Suppose $\mathcal{A}$ is a quasi coherent sheaf of algebras over a group scheme $\mathcal{G}$. Suppose it is generated by global section. Then , what can we say about the external tensor product $\mathcal{A}\boxtimes\mathcal{A}$? Will this sheaf also be generated by the tensor product of the global section with itself? Or is it bigger than this?
[Math] External tensor product of sheaves
tensor-products
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I am not sure whether this helps, but along the lines of the comment of Theo Johnson-Freyd, one can say the following:
If $A,B,C$ are von Neumann algebras and $Mod(A),Mod(B),Mod(C)$ their categories of representations, then there exists a natural isomorphism between morphisms $A \to B \bar{\otimes} C$ and $Mod(B)\times Mod(C) \to Mod(A)$. Here, the representation categories are viewed as categories ${\cal D}$ fibered over (equipped with forgetful functor $U_\cal D$ to) the category of Hilbert spaces, and the product ${\cal D} \times {\cal E}$ of two such fibered categories is the product category, equipped with the forgetful functor $U_{\cal D} \times U_{\cal E}$ followed by the Hilbert space tensor product functor.
This statement depends on the fact that the categories of representations have generators and extends from the spatial tensor product to fiber products of von Neumann algebras relative to some subalgebra. Categorically, this has to be made more precise and may not be quite what you're looking for. Details for the fiber product (which reduces to the spatial tensor product if the subalgebra is just complex numbers) may be found in Section 4.3 of link text
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.
Best Answer
What do you mean by $A\boxtimes A$? If this is the sheaf $p_1^*A\otimes p_2^*A$ on $G\times G$ then certainly it is globally generated by the tensor square of global sections. It follows easily from right-exactness of the pullback and of the tensor product --- let $V$ be the space of global sections, then $V\otimes O_G \to A$ is surjective, hence $V\otimes O_{G\times G} \to p_i^*A$ is surjective, hence $V\otimes V\otimes O_{G\times G} \to A\boxtimes A$ is surjective.