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?



Counterexamples to Kunneth formula in algebraic K-theory
**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.


