The n-Category Café

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.

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?

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.)

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

So somehow Mod_R 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 Mod_R, 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 |-> Mod_R, a monoidal category.
Then a Schur functor should be a (special) natural transformation Mod -> Mod. So any commutative ring R is handed a functor Mod_R -> Mod_R in a natural way.

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!

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.

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

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!

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.

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!

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?

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.

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?

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.

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.