In the context of algebraic (equivariant) K-theory (more specifically, in the context of Chriss and Ginzburg's book representation theory and complex geometry) I would like to know if I have the correct intuition for the following.

Let $N$ be a smooth variety and $\pi:M\to N$ a vector bundle (or really $M$ a smooth variety and $\pi$ any flat map). Let $\mathcal{F},\mathcal{G}$ be two locally free sheaves on $N$, and write $[\mathcal{F}]$ for the element represented by $\mathcal{F}$ in the (zero-th) K-group. Is it true that $\pi^*([\mathcal{F}]\otimes[\mathcal{G}]) = [\pi^* (\mathcal{F})]\otimes[\pi^* (\mathcal{G})]$ in $K(M)$, the zero-th K-group?

To what extent is this true, provided that we must keep $M,N$ smooth so that the tensor products are defined? Specifically is it true provided only one of the sheaves is locally free?

EDIT: We must be careful here, as the definition of tensor product of modules isn't quite sufficient to define tensor product in K-theory. Perhaps we should expand the question to a question about the notion of pull back in $K$-theory.

Essentially we build the tensor product of modules in 2 steps. Fist we have the exact construction, $\mathcal{F},\mathcal{G}$ modules over $X,Y$ reps., then we can form the outer tensor product of $\mathcal{F},\mathcal{G}$ on $X\times Y$. To do this let $p_i$ be the $i$-th projection and set, $$ \mathcal{F}\boxtimes \mathcal{G} = p_1^*\mathcal{F}\otimes_{\mathcal{O}_{X\times Y}}p_2^*\mathcal{G} $$ As the projection is a flat map the pullback functor $p_i^*$ is exact, and hence the above formula gives rise to a well defined map $K(X)\times K(Y)\to K(X\times Y)$.

In the case $X=Y$ we also have the diagonal embedding, $\Delta:X_\Delta\to X\times X$, and hence we can pull back the outer product of two sheaves to the diagonal, $\mathcal{F}\otimes\mathcal{G}:=?\Delta^*(p_1^*\mathcal{F}\otimes_{\mathcal{O}_{X\times Y}}p_2^*\mathcal{G})$.

BUT $\Delta^*$ need not be exact! In fact it is only a closed immersion even if we assume $X$ is smooth, so even then the diagonal map is not flat! To make matters worse, if $X$ is singular the higher derived functors $R^i(\Delta^*)$ need not vanish for $i>>0$, so the typical trick of taking as the definition of $\Delta^*$ an alternating sum of the higher derived functors doesn't work.

Thus Chriss and Ginzburg's definition of the pull back under a closed immersion is somewhat ad hoc. Perhaps the better question to ask here is for someone do describe the geometry of this construction, and give some intuition for it in simple cases. (i.e. does transversality simplify things some?)

Their definition goes as follows. Let $f:Y\hookrightarrow X$ be a closed immersion of smooth quasi-projective varieties. For any coherent sheaf $\mathcal{F}$ on $X$ there is a finite locally free resolution $F^\bullet$ of $\mathcal{F}$ (Hilbert's syzygy theorem), $$ \cdots \to F^1\to F^0 \to \mathcal{F}\to 0. $$ (For me, the following is the strange part) The sheaf $f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^i$ is a coherent sheaf of $\mathcal{O}_Y \cong f_*(\mathcal{O}_Y)$-modules (i.e. it is annihilated by the ideal $\mathcal{I}_Y$). Thus, the homology sheaves of the complex, $$ \cdots \to f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^1\to f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^0\to 0 $$

are coherent sheaves of $\mathcal{O}_Y$-modules. Finally(!) we define, $$ f^*([\mathcal{F}]) = \sum (-1)^i[\mathcal{H}_i(f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^\bullet)] $$

It seems clear (but I have a hard time trusting) that for $\mathcal{F}$ locally free we have an equality with the old notion of pullback, $$ f^*([\mathcal{F}]) = [\mathcal{O}_Y\otimes_{f^{-1}(\mathcal{O}_X)}f^{-1}(\mathcal{F})], $$ so the intuition for the the pull back of tensor products problem mentioned at the beginning is that we can simply use the composition of pull backs is the pull back of the composition for locally free sheaves. Is this the right intuition?

In practice the resolution construction might be overkill. Suppose $i:Z\hookrightarrow X$ is another closed immersion of smooth quasi-projective varieties, and suppose that $Z$ and $Y$ intersect transversally. Then can we say something without using resolutions about the pullback of the structure sheaf $f^*([i_*\mathcal{O}_Z])$ to $Y$? (i.e. is it the obvious thing, $[\mathcal{O}_{Y\cap Z}]$?)

  • $\begingroup$ Why should tensor products require smoothness? $\endgroup$ – Will Sawin Jun 15 '12 at 0:53
  • $\begingroup$ What's $\mathcal H_i$? $\endgroup$ – Will Sawin Jun 15 '12 at 20:52
  • $\begingroup$ In the formula above, $\mathcal{H}_i(f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^\bullet)$ stands for the $i$-th homology group of the complex. Explicitly, $$ \mathcal{H}_i(f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^\bullet) = ker(f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^i\to f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^{i-1})/image(f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^{i-1}\to f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^i) $$ $\endgroup$ – Rob Denomme Jun 15 '12 at 21:27
  • $\begingroup$ oops, make that second $F^{i-1}$ an $F^{i+1}$ $\endgroup$ – Rob Denomme Jun 15 '12 at 21:37
  • $\begingroup$ @Rob, I'm not sure my response completely answers your question, but this is partly because it's a little hard to parse what you're asking. Could you tighten up the question a little to clarify? Thanks! $\endgroup$ – Dave Anderson Jun 15 '12 at 23:00

There are a couple issues here. I think the main confusion is between K-theory of vector bundles ($K^\circ X$) and K-theory of coherent sheaves ($K_\circ X$). In Chriss-Ginzburg, I think they often assume $X$ is smooth, in which case the two are isomorphic. The isomorphism comes from resolving a coherent sheaf by vector bundles, and taking the alternating sum.

The former is always a ring under tensor product, and pullback is indeed the naive one. As you say, pullback is not exact in general, but since tensor product with a locally free sheaf is exact, pullback on K-theory of vector bundles is well-defined, for any morphism and any spaces.

For coherent sheaves, there are pullbacks for some classes of morphisms. The easiest case is when $X$ and $Y$ are smooth, so that there's an isomorphism with K-theory of vector bundles, and there's a pullback for any morphism $f: X \to Y$. More generally, the intuition is that a pullback $f^*: K_\circ Y \to K_\circ X$ should be defined by $$f^*[\mathcal{F}] = \sum (-1)^i [Tor^Y_i(\mathcal{O}_X,\mathcal{F})].$$ This makes sense whenever $f$ is a perfect morphism, i.e., $\mathcal{O}_X$ has a finite resolution by flat $f^{-1}\mathcal{O}_Y$-modules, because in that case the Tor sheaves are zero all but finitely many $i$. In particular, it does work when $f$ is flat, or a regular embedding.

(A word of warning: these Tor sheaves are not computed using the pushforward $f_*\mathcal{O}_X$ in general, although that does work when $f$ is a closed embedding. The correct definition is to cover $Y$ and $X$ by affines, construct the Tor locally, and glue --- see EGA III.6.)

The answer to your last question is yes: if $Y$ and $Z$ are transversally intersecting subvarieties of a smooth variety $X$, with $Y\cap Z = W$, then $[\mathcal{O}_Y]\cdot [\mathcal{O}_Z] = [\mathcal{O}_W]$, because transversality implies vanishing of the higher Tor's.


We can say more. These are naturally isomorphic as sheaves, with no conditions on $M$ and $N$.

It is easiest to see this in the affine case, $N=\textrm{Spec} A$, $M=\textrm{Spec} B$, $\mathcal F$ and $\mathcal G$ the sheafification of the $A$-modules $F$ and $G$. Then your equation translates to

$ B \otimes_A (F \otimes_A G) = (B \otimes_A F)\otimes_B (B\otimes_A G)$

which is true because tensor product is commutative and associative and $B \otimes_B B=B$.

This gives us a local isomorphism between the two sheaves on every affine open set which clearly restrict to the same isomorphism on their intersection, meaning that they glue together into a global isomorphism of sheaves.

  • $\begingroup$ Thanks for the response, this is exactly the intuition one wants to use, pull back of composition is the composition of the pull backs. The tensor product in algebraic K-theory isn't so easily constructed due to the failure of diagonal embedding to be flat, see the discussion added above for more details, and so I wonder how well this intuition is suited to the problem after this consideration is made. $\endgroup$ – Rob Denomme Jun 15 '12 at 20:15
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    $\begingroup$ Ah. I don't know much about K-theory, so I didn't see the difficulty with exactness. I will think about the discussion and see if I can figure out the answer. $\endgroup$ – Will Sawin Jun 15 '12 at 20:51

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