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A linear complex structure on $\mathbb R^{2n}$ is an endomorphism $J: \mathbb R^{2n} \to \mathbb R^{2n}$ such that $J^2 = -Id$. (Then $J$ is necessarily an isomorphism.) We have an action of $GL(2n,\mathbb R)$ on the set of all complex structures $J$ by conjugation ($g\cdot J := g J g^{-1}$) and I want to show that this action is transitive. In order to do so, I thought to take some linearly independent set $(v_1,...,v_n)$ of vectors such that, if $V$ denotes their $n$-dimensional span, we have $\mathbb R^{2n} = V \oplus J(V)$. From there I would show that I can move any $J$ to a 'standard'-$J_0$, e.g. one of the form $J_0 = \begin {pmatrix} 0_n & -I_n\\I_n & 0_n\end{pmatrix}$. To find such a $V$, I would start from any $v_1 \neq 0$. Then $J v_1$ is not a multiple of $v_1$ and in the next step I pick some $v_2$ that does not lie in the span of $\{v_1, J v_1\}$ and observe that $J v_2$ does also not lie in the span of $\{v_1, J v_1\}$ and then pick some $v_3$ etc. Is my understanding correct, that these $v_j$'s cannot be choosen explicitely? Conversely, if I have given a $\mathbb C$-basis $\{v_j\}$ of $(\mathbb R^{2n},J)$ I find an explicit $\mathbb R$-basis of $\mathbb R^{2n}$ as $\{v_j\}_j \cup \{J(v_j)\}_j$.

Is this ok so far?

My actual question is, how do I see that the stabilizer of $J_0$ can be identified with $GL(n,\mathbb C)$? My idea is to identify the stabilizer with the image of the inclusion $M(n,\mathbb C) \to M(2n,\mathbb R)$ given by $g = Re g + i Im g \mapsto \begin {pmatrix} Re g & -Im g\\Im g & Re g\end{pmatrix}$, restricted to invertible matrices. How do I formalize that this inclusion is 'induced' by $J_0$?. I see that this inclusion is an algebra homomorphism. How do I see that it respects invertibility?

Finally a stupid question, is any $J$ skew-symmetric as is $J_0$? Is it the case, that the matrix $g$ that conjugates $J$ to $J_0$, is orthogonal? If so, why?

Mekanik
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1 Answers1

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  1. A linear complex structure on $\mathbb{R}^{2n}$ is the structure of a complex vector space on it compatible with its real vector space structure. $J$ is multiplication by $i$. Since there is only one $n$-dimensional complex vector space up to isomorphism, any two such complex structures give rise to two complex vector spaces $V, V'$ such that there must be some isomorphism $f : V \to V'$ between them. Since $V$ and $V'$ have the same underlying real vector space this isomorphism can be identified with an element of $GL_{2n}(\mathbb{R})$.

  2. The stabilizer of the standard complex structure is the automorphism group of $\mathbb{C}^n$ as a complex vector space.

  3. $J_0$ is skew-symmetric because $e^{J_0 t}$ is a rotation; equivalently, it preserves the standard inner product on $\mathbb{R}^{2n}$. But there's no reason for an arbitrary complex structure $J$ to have this property; it needs to have a compatible complex inner product which restricts to the standard inner product. This is precisely the condition that $J$ and $J_0$ are conjugate by an element of the orthogonal group.

Qiaochu Yuan
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