# Let $gl_S(n,F) = \{x \in gl(n,F) : x^tS = -Sx \}$ and $T = P^tSP$. Show $gl_S(n,F) \cong gl_T(n,F)$

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.

• You got a great answer here, but note that its upshot was mentioned at the beginning of step 2 of my answer math.stackexchange.com/a/3708980/96384, whose link I already posted under your previous question about this $gl_S$ construction. Sep 11, 2020 at 3:05

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: $$\begin{eqnarray*}g(y)^tS&=&(PyP^{-1})^t(P^t)^{-1}TP^{-1}\\ &=&(P^t)^{-1}y^t P^t(P^t)^{-1}TP^{-1}\\ &=&(P^t)^{-1}y^t TP^{-1}\\ &=&-(P^t)^{-1}TyP^{-1}\\ &=&-(P^t)^{-1}TP^{-1} PyP^{-1}\\ &=&-Sg(y). \end{eqnarray*}$$