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 $$w_1\wedge\cdots w_n\wedge Jw_1\wedge Jw_n = {\det} g\, v_1\wedge\cdots v_n\wedge Jv_1\wedge Jv_n.$$ 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 $$w^+_1\wedge\cdots \wedge w^+_n\wedge w^-_1\wedge\cdots \wedge w^-_n = d\bar d v_1^+\wedge\cdots \wedge v_n^+\wedge v_1^-\wedge\cdots v_n^-.$$ 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

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 $\{X^1,\cdots,X^n,JX^1,\cdots,JX^n\}$.