Suppose $X$ is a smooth scheme, $E=O\_X^{\oplus n}$ and $\varphi\in SL\_n(E)$, i.e. $\varphi$ has trivial determinant and is an isomorphism. Is the morphism
$$\mathbb{P}(\varphi)^\*:CH^{\bullet}(\mathbb{P}(E))\longrightarrow CH^{\bullet}(\mathbb{P}(E))$$
equal to the identity map?



Invariance of Chow groups of projective bundles under automorphisms of bundles
This is true for any smooth variety $X$ (i.e. separated scheme of finite type over a field) and any vector bundle $E$ if and only if $\mathbb{P}(\varphi)^\*(\mathcal{O}\_{\mathbb{P}(E)}(1))=\mathcal{O}\_{\mathbb{P}(E)}(1)$.


For one direction, note that if this equality doesn't hold, then the automorphism induces a morphism that's different on the Picard group which is a subgroup of the Chow ring. Below, I'll show the converse.


**Claim 1**: Let $X$ be any variety with irreducible components $X\_i$. Let $\varphi: E\rightarrow E$ be any $X$-automorphism of a vector bundle $E$ over $X$. Let $\varphi\_i$ be the restriction of $\varphi$ to $E|\_{X\_i}$. If $\mathbb{P}(\varphi\_i)^\*$ is the identity for all $i$, then $\mathbb{P}(\varphi)^\*$ is the identity.


*Proof.* Let $\rho\_i:\mathbb{P}(E|\_{X\_i})\rightarrow \mathbb{P}(E)$ be the inclusion. Then $\rho\_{i\*}\circ \mathbb{P}(\varphi\_i)^\*= \mathbb{P}(\varphi^\*)\circ \rho\_{i\*}$ by Proposition 1.7 of Fulton. The result follows since the images of $\rho\_{i\*}$ generate $\mathrm{CH}(\mathbb{P}(E))$. $\square$


**Claim 2**: Let $X$ be any variety and let $\varphi:E\rightarrow E$ be any $X$-automorphism of a vector bundle $E$ over $X$. If $\mathbb{P}(\varphi)^\*(\mathcal{O}\_{\mathbb{P}(E)}(1))=\mathcal{O}\_{\mathbb{P}(E)}(1)$, then $\mathbb{P}(\varphi)^\*$ is the identity.


*Proof.* We'll proceed by Noetherian induction on the dimension of the irreducible components of $X$. For this purpose, let $X$ be any variety with irreducible components $X\_i$ having dimension $\max\_i\{ \mathrm{dim}(X\_i)\}\leq n$.


For the base case, we observe that if $X$ is a point, then any automorphism of $\mathbb{P}^m\_X$ has to take $\mathcal{O}(1)$ to itself in $\mathrm{Pic}(X)$ and this generates the ring $\mathrm{CH}^\*(\mathbb{P}^m\_X)$.


Assume that the result is true for all $X$ with irreducible components $X\_i$ with $\max\_i\{\mathrm{dim}(X\_i)\}\leq n-1$. By Claim 1, it suffices to show that this implies the result when $X$ is irreducible. Then for any open $U\subset X$ there is a commuting ladder with exact rows


$$\begin{matrix} \mathrm{CH}(\mathbb{P}(E|\_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|\_U))&\rightarrow & 0 \\ \downarrow && \downarrow && \downarrow\\ \mathrm{CH}(\mathbb{P}(E|\_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|\_U))&\rightarrow & 0\end{matrix}$$ where the vertical morphisms are induced by $\mathbb{P}(\varphi)^\*$. By our induction hypothesis, the left-most arrow is the identity whenever $U$ is nonempty. Taking a limit over the inductive system of closed subschemes $Z\subset X$, we get a commutative ladder with exact rows given by localization
$$\begin{matrix} \varinjlim\_Z \mathrm{CH}(\mathbb{P}(E|\_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|\_{\eta}))&\rightarrow & 0 \\ \downarrow && \downarrow && \downarrow\\ \varinjlim\_Z \mathrm{CH}(\mathbb{P}(E|\_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|\_{\eta}))&\rightarrow & 0\end{matrix}$$ where $\eta$ is the generic point of $X$. By our induction, the rightmost vertical arrow is the identity. These sequences are (canonically) split by Claim 3 below. A diagram chase finishes the proof. $\square$


**Claim 3**: Let $L=\mathcal{O}\_{\mathbb{P}(E)}(1)$. If $\mathbb{P}(\varphi)^\*(L)=L$, then the map $p^\*:\mathrm{CH}\_i(\mathbb{P}(E))\rightarrow \mathrm{CH}\_{i-\mathrm{dim}(X)}(\mathbb{P}(E|\_\eta))$ is right-split by the map $s:\mathrm{CH}\_{i-\mathrm{dim}(X)}(\mathbb{P}(E|\_\eta))\rightarrow \mathrm{CH}\_i(\mathbb{P}(E))$ sending $c\_1(L|\_\eta)^j$ to $c\_1(L)^j$. Moreover, there is a commutative square $s\circ\mathbb{P}(\varphi|\_\eta)^\*=\mathbb{P}(\varphi)^\*\circ s$.


*Proof*. Since the group $\mathrm{CH}\_i(\mathbb{P}(E|\_\eta))=\mathbb{Z}$, for the first claim it suffices to see that $p^\*c\_1(L)=c\_1(L|\_\eta)$ which is true by functorality of the pullbacks. To see the second claim, we need to check
$$s\circ\mathbb{P}(\varphi|\_\eta)^\*(c\_1(L|\_\eta)^j)=\mathbb{P}(\varphi)^\*\circ s(c\_1(L|\_\eta)^j).$$ The left hand side is equivalent to
$$s\circ\mathbb{P}(\varphi|\_\eta)^\*(c\_1(L|\_\eta)^j)=s(c\_1(\mathbb{P}(\varphi)^\*(L|\_\eta))^j)=s(c\_1(L|\_\eta)^j)=c\_1(L)^j$$ while the right hand side is equivalent to $$\mathbb{P}(\varphi)^\*\circ s(c\_1(L|\_\eta)^j)=\mathbb{P}(\varphi)^\*(c\_1(L)^j))=c\_1(\mathbb{P}(\varphi)^\*(L))^j$$ by functorality. It suffices then to observe that $\mathbb{P}(\varphi)^\*(L)=L$. $\square$


Let $\pi\_X:\mathbb{P}(E)\rightarrow X$ be the structure map and let $\mathbb{P}(\varphi):\mathbb{P}(E)\rightarrow \mathbb{P}(E)$ be an $X$-isomorphism. If $X$ is smooth, then by the projective bundle formula (and the relation between the divisor class group and the Picard group) we have $$\mathrm{Pic}(\mathbb{P}(E))=\pi\_X^\*\mathrm{Pic}(X)\oplus \mathbb{Z}$$ where $L$ generates the copy of $\mathbb{Z}$. The projection formula then yields
\begin{align\*}\pi\_{X\*}\mathbb{P}(\varphi)^\*(L) &=\pi\_{X\*}(L\otimes \pi\_X^\*(M))\\ &= \pi\_{X\*}(L)\otimes M\\ &=E\otimes M\end{align\*} with $M$ a line bundle on $X$.


**Claim 4**. If $E=\mathcal{O}\_X^{\oplus n+1}$, then $\mathbb{P}(\varphi)^\*$ is the identity.


*Proof*. The composition of a section $\rho: X\rightarrow X\times \mathbb{P}^n$ with $\varphi$ and the projection $\psi:X\times \mathbb{P}^n\rightarrow \mathbb{P}^n$ shows that $$ (\rho\circ\varphi\circ\psi)^\*(\mathcal{O}(1))=M.$$ But we also have that the composition $X\rightarrow \mathbb{P}^n$ factors through a point in $\mathbb{P}^n$ so $M=\mathcal{O}\_X$.$\square$


