1
$\begingroup$

Let $X$ be a smooth projective variety with an action of linear algebraic group $G$. Theorem 5.6.1 in Criss/Ginzburg (Representation Theory and Complex Geometry) lists a bunch of equivalent conditions for when this space satisfies a "Kunneth formula:"

(a) for any $Y$ with an action of $G$, the exterior tensor map $\pi: K^G(X) \otimes_{K^G(\text{pt})} K^G(Y) \simeq K^G(X \times Y)$ is an isomorphism.

(b) the diagonal $\mathcal{O}_X \in K^G(X \times X)$ is in the image of $\pi$ for $Y = X$ as above

(c) The convolution in K-theory map $$K^G(X \times Y) \rightarrow \text{Hom}_{K^G(\text{pt})}(K^G(X), K^G(Y))$$ is an isomorphism, and $K^G(X)$ is projective over $K^G(\text{pt})$.

Are there known examples of smooth projective varieties $X$ such that the above statements do not hold?

Edit: In light of the counterexample provided by Jason in the comments, is there a counterexample where $G$ acts on $X$ by finitely many orbits or less strictly, finitely many closed orbits?

$\endgroup$
5
  • 4
    $\begingroup$ For $G$ equal to a trivial group and for $X$ any smooth, projective curve of genus $g>0$, $K^G(X\times X)$ is strictly larger than $K^G(X)\otimes K^G(X)$. In fact, $\text{Pic}(X\times X)$ is strictly larger than $\text{Pic}(X)\times \text{Pic}(X)$. $\endgroup$ Commented Jul 20, 2016 at 19:55
  • $\begingroup$ Thanks! Followup: are there examples where G acts on X by finitely many orbits, or maybe less strictly, X has only finitely many closed orbits? Also, maybe for future passers-by, this MO post discusses that counterexample in more detail and references the relevant Hartshorne exercises: mathoverflow.net/questions/244720/… $\endgroup$ Commented Jul 20, 2016 at 20:23
  • 3
    $\begingroup$ "... is there a counterexample where $G$ acts on $X$ by finitely many orbits or less strictly, finitely many closed orbits?" Let $G$ be $\mathbf{SL}_n$, $n\geq 2$, and let $X=Y=G$. Then $G$ acts freely on $X$ and $Y$ with quotient a point, i.e., there is a single orbit. So $\text{Pic}(X/G) = \text{Pic}(Y/G) = \{0\}$. On the other hand, for the diagonal action of $G$ on $X\times Y = G\times G$, the quotient is isomorphic to $G$. Since $\text{Pic}(\mathbf{SL}_n) \cong \mathbb{Z}/n\mathbb{Z},$ the map $\text{Pic}(X/G)\times \text{Pic}(Y/G) \to \text{Pic}((X\times Y)/G)$ is not surjective. $\endgroup$ Commented Jul 20, 2016 at 23:04
  • $\begingroup$ Oops. Please replace $\textbf{SL}_n$ by $\textbf{PGL}_n$ in my last comment. $\endgroup$ Commented Jul 20, 2016 at 23:19
  • 1
    $\begingroup$ Oops again. My example of $X$ is not projective. $\endgroup$ Commented Jul 21, 2016 at 0:41

1 Answer 1

1
$\begingroup$

Edit. User hic points out that with the first action I specified, there are infinitely many orbits. So please change the group to $\textbf{PGL}_2\times \textbf{SL}_2$ with its left-right action.

The examples in my comment are only quasi-projective, not projective. However, there are also projective examples. Begin with the $4$-dimensional vector space $\text{Mat}_{2\times 2}$ of $2\times 2$-matrices, with its usual matrix product. Let $\mathbb{P}\text{Mat}_{2\times 2}$ denote the associated projective $3$-space. Denote by $\textbf{PGL}_2\subset \mathbb{P}\text{Mat}_{2\times 2}$ denote the open subset of invertible $2\times 2$ matrices considered up to scaling. Edit. Also let $\textbf{SL}_2\subset \text{Mat}_{2\times 2}$ denote the degree $2$ cover of $\textbf{PGL}_2$; the group of $2\times 2$ matrices with determinant $1$. Let $G$ be $\textbf{PGL}_2\times \text{SL}_2$. For every action of $G$ on a scheme $Z$, since the character group of $G$ is trivial, the natural homomorphism $$\text{Pic}^{G}(Z) \to \text{Pic}(Z)$$ is injective. This raises the question: what is the image?

The left action of $\textbf{PGL}_2$ on itself extends to a left action of $\textbf{PGL}_2$ on $\mathbb{P}\text{Mat}_{2\times 2}$. Edit.The right action of $\textbf{SL}_2$ on $\textbf{PGL}_2$ extends to a right action of $\textbf{SL}_2$ on $\mathbb{P}\text{Mat}_{2\times 2}$. Together this defines an action of $G$ on $\mathbb{P}\text{Mat}_{2\times 2}$, $([g],h)\cdot [A] = [g\cdot A\cdot h^{-1}]$. This action has two orbits, the open orbit $\textbf{PGL}_2$ and the closed orbit is the zero locus of the determinant. In particular, the closed orbit is a $G$-invariant Cartier divisor whose associated invertible sheaf is isomorphic to $\mathcal{O}(2)$ with a natural $G$-linearization. The image of the natural group homomorphism $$\text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}) \to \text{Pic}(\mathbb{P}\text{Mat}_{2\times 2}) $$ is the index $2$ subgroup generated by the class $[\mathcal{O}(2)]$. Now set $X=Y=\mathbb{P}\text{Mat}_{2\times 2}$. The Künneth homomorphism $$\text{Pic}(\mathbb{P}\text{Mat}_{2\times 2}) \times \text{Pic}(\mathbb{P}\text{Mat}_{2\times 2}) \to \text{Pic}(\mathbb{P}\text{Mat}_{2\times 2} \times \mathbb{P}\text{Mat}_{2\times 2} ) $$ is an isomorphism. Thus the image of the injective group homomorphism $$\text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}) \times \text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}) \to \text{Pic}(\mathbb{P}\text{Mat}_{2\times 2} \times \mathbb{P}\text{Mat}_{2\times 2} ) $$ is the subgroup generated by $([\mathcal{O}(2)],0)$ and $(0,[\mathcal{O}(2)])$.

Now for elements $([A_1],[A_2]) \in \mathbb{P}\text{Mat}_{2\times 2}\times \mathbb{P}\text{Mat}_{2\times 2}$ with coordinates $$A_1 = \left[ \begin{array}{cc} a_1 & b_1 \\ c_1 & d_1 \end{array} \right], \ A_2 = \left[ \begin{array}{cc} a_2 & b_2 \\ c_2 & d_2 \end{array} \right],$$ consider the matrix product of the adjugate of $A_2$ and $A_1$, $$ B = \text{adj}(A_2)\cdot A_1 = \left[ \begin{array}{cc} d_2 & -b_2 \\ -c_2 & a_2 \end{array} \right] \cdot \left[ \begin{array}{cc} a_1 & b_1 \\ c_1 & d_1 \end{array} \right] = \left[ \begin{array}{cc} a_1d_2-b_2c_1 & b_1d_2-b_2d_1 \\ a_2c_1-a_1c_2 & a_2d_1-b_1c_2 \end{array} \right].$$ Then entries of the matrix $B$ generate a subspace $U$ of the space $$ V=H^0(\mathbb{P}\text{Mat}_{2\times 2} \times \mathbb{P}\text{Mat}_{2\times 2}, \text{pr}_1^*\mathcal{O}(1)\otimes \text{pr}_2^* \mathcal{O}(1)).$$ The left, resp. right, diagonal action of $\textbf{SL}_2$ on $\mathbb{P}\text{Mat}_{2\times 2} \times \mathbb{P}\text{Mat}_{2\times 2}$ lifts to a unique linearization of $\text{pr}_1^*\mathcal{O}(1)\otimes \text{pr}_2^* \mathcal{O}(1)$. The subspace $U$ is a left-right $\textbf{SL}_2\times \textbf{SL}_2$-subrepresentation of $V$ that is left trivial, i.e., for every $g\in \textbf{SL}_2$, $$\text{adj}(g\cdot A_2)\cdot (g\cdot A_1) = \left( \text{adj}(A_2)\text{adj}(g)\right)\cdot (g\cdot A_1) = $$ $$\text{adj}(A_2)\left( \text{adj}(g)g \right) A_1 = \text{adj}(A_2)\left( \text{Id}_{2\times 2}\right) A_1= \text{adj}(A_2)A_1.$$ Edit.Thus, the left-right action of $\textbf{SL}_2\times \textbf{SL}_2$ factors through an action of $\textbf{PGL}_2\times \textbf{SL}_2$. Edit.In particular, the nonzero element in $U$ corresponding to the trace of $B$ is invariant under $\textbf{PGL}_2\times \textbf{SL}_2$. The corresponding zero divisor of this section of $\text{pr}_1^*\mathcal{O}(1)\times \text{pr}_2^*\mathcal{O}(1)$ is a $G$-invariant divisor. Thus, the invertible sheaf of this divisor has a natural $G$-linearization. Therefore $([\mathcal{O}(1)],[\mathcal{O}(1)])$ is an element in the image of $\text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}\times \mathbb{P}\text{Mat}_{2\times 2})$ that is not in the image of $\text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}) \times \text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}).$ Thus the Künneth homomorphism $$\text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}) \times \text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2}) \to \text{Pic}^{G}(\mathbb{P}\text{Mat}_{2\times 2} \times \mathbb{P}\text{Mat}_{2\times 2} ) $$ is not surjective.

$\endgroup$
2
  • $\begingroup$ I like this example! However, if I'm not making a mistake, I think $\mathbb{P}^3$ has infinitely many closed orbits "at infinity" (i.e. in the complement of $PGL_2$) i.e. there is a (trivial) family of orbits isomorphic to $\mathbb{P}^1$ indexed by $\mathbb{P}^1$ (each orbit has representative a row echelon form matrix modulo scalars, and the singular orbits can be reduced to have zeroes in the second row and $\mathbb{P}^1$ worth of choices in the first row). In any case this is still good to know, so thanks! $\endgroup$ Commented Jul 22, 2016 at 7:22
  • 1
    $\begingroup$ @hic. Thank you for pointing out the mistake. If I "double" the group and consider its left-right action (instead of just the left action), then there are only two orbits. I have edited the answer accordingly. $\endgroup$ Commented Jul 22, 2016 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.