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?

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    $\begingroup$ 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
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    $\begingroup$ "... 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
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    $\begingroup$ Oops again. My example of $X$ is not projective. $\endgroup$ Commented Jul 21, 2016 at 0:41

1 Answer 1


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.

  • $\begingroup$ 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
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    $\begingroup$ @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

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