)$. And this means we can only get representations of form $g \to det(g)^k$ with $k$ nonnegative.

But all powers of the determinant representation of $GL(n,\mathbb{C})$ can be described by Young diagrams: tall skinny Young diagrams with $n k$ boxes.

So, every endofunctor

$FinVect_{\mathbb{C}} \to FinVect_{\mathbb{C}}$

gives us, for each $n$, a representation of $GL(n,\mathbb{C})$ which is a direct sum of finitely many representations described by Young diagrams.

But as mentioned, the representations for different $n$ need to ‘fit together’ in a very tight way, thanks to the standard inclusions $\mathbb{C}^n \hookrightarrow \mathbb{C}^{n+1}$ and also, for that matter, the projections $\mathbb{C}^{n+1} \hookrightarrow \mathbb{C}^{n}$.

I hope these are most of the raw ingredients of the proof.

For various reasons (which I’m coming to), I decided to play it safe by restricting attention to symmetric monoidal categories enriched in vector spaces over $\mathbb{Q}$ (with some exactness assumptions, pretty mild ones will do I think).

That seems like a great place to start. If we can’t do that, we probably can’t do the fancier stuff.

… I wonder whether John meant the 1-cells of $C$ to be symmetric monoidal (right exact) functors. I think of the 1-cells in $C$ as being suitable “change of base” functors which should preserve the construction of the basic gadgets $S_\lambda$, which involves tensor products and tensor powers.

Yes, Schur functors are supposed to be ‘constructions preserved by change of base’, so I should indeed have said symmetric monoidal right exact functors. And that makes me think the 2-cells should be symmetric monoidal natural transformations.

Now, split coequalizers are preserved by any functor!

Oh! Cute!

It seems to me there may be some point to weakening the abelian hypothesis on 0-cells of $C$ to just Cauchy completeness (along the lines of what Noah said in his comment), and perhaps strengthening the hypothesis on the 1-cells to include preservation of direct sums.

That sounds right.

Posted by: John Baez on October 25, 2009 5:52 PM | Permalink | Reply to this

### Re: Schur Functors

John wrote:

I’d like us to publish a — short and sweet — paper about it.

But if you prefer the more communal approach of putting stuff on the $n$Lab and talking about it with folks there, don’t hesitate to say so. Maybe that style works better for you than ‘publishing papers’. I’m discovering that the opposite is true for me. Having written tons of papers, I know how that works. I don’t know what happens to really interesting new results that are discovered on the $n$Lab.

The reason I’m acting a bit possessive about this conjecture I have some dreams of redoing old-fashioned representation theory — for example, Weyl’s The Classical Groups: Their Invariant and Representations — in 21st-century language. My first step in that direction was characterizing the categories of unitary representations of the compact classical groups. But this conjecture would be a big step forwards.

Unfortunately, I have a bunch of dreams, and not enough time or skill to pursue most of them…

Posted by: John Baez on October 25, 2009 7:19 PM | Permalink | Reply to this

### Re: Schur Functors

I don’t know what happens to really interesting new results that are discovered on the nLab.

At least until the revolution is complete, I would hope that they eventually get published in a paper. I don’t see why working on a really interesting new result at the nLab would prevent it from then being published, any more than discussing it on a blog or putting it on the arxiv would. Remember the “lab book” analogy—the lab book doesn’t get published, but the interesting things that are written in it do eventually (one hopes) get published.

Posted by: Mike Shulman on October 25, 2009 10:13 PM | Permalink | PGP Sig | Reply to this

### Re: Schur Functors

Mike wrote:

I don’t see why working on a really interesting new result at the nLab would prevent it from then being published…

I don’t either. At present, journals don’t give a darn about whether their results previously appeared on the $n$Lab. I was worrying about something else: how do we decide who are the authors?

Suppose Todd takes a conjecture of mine, fixes some mistakes in it, and throws it onto the $n$Lab. Then suppose he and a bunch of other people work on it and finally it gets proved. Then someone decides it’s so nice that it should be published. Who gets to be a coauthor?

You could say “Don’t worry too much; we’ll all friends, so we’ll sort something out.” And maybe that’s true. But one thing I think I’ve learned from publishing dozens of papers is that it really pays to decide early on who is a coauthor, and stick to that decision. Otherwise, people’s feelings can be hurt when they discover that everyone else thinks they haven’t done enough work to be a coauthor. It really happens. Indeeed, I recently violated this rule once, thinking the circumstances warranted it. But I rediscovered that it was a good rule, and I shouldn’t have violated it.

Anyway, maybe this is a good conversation to have on the $n$-Forum. The issue already came up — see Andrew’s comment about being a ‘professional mathematician’ — and I wanted to join in, but I was too stupid to remember my password. Now Andrew has given me a new one, and I think I can even remember it.

Posted by: John Baez on October 26, 2009 4:40 AM | Permalink | Reply to this

### Re: Schur Functors

Who gets to be a coauthor?

Not it!
But I would appreciate if my role as instigator was mentioned somewhere.

Posted by: Allen Knutson on October 26, 2009 7:49 PM | Permalink | Reply to this

### Re: Schur Functors

Yes, you’ll certainly be cited as the Primary Instigator… and whoever winds up writing a paper will certainly make sure to thank everyone who helped out.

Posted by: John Baez on October 26, 2009 7:57 PM | Permalink | Reply to this

### Re: Schur Functors

Mike wrote:

I don’t see why working on a really interesting new result at the nLab would prevent it from then being published…

I don’t either. At present, journals don’t give a darn about whether their results previously appeared on the $n$Lab. I was worrying about something else: how do we decide who are the authors?

Suppose Todd takes a conjecture of mine, fixes some mistakes in it, and throws it onto the $n$Lab. Then suppose he and a bunch of other people work on it and finally it gets proved. Then someone decides it’s so nice that it should be published. Who gets to be a coauthor?

You could say “Don’t worry too much; we’ll all friends, so we’ll sort something out.” And maybe that’s true. But one rule I’ve learned from publishing dozens of papers is that it really pays to decide early on who is a coauthor, and stick to that decision. Otherwise, people’s feelings can be hurt when they discover that everyone else thinks they haven’t done enough work to be a coauthor.

(These problems really happen. Indeeed, I recently violated my rule, thinking the circumstances warranted it. But I rediscovered that it was a good rule, and I shouldn’t have violated it.)

Anyway, maybe this is a good conversation to have on the $n$-Forum. The issue already came up — see Andrew’s comment about being a ‘professional mathematician’ — and I wanted to join in, but I was too stupid to remember my password. Now Andrew has given me a new one, and I think I can even remember it.

Posted by: John Baez on October 26, 2009 4:54 AM | Permalink | Reply to this

### Re: Schur Functors

I just got back from a nice productive meeting with Alex Hoffnung, whom I hadn’t met before; hence the delay in responding.

I think a short paper on this would be fun! I would also like to get back to unitary representations of the Poincaré group, which might fit in well with the larger dreams you have of updating classical representation theory. (I tried to get a start on that in the nLab as well. In particular, any comments you have on the Poincaré group article would be appreciated, and there are a few other related articles which I wouldn’t mind your looking at, if you have some time.)

I’ll think about the mathematics in your response… thanks for the hints about the direct sum decomposition into $S_\lambda$’s; I think I get the idea now.

Posted by: Todd Trimble on October 25, 2009 10:19 PM | Permalink | Reply to this

### Re: Schur Functors

Todd wrote:

In particular, any comments you have on the Poincaré group article would be appreciated…

It looks great so far. It’s unusually self-contained for an $n$Lab article: for example, it says a bit about what a unitary representation is and why we might care, instead of referring to some other $n$Lab article for that. But that’s a good thing.

I’ll have a lot more to say when this article starts classifying Poincaré representations. I can probably say a bit about stuff like ‘mass’, ‘spin’ and ‘helicity’ and what they mean to physicists.

Posted by: John Baez on October 26, 2009 7:02 AM | Permalink | Reply to this

### Re: Schur Functors

Every (finite-dimensional) representation of $GL(n,\mathbb{C})$ is a finite direct sum of irreducibles. Every irreducible is a tensor product of 1) a representation coming from a Young diagram with $\leq n$ boxes and 2) a 1-dimensional representation of the form $g \to \det(g)^k$ for some $k \in \mathbb{Z}$.

This may be related to Todd's doubts as to whether every endofunctor of $Vect_{fd}$ is a direct sum of the corresponding Schur functors, because what you've just written is not quite true. Besides these, there are other $1$-dimensional representations: $g \mapsto {|\det g|}^a$ for any $a \in \mathbb{R}$ and $g \mapsto \overline{\det g}$, for example. If you believe in nontrivial multiplicative automorphisms of the real line, then there are even more.

As explained at the SBS, you can restrict to the representations that you want by insisting on algebraic or (equivalently) holomorphic representations. I expect (although I don't have a reference or proof for this) that you could get my absolute-value-based and conjugation-based counterexamples while still ruling out the nonconstructive ones, if you accept all continuous (or even all measurable) representations.

Now here’s the cute part: the only representations of $GL(n,\mathbb{C})$ we can get are those that extend to representations of the monoid $End(\mathbb{C}^n)$. And this means we can only get representations of form $g \to \det(g)^k$ with $k$ nonnegative.

Again, for this conclusion, you need to insist that your representations are at least continuous.

Posted by: Toby Bartels on October 25, 2009 10:49 PM | Permalink | Reply to this

### Re: Schur Functors

Whoops. Good point, Toby!

I wanted only the algebraic representations of $GL(n,\mathbb{C})$ to show up when we restrict a functor $F: FinVect_{\mathbb{C}} \to FinVect_{\mathbb{C}}$ to the subcategory $GL(n,\mathbb{C})$. And I’m still hoping it’s almost true. But my argument was certainly wrong, because I forgot about all the nonalgebraic representations of the multiplicative group $\mathbb{C}^*$.

Why ‘almost’ true? Well, I realize now that the representation $\overline{det(g)}$ does arise from a functor $F: FinVect_{\mathbb{C}} \to FinVect_{\mathbb{C}}$. Why? Just take the endofunctor sending each complex vector space to its complex conjugate!

More generally: the group of automorphisms of a field $k$ acts as endofunctors $F: FinVect_{k} \to FinVect_{k}$. And these are not Schur functors, except for the identity.

So, my claim that all endofunctors of $FinVect_{\mathbb{C}}$ are Schur functors is definitely false. But this is not the sort of thing I’d have the temerity to claim if I hadn’t read something like it somewhere! So I still hope it’s almost true: maybe I just left out a hypothesis.

And I don’t think that hypothesis would mention ‘continuity’ or ‘measurability’ — it would have been algebraic in nature. Why? Because whoever claimed this would have been an algebraist of the sort that avoids analysis.

For example: maybe I read that all endofunctors of $FinVect_{\mathbb{Q}}$ are Schur functors. Switching to $\mathbb{Q}$ would get rid of complex conjugation and all those nasty nonmeasurable automorphisms of the field $\mathbb{C}$. So $\mathbb{Q}$ should make things easier, though Toby’s points about $|det(g)|$ and $det(g)^k$ with $k$ negative still need to be reckoned with.

By the way: all this stuff makes me like even more my hoped-for ‘field-independent’ description of Schur functors. All this annoying stuff that Toby just pointed out is very ‘field-dependent’, and I think it would get washed out if we demanded constructions that work uniformly for all symmetric monoidal abelian categories (or something like that).

Posted by: John Baez on October 26, 2009 4:12 AM | Permalink | Reply to this

### Re: Schur Functors

My point above is that ‘field-independent’ representations of $GL(n)$ should be the same as ‘algebraic’ ones. and that ‘field-independence’ can be formalized using the pseudonaturality condition described here.

Posted by: John Baez on October 26, 2009 7:04 AM | Permalink | Reply to this

### Re: Schur Functors

So I still hope it’s almost true: maybe I just left out a hypothesis.

And I don’t think that hypothesis would mention ‘continuity’ or ‘measurability’

Those wouldn't be strong enough anyway; they wouldn't rule out complex conjugation.

Presumably there's some way to say that we want only algebraic functors in a sense that mirrors the algebraicity condition at the SBS. (And for $\mathbb{C}$, there might be an equivalent way to define holomorphic functors, but that won't make sense for other fields.) It would be very nice if that followed from a base-change-preservation (field independence) condition as well.

As for the question in the OP about other characteristic, note that the SBS post claimed that its classification of algebraic representations was correct over any field of characteristic $0$.

Posted by: Toby Bartels on October 26, 2009 7:32 AM | Permalink | Reply to this

### Re: Schur Functors

If I understand what you want, I think the issues of algebraicity and field independence can be handed by enriching everything over algebraic geometry, which is probably not a surprise. If $R$ is a ring, then the category $C$ of finite free $R$-modules is enriched in $R$-schemes. This is just a fancy way of saying that the set of $m\times n$ matrices with entries in $R$ is the set of points of a scheme (affine $m n$-space) with coordinates in $R$ and that composition is a scheme map. Then define an algebraic structure on a functor $C\to C$ to be an enrichment in $R$-schemes. This handles algebraicity and uniformity across fields which are $R$-algebras. To get all fields, let $R$ be the ring of integers.

There are much more down to earth ways of saying the same thing. If $R$ is $\mathbf{Z}$ or $\mathbf{Q}$, then such an enrichment is unique if it exists. This is because the points in affine $n$-space with coordinates in $\mathbf{Z}$ or $\mathbf{Q}$ are Zariski dense (unlike the case with finite fields). So, the algebraicity condition on the functor is the same as saying that the maps between Hom sets can be described by polynomials in the entries of the matrices in some basis (or equivalently, in any basis).

Posted by: James on October 26, 2009 9:11 AM | Permalink | Reply to this

### Re: Schur Functors

So when Todd writes that he'll work in a sufficiently exact symmetric monoidal $Vect_{\mathbb{Q}}$-enriched category, he really should move from $Vect_{\mathbb{Q}}$ to $Sch_{\mathbb{Q}}$? That will cover all fields of characteristic zero. Then when we're ready to worry about non-zero characteristics or even non-fields, we move from $Sch_{\mathbb{Q}}$ to $Sch$.

Posted by: Toby Bartels on October 26, 2009 6:04 PM | Permalink | Reply to this

### Re: Schur Functors

James wrote:

If I understand what you want, I think the issues of algebraicity and field independence can be handed by enriching everything over algebraic geometry, which is probably not a surprise.

It’s not, but it’s good to hear how it works!

There are much more down to earth ways of saying the same thing.

I’m ultimately aiming for much less down to earth ways of saying the same thing — but again, this is very good to hear.

Posted by: John Baez on October 26, 2009 3:56 PM | Permalink | Reply to this

### Re: Schur Functors

One more thing. I bet it’s not necessary to enrich over schemes per se, it’s probably enough just to enrich over the category of functors from rings to sets, which contains the category of schemes as a full subcategory. People around these parts might prefer to do it this way.

Posted by: James on October 27, 2009 6:54 AM | Permalink | Reply to this

### Re: Schur Functors

In the other direction, it is probably also enough to enrich over Rings^{op}, which is a subcategory of schemes.

Posted by: David Speyer on October 27, 2009 2:24 PM | Permalink | Reply to this

### Re: Schur Functors

James wrote:

One more thing. I bet it’s not necessary to enrich over schemes per se, it’s probably enough just to enrich over the category of functors from rings to sets, which contains the category of schemes as a full subcategory. People around these parts might prefer to do it this way.

One reason this approach would be nice here is that ultimately we’re aiming for a description of Schur functors in terms of some 2-category $C$ like

• symmetric monoidal abelian categories,
• right exact symmetric monoidal functors,
• symmetric monoidal natural transformations.

and there’s a functor

$Mod : CommRing \to C$

sending each commutative ring $R$ to the category $R Mod$. So, the sort of ‘ring-independence’ you’re expressing with the language of schemes, or functors from $CommRing$ to $Set$, should be directly related to the ‘ring-independence’ that I’m trying to express by describing Schur functors with the help of the 2-category $C$.

Basically I’m trying to drop the reference to rings and work directly with their categories of modules — or more general, any sort of category upon which Schur functors know how to act. And that’s why $C$ consists of ‘symmetric monoidal abelian categories’, or something a bit more general, like Todd’s ‘symmetric monoidal Cauchy complete categories with direct sums’.

Posted by: John Baez on October 27, 2009 6:47 PM | Permalink | Reply to this

### Re: Schur Functors

In the current nLab entry, I’ve written out John’s conjecture (leaving some details like exactness conditions up in the air), and wrote that it’s easy to see that symmetric monoidal linear functors $Mod_R \to Mod_S$ are monoidally equivalent to ones of the form

$S \otimes_R -: Mod_R \to Mod_S$

for some unique $R$-algebra structure $f: R \to S$. (This I’m sure can be beefed up to a 2-categorical statement, but I’ll let that pass for now.) So for $F$ a Schur functor according to John’s conjectured definition, we have (coherent) strength isomorphisms of the form

$S \otimes_R F_R(M) \cong F_S(S \otimes_R M)$

for any $R$-module $M$ and $R$-algebra $f: R \to S$, where I am using $F_R$ to denote the component of the Schur functor as endofunctor on $Mod_R$.

Let’s see if we can make a connection to what James was saying. Instead of his category $C$ of finite free $R$-modules, it actually seemed more appropriate to me to consider the category of finite projective $R$-modules. Whatever – I’ll call it $C$ again. For any two finite projectives $P$, $Q$, we have that $hom(P, Q)$ is an $R$-scheme. In fact, echoing David, it’s easy to see that $hom(P, Q)$ is an affine scheme (essentially because it’s a retract of an affine scheme or a spectrum of a ring, and rings admit retracts = splittings of idempotents). Anyway, the enrichment means we have appropriate maps of affine schemes

$hom(P, Q) \to hom(F_R(P), F_R(Q))$

Hm, it’s getting late and I’m a bit bleary-eyed, but I’ll just throw this out there and see if it sticks. We’re trying to check (we’re hoping) that being Schur in John’s sense implies enrichment or algebraicity in James’s sense. Now, it’s well-known that there is a close connection between strengths and enrichments. Throwing some caution to the wind, I think the formal calculation we’re after is that we want to produce an $R$-scheme map

$hom(P, Q) \times F_R(P) \to F_R(Q)$

(even if schemes aren’t cartesian closed, I think maybe $F_R(P)$ is exponentiable – something to check later). If luck is with us, the strength above would give us an $R$-scheme map

$hom(P, Q) \times F_R(P) \to F_R(hom(P, Q) \times P)$

which we then compose with

$F_R(ev): F_R(hom(P, Q) \times P) \to F_R(Q)$

to get what we want.

I’m pretty confused about aspects of this; it might not bear scrutiny in the light of day.

Posted by: Todd Trimble on October 28, 2009 9:16 AM | Permalink | Reply to this

### Re: Schur Functors

Let me ask a question about something that was being said back here:

First, to me, these Schur functors were always the Schur functors $S_\lambda$ attached to Young diagrams, and I didn’t know that every endofunctor on $Vect_{fd}$ (finite-dimensional vector spaces over $\mathbb{C}$; let me stick to the classical case for a minute) was a direct sum of $S_\lambda$’s.

I’d like to understand that a bit better. First: is it literally every endofunctor? At first glance, that’s surprising to me.

Yeah, it’s surprising that you can classify them so neatly, but I seem to recall that it’s a well-known fact.

Is there a nice way of seeing that? Is it that every endofunctor is a finite direct sum of irreducibles?

Let’s see. Either I don’t understand this situation (quite possible) or your guess is right: the endofunctor category $[FinVect_{\mathbb{C}};,FinVect_{\mathbb{C}}]$ is an abelian category where every object is a finite direct sum of simple objects.

I’m a little worried about the statement that every endofunctor

$F: Vect_{fd} \to Vect_{fd}$

that is algebraic (in James’s sense) must be a finite direct sum of Schur functors $S_\lambda$.

Here’s an example: let $F$ be the total space of the exterior algebra:

$F(V) = \Lambda(V) = \sum_{n \geq 0} \Lambda^n(V)$

Then $F(V)$ is finite-dimensional for each finite-dimensional $V$, and so defines an endofunctor on $Vect_{fd}$. Unless I’m missing a subtlety, it seems to me that $F$ is algebraic in James’s sense, e.g.,

$hom(R^m, R^n) \to hom(F(R^m), F(R^n))$

(say for $R = \mathbb{C}$) is defined by polynomial mappings for each pair $m, n$. But $F$ seems to be an infinite sum of Schur functors $\Lambda^n$. Am I missing something?

Posted by: Todd Trimble on October 28, 2009 8:47 PM | Permalink | Reply to this

### Re: Schur Functors

I wish you were missing something, but I can’t see what it would be.

But if this is as bad as it gets for endofunctors of $FinVect_{\mathbb{Q}}$, I won’t be too upset. We should try to understand all these endofunctors, just because it seems like an interesting question. My fallback position, obviously, is that they’re infinite direct sums of Schur functors any one of which vanishes for vector spaces of sufficiently high dimension!

Can you think of a symmetric monoidal abelian category $C$ for which this endofunctor $X \mapsto \bigoplus_n \Lambda^n X$ is ill-defined because the infinite direct sum fails to exist? If so, the ‘polymorphic’ approach to Schur functors might rule out this pathology.

By the way, on another front: I’ve been meaning to post something about why I hope nonalgebraic representations of $GL(n,\mathbb{Q})$ don’t arise from endofunctors of $FinVect_{\mathbb{Q}}$. But I need time to do a few more calculations.

Posted by: John Baez on October 28, 2009 9:44 PM | Permalink | Reply to this

### Re: Schur Functors

John wrote:

Can you think of a symmetric monoidal abelian category C for which this endofunctor $X \mapsto \bigoplus_n \Lambda^n X$ is ill-defined because the infinite direct sum fails to exist?

Oh, duh. Just use the category of finite-dimensional super-vector spaces and take $X$ to be a purely odd super-vector space. Then $\bigoplus_n \Lambda^n X$ is really the symmetric algebra of $X$, and this will typically be infinite-dimensional — hence not in our category.

Posted by: John Baez on October 28, 2009 10:02 PM | Permalink | Reply to this

### Re: Schur Functors

Okay, good. I am reasonably confident that the same kind of trick could be used to handle any potentially infinite direct sum, so I think we’re safe for the time being. :-)

Posted by: Todd Trimble on October 30, 2009 5:20 PM | Permalink | Reply to this
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### Re: Schur Functors

On the $n$Lab page on Schur functors, it says “Recall that the group algebra $\mathbb{Q}[S_n]$ decomposes as a direct sum $\sum V_\lambda$ where $\lambda$ ranges over isomorphism classes of partitions”. I think this is a bit misleading — surely each isomorphism class should be counted once for each possible numbering, as I asked about in my earlier comment.

Also, I’m confused by the phrase “$\mathbb{Q}[S_n]$ acts diagonally on the object $V_\lambda \otimes X ^{\otimes n}$”. Is the idea that we’ve identified $V_\lambda$ as an object in the category? How?

And what’s this “acts diagonally” business? We’re taking the product of two representations — does it just mean $\mathbb{Q}[S_n]$ acts in the obvious way, taking the tensor products of the two individual actions?

Posted by: Jamie Vicary on October 26, 2009 11:48 PM | Permalink | Reply to this

### Re: Schur Functors

Yes, the way the decomposition was written was a slip on my part. There are various ways to fix it: either by Young tableaux as you suggest, or by expressing $\mathbb{Q}[S_n]$ as a sum of matrix algebras:

$\mathbb{Q}[S_n] \cong \sum_{\lambda} hom(V_\lambda, V_\lambda)$

I think I’ll mention both.

Yes, in the set-up as I was originally writing it, I was employing a mild abuse of notation: $V_\lambda$ could denote either a rational vector space representation of $S_n$, i.e., a functor

$V_{\lambda}: S_n \to Vect_{\mathbb{Q}}$

(the category on the right is supposed to be finite-dimensional vector spaces), or it could denote a representation of $S_n$ in a general symmetric monoidal additive category $C$ with rational vector space homs, via a (symmetric monoidal) change of base functor

$Vect_{\mathbb{Q}} \to C: \mathbb{Q}^n \mapsto I^n$

($I$ the monoidal unit of $C$) applied to $V_{\lambda}$ as given above. Unless someone erased it, this was discussed in the draft article.

Yes, I mean that $\mathbb{Q}[S_n]$ acts in the obvious way on the tensor product. The “acts diagonally business” just referred to the fact that the category of modules over a bialgebra such as $\mathbb{Q}[S_n]$ is a monoidal category; here the comultiplication

$\mathbb{Q}[S_n] \to \mathbb{Q}[S_n] \otimes \mathbb{Q}[S_n]$

comes from the diagonal map

$S_n \to S_n \times S_n.$

In other words, you’re essentially taking the tensor product of the two individual actions as you said, and then pulling back along the comultiplication.

Posted by: Todd Trimble on October 28, 2009 3:13 AM | Permalink | Reply to this

### Re: Schur Functors

I’ve already gone over to the $n$Lab and fixed these problems in a minimal sort of way, and made a few other changes too. Of course Todd (and maybe other people) can improve things further.

Posted by: John Baez on October 28, 2009 3:23 AM | Permalink | Reply to this

### Re: Schur Functors

Todd, thanks very much! That’s very clear now, and makes a lot of sense to me.

I now realist that the reason I was banging on so much about Young tableaux was that I wanted to see the decomposition $X^n \simeq \bigoplus_{\lambda'} V_{\lambda'}$ being directly relevant for the Schur functors — for example, if $F$ is a Schur functor on a category C corresponding to a Young diagram for $S_3$, then for each object $X$ in C, I wanted the object $F(X)$ to come equipped with a canonical embedding into $X\otimes X \otimes X$. I now see that this isn’t necessary, and isn’t the case.

In particular, this won’t come out from the coequalizer that you’re using, as it doesn’t directly give rise to a subobject of $X ^{\otimes n}$, but rather of $V_\lambda \otimes X^{\otimes n}$.

Posted by: Jamie Vicary on October 28, 2009 11:09 AM | Permalink | Reply to this

### Re: Schur Functors

Here’s another Schur functor question. In sufficiently nice situations, for any faithful representation $R$ of a group $\mathbf{G}$, every representation of $\mathbf{G}$ is a subrepresentation of $R ^{\otimes n}$ for some integer $n$.

Some of these subrepresentations will be given by Schur functors applied to $R$. Can you, in fact, construct any representation by applying Schur functors to a given faithful representation? Let’s say you’re also allowed to have access to the conjugate representation $\overline{R}$, so you can form representations like $R \otimes \overline{R} \oplus 1$ and $\Lambda^2 (R \otimes R \otimes \overline{R})$.

Let’s say the groups are finite, we’re working over the complex numbers, and all representations are unitary and finite-dimensional. It’s difficult to play around with Schur functors to test these things out, but I’m pretty sure that there aren’t any completely obvious counterexamples to this suggestion. At the very least, the representations obtainable in this way from $(R,\overline{R})$ must form an interesting class.

Posted by: Jamie Vicary on November 30, 2009 1:28 PM | Permalink | Reply to this

### Re: Schur Functors

I’ve been making too many mistakes on this thread to want to fire off guesses… but your question is too interesting to resist saying anything about it!

In sufficiently nice situations, for any faithful representation $R$ of a group $G$, every representation of $G$ is a subrepresentation of $R^{\otimes n}$ for some integer $n$.

Can you remind me what counts as ‘sufficiently nice’? Maybe a compact Lie group, in the context where ‘representation’ means ‘finite-dimensional complex representation’?

Some of these subrepresentations will be given by Schur functors applied to $R$. Can you, in fact, construct any representation by applying Schur functors to a given faithful representation?

It would be fun to try to disprove this. For that, you might try the group $GL(n,\mathbb{F}_q)$ where $q$ is a prime power. There’s a lot of interesting material on this group in Daniel Bump’s book Lie Groups. As $q\to 1$ the representation theory of this group more and more resembles that of the symmetric group $S_n$. Indeed, every irrep of $S_n$ corresponds (in some way) to a certain ‘bland’ irrep of $GL(n,\mathbb{F}_q)$. But as $q \to \infty$ the number of irreps of $GL(n,\mathbb{F}_q)$ keeps increasing. The most exciting new ones are called the ‘cuspidal’ ones. Maybe these can’t be built using Schur functors from the bland irreps and their duals.

On the other hand, it would be fun to try to prove your statement by starting with a faithful irrep $R$ and its dual $\overline{R}$ and building up a subcategory of reps by tensoring, direct sums, and Schur functors. Say the only morphisms we allow are the ‘obvious’ ones associated to tensor products, direct sums and Schur functors. If this subcategory meets the conditions of the Doplicher-Roberts theorem then it’s the category of representations of some group. And how is this group related to our original group?

Just some vague thoughts. Your question is almost well-formed enough to throw at mathoverflow! But anyway, you can’t lose by looking at Daniel Bump’s book. The answer to your question won’t be sitting there staring you in the face — but it’s a great book.

Posted by: John Baez on November 30, 2009 5:40 PM | Permalink | Reply to this

### Re: Schur Functors

Can you remind me what counts as ‘sufficiently nice’? Maybe a compact Lie group, in the context where ‘representation’ means ‘finite-dimensional complex representation’?

I wish I knew. Unfortunately, I don’t know how to prove this important result, even in the case of finite groups. Surely it seems likely for the case you mention.

It would be fun to try to disprove this. For that, you might try the group $GL(n,F_q)$ where $q$ is a prime power.

Thanks for this suggestion! (Un)fortunately I’m most interested in this question in the finite group case, so I am hoping to be able to get away without such cutting-edge representation theory. I will certainly have a look at the book, I need as many good group theory books as I can get my hands on.

On the other hand, it would be fun to try to prove your statement by starting with a faithful irrep R and its dual R¯ and building up a subcategory of reps by tensoring, direct sums, and Schur functors … If this subcategory meets the conditions of the Doplicher-Roberts theorem then it’s the category of representations of some group. And how is this group related to our original group?

By cheating a little bit, you can use the Doplicher-Roberts duality, along with some of your HDA2 results, to ‘prove’ the answer to this. Forget the condition that $R$ is faithful: then the generated category is the category of representations of of the image of the representation $R$, seen as a group homomorphism $R:G \to U(n)$.

You probably see it already, but here’s why. The objects in $\mathrm{Rep}(G)$ that we can build from $R$, $R^*$ and Schur functors are exactly the objects in the image of $\mathrm{Rep}(R):\mathrm{Rep}(U(n))\to\mathrm{Rep}(G)$, up to isomorphism, where Rep is now standing for the Doplicher-Roberts contravariant equivalence between the 2-category of groupoids, and the 2-category of symmetric even 2-H*-algebras.

If the objects in the image of $\mathrm{Rep}(R)$ form a nice symmetric 2-H*-algebra, then this should certainly embed nicely into $\mathrm{Rep}(G)$. So we should get a pair of symmetric monoidal functors $\mathrm{Rep}(U(n)) \to \mathrm{Rep}(?) \to \mathrm{Rep}(G)$ which compose to give $\mathrm{Rep}(R)$, where the first is onto and the second into. Under Doplicher-Roberts equivalence, this is the same as a factorization $G \to ? \to U(n)$ where again, the first is onto and the second into. By thinking about some edge cases, it’s pretty clear that the only candidate for this is $?=\mathrm{Im}(R)$, the image of the representation.

This works out for all the simple cases I can actually calculate (trivial cases, and less trivial cases where $G=S_3$.) It took me a long time to get around to this point of view!

It’s cheating because it’s not a proof; it’s just a different perspective from which it seems very clear that there’s only one possible answer. And note that it doesn’t actually establish that, for the case that $R$ is faithful, we can hit every representation of $G$ by applying $R$, $R^*$, Schur functors and categorical operations. But it does let us reformulate this question: if $R:G \to U(n)$ is a faithful representation of a finite group, then does every representation of $G$ arise by restricting some representation of $U(n)$? Surely someone out there knows the answer!

If this subcategory meets the conditions of the Doplicher-Roberts theorem then it’s the category of representations of some group.

The problem with this is that there’s no natural monoidal structure to put on the subcategory that gets generated. More generally, there’s no theorem that I can prove saying that a monoidal functor $F:C \to D$ between abelian monoidal categories has an onto-into factorization. I’m not even sure what the appropriately weak definitions of ‘onto’ and ‘into’ should be. It’s pretty surprising to me that this seems so difficult, since the equivalent proof for ordinary algebras is very straightforward, so maybe I’m missing something obvious. This is a very general question that many category theorists might have thought about — any ideas out there?

But anyway, you can’t lose by looking at Daniel Bump’s book. The answer to your question won’t be sitting there staring you in the face — but it’s a great book.

Thanks! Being a physicist, I am constantly stymied by my lack of basic knowledge about these things. I’ll check it out.

Posted by: Jamie Vicary on November 30, 2009 10:39 PM | Permalink | Reply to this

### Re: Schur Functors

John wrote:

It would be fun to try to disprove this. For that, you might try the group $GL(n,F_q)$ where $q$ is a prime power.

Jamie wrote:

Thanks for this suggestion! (Un)fortunately I’m most interested in this question in the finite group case, so I am hoping to be able to get away without such cutting-edge representation theory.

Well, $GL(n,F_q)$ is a finite group: invertible $n \times n$ matrices with coefficients in the field with $q$ elements. And it’s the simplest example where I have a feeling I might be able show that not every irrep is built from some obvious faithful rep by Schur functors and taking duals. The reason is that its representation theory is very much like that of $S_n$, but ‘bigger’.

Gotta run… good luck!

Posted by: John Baez on November 30, 2009 11:30 PM | Permalink | Reply to this

### Re: Schur Functors

Oh God, sorry, obviously it’s finite.

Posted by: Jamie Vicary on December 1, 2009 9:48 AM | Permalink | Reply to this

### Re: Schur Functors

John wrote:

If this subcategory meets the conditions of the Doplicher-Roberts theorem then it’s the category of representations of some group.

Jamie wrote:

The problem with this is that there’s no natural monoidal structure to put on the subcategory that gets generated.

Really? I thought you were letting us take tensor products — at least of objects, and I assumed of morphisms.

I thought the problem was that you weren’t letting us take cokernels of arbitary morphisms.

Once you give me a faithful representation $R$ of $G$, I’m willing to believe every irreducible representation of $G$ sits inside some tensor product of copies of $R$ and $R^*$, or maybe even just $R$. I used to know the precise theorem.

But how do we ‘extract’ these irreducible representations? We can do it if we can take arbitrary cokernels (or kernels, if you prefer). And the ability to do this is one of the assumptions built into the hypotheses of the Tannaka–Krein and Doplicher–Roberts theorems.

But I thought the point of your question was that you were denying me the ability to take arbitrary cokernels of morphisms. I thought you were only letting me take those cokernels given to me by Schur functors.

(A Schur functor coming from an $n$-box Young diagram sends a representation $R$ to the cokernel of a specific morphism $R^{\otimes n} \to R^{\otimes n}$.)

Posted by: John Baez on December 1, 2009 9:07 PM | Permalink | Reply to this

### Re: Schur Functors

I agree with everything you’re saying; I am indeed only allowing cokernels corresponding to Schur functors. Saying there wasn’t a good monoidal structure was misleading, sorry.

Here’s the problem as I see it. Let a ‘generalized Schur functor’ be a gadget which turns representations into new ones by applying conjugation, tensor product and ordinary Schur functors a finite number of times. Let’s say that the images of any particular representation under the family of all these generalized Schur functors — along with morphisms given by the associators, unitors, multiplication by complex numbers, and injections and projections into the subspaces defined by these generalized Schur functors — generate a well-behaved monoidal category to which the Doplicher-Roberts theorem applies; I think this is reasonable.

The problem lies in proving that this isn’t an evil construction. Let $S$ and $T$ be two different generalized Schur functors on $\mathrm{Rep}(G)$, and $R$ a representation of $G$ such that $S(R) \simeq T(R)$ in $\mathrm{Rep}(G)$. Are $S(R)$ and $T(R)$ also isomorphic in the generated category? It seems difficult to prove that they would be, although you might guess that they are. But now pass to a monoidally-equivalent skeletalization of $\mathrm{Rep}(G)$ before you start: then $S(R)$ and $T(R)$ are certainly isomorphic in the generated category, because they’re equal!

Like all evilness problems, it’s sort of funny to think about — but it really gets in the way of actually using this construction for anything!

Posted by: Jamie Vicary on December 2, 2009 6:16 PM | Permalink | Reply to this

### Re: Schur Functors

The theorem is “the representations of any compact group over a field of characteristic 0 is generated by any faithful representation.”

The point is that if $\chi$ is a character, the absolute value of its value at an non-central element is either less than dim V, or a root of unity times it (and this only happens at a central element). As one considers higher and higher powers of $V$, the value of $\chi^m$ at any non-central point becomes small compared to $\chi(m)^1$ (if you can non-discrete compact groups, you have to look at small neighborhoods of central points), so eventually, the inner product of any irreducible rep $W$ with $\chi^m$ gets close enough to the inner product of the restriction to the center to guarantee the inner product is non-zero if its restriction to the center is.

Thus, this reduces to the abelian case, which follows from the fact that a representation is faithful if and only if its summands generate the Pontryagin dual.

Posted by: Ben Webster on December 2, 2009 12:00 AM | Permalink | Reply to this

### Re: Schur Functors

Ben wrote:

The theorem is “the representations of any compact group over a field of characteristic 0 is generated by any faithful representation.”

Since you’re talking about ‘compact’ groups, that makes me guess you’re talking about continuous representations, which in turn makes me want to equip my field with a topology…

But I was perfectly happy just taking the field to be $\mathbb{C}$.

I guess the theorem I was thinking about goes sort of like this: suppose $G$ is a compact Hausdorff group, and by a ‘representation’ of $G$ let’s mean a finite-dimensional continuous complex representation. Then if $R$ is a faithful representation, every irreducible representation is a subrepresentation of $R^{\otimes n} \otimes (R^*)^{\otimes m}$ for some $n, m \ge 0$.

And I guess the proof I once knew goes something like this. Since $R$ is faithful, its matrix entries are continuous functions on $G$ that separate points. Linear combinations of matrix entries of the representations $R^{\otimes n} \otimes (R^*)^{\otimes m}$ form a $*$-algebra of continuous functions on $G$. By Stone-Weierstrass, this $*$-algebra is dense in the continuous functions on $G$. If some irrep of $G$ did not appear as a subrepresentation of any of these reps $R^{\otimes n} \otimes (R^*)^{\otimes m}$, its matrix entries would be orthogonal to everything in this $*$-algebra. But that’s a contradiction, since being dense in the continuous functions on our compact group $G$ implies being dense in $L^2(G)$.

I’d forgotten whether we needed to include tensor powers of $R^*$ in this game, but we clearly do.

If $R$ is the obvious representation of $\U(1)$, it’s faithful, but $R^*$ doesn’t appear in any of the tensor powers $R^{\otimes n}$ ($n \ge 0$). The point is that their matrix entries form an algebra of continuous functions on $\U(1)$ that seperate points… but not a $*$-algebra, so Stone-Weierstrass doesn’t apply!

Posted by: John Baez on December 2, 2009 6:26 AM | Permalink | Reply to this

### Re: Schur Functors

Wow! I never knew about the Stone-Weierstrass theorem before, it’s a good ‘un.

Posted by: Jamie Vicary on December 2, 2009 6:25 PM | Permalink | Reply to this

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