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 $S^{\mathrm{1,1}}$. 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 $S^{\mathrm{1,1}}(V)$ 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 $S^{\mathrm{1,1}}$ 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 $V^{\otimes 2}\to S^{\mathrm{1,1}}(V)$ and let $S_{\mathrm{1,1}}(V)$ be the kernel of $V^{\otimes 2}\to S^2(V)$. $S_2$ and $S_{\mathrm{1,1}}$ 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 $ℂ$ 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, ${\mathrm{Sym}}^2$ resp. ${\mathrm{Alt}}^2$, 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

$$F:C\to C^{\prime }$$

Then we should get a square of functors

$$\begin{array}{ccc}C& \stackrel{F}{\to }& C^{\prime }\\ Y_C\downarrow & & \downarrow Y_{C^{\prime }}\\ C& \stackrel{F}{\to }& C^{\prime }\end{array}$$

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

$$F:{\mathrm{Mod}}_R\to {\mathrm{Mod}}_{R^{\prime }}$$

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!