# Eigenvalues and eigenspaces of almost complex structures under each other [closed]

I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier

I have several questions on the concepts of almost complex structures and complexification. Here are some:

Let $$L$$ be $$\mathbb C$$-vector space. Let $$L_{\mathbb R}$$ be its realification, and let the $$(L_{\mathbb R})^{\mathbb C} = (L_{\mathbb R}^2,J)$$ be the complexification of its realification with almost complex structure $$J(l,m):=(-m,l)$$ on $$L_{\mathbb R}^2$$. For every almost complex structure $$K$$ on $$L_{\mathbb R}$$, $$K \oplus K$$ is an almost complex structure on $$L_{\mathbb R}^2$$. Then $$K^{\mathbb C} := (K \oplus K)^J$$ (see notation and definitions here, in particular the bullet below 'Definition 4') is $$\mathbb C$$-linear, i.e. $$K \oplus K$$ and $$J$$ commute.

Based on this question, it appears we have that for $$K=i^{\sharp}$$, we have that $$(K \oplus K)^J$$ has the same eigenvalues as $$J^{K \oplus K}$$

Question 1. For any almost complex structure $$K$$ on $$L_{\mathbb R}$$, does $$(K \oplus K)^J$$ always have the same eigenvalues as $$J^{K \oplus K}$$?

Question 2. For any eigenvalues $$(K \oplus K)^J$$ and $$J^{K \oplus K}$$ have in common, do the corresponding eigenspaces have the same underlying sets?

I think the answer to both questions is yes and that this need not be only for the case where we have an almost complex structure on $$L_{\mathbb R}^2$$ that is the realification of a complexification of a map on $$L_{\mathbb R}$$ (such map must, I think, be an almost complex structure on $$L_{\mathbb R}$$):

Question 3. For any almost complex structure $$H$$ on $$L_{\mathbb R}^2$$ (not necessarily the realification of a complexification of a map on $$L_{\mathbb R}$$) such that $$H$$ and $$J$$ commute, does $$H^J$$ always have the same eigenvalues as $$J^H$$?

Question 4. For any eigenvalues $$H^J$$ and $$J^H$$ have in common, do the corresponding eigenspaces have the same underlying sets?

Question 5. For any almost complex structures $$K$$ and $$M$$ on $$L_{\mathbb R}^2$$ that commute, are the eigenvalues of $$K^M$$ a subset of $$\{ \pm i\}$$?

Question 6. If yes to Question 5, then is it that $$K^K$$ has $$i$$ as its only eigenvalue if $$L \ne 0$$ and has no eigenvalues if $$L=0$$? (I assume $$L=0$$ iff $$L_{\mathbb R} = 0$$ iff $$(L_{\mathbb R})^{\mathbb C} = 0$$ iff $$L_{\mathbb R}^2 = 0$$)

• Arturo Magidin, this isn't my only question like this. See many of my questions about almost-complex or complexification. Why don't you close them all like in pokemon (gotta close 'em all)? Oct 27 at 15:49

The answer to all your questions is yes and has nothing to do with complexification. Let $$V$$ be a real vector space and let $$J,H$$ be two commuting linear complex structures on $$V$$ ($$J^2 = H^2 = -\operatorname{id}_W$$ and $$JH = HJ$$).

You can consider $$V$$ as a complex vector space with respect to $$H$$. Then, since $$J$$ commutes with $$H$$, the map $$J$$ is actually $$\mathbb{C}$$-linear as a map $$J^H \colon (V,H) \rightarrow (V,H)$$. As a real map, $$J \colon V \rightarrow V$$ doesn't have any eigenvalues since if $$Jv = \lambda v$$ then $$J^2v = \lambda^2 v = -v$$ which implies that $$\lambda^2 = -1$$. Considering $$J^H$$ as a complex linear map, the above calculation shows that the only possible eigenvalues of $$J^H$$ are $$\pm i$$. We also have a direct sum decomposition

$$V = \{ v \in V \, | \, J^H v = iv \ \iff Jv = Hv \} \oplus \{ v \in V \, | \, J^Hv = -iv \iff Jv = -Hv\}$$

where the first factor is the "eigenspace" of $$J^H$$ corresponding to the eigenvalue $$i$$ and the second is the "eigenspace" of $$J^H$$ corresponding to the eigenvalue $$-i$$. The only caveat is that one of the factors might be trivial so $$J^H$$ won't necessarily have both $$\pm i$$ as eigenvalues.

Similarly, you can consider $$V$$ as a complex vector space with respect to $$J$$ and then $$H^J \colon (V,J) \rightarrow (V,J)$$ is $$\mathbb{C}$$-linear with the only possible eigenvalues being $$\pm i$$ and you get a direct sum decomposition

$$V = \{ v \in V \, | \, H^Jv = iv \iff Hv = Jv \} \oplus \{ v \in V \, | \, H^Jv = -iv \iff Hv = -Jv \}$$ where the first factor is the "eigenspace" of $$H^J$$ corresponding to the eigenvalue $$i$$ and the second is the "eigenspace" of $$H^J$$ corresponding to the eigenvalue $$-i$$. This shows that $$J^H$$ and $$H^J$$ have the same eigenvalues and the same eigenspaces.

Finally, the map $$J^J \colon (V,J) \rightarrow (V,J)$$ is also complex linear and is just given by multiplication by $$i$$ so it has only $$i$$ as an eigenvalue (at least as long as $$V \neq \{ 0 \}$$).

• Thank you, levap. Feb 3 '20 at 11:42
• levap do you think the question should be closed? Oct 27 at 15:16
• @JohnSmithKyon: While I agree with the criticism that this question would benefit from "more focus" two years ago, I don't see the point in closing a question that was asked so long ago and was given an accepted answer. For all practical reasons, this question was already "closed". Oct 28 at 21:41