Ubiquity of the push-pull formula - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:28:42Z http://mathoverflow.net/feeds/question/18799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula Ubiquity of the push-pull formula Andrea Ferretti 2010-03-19T22:09:08Z 2012-07-13T17:42:55Z <p>The push-pull formula appears in several different incarnations. There are, at least, the following:</p> <p>1) If $f \colon X \to Y$ is a continous map, then for sheaves $\mathcal{F}$ on $X$ and $\mathcal{G}$ on $Y$ we have <code>$f_{*} (\mathcal{F} \otimes f^{*} \mathcal{G}) \cong f_{*} (\mathcal{F}) \otimes \mathcal{G}$</code>.</p> <p>A similar formula holds for the derived functors and for $f^{!}$.</p> <p>2) If $f \colon X \to Y$ is a proper map of schemes, with $Y$ smooth, both <code>$f^{*}$</code> and <code>$f_{*}$</code> are defined on the Chow groups, and <code>$f_{*}(\alpha \cdot f^{*} \beta) = f_{*} \alpha \cdot \beta$</code> for classes <code>$\alpha \in CH^{*}(X)$</code> and <code>$\beta \in CH^{*}(X)$</code>.</p> <p>Of course a similar results holds in cohomology if $f$ is a proper map of smooth manifolds, using Gysyn map for push-forward.</p> <p>3) If $H &lt; G$ are finite groups, we have two functors <code>$\mathop{Ind}_{H}^{G}$</code> and <code>$\mathop{Res}_{H}^{G}$</code>, which can be seen as pull-back and push-forward maps between the representations rings $R(G)$ and $R(H)$. Again we have <code>$\mathop{Ind}(U \otimes \mathop{Res} V) \cong \mathop{Ind} U \otimes V$</code>.</p> <p>Edit: one more example appears in the book linked in Peter's answer. It is a bit complicated to state, but basically (if I understand well)</p> <p>4) for a compactly generated topological group $G$ and for $G$-spaces $A$ and $B$ one considers the category <code>$G \mathcal{K}_A$</code> of $G$-spaces over $A$ with equivariant maps (up to homotopy). Then for a $G$-map $f \colon A \to B$ one has functors <code>$f^{*} \colon G\mathcal{K}_B \to G\mathcal{K}_A$</code> and <code>$f_{!} \colon G\mathcal{K}_A \to G\mathcal{K}_B$</code> satisfying <code>$f_{!}(f^{*}Y \wedge_A X) \cong Y \wedge_B f_{!} X$</code>.</p> <p>There are probably several other variations which now I fail to recall. I should mention that in some situations 2) can be obtained by 1), but not always, as far as I know.</p> <blockquote> <p>Is there a unifying principle (even informal) which explains why in these diverse settings we should always have the same formula?</p> </blockquote> http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula/18807#18807 Answer by Kevin Lin for Ubiquity of the push-pull formula Kevin Lin 2010-03-19T23:10:24Z 2010-03-20T00:19:57Z <p>To connect (1) and (3), note that a representation of a group G is the same as a vector bundle over the stack pt/G.</p> <p>I am not sure, but I would guess that (some (but not all?) cases of) (2) follow from (1) by taking (appropriate) Chern classes?</p> http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula/18808#18808 Answer by darij grinberg for Ubiquity of the push-pull formula darij grinberg 2010-03-19T23:12:53Z 2010-03-19T23:12:53Z <p>The explanation for 3) is that $\mathrm{Ind}$ is some kind of tensoring from the left: $\mathrm{Ind}^G_H V\cong k\left[G\right]\otimes_{k\left[H\right]} V$ as $k\left[G\right]$-modules. Now the isomorphism $\mathrm{Ind}\left(U\otimes \mathrm{Res} V\right)\cong \mathrm{Ind}U\otimes V$ takes the more obviously-looking form $k\left[G\right]\otimes_{k\left[H\right]} \left(U\otimes V\right)\cong \left(k\left[G\right]\otimes_{k\left[H\right]} U\right)\otimes V$. Of course, this seems to trivially follow from associativity of the tensor product, but this is not that easy: the tensor product sign $\otimes$ (without $k\left[H\right]$ below) is <em>not</em> a tensor product of modules. However, we can save the approach by making $V$ a <em>right</em> $k\left[G\right]$-module by $vg=g^{-1}v$, and then applying associativity of the tensor product. We even get a stronger assertion this way: that $k\left[G\right]\otimes_{k\left[H\right]} \left(U\otimes V\right)\cong \left(k\left[G\right]\otimes_{k\left[H\right]} U\right)\otimes V$ as $\left(k\left[G\right],k\left[G\right]\right)$-bimodules.</p> <p>Unfortunately, I have absolutely no intuition for sheaves and schemes, but maybe it's this kind of argument that you should be looking for: writing the functor as a tensoring with something from the left, and applying associativity, possibly after rescuing some structure to the right which would otherwise be damaged by tensoring.</p> http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula/18809#18809 Answer by Graham Leuschke for Ubiquity of the push-pull formula Graham Leuschke 2010-03-20T00:05:14Z 2010-03-20T00:05:14Z <p>This is so transparent in commutative algebra that it doesn't even merit a name. Reversing the arrows apparently gives it some kind of status, usually called "the projection formula". Given a ring map $R \longrightarrow S$, an $R$-module $M$, and an $S$-module $N$, you can either (a) consider $N$ as an $R$-module and tensor it with $M$, getting $M \otimes_R N$, or (b) tensor $M$ up to $S$, tensor there with $N$, and consider the result as an $R$-module, getting ${}_R(M \otimes_R S \otimes_S N)$. The two are identified via the associativity of the tensor product. </p> http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula/18814#18814 Answer by Peter Arndt for Ubiquity of the push-pull formula Peter Arndt 2010-03-20T01:06:27Z 2010-03-20T03:18:37Z <p><a href="http://www.math.uiuc.edu/K-theory/0573/" rel="nofollow">This paper</a> by Fausk, Hu and May does not exactly tell you why those maps should be isomorphisms in more concrete situations, but it cleanly explains the abstract settings in which they arise - look e.g. at Propositions 2.4 and 2.8 for equivalent formulations of projection formulas.</p> <p>For an example of a projection formula that is not on the list in your question see equation 2.2.5 in <a href="http://www.math.uiuc.edu/K-theory/0716/" rel="nofollow">this book</a> by May and Sigurdsson - it is an example for the abstract "Wirthmüller context" from the paper above, which, I think, inspired the authors to do the abstract analysis in the first place.</p> http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula/18836#18836 Answer by David Corfield for Ubiquity of the push-pull formula David Corfield 2010-03-20T13:30:34Z 2010-03-20T13:30:34Z <p>Isn't this about the <a href="http://ncatlab.org/nlab/show/Beck-Chevalley+condition" rel="nofollow">Beck-Chevalley condition</a>? Some of its manifestations were discussed in this <a href="http://golem.ph.utexas.edu/category/2007/10/concrete_groups_and_axiomatic.html" rel="nofollow">thread</a>.</p> http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula/86147#86147 Answer by Drew for Ubiquity of the push-pull formula Drew 2012-01-19T23:29:53Z 2012-01-19T23:56:04Z <p>I've been reading <a href="http://math.stanford.edu/~vakil/216blog/" rel="nofollow">Ravi Vakil's Albegraic Geometry notes</a> , and he has you use the FHHF Theorem to prove the projection formula for $R^i\pi_i$ (ex 20.7.E in the Jan. 2011 version), when we are dealing with schemes and sheaves of modules. </p> <p>The FHHF theorem says that if you have a functor $F: A \rightarrow B$ and a complex $C^\bullet$ in $A$, and $H$ is right exact, you have a map $FH(C^\bullet) \rightarrow HF(C^{\bullet})$ (where $H$ means take cohomology). If instead $H$ is left exact, the map goes the other way, and if it is exact, there is an isomorphism.</p> <p>I'm not familiar with the other examples, but I would suspect that they would use the FHHF theorem as well.</p> http://mathoverflow.net/questions/18799/ubiquity-of-the-push-pull-formula/86504#86504 Answer by Ryan Reich for Ubiquity of the push-pull formula Ryan Reich 2012-01-24T01:49:30Z 2012-07-13T17:42:55Z <p>Warning! Enormous and possibly unreadable answer follows. But you should read it, since I believe it addresses your question. There's a section about halfway through where the actual answer starts.</p> <p>$\def\sh{\mathcal}\def\on{\operatorname}\def\id{\mathrm{id}}$All your examples are, as others have described them, examples of "the" projection formula. In the context of the six functors, I know a very formal way of proving this isomorphism as a consequence of base change. Suppose you have a map $f \colon X \to Y$ of "spaces" (schemes, whatever) and you want to prove that for sheaves $\sh{F}$ and $\sh{G}$ on $X$ and $Y$ respectively, there is an isomorphism <code>$$f_!(\sh{F} \otimes f^* \sh{G}) \cong f_! \sh{F} \otimes \sh{G}.$$</code> First, you rewrite it in terms of the "external tensor product" $\sh{F} \boxtimes \sh{G}$ on $X \times Y$. This combination has three important properties:</p> <ol> <li><p>If we have other maps $g \colon W \to X$ and $h \colon Z \to Y$, then on $W \times Z$: <code>$$(g \times h)^* (\sh{F} \boxtimes \sh{G}) \cong g^* \sh{F} \boxtimes h^* \sh{G}.$$</code></p></li> <li><p>Likewise, if the maps are $g \colon X \to W$ and $h \colon Y \to Z$, then on $W \times Z$ we have <code>$$(g \times h)_! (\sh{F} \boxtimes \sh{G}) \cong g_! \sh{F} \boxtimes h_! \sh{G}.$$</code></p></li> <li><p>Finally, if $X = Y$ and $\Delta_X \colon X \to X \times X$ is the diagonal map, then we have <code>$$\sh{F} \otimes \sh{G} \cong \Delta_X^* (\sh{F} \boxtimes \sh{G}).$$</code></p></li> </ol> <p>It's best to think of property 3 as defining the usual tensor product, rather than the equation <code>$$\sh{F} \boxtimes \sh{G} = \on{pr}_X^* \sh{F} \otimes \on{pr}_Y^* \sh{G}$$</code> defining the external tensor product. This is especially valid for representations: if $V$ is a representation of a group $G$ and $W$ of a group $H$, then the vector space $V \otimes W$ is naturally a representation of $G \times H$, even if $G = H$ when you are allowed to restrict to the diagonal and produce the usual tensor product representation.</p> <p>Anyway, using this, you want an isomorphism: <code>$$f_! (\id, f)^* (\sh{F} \boxtimes \sh{G}) \cong \Delta_X^* (f \times \id)_! (\sh{F} \boxtimes \sh{G}).$$</code> (The left-hand side is obtained by writing $(\id, f) = (\id \times f) \Delta_Y$ and using property 1.) Obviously, it is more productive to just remove the sheaves and prove the natural isomorphism of functors. The combination $(\id, f) \colon X \to X \times Y$ is usually called $\Gamma_f$ and is the graph of $f$; it is the base change of the diagonal map $\Delta_X$: <code>$$\begin{matrix} X &amp; \xrightarrow{\Gamma_f} &amp; X \times Y \\ {\scriptstyle f} \downarrow &amp; &amp; \downarrow{\scriptstyle\id \times f} \\ Y &amp; \xrightarrow{\Delta_X} &amp; Y \times Y \end{matrix}$$</code> Therefore, it follows (effectively from proper base change, though it does not matter that $\Delta_Y$ is proper since we are using the $!$ pushforward) that <code>$$f_! \Gamma_f^* \cong \Delta_Y^* (f \times \id)_!$$</code> and that's the projection formula.</p> <h2>Connection with the examples</h2> <p>If you want to establish a projection formula in a more general context, you need the following ingredients:</p> <ol> <li><p>Some kind of external tensor product (and, correspondingly, some kind of product of "spaces") satisfying point 3 above with respect to whatever functor you are calling "pullback".</p></li> <li><p>Points 1 and 2 above for whatever you are calling "pullback" and "pushforward".</p></li> <li><p>A base change isomorphism for fiber products of spaces.</p></li> </ol> <p>Let me address these points one-by-one for your examples.</p> <ul> <li><p>Abelian sheaves and continuous pushforward/pullback. I am embarrassingly ignorant of "easy" examples of cohomology, which is to say that I do not actually know what is true of topological spaces. However, it is implied in Brian Conrad's <a href="http://math.stanford.edu/~conrad/248BPage/handouts/basechange.pdf" rel="nofollow">notes</a> (above Proposition 3.1) that <em>any</em> continuous map of topological spaces ringed by the constant sheaf $\mathbb{Z}$ is "flat" (well, it obviously satisfies the algebraic criterion anyway) and if that is the case, then the base change map is an isomorphism because of "flat base change" rather than proper. Points 1&ndash;3 are obviously satisfied here just as in the six functors formalism, so you get your projection formula.</p></li> <li><p>Algebraic cycles and pullback/pushforward in the Chow ring. Take your "sheaves" to be closed subschemes in $X$ or $Y$, with pullback being flat base change and pushforward being proper scheme-theoretic image. The "external tensor product" is just the obvious <em>product of subschemes</em> in $X \times Y$, whose restriction to the diagonal is of course the intersection, so the formalism of point 1 is satisfied. Note that the product in the Chow ring is generically the intersection product. The formalism of point 2 is also satisfied, since you can work independently with each coordinate. As for the base-change isomorphism, it is certainly true set-theoretically, and I believe it will work scheme-theoretically also since for affine schemes, the scheme-theoretic image of a map $A \to B$ is the ring generated in $B$ by $A$, and this is stable under taking a tensor product with some extension of $A$. I admit I'm a little fatigued (I'm writing this paragraph last) so I am willing to accept criticism for being so vague.</p></li> <li><p>Finite groups with induction (= pushforward)/restriction (= pullback). Here, obviously, the product of "spaces" is the product of groups and the external product of "sheaves" is the tensor product $V \otimes W$ of representations of two groups $G$ and $H$, considered in the natural way as a representation of $G$ and $H$. Point 1 is satisfied since $G$ and $H$ just act separately on each factor. Point 2 is obvious for restriction and for induction, we use the fact that darij gave in his answer: $\def\Ind{\operatorname{Ind}}\Ind_H^G$ is tensoring up with $k[G]$, and <code>$$(V \otimes_{k[G]} k[G']) \otimes_k (W \otimes_{k[H]} k[H']) = (V \otimes_k W) \otimes_{k[G \times H]} k[G' \times H'].$$</code> So point 2 is satisfied too. The base change isomorphism is tricky; it is not actually true in general that for subgroups $H, K \subset G$ we have <code>$\Ind_H^G(V)|_K \cong \Ind_{K \cap H}^K(V|_{K \cap H})$</code>, though we do have a map (the base change map, of course): <code>$$\Ind_{K \cap H}^K(V |_{K \cap H}) = V \otimes_{k [K \cap H]} k[K] \to V \otimes_{k[H]} k[G]|_K = \Ind_H^G(V)|_K$$</code> sending elements of $K \subset G$ into $G$, which is obviously $k[K \cap H]$-linear. The two sides have dimensions (times $\dim V$), respectively, $[K : K \cap H]$ and $[G : H]$, where the former is also equal to <code>$\#(K.H)/H$</code>. So these indices are the same if and only if $G = K.H$, and this actually happens in the particular situation of the projection formula, where $K = G \subset G \times G$ and "$H$" is $G \times H \subset G \times G$. Then the base change map is an isomorphism by counting the (finite!) dimensions, and you get your projection formula again.<br/><br/> It is perhaps worth noting that the projection formula for group representations is very closely related to that for sheaves, since for any algebraic (for example, finite) group $G$, we can form the algebraic stack $*/G$, on which quasi-coherent sheaves are identified with representations of $G$. So, modulo a satisfactory theory of base change for morphisms of stacks, this example is actually sort of the same as example 1.</p></li> <li><p>Spaces with topological group actions and restriction (= pullback)/??? (= pushforward). Alas, I cannot comment on this example, since I do not understand modern (or even somewhat dated) homotopy theory. However, if I recall what I once knew about smash products, we do have for spaces $X$ and $Y$ over a space $A$ having distinguished sections from $A$ that (in sort of informal quotient notation) <code>$$X \wedge_A Y = X \times_A Y / (A \times_A Y = A, X \times_A A = A),$$</code> the "equalities" being that the $Y$-coordinate or $X$-coordinate is identified with whatever point of $A$ it is mapped to via the structure as $A$-spaces. Therefore, if we have <em>two</em> base spaces $A$ and $B$, and spaces $X$ and $Y$ over them respectively, we have an obvious "external smash product" over $A \times B$ (I apologize for the hacky symbol): <code>$$X \mathop{\fbox{\wedge}} Y = X \times Y / (A \times Y = A \times B, X \times A = A \times B)$$</code> in the same way. And if $A = B$, this base-changes correctly to the diagonal copy of $A$, so at least the formalism of point 1 is satisfied, along with the restriction part of point 2. I can't say anything about the pushforward parts, alas, but perhaps you (if you are still following this at all!) can fill in the details now.</p></li> </ul>