For fixed $C \in \mathrm{\mathbf{GL}}_n(\mathbb{R})$, we defined $\varphi, \psi: \mathrm{\mathbf{SL}}_n(\mathbb{R}) \rightarrow \mathrm{\mathbf{SL}}_n(\mathbb{R})$ as $$\varphi(A) = CAC^{-1}, \quad \psi(A) = (A^{\mathrm{\mathbf{t}}})^{-1}, \qquad (A\in \mathrm{\mathbf{SL}}_n(\mathbb{R}))$$

My question is, for the case where $C \notin \mathrm{\mathbf{SL}}_n(\mathbb{R})$,

Is $\varphi, \psi$ an outer automorphism of $\mathrm{\mathbf{SL}}_n(\mathbb{R})$?

I can see that $\varphi, \psi \in \mathrm{Aut}(\mathrm{\mathbf{SL}}_n(\mathbb{R}))$, so what I tried was the following.

If $\varphi, \psi$ is an inner automorphism of $\mathrm{\mathbf{SL}}_n(\mathbb{R})$, there exists $B\in \mathrm{\mathbf{SL}}_n(\mathbb{R})$, such that $$BAB^{-1}=\varphi(A)=CAC^{-1}, \quad BAB^{-1} = \psi(A) = (A^{\mathrm{\mathbf{t}}})^{-1}$$ But from here, I cannot get any further.

It seems like $\varphi \in \mathrm{Inn}(\mathrm{\mathbf{SL}}_n(\mathbb{R}))$ and $\psi \in \mathrm{Out}(\mathrm{\mathbf{SL}}_n(\mathbb{R}))$ but I cannot prove it. Any help would be greatly appreciated. Thank you.

Note: I have read an answer from Transpose inverse automorphism is not inner, but I couldn't understand the answer.

  • 1
    $\begingroup$ Note for the first one : $\varphi$ is invariant under replacing $C$ by $\gamma C$ for some $\gamma\in \mathbb{R}$. Thus at least for odd $n$ or $C$ with positive determinant you can always assume that $C\in SL_n(\Bbb R)$. $\endgroup$ – Arnaud D. Oct 25 '18 at 12:17
  • 2
    $\begingroup$ As for $\psi$, there is also this question : math.stackexchange.com/questions/98378/… $\endgroup$ – Arnaud D. Oct 25 '18 at 12:21
  • $\begingroup$ @ArnaudD. That makes sense. It would work for $\varphi$. And as for $\psi$, thank you! $\endgroup$ – zxcvber Oct 25 '18 at 12:30

The case $n$ is odd or $\det C>0$, Arnaud D showed in the comment that $\varphi$ is an inner automorphism. However, $\varphi$ is not necessarily inner if $n$ is even and $\det C<0$.

Let $n$ be an even positive integer. WLOG, consider $n=2$. Otherwise, you can look at the subgroup of $\mathbf{SL}_n(\Bbb R)$ consisting of $\begin{pmatrix} X&0\\0&I\end{pmatrix}$ with $X\in\mathbf{SL}_2(\Bbb R)$ and look at the action of $\varphi$ on such this subgroup. To make $\varphi$ fix this subgroup, we can set $C=\begin{pmatrix} Z&0\\0&I\end{pmatrix}$ for some $Z\in\mathbf{GL}_2(\Bbb R)$.

Now, for $n=2$, we can take $C=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Prove that $\varphi$ is not inner with this $C$.

For $\psi$, it is also not inner except for the trivial case $n=1$. For $n=2$, it is not difficult to show that the only matrix $X$ such that $(A^t)^{-1}X=XA$ for all $A\in\mathbf{SL}_2(\mathbb{R})$ is $X=0$. (This is not true of $\mathbb{R}$ is replaced by a field of characteristic $2$. The map $\psi$ is an inner automorphism when $n=2$ and the field has characteristic $2$.)

For $n>2$, note that $$B=\begin{pmatrix}1&1&1\\1&1&0\\-1&0&0\end{pmatrix}$$ is in $\mathbf{SL}_3(\mathbb{R})$. Let $A$ be the matrix $\begin{pmatrix}B&0\\0&I\end{pmatrix}$ in $\mathbf{SL}_n(\mathbb{R})$. Then, $\psi(A)=\begin{pmatrix}\psi(B)&0\\0&I\end{pmatrix}$, where $$\psi(B)=\begin{pmatrix}0&0&1\\0&1&-1\\-1&1&0\end{pmatrix}.$$ Thus, $\operatorname{tr}(A)=\operatorname{tr}(B)+(n-3)=n-1$ and $\operatorname{tr}\big(\psi(A)\big)=\operatorname{tr}\big(\psi(B)\big)+(n-3)=n-2$. Therefore, $\psi$ cannot be an inner automorphism, since inner automorphisms preserve the trace of each matrix. (This argument also shows that $\psi$ is not an inner automorphism for every base field, if $n\geq 3$.)

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  • $\begingroup$ Thank you very much, it helped a lot! $\endgroup$ – zxcvber Oct 30 '18 at 8:19

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