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YCor
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Algebraic K theory-theory of a scheme with group action of a semi-directsemidirect product

Given $A$ and $H$ linear algebraic groups over a field $k$ and a homomorphism $\phi\colon H \rightarrow Aut(A)$$\phi\colon H \rightarrow \mathrm{Aut}(A)$ we can form the semi-direct product $G = A \rtimes_{\phi} H$.

Suppose that $G$ acts on a scheme $X$. The splitting $H\rightarrow G$ gives the restriction map $K^{0}(G,X)\rightarrow K^{0}(H,X)$ in equivariant algebraic K-theory.

$\textbf{Question:}$ Is this map an isomorphism of groups if we assume that $A$ is unipotent?

If $X=Spec(k)$$X=\mathrm{Spec}(k)$ then $K^{0}(G,X)=R(G)$ is the group completion of $Rep(G)$$\mathrm{Rep}(G)$. I think that restriction $R(G)\rightarrow R(H)$ is an isomorphism because the only finite dimensional irreducible representation of $A$ is the trivial one, so $R(A)\simeq \mathbb{Z}$. The inverse is given by precomposition by the projection $\pi \colon G \rightarrow H$.

If $H$ is trivial then $G \simeq A$ is unipotent and we get that $K^{0}(G,X)\rightarrow K^{0}(X)$ is an isomorphism, according to Thomason.

In the setting of Chow rings an analogous statement (with rational coefficients) is Lemma 1.4.7 of Brokemper 'On the Chow ring of the stack of truncated Barsotti-Tate groups', "On the Chow ring of the stack of truncated Barsotti-Tate groups".

Algebraic K theory of a scheme with group action of a semi-direct product

Given $A$ and $H$ linear algebraic groups over a field $k$ and a homomorphism $\phi\colon H \rightarrow Aut(A)$ we can form the semi-direct product $G = A \rtimes_{\phi} H$.

Suppose that $G$ acts on a scheme $X$. The splitting $H\rightarrow G$ gives the restriction map $K^{0}(G,X)\rightarrow K^{0}(H,X)$ in equivariant algebraic K-theory.

$\textbf{Question:}$ Is this map an isomorphism of groups if we assume that $A$ is unipotent?

If $X=Spec(k)$ then $K^{0}(G,X)=R(G)$ is the group completion of $Rep(G)$. I think that restriction $R(G)\rightarrow R(H)$ is an isomorphism because the only finite dimensional irreducible representation of $A$ is the trivial one, so $R(A)\simeq \mathbb{Z}$. The inverse is given by precomposition by the projection $\pi \colon G \rightarrow H$.

If $H$ is trivial then $G \simeq A$ is unipotent and we get that $K^{0}(G,X)\rightarrow K^{0}(X)$ is an isomorphism, according to Thomason.

In the setting of Chow rings an analogous statement (with rational coefficients) is Lemma 1.4.7 of Brokemper 'On the Chow ring of the stack of truncated Barsotti-Tate groups'.

Algebraic K-theory of a scheme with group action of a semidirect product

Given $A$ and $H$ linear algebraic groups over a field $k$ and a homomorphism $\phi\colon H \rightarrow \mathrm{Aut}(A)$ we can form the semi-direct product $G = A \rtimes_{\phi} H$.

Suppose that $G$ acts on a scheme $X$. The splitting $H\rightarrow G$ gives the restriction map $K^{0}(G,X)\rightarrow K^{0}(H,X)$ in equivariant algebraic K-theory.

$\textbf{Question:}$ Is this map an isomorphism of groups if we assume that $A$ is unipotent?

If $X=\mathrm{Spec}(k)$ then $K^{0}(G,X)=R(G)$ is the group completion of $\mathrm{Rep}(G)$. I think that restriction $R(G)\rightarrow R(H)$ is an isomorphism because the only finite dimensional irreducible representation of $A$ is the trivial one, so $R(A)\simeq \mathbb{Z}$. The inverse is given by precomposition by the projection $\pi \colon G \rightarrow H$.

If $H$ is trivial then $G \simeq A$ is unipotent and we get that $K^{0}(G,X)\rightarrow K^{0}(X)$ is an isomorphism, according to Thomason.

In the setting of Chow rings an analogous statement (with rational coefficients) is Lemma 1.4.7 of Brokemper, "On the Chow ring of the stack of truncated Barsotti-Tate groups".

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Simon Cooper
Simon Cooper
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Algebraic K theory of a scheme with group action of a semi-direct product

Given $A$ and $H$ linear algebraic groups over a field $k$ and a homomorphism $\phi\colon H \rightarrow Aut(A)$ we can form the semi-direct product $G = A \rtimes_{\phi} H$.

Suppose that $G$ acts on a scheme $X$. The splitting $H\rightarrow G$ gives the restriction map $K^{0}(G,X)\rightarrow K^{0}(H,X)$ in equivariant algebraic K-theory.

$\textbf{Question:}$ Is this map an isomorphism of groups if we assume that $A$ is unipotent?

If $X=Spec(k)$ then $K^{0}(G,X)=R(G)$ is the group completion of $Rep(G)$. I think that restriction $R(G)\rightarrow R(H)$ is an isomorphism because the only finite dimensional irreducible representation of $A$ is the trivial one, so $R(A)\simeq \mathbb{Z}$. The inverse is given by precomposition by the projection $\pi \colon G \rightarrow H$.

If $H$ is trivial then $G \simeq A$ is unipotent and we get that $K^{0}(G,X)\rightarrow K^{0}(X)$ is an isomorphism, according to Thomason.

In the setting of Chow rings an analogous statement (with rational coefficients) is Lemma 1.4.7 of Brokemper 'On the Chow ring of the stack of truncated Barsotti-Tate groups'.