The set of real structures on $\mathbb{C}^n$ is isomorphic to $GL(n,\mathbb{C})/GL(n,\mathbb{R})$ as a $GL(n,\mathbb{C})$-space.

Question

In representation theory a real structure on a $G$-module $V$ (finite dimensional complex vector space, in which the group $G$ has a linear action on it) can be definide as a conjugate-linear $G$-map $J:V\to V$ (that is, $J(g.v)=g.J(v)$) s.t. $J^2=id$.

Let $X$ be the set of all real structures on $\mathbb{C}^n$. I want to show that $X$ is isomorphic to $GL(n,\mathbb{C})/GL(n,\mathbb{R})$ as a $GL(n,\mathbb{C})$-space.

For this porpose I know that $GL(n,\mathbb{C})$ acts on $X$ by $(A,J)\mapsto AJA^{-1}$, as $(AJA^{-1})^2=id$.

But I cannot go any further. For example, given a $B\in GL(n,\mathbb{R})$ i cannot see why $BJ=id$ for every $J\in X$.

Answer 1

You shouldn't expect $BJ=\mathrm{Id}$ for all $B\in GL(n,\mathbb{R})$.

Try this: Let $V=\mathbb{C}^n$ and $J:V\to V$ be the conjugate linear map $J(v)=\bar{v}$, where $$v=\begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}\;\Rightarrow\;\bar{v}=\begin{pmatrix}\bar{a_1}\\\vdots\\\bar{a_n}\end{pmatrix}.$$ Define a map $$f:GL(n,\mathbb{C})\to X\;\;\;\mbox{by}\;\;\;f(A)=AJA^{-1}.$$ You need to show two things:

1) $f$ is surjective, and

2) $f(A)=f(B)$ if, and only if, $B^{-1}A\in GL(n,\mathbb{R})$.

These two facts together imply that there is a well defined bijection $$\bar{f}:GL(n,\mathbb{C})/GL(n,\mathbb{R})\to X$$ given by $\overline{f}(A\, GL(n,\mathbb{R}))=f(A)$.

To prove 2, note that $f(B)=J$ if, and only if $B\in GL(n,\mathbb{R})$. Indeed, \begin{align} (BJB^{-1})(v)=J(v)\;\forall v\in V&\iff B\bar{B}^{-1}(\bar{v})=\bar{v}\;\forall v\in V\\ &\iff B\bar{B}^{-1}=\mathrm{Id}_V\\ &\iff B=\bar{B}\\ &\iff B\in GL(n,\mathbb{R}). \end{align} Now, \begin{align} f(A)=f(B)&\iff AJA^{-1}=BJB^{-1}\\ &\iff J=B^{-1}AJA^{-1}B=B^{-1}AJ(B^{-1}A)^{-1}\\ &\iff B^{-1}A\in GL(n,\mathbb{R}). \end{align} This proves 2.

To prove 1, let $K\in X$ and let $V^K=\{v\in V\mid K(v)=v\}$ be the set of $K$-fixed points. Let $W=\mathrm{span}_{\mathbb{C}}V^K$. We claim that $W=V$. If not, $V=W\oplus U$ for some complementary subspace $U$. If $u\in U\backslash\{0\}$, then $K(u)=w+u'$ for some $w\in W$ and $u'\in U$. Since $u+K(u)=w+(u+u')\in W$ we must have $u'=-u$. Now, consider the element $2iu-iw\in V\backslash W$. We have $$K(2iu-iw)=-2i(w-u)+iw=2iu-iw.$$ Hence $2iu-iw\in W$, a contradiction. Hence $W=V$ as desired.

Finally, let $\beta=\{v_1,\ldots,v_n\}\subset V^K$ be a basis for $V$. The map $A:V\to V$ given by $A(e_i)=v_i$ ($e_i$ the $i$th coordinate vector) satisfies $f(A)=AJA^{-1}=K$, proving 1.

Now, $GL(n,\mathbb{C})$ acts on $GL(n,\mathbb{C}/GL(n,\mathbb{R})$ by left multipication: $$A.(B\,GL(n,\mathbb{R})=(AB)\,GL(n,\mathbb{R}),$$ and $GL(n,\mathbb{C})$ acts on $X$ by conjugation: $$A.K=AKA^{-1}.$$ Now, observe that \begin{align} \overline{f}(A.(B\,GL(n,\mathbb{R}))&=\overline{f}((AB)\,GL(n,\mathbb{R}))\\ &=f(AB)\\ &=(AB)J(AB)^{-1}\\ &=A(BJB^{-1})A^{-1}\\ &=A.(BJB^{-1})\\ &=A.f(B)\\ &=A.\overline{f}(B\,GL(n,\mathbb{R})). \end{align} Therefore, $\overline{f}$ is a map of $GL(n,\mathbb{C})$-spaces.