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, $[\mathcal{O}_{Y\cap Z}]$?)

`$\mathcal{H}_i(f_*(\mathcal{O}_Y)\otimes_{\mathcal{O}_X}F^\bullet)$`

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`$F^{i-1}$`

an`$F^{i+1}$`

$\endgroup$ – Rob Denomme Jun 15 '12 at 21:37