Does the tangent bundle of an almost complex manifold split into the direct sum of $n$ two-dimensional $J$-invariant subbundles?

Question

Is it true that for a manifold (of dimension $2n$) with an almost complex structure $J$, the tangent bundle splits as sum of $n$ $2$-dimensional invariant sub bundles? At least is it true that for a linear map $J$ with $J^2=-I$, the vector space splits as a sum of $n$ invariant $2$-dimensional subspaces?

Answer 1

The tangent bundle of an almost complex manifold $(M, J)$ can be viewed as a complex vector bundle by defining $(a + bi)\cdot v = av + bJ(v)$. A real subbundle $E \subseteq TM$ is $J$-invariant if and only if $E$ is a complex subbundle. Therefore, your first question is equivalent to the following:

For an almost complex manifold $M$, is $TM$ isomorphic to the direct sum of $n$ complex line bundles?

The answer is no. For example, consider the tangent bundle of $\mathbb{CP}^2$. Suppose $T\mathbb{CP}^2 \cong L_1\oplus L_2$, then

$$c(T\mathbb{CP}^2) = c(L_1)c(L_2) = (1 + ax)(1+bx) = 1 + (a + b)x + abx^2$$

where $H^*(\mathbb{CP}^2; \mathbb{Z}) \cong \mathbb{Z}[x]/(x^3)$, and $c_1(L_1) = ax$, $c_1(L_2) = bx$. On the other hand $c(T\mathbb{CP}^2) = 1 + 3x + 3x^2$, so $a + b = 3$ and $ab = 3$. This is impossible as $a, b \in \mathbb{Z}$.

As for the second question, note that a real $2n$-dimensional vector space $V$ with a linear map $J : V \to V$ with $J^2 = -\operatorname{id}_V$ can be viewed as an $n$-dimensional complex vector space by defining $(a + bi)\cdot v = av + bJ(v)$. A real subspace $W \subseteq V$ is $J$-invariant if and only if $W$ is a complex subspace. Therefore, your second question is equivalent to the following:

Is an $n$-dimensional complex vector space $V$ isomorphic to the direct sum of $n$ one-dimensional complex vector spaces?

The answer is yes because, after choosing a complex basis for $V$, we have $V \cong \mathbb{C}^n \cong \mathbb{C}\oplus\dots\oplus\mathbb{C}$.