6

The set of almost complex structures on $\mathbb R^{2n}$ is given by $$ M_n = \frac{GL(2n,\mathbb R)}{GL(n,\mathbb C)} = \mathcal C_+ \sqcup \mathcal C_-,$$ taking into account that $\det = \pm 1$ gives two disjoint sets.

How do we show that $M_2=S^2 \sqcup S^2$? Moreover, I'm a bit confused by dimensions: how does $\dim_{\mathbb R}M_2=8$ relates with the latter decomposition?

This is exercise 1.2.1 from the book Complex geometry by Huybrechts.

jj_p
  • 2,500
  • 1
  • 15
  • 31
  • I am a beginner and I am confused. Is the decomposition $\mathcal C_+ \sqcup \mathcal C_-$ really because of $\det = \pm 1$? In the case of $\mathbb R^2$, the two complex structures are given by $J_1=\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ and $J_2=\begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}$. But both of these have $\det=1$. What am I missing? – Mohith Nagaraju Oct 20 '23 at 04:43
  • 1
    @MohithNagaraju: Since $GL(2, \mathbb{R})$ acts transitively, there is $A \in GL(2, \mathbb{R})$ such that $J_2 = A^{-1}J_1A$. If $A$ has positive determinant, then $J_1$ and $J_2$ are in the same connected component, while if $A$ has negative determinant, then $J_1$ and $J_2$ are in different connected components. Note, $A$ is not unique and if $J_2 = B^{-1}J_1B$, then $A = CB$ where $C \in GL(1, \mathbb{C})$ - note that $\det C > 0$, so $\det A$ and $\det B$ have the same sign. In the (left) quotient $GL(2, \mathbb{R})/GL(1, \mathbb{C})$, the equivalence class $[A]$ corresponds to $J_2$. – Michael Albanese Jan 22 '24 at 03:48

1 Answers1

3

As you noted, $M$ is not diffeomorphic to $S^2\coprod S^2$ for dimension reasons.

On the other hand, what is true is $M$ is homotopy equivalent to $S^2 \coprod S^2$.

(The following argument is partly adapted from a paper of Montgomery)

To see this, it's enough to show that $Gl^+(4,\mathbb{R})/Gl(2,\mathbb{C})$ is homotopy equivalent to $S^2$, where $Gl^+$ denotes those matrices of positive determinant.

Now, consider the subgroups $U(2)\subseteq Gl(2,\mathbb{C})$ and $SO(4)\subseteq Gl^+(4,\mathbb{R})$.

It's relatively well known that $Gl(2,\mathbb{C})$ is diffeomorphic to $U(2)\times \mathbb{R}^4$ and that $Gl^+(4,\mathbb{R})$ is diffeomorphic to $SO(4)\times \mathbb{R}^{10}$. Further, in the usual inclusion $Gl(2,\mathbb{C})\rightarrow Gl^+(4,\mathbb{R})$, $U(2)$ becomes a subgroup of $SO(4)$.

Now, the chain of subgroups $U(2)\subseteq SO(4)\subseteq Gl^+(4,\mathbb{R})$ gives rise to a homogeneous fibration $$SO(4)/U(2)\rightarrow Gl^+(4,\mathbb{R})/U(2)\rightarrow Gl^+(4,\mathbb{R})/SO(4).$$ In light of the above diffeomorphisms, $Gl^+(4,\mathbb{R})/SO(4)$ is diffeomorphic to $\mathbb{R}^{10}$. Since Euclidean spaces are contractible, it follows that the fibration is trivial, so $Gl^+(4,\mathbb{R})/U(2)$ is diffeomorphic to $SO(4)/U(2)\times \mathbb{R}^{10}$. In particular, $SO(4)/U(2)$ is homotopy equivalent to $Gl^+(4,\mathbb{R})/U(2)$.

Now, consider the chain of subgroups $U(2)\subseteq Gl(2,\mathbb{C})\subseteq Gl^+(4,\mathbb{R})$. This gives rise to a homogeneous fibration $$Gl(2,\mathbb{C})/U(2)\rightarrow Gl^+(4,\mathbb{R})/U(2) \rightarrow Gl^+(4,\mathbb{R})/Gl(2,\mathbb{C}).$$ In this case, the fiber is diffeomoprhic to $\mathbb{R}^4$, which immediately implies that $Gl^+(4,\mathbb{R})/U(2)$ is homotopy equivalent to $Gl^+(4,\mathbb{R})/Gl(2,\mathbb{C})$.

Putting the last two paragraphs together, we now know that $SO(4)/U(2)$ is homotopy equivalent to $Gl^+(4,\mathbb{R})/Gl(2,\mathbb{C})$.

To finish off the argument, we need to show that $SO(4)/U(2)$ is diffeomorphic to $S^2$.
To see this, first note that $U(2)$ intersects the center $Z(SO(4)) = \{\pm I\}$ of $SO(4)$. It follows that $$SO(4)/U(2) \cong [SO(4)/Z(SO(4)]/[U(2)/(Z(SO(4))\cap U(2))].$$

But $SO(4)/Z(SO(4))\cong SO(3)\times SO(3)$ and $U(2)/(Z(SO(4))\cap U(2)) \cong SO(3)\times S^1$. So, $SO(4)/U(2)\cong (SO(3)\times SO(3))/(SO(3)\times S^1)\cong SO(3)/S^1$.

But the standard action of $SO(3)$ on $S^2$ is transitive with stabilizer $S^1$, so $SO(3)/S^1 \cong S^2$.

Jason DeVito - on hiatus
  • 51,620
  • 2
    This argument works for any $n$ to show $Gl^+(2n,\mathbb{R})/Gl(n,\mathbb{C})$ is always homotopy equivalent to $SO(2n)/U(n)$. I don't think $SO(2n)/U(n)$ has a nicer description for most $n$. But, when $n=1$, it's trivial. When $n=2$, it's $S^2$, and when $n=3$, it's $SU(4)/U(3) = \mathbb{C}P^3$. – Jason DeVito - on hiatus Sep 08 '14 at 18:20
  • Thanks for the nice answer; I also found an alternative way of looking at this fact, though I'm not able to fully understand it. It goes as follows: a matrix in $GL^+(2n,\mathbb R)/GL(n,\mathbb C)$ has polar decomposition $J=SO=-O^{-1}S^{-1}=(O^tS^{-1}O)(-O^{-1})$. This is OK, and implies that $O=-O^{-1}=P^{-1}J_0P$ defines an element $U(n)P \in SO(2n)/U(n)$: what I don't understand of this last statement is how the properties of $O$ imply that $P \in GL(2n,\mathbb R)$ is also in $SO(2n)$. Then one could conclude observing that $SO(4)/U(2)$ is $S^2$. Could you help? – jj_p Sep 12 '14 at 16:45
  • Here $J_0$ is the standard symplectic form – jj_p Sep 12 '14 at 16:46
  • @jj_p: I'm not sure I understand all the notation. For example, why is $SO = -O^{-1} S$? Is $J$ a specific matrix? And are you claiming that any orthogonal matrix can be obtained from $J_0$ by conjugating? – Jason DeVito - on hiatus Sep 12 '14 at 18:29
  • I try to explain better: $J$ is a generic complex structure, s.t. $J^2=-1$, and the polar decomposition is given by a symmetric matrix $S=S^t$ times an orthogonal matrix $O^t=O^{-1}$. So the equality you cite, follows from $J=-J^{-1}$. The other property one uses to prove the second set of equalities is uniqueness of $O$ in the polar decomposition. Does these clarify notation? – jj_p Sep 12 '14 at 19:58
  • 1
    @jj_p: Sorry for the late response. The farthest I've been able to get is that $(PP^t) J_0 (PP^t)^{-1} = J_0$. So, we conjugate $J_0$ by a symmetric matrix and get $J_0$ back. I've been unable to show that this implies $PP^t = I$, and I'd have no idea how to get to showing $\det P = 1$ from there. Sorry! – Jason DeVito - on hiatus Sep 17 '14 at 17:54