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Problem: Let $\mathfrak{g}$ be an orthogonal Lie algebra (type $B_l$ or $D_l$). If $Y$ is an orthogonal matrix, that is, $Y$ is invertible such that $Y^TSY = S$, prove that the function $X\mapsto YXY^{-1}$ is an automorphism of $ \mathfrak{g}$.

Attempt: Let the function:

$$\begin{align*} \varphi:\mathfrak{g}&\longrightarrow\mathfrak{g} \\[0.5em] X&\longmapsto YXY^{-1} \end{align*}$$

Let's see that $\varphi$ is an automorphism of $\mathfrak{g}$, that is, $\varphi$ is a homomorphism of $\mathfrak{g}$ to $\mathfrak{g}$.

Let $X_1,X_2\in\mathfrak{g}$, then:

$$\begin{align*} \varphi([X_1,X_2]) &= Y[X_1,X_2]Y^{-1} \\[0.5em] &= Y(X_1X_2 - X_2X_1)Y^{-1} \\[0.5em] &= Y(X_1X_2)Y^{-1} - Y(X_2X_1)Y^{-1} \\[0.5em] &= (YX_1Y^{-1})(YX_2Y^{-1}) - (YX_2Y^{-1})(YX_1Y^{-1}) \\[0.5em] &= \varphi(X_1)\varphi(X_2) - \varphi(X_2)\varphi(X_1) \\[0.5em] &= [\varphi(X_1),\varphi(X_2)] \end{align*}$$

So $\varphi$ is an automorphism of $\mathfrak{g}$. $\blacksquare$

Question: Is this proof correct? I think I'm missing something about the properties of $\varphi$ because I didn't use the hypothesis ($Y$ is invertible such that $Y^TSY = S$). I think I actually used it in the proof but I don't know where specifically. If the proof is correct I want to make it more rigorous.

Tryncha
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    $Y$ should not be viewed as an element of $\mathfrak g$, but rather as an element of the Lie group. Indeed, the Lie group acts on the Lie algebra by conjugation, but the Lie algebra doesn't. – Kenta S Mar 09 '23 at 01:17
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    Your calculation shows it is a Lie algebra homomorphism. But probably you want to use the properties of $Y$ to show that the map $\phi$ is well-defined in the first place. That is, if $X$ is skew-symmetric and $Y$ is orthogonal, then $YXY^{-1}$ is also skew-symmetric. – Nick Mar 09 '23 at 01:23
  • @Nick This calculation proves that $\varphi$ is an automorphism of $\mathfrak{g}$. But I must show first that $\varphi$ is well-defined. Did I understand it right? – Tryncha Mar 09 '23 at 02:16
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    Yes, I'm just saying that in your calculation, you are implicitly assuming that $YXY^{-1} \in \mathfrak{g}$, but maybe this isn't obvious. And I suspect that showing $YXY^{-1} \in \mathfrak{g}$ is where you might use the fact that $Y$ is orthogonal. – Nick Mar 09 '23 at 02:59
  • @Nick Oh, ok. I understand. Thank you. – Tryncha Mar 09 '23 at 03:04
  • Nick is correct. Indeed the map $X \mapsto YXY^{-1}$ is a Lie algebra automorphism of $\mathfrak{gl}_n$ for any $Y\in GL_n$ so orthogonality doesn't play a role in preserving the Lie bracket and instead is about preserving $\mathfrak{g}\leq \mathfrak{gl}_n$. However I will note additionally that you only show it is a homomorphism and you need it to be an isomorphism in order for it to be an automorphism. That is, you need to show it is bijective as well. Since it is linear, injectivity will do, so we can simply note that $YXY^{-1} =0$ implies $X = 0$. – Callum Mar 09 '23 at 22:12

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