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Let $$X:=\{ g\in \mathrm{GL}(2n,\mathbb{R})\mid g^2=(-I)\}\subset \mathrm{GL}(2n,\mathbb{R}).$$

I want to show that it is a submanifold, and find out the number of its connected components as real smooth manifold. But it seems that it is not wise to use the inverse function theorem directly (for $\mathrm{GL}(2n,\mathbb{R})\rightarrow \mathrm{M}(2n,\mathbb{R})$, $g\mapsto g^2+I$). Is there an easier way for this specific example (I guess there is some background related to geometry or complex structure).

I know that $\mathrm{GL}(n,\mathbb{R})$ has two connected components (How many connected components does $\mathrm{GL}_n(\mathbb R)$ have? ). Is there a variant argument for the example here?

Thanks a lot in advance for any explanation and help!

youknowwho
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  • I guess that are not many such $g$. Think of the Jordan block decomposition of $g$, $g^2 =I$ implies there's not many such $g$. – Arctic Char May 16 '24 at 05:45
  • As @ArcticChar says, look at the Jordan normal form. It must only contain blocks of the form$\begin{pmatrix}0&1\ -1 & 0\end{pmatrix}$ (or its transpose but that results in a similar matrix). So $X$ is the orbit under conjugation of that Jordan normal form and as such has only one component and must be an immersed submanifold. – Callum May 16 '24 at 14:12
  • @Callum It does not follow that there is only one component. In fact, it is not too hard to see that (choosing one specific $g$ to start with) $X$ splits into two disjoint sets, coming from the conjugates of $g$ by the orientation preserving (resp. reversing) operators. Each of these is clearly connected (as a continuous image from a component of $\mathrm{GL}_n$). As such, $X$ is the disjoint union of two homeomorphic connected sets. (I don't know, if we can conclude from this that these must be the components here, maybe that follows by some machinery.) – Tim Seifert May 17 '24 at 09:33
  • @TimSeifert I don't think you are correct. That would be true for the action of left multiplication for example but conjugation is different. Conjugation by orientation reversing operators doesn't result in a disjoint component. Consider simply $\begin{pmatrix}-1&0 \0&1\end{pmatrix}\begin{pmatrix}0&1 \-1&0\end{pmatrix}\begin{pmatrix}-1&0 \0&1\end{pmatrix} = \begin{pmatrix}0&-1 \1&0\end{pmatrix}$ which is clearly path connected to $\begin{pmatrix}0&1 \-1&0\end{pmatrix}$ – Callum May 17 '24 at 10:24
  • Is it path connected in $X$ though? Maybe it is, then I was mistaken. But I don't think this is so clear – Tim Seifert May 17 '24 at 10:28
  • I did some explicit calculations and it turns out that in the $2\times 2$-case $X$ is homeomorphic to the plane without a line, hence disconnected. This is of course no guarantee for the general case, but it seems that these two subsets I mentioned may always be separated and thus form the two components – Tim Seifert May 17 '24 at 10:56
  • @TimSeifert Ah yes my mistake. I though you could take the obvious path in $\operatorname{GL}(2n,\mathbb{R})$ and adjust it to lie in $X$ but that is not the case. I suspect the general case might be $n+1$ connected components by a Sylester's law of inertia style argument then. – Callum May 17 '24 at 12:45
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    It is a submanifold (known as the space of linear almost complex structures on $\mathbb R^{2n}$), diffeomorphic to $GL(2n,\mathbb R)/GL(n,\mathbb C)$, hence, has two connected components. See here for details. – Moishe Kohan May 17 '24 at 13:27

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To convert my comment to an answer:

A matrix $J\in GL(2n, \mathbb R)$ satisfying $J^2=-I$ is called a linear almost complex structure on $\mathbb R^{2n}$. Basic properties are discussed in detail in Chapter IX, section 1 of

Kobayashi, Shoshichi; Nomizu, Katsumi, Foundations of differential geometry. Vol. II, Interscience Tracts in Pure and Applied Mathematics 15. New York-London-Sydney: Interscience Publishers a division of John Wiley and Sons (1969). ZBL0175.48504.

See also this Wikipedia article and my answer here.

In particular, all such matrices $J$ are conjugate by elements of $GL(2n, \mathbb R)$. The centralizer of such $J$ (the stabilizer with respect to the action of $GL(2n, \mathbb R)$ by conjugation) is isomorphic to $GL(n, \mathbb C)$. In particular, the subset of linear almost complex structures is a submanifold of $GL(2n, \mathbb R)$ naturally diffeomorphic to the quotient $M=GL(2n, \mathbb R)/GL(n, \mathbb C)$, which is a manifold of real dimension $(2n)^2 - 2n^2=2n^2$. Since $GL(n, \mathbb C)$ is connected, $M$ has two connected components (one for each component of $GL(2n, \mathbb R)$, i.e. one for each orientation on $\mathbb R^{2n}$).

Moishe Kohan
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