Almost complex manifolds are orientable

Question

I want to verify the fact that

every almost complex manifold is orientable.

By definition, an almost complex manifold is an even-dimensional smooth manifold $M^{2n}$ with a complex structure, i.e., a bundle isomorphism $J\colon TM\to TM$ such that $J^2=-I$, where $I$ is the identity.

For every tangent space $T_pM$, there exists tangent vectors $v_1,\cdots, v_n$ such that $(v_1,\cdots, v_n,Jv_1,\cdots,Jv_n)$ is a real ordered basis for $T_pM$. I've checked that any two such bases are related by a matrix with positive determinant.

So for each tangent space $T_pM$ I fix such a basis. Hence, I've obtained a possibly discontinuous global frame $X_1,\cdots,X_n,JX_1,\cdots,JX_n$. To show this frame determine an orientation on $M$, I've been trying to construct for each $p\in M$ a coordinate neighborhood $(U,x^1,\cdots,x^{2n})$ such that

$$(dx^1\wedge\cdots dx^{2n})(X_1,\cdots,X_n,JX_1,\cdots JX_n)>0$$ for every point in $U$.

How do I show that? Thanks!

Answer 1

Here's a slightly different way of looking at it. First, note that we can choose a Riemannian metric $g$ on $M$ that satisfies $g(X,Y) = g(JX, JY)$ (start with any Riemannian metric $h$ and then define $g(X,Y) = h(X,Y) + h(JX, JY)$). You can then show that $\omega(X,Y) = g(X,JY)$ is skew-symmetric:

\begin{align} \omega(X,Y) & = g(X,JY) = g(JY, X) \\& = g(J^2Y, JX) = -g(Y, JX) = -\omega(Y,X) \end{align}

and is therefore a 2-form. $\omega$ is also non-degenerate, since $\omega(X, -) = - g(JX, -)$ and a metric is non-degenerate by definition. Then since $\omega$ is non-degenerate, its top exterior power $\omega^n = \omega\wedge \dots \wedge \omega$ is nowhere zero. A nowhere vanishing top form defines an orientation on $M$, so we are done.

Answer 2

I see you have added an answer - and Charlie's is (more than) fine too. Here is another way...

First of all - you have checked this, but for completeness - if we endow a real vector space $V$ of dimension $2n$ with a complex structure (i.e.,view it as a $C=\mathbb R[J]$ module), then we have chosen an orientation. For, suppose $v_1,\cdots, v_n$ and $w_1,\cdots, w_n$ are $C$-bases, and $g $ is the transition matrix ($g$ has entries in $C$, or $g = a + Jb$, with $a$, $b$ real) between the two, with determinant $d\in C$. Then $g,$ viewed as a transformation over the reals, has determinant $\det_{\mathbb R}g =d\bar d$, and hence is positive.

Proof: [Updated in 2025, in response to a very valid comment below.]

Viewing $$g\colon V\to V$$ as a $\mathbb R$-linear map, we get a $\mathbb R$-linear map $$\det g\colon \det V \to \det V,$$ where $\det V$ is the real, one-dimensional, vector space $\Lambda^{2n}V$. Since $\det V$ is one-dimensional, one can identify $\det g$ with a real number once a choice of bases is made: here, we wish to calculate the real number $\mathop{\det} g$ such that $$w_1\wedge\cdots w_n\wedge Jw_1\wedge\cdots\wedge Jw_n = {\det} g\, v_1\wedge\cdots v_n\wedge Jv_1\wedge\cdots\wedge Jv_n.$$ Tensor $V$ with $\mathbb C$ - that is, extend scalars. The determinant 'does not change:'$$({\det}_{\mathbb R}g )\otimes 1 = {\det}_{\mathbb C} (g\otimes 1).$$

Now, $g$ commutes with $J$, so we can calculate $\det_{\mathbb C} (g\otimes 1)$ making use of the fact that $g\otimes 1$ preserves $V_{\pm i}$, the $\pm i$-eigen-spaces of $J$:

For $v\in V$ (or $\in V\otimes{\mathbb C}$), write $$ v = v^+ + v^-,$$ with $v^{\pm}\in V_{\pm i}.$ [Explicitly, $$v^{\pm} = {1\over 2}\Big(1\pm {J\over i}\Big)v,$$ if one cares.]

Now, $g\otimes 1$, acting on $V_{\pm i}$, is the transition matrix $a\pm ib$ taking $$v_1^{\pm},\cdots, v^{\pm}_n\ \text{to}\ w_1^{\pm},\cdots,w_n^{\pm}.$$ Therefore, $$w_1^\pm\wedge \cdots \wedge w^\pm_n = \det (a \pm i b)\, v_1^\pm\wedge \cdots \wedge v^\pm_n,$$ so that $$w^+_1\wedge\cdots \wedge w^+_n\wedge w^-_1\wedge\cdots \wedge w^-_n = d\bar d v_1^+\wedge\cdots \wedge v_n^+\wedge v_1^-\wedge\cdots\wedge v_n^-.$$ Now, the non-zero constant $K$ such that $$v_1\wedge \cdots\wedge v_n\wedge Jv_1\wedge\cdots\wedge Jv_n= K\, v_1^+\wedge\cdots \wedge v_n^+\wedge v_1^-\wedge\cdots \wedge v_n^-$$ is (of course?) the same that would appear above with the $w_k$ in place of the $v_k$. [Explicitly, $K= (-2i)^n$, if I haven't goofed up.]

Therefore, we also get $$ w_1\wedge\cdots \wedge w_n \wedge J w_1\wedge\cdots \wedge J w_n = d\bar d\ v_1\wedge\cdots \wedge v_n \wedge J v_1\wedge\cdots \wedge J v_n.$$ The real number $ d\bar d$ is positive, and the above orientation on $V$ does not depend on choice of $C$-basis.

To answer the question proper...

By assumption, $J$ is a global tensor, and thus each $T_p(M)$ is endowed with a $J$ structure, and hence (by the previous argument), an orientation, independent of charts. To bring this in line with the original formulation in your question: suppose $$\phi: U\subset M\to \mathbb R^{2n},$$ is a chart, and $p\in U$. Then $\partial/\partial x_1|_p,\cdots ,\partial/ \partial x_{2n}|_p$ - obtained from $\phi$ (or $\phi^{-1}$), is an oriented basis. If the orientation fails to match the $J$ orientation, modify $\phi$ by swapping the first two coordinates. By continuity of $J$, the (modified, if needed) chart will give you an oriented frame on all of $U$ that matches that of $J$. The manifold $M$ is therefore orientable - that is the transition function jacobian determinants will be positive.

Answer 3

I finally figured this out, so I want to post my own answer, since I couldn't find a satisfactory answer myself online.

It suffices to prove that for every $p\in M$, there exists a local frame $(\sigma_1,\cdots,\sigma_n,J\sigma_1,\cdots,J\sigma_n)$ in some open neighborhood $U$ of $p$.

Fix $p\in M$ and choose an ordered basis $(v_1,\cdots,v_n,Jv_1,\cdots,Jv_n)$. Choose sections $\sigma_1,\cdots,\sigma_n$ such that $\sigma_i(p)=v_i$ for each $1\leq i\leq n$. Since $(\sigma_1|_p,\cdots,\sigma_n|_p,J\sigma_1|_p,\cdots,J\sigma_n|_p)$ is linearly independent,

$$\omega=\sigma_1\wedge\cdots\sigma_n\wedge(J\sigma_1)\wedge\cdots\wedge(J\sigma_n)\neq0$$

at $p$. Since $\omega$ is continuous, it is nonzero for some neighborhood $U$ of $p$. In other words,$(\sigma_1,\cdots,\sigma_n,J\sigma_1,\cdots,J\sigma_n)$ is linearly independent in $U$, so that it is a local frame.

Answer 4

This answer summarizes Charlie Cifarelli's nice answer using only words:

1. A paracompact manifold can be endowed with a Riemannian metric.

2. A paracompact, almost complex manifold can be endowed with an almost Hermitian metric.

3. An almost Hermitian manifold is an almost symplectic manifold.

4. An almost symplectic manifold has a canonical volume form, and is hence orientable.

Answer 5

A complex vector space is a real vector space $V$ endowed with an automorphism $J:V\rightarrow V$ such that $J^2=-I$. For example, on the real tangent space at a point on a complex manifold one gets such a $J$ by fixing a chart homeomorphic to $\mathbb{C}^n$ and transferring the action of $i$. A complex morphism $A:V\rightarrow V$ is a real linear map such that $AJ=JA$. If one fixes a basis of type $\{v_1,Jv_1,\ldots,v_n,Jv_n\}$ on $V$, which one can always do, then with respect to the decomposition $V=W_+\oplus W_-$ where $W_+:=\{v_1,\ldots,v_n\}$ and $W_-:=\{Jv_1,\ldots,Jv_n\}$ the morphism $A$ has the block decomposition $\left(\begin{array}{cc} B& C\\ -C& B\end{array}\right)$. Then $UAU^*=(B+iC)\oplus (B-iC)$, where $U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1&i \\ i&1\end{array}\right)$. So $$\det(A)=\det(UAU^*)=\det(B+iC)\det(B-iC)>0.$$ This works to rapidly prove that a cx manifold is oriented as the determinant of the (differential of the holomorphic) transition mappings will be positive by the previous reasoning.

For an almost cx manifold, the above reasoning does not work unless you can guarantee that you have charts that take $J$ to some canonical form. But orientability is simpler than that. You want to prove that you can construct a global continuous section of the orientation double cover. The idea of looking for basis of type $v_1,v_2,\ldots, v_n,Jv_1,\ldots,Jv_n$ is good. The way to make this argument correct is as follows. Just as in the previous paragraph you want to choose splittings $T_xM=W_x\oplus J_xW_x$ which are coherent (read continuous) with $x$. You can do this locally since the condition $W\cap JW=\{0\}$ is open in the Grassmannian $G(n,2n)$. So roughly, the same subspace $W\subset \mathbb{R}^{2n}$ transferred via the differential of the same chart map will do the trick. So a basis $w_1,\ldots, w_n$ of $W$ will give a basis $v_1(y),\ldots,v_n(y),Jv_1(y),\ldots Jv_n(y)$ for every $y$ in the neighborhood of $x$ and the dependence on $y$ is continuous. While you can't guarantee the choice of such basis globally, note that the matrix of change of basis between two such basis has $\det>0$ since if you choose another splitting $T_xM=W_x'\oplus J_xW_x'$ with $W_x'$ generated by $w_1',\ldots,w_n'$, then the matrix of change of basis between $\{w_1,\ldots,w_n,J_xw_1,\ldots,J_xw_n\}$ and $\{w_1',\ldots,w_n',J_xw_1',\ldots, J_xw_n'\}$ will be of type $A\oplus J_xAJ_x^{-1}$.So you end up in the same orientation class. This guarantees a continuous section of the orientation double cover.

Answer 6

For each point $x\in M$, $(T_xM,J_x)$ is a real vector space with complex structure $J_x$, there exists a collection of vectors $\{X^1_x,\cdots,X^n_x,J_xX^1_x,\cdots,J_xX^n_x\}$ forms a basis of $T_xM$. We can extend each vector $X^k_x$ to be a local vector field $X^k$ such that the vector fields $\{X^1,X^2,\cdots,X^n\}$ are linearly independent on an open set $U'$. On the other hand, becasue of the smoothness of $J$, there exists an open set $U\subseteq U'$ such that vector fields $\{X^1,\cdots,X^n,JX^1,\cdots,JX^n\}$ are linearly independent on $U$. Namely, there is an open covering $(U_i)$ of $M$ such that over each $U_i$, we have a frame of form $\{X^1,\cdots,X^n,JX^1,\cdots,JX^n\}$.