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Let $gl_S(n,F) = \{x \in gl(n,F) : x^tS = -Sx \}$ where $gl(n,F)$ is the lie algebra of $GL(n,F)$.

Show that if $P$ is an $n \times n$ invertible matrix and $T = P^tSP$, then

$$gl_T(n,F) = \{x \in gl(n,F) : x^tT = -Tx \} \cong gl_S(n,F)$$

my thoughts:

So at first I was thinking that $T$ was aquired from $S$ by a change of basis, then I realized we were taking the transpose and not the inverse.

Looking at $gl_T(n,F) = \{x \in gl(n,F) : x^tP^tSP = -P^tSPx \}$, I don't see any obvious way to make the $P$'s inconsequential.

I appreciate your time.

Eugene
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    You got a great answer here, but note that its upshot was mentioned at the beginning of step 2 of my answer https://math.stackexchange.com/a/3708980/96384, whose link I already posted under your previous question about this $gl_S$ construction. – Torsten Schoeneberg Sep 11 '20 at 03:05

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We have an isomorphism $f\colon gl_S(n,F)\to gl_T(n,F)$ is given by $f\colon x\mapsto P^{-1}xP$.

First note that if $x^tS=-Sx$, then: \begin{eqnarray*}f(x)^tT&=&(P^{-1}xP)^tP^tSP\\ &=&P^tx^t(P^t)^{-1}P^tSP\\ &=&P^tx^tSP\\ &=&-P^tSxP\\ &=&-P^tSPP^{-1}xP\\ &=&-Tf(x), \end{eqnarray*} so $f$ is well defined.

Secondly note that $f$ is a Lie algebra homomorphism:$$[f(x),f(y)]=P^{-1}xyP-P^{-1}yxP=f([x,y]).$$

Finally note that $f$ has inverse $g\colon gl_T(n,F)\to gl_S(n,F)$, given by $g\colon y\mapsto PyP^{-1}$.

For completeness we should verify that $g$ is well defined. If $y^tT=-Ty$ then: \begin{eqnarray*}g(y)^tS&=&(PyP^{-1})^t(P^t)^{-1}TP^{-1}\\ &=&(P^t)^{-1}y^t P^t(P^t)^{-1}TP^{-1}\\ &=&(P^t)^{-1}y^t TP^{-1}\\ &=&-(P^t)^{-1}TyP^{-1}\\ &=&-(P^t)^{-1}TP^{-1} PyP^{-1}\\ &=&-Sg(y). \end{eqnarray*}

tkf
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