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The kind of question I'm interested in has the following flavor: having two moduli stacks with one being an enhanced (i.e more data) version of the other with the natural "forgetting" map between them can one describe the pushforward (when it makes sense) of the universal object? In general, the pullback of the universal object is the universal object, but I couldn't find anything about the pushforward.

Here are the examples that led me to this question (the pushforward is derived):

I'm considering moduli stacks of the type $BGL_n$ or $Bun_n(X)$ where $X$ is some variety. These stacks all come with some universal object (bundle, vector bundle, coherent sheaf, etc.) on them (or crossed with X).

For example, consider $BGL_n$ and pick a parabolic $P_{n-d,d}$ in $GL_n$. We have a canonical map $p:BP_{n-d,d}\to BGL_n$.

The stack $BGL_n$ has a universal vector bundle $\mathcal{V}_n$ of rank $n$ (I identify vector bundles with $GL_n$-bundles through the natural representation of $GL_n$). Also the stack $BP_{n-d,d}$ has a universal object which is a vector bundle $\mathcal{W}_n$ equipped with a subvector bundle $\mathcal{W}_{n-d}\subseteq\mathcal{W}_n$.

We know that to give a map from a stack to $BGL_n$ is the same as to give a vector bundle on the stack. The map $p:BP_{n-d,d}\to BGL_n$ is given precisely by $\mathcal{W}_n$ and $\mathcal{W}_n=p^*(\mathcal{V}_n)$.

My question is: can we describe $p_*(\mathcal{W}_n)$ or $p_*(c_i(\mathcal{W}_n))$ ($c_i$ Chern classes) in a nice way? I'm happy if it's in the Grothendieck group. (I know about Riemann-Roch but it doesn't seem to help here).

More generally, if one has a map from $Y$ (a variet/stack) into $BGL_n$ how can one describe the pushforward of the corresponding vector bundle? (or it's Chern classes).

The same question can be asked about morphisms of the type $Bun_{n-d,d}(X)\to Bun_n$ or $Coh_{n-d,d}(X) \to Coh_n(X)$.

Do you know of any similar situation when this kind of question is considered or any relevant computations?

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  • $\begingroup$ As Qiaochu says, the pushforward is the induction from $P$ to $G$, which can be written as the space of sections of the vector bundle $G\times_P \mathbf{C}^n\rightarrow G/P$ (or as a cotensor product $\mathbf{C}^n\otimes^{\mathcal{O}(P)}\mathcal{O}(G)$). For instance, for $G=SL_2$ and $P=B$ the vector bundle is, I think, the trivial rank 2 bundle, so you get the trivial rank 2 representation. $\endgroup$ – Pavel Safronov Feb 4 '15 at 13:38
  • $\begingroup$ Pavel, I'm a bit confused: shouldn't it be the natural representation of $SL_2$ ? Also, when having an equivariant vector bundle, even if it's trivial (when one forgets the equivariant structure) the global sections could be a non-trivial representation. $\endgroup$ – Dragos Fratila Feb 5 '15 at 13:48
  • $\begingroup$ You're totally right on both counts. The bundle is trivial rank 2 and the representation is the defining 2-dimensional one. $\endgroup$ – Pavel Safronov Feb 5 '15 at 16:26
  • $\begingroup$ In this case isn't $p_*(W_n)=p_*(p^*V_n)$ just isomorphic to $V_n$ itself? I'm not sure there is a projection formula for Artin stacks that doesn't involve derived categories, but in this case the map is representable and flat.. $\endgroup$ – Mattia Talpo Feb 12 '15 at 23:54
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This is long and may not at all address your question, but here's a claim I'd like to defend:

taking pullbacks of universal things is an abstract-nonsense operation, but taking pushforwards is not.

So you can hope for abstract-nonsense answers to what pullbacks of universal things are, but my claim is you shouldn't hope for abstract-nonsense answers to what pushforwards of universal things are.

The starting point is the observation that we can make sense of the abstract-nonsense yoga of classifying spaces and pullbacks in a very wide variety of contexts (e.g. algebraic geometry and algebraic topology). In any category of "spaces," whatever "spaces" means to you, any reasonable notion of "families of foos over a space $X$" should at the very least be a contravariant functor $F$ in $X$; you should be able to restrict families to subspaces, take the stalk of a family at a point, and so forth.

Let's not worry too much about the target of this functor; the point is that if you believe in the Yoneda embedding hard enough, any contravariant functor $F$ on a category of "spaces" is already morally a "generalized space" (the classifying space of foos). In particular, if $F$ is representable, so that $F(X) \cong [X, M]$ for some actual "space" $M$, then $M$ is the "moduli / classifying space" of foos and pullback of families of $Y$-objects is just precomposition and everything is as nice as possible.

Okay, so what about pushforwards? One intuition is that pushforwards involve "integration over fibers," and in particular they require a choice of "measure" to make sense. That is in general not something we get for free. Consider the following terrible situation: let our category of "spaces" be sets, and let "families of foos" be real-valued functions

$$F(X) = \text{Hom}(X, \mathbb{R}).$$

Pulling back real-valued functions is a formal operation; it doesn't even require any extra structure on $\mathbb{R}$. What does it mean to push forward real-valued functions? If a map $f : X \to Y$ has finite fibers, then a pretty good candidate for the pushforward of a real-valued function $r : X \to \mathbb{R}$ might be

$$f_{\ast}(r)(y) = \sum_{x \in f^{-1}(y)} f(x).$$

This is just "integration over fibers" with respect to the counting measure. But if $f$ doesn't have finite fibers then we're just stuck unless someone gives us a family of measures on each fiber, and even then there are restrictions on what kinds of functions we can push forward.

Let's switch to something that looks more like the examples in the OP. One big difference between algebraic topology and algebraic geometry is that in algebraic topology there is in general no reasonable way to take the pushforward of a vector bundle and get another vector bundle. We might settle for a way to take the pushforward of a $K$-theory class and get another $K$-theory class. In general, if $f : X \to Y$ is a map, we might ask under what conditions we get a reasonable pushforward map

$$f_{\ast} : E^{\bullet}(X) \to E^{\bullet \pm ?}(Y)$$

in some cohomology theory $E$, having in mind the special case of the pushforward map available for a map between closed oriented manifolds that we get from Poincare duality.

There is a general machine that tells us the extra structure that $f$ needs to be equipped with, which is called an $E$-orientation; for example, if $E$ is integral cohomology $H \mathbb{Z}$ then, roughly speaking, $f$ needs to be a map whose fibers are closed oriented manifolds, and "integration over fibers" is given by fiberwise pairing with the fundamental class. The choice of "measure" here is a fiberwise choice of orientations.

If $E$ is real resp. complex $K$-theory, then, roughly speaking, $f$ needs to be a map whose fibers are closed spun (resp. complex spun) (by "spun" I mean equipped with a choice of spin structure, as opposed to "spinnable" which means there exists a spin structure) manifolds, and "integration over fibers" has a description in terms of fiberwise Dirac operators, which give fundamental classes in real resp. complex K-theory. The choice of "measure" here is a fiberwise choice of spin structures.

Let's look in particular at the special case that $f : X \to 1$ is the canonical map to a point. The pullback along this map is a formal operation: given a vector space (or any kind of object really) it returns the constant vector bundle (or constant family of objects) over $X$ with fiber that vector space. But if $X$ is a closed spun resp. complex spun manifold, then pushforward to a point in real resp. complex K-theory recovers the $\hat{A}$-genus resp. the Todd genus of $X$. That's not abstract nonsense!

I can't resist making one last remark. Suppose our families-of-foos functor $F$ takes values, not in sets or abelian groups, but in categories. Then if $f : X \to Y$ induces a pullback functor

$$f^{\ast} : F(Y) \to F(X)$$

we can ask if the pullback functor has left or right adjoints. That certainly seems like an abstract-nonsense way to arrive at pushforwards. But my claim is that neither the left nor the right adjoint will behave how you expect a pushforward to unless you have some extra structure, such as a compatible family of isomorphisms between the left and right adjoints. (I don't understand this as well as I'd like to, and in particular I don't understand the significance of Wirthmüller contexts vs. Grothendieck contexts here.) This extra structure is the choice of "measure."

For example, if $f : H \to G$ is the inclusion of a subgroup $H$ into a group $G$, then the pullback functor

$$f^{\ast} : \text{Rep}(G) \to \text{Rep}(H)$$

on linear representations has a left and a right adjoint, induction and coinduction. They are naturally isomorphic iff $H$ has finite index in $G$, and I believe that the choice of such a natural isomorphism amounts to the choice of a measure of some kind on $G/H$. Here it might be useful to think of $\text{Rep}(G)$ as $\text{QCoh}(BG)$ and of $G/H$ as the fiber of the map $BH \to BG$.

Sometimes, if you're lucky, there are canonical "measures" (e.g. smooth projective complex varieties are automatically endowed with orientations and in fact are automatically endowed with complex spin structures). But this shouldn't be the general situation.

Anyway, let me try to get all the way back to your question. If $H \to G$ is a nice inclusion of algebraic groups you like and you'd like to push vector bundles and so forth along the map $BH \to BG$ of stacks, my guess is that figuring out what this pushforward does should be at least as hard as computing the Todd genus of the fiber $G/H$.

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  • $\begingroup$ Interesting comment Qiaochu. Thank you! However, as you say in the last sentence, I'm happy to assume that we know the cohomology of the spaces involved. Also, I'm not necessarily hoping to get the most general situation possible of pushforward, but I was thinking that given the "canonicity" (or naturality) of the examples that I mentioned, there should exist some nice formula (again, I'm happy to assume the cohomologies known, the Todd class, cotangent complex [I think they can be calculated without much trouble in the examples that I mentioned]). $\endgroup$ – Dragos Fratila Feb 4 '15 at 22:19

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