# How to show that an automorphism is inner/outer

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.

• 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 $C\in SL_n(\Bbb R)$. – Arnaud D. Oct 25 '18 at 12:17
• As for $\psi$, there is also this question : math.stackexchange.com/questions/98378/… – Arnaud D. Oct 25 '18 at 12:21
• @ArnaudD. That makes sense. It would work for $\varphi$. And as for $\psi$, thank you! – 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 $$n\geq 3$$.)