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I came across this post that says $\frac{\mathrm{GL}(2n,\mathbb{R})}{\mathrm{GL}(n,\mathbb{C})}$ is the set of all almost complex structures. However, I do not understand why. Can any point out an reference to this?

Also, any geometric/physical interpretation is warmly welcomed!

taper
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    $GL_n$ is not a vector space, and half of your "basis" elements aren't even invertible. – Max Sep 15 '16 at 04:21

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An almost-complex structure is a matrix $J$ such that $J^2 = -I$ is the negative identity. As you said, one example of such a matrix $J$ is $$\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}.$$

Interpreting $GL(n,\mathbb{C})$ as a subgroup of $GL(2n;\mathbb{R})$ depends on having fixed such an almost-complex structure. Once we have a matrix $J$, we can call a matrix $A \in GL(2n;\mathbb{R})$ complex-linear if it commutes with $J$, i.e. $AJA^{-1} = J.$

(The idea is that $\mathbb{C}$-linear maps $T$ are just real linear maps with the additional property that $T(iv) = i T(v)$ for all vectors $v$)

Given any matrix $A \in GL(2n;\mathbb{R})$, we get another almost-complex structure $AJA^{-1}$. This is the same almost-complex structure $J$ if and only if $A \in GL(n;\mathbb{C}).$ On the other hand, all almost-complex structures are similar (although it may take some work to be convincing that they are similar over $\mathbb{R}$ and not only $\mathbb{C}$) since they are diagonalizable with the same eigenvalues $\pm i$. That gives you a bijection $$GL(2n;\mathbb{R}) / GL(n;\mathbb{C}) \longrightarrow \{\mathrm{almost}-\mathrm{complex}\; \mathrm{structures}\}$$ under which a class $A \cdot GL(n;\mathbb{C})$ corresponds to the almost-complex structure $AJA^{-1}.$

  • Thanks for this. Could you add some references to harness some claims you made in your argument? – taper Sep 15 '16 at 05:24
  • What claim is unclear? –  Sep 15 '16 at 06:06
  • This two points: 1.Why $AJA^{-1}$ is the same almost-complex structure $J$ if and only if $A \in GL(n;\mathbb{C}).$. 2. Why all almost-complex structures are similar over $\mathbb{R}$. – taper Sep 15 '16 at 06:18
  • But I think they might be too trivial. So any reference on this topic is welcomed~~ :P – taper Sep 15 '16 at 06:25
  • is the definition of $GL(n;\mathbb{C})$ as matrices $A$ with $AJA^{-1} = J.$ 2. comes from the fact that any real matrices that are similar over $\mathbb{C}$ are already similar over $\mathbb{R}$. This isn't trivial but it has been asked and answered many times on this site: here is one reference http://math.stackexchange.com/questions/57242/similar-matrices-and-field-extensions?noredirect=1&lq=1
  • –  Sep 15 '16 at 06:58