# On the vector bundles associated to the eigenspaces of an almost complex structure.

Let $(M,J)$ be an almost complex manifold, that is $M$ is a smooth differentiable manifold of even dimension and $J$ is an endomorphism of the tangent vector bundle $TM$ such that $J^2=-\textrm{id}_{TM}$.

Let $T^{\mathbb{C}}M$ be the complexification of $TM$, that is: $$T^{\mathbb{C}}M:=TM\otimes_{\mathbb{R}}\mathbb{C}=TM\oplus iTM,$$ where the tensor product is done fiber by fiber. Then, the endomorphism $J$ is diagonalizable on each fiber. Let $T^{(1,0)}M$, respectively $T^{(0,1)}M$ be the disjoint union of eigenspaces associated with $+i$, respectively $-i$. For example, one has: $$T^{(1,0)}M:=\coprod_{x\in M}E_i(J_{\vert T_x^{\mathbb{C}}M}).$$

Why are $T^{(1,0)}M$ and $T^{(0,1)}M$ vector bundles?

Roughly speaking, why the vector space structure of $E_{\pm i}(J_{\vert T_x^\mathbb{C}M})$ depends continuously on $x$?

More generally the following question arises:

Question. Let $\xi$ be a vector bundle and let $F$ be an endomorphism of $\xi$, when is $\ker(F)$ a vector bundle?

It seems that the question is non-empty since the dimension of $\ker(F_{\vert E_b})$ can change with the point $b\in B$, where $p\colon E\rightarrow B$ is the projection of $\xi$, which obviously prevents $\ker(F)$ from being a vector bundle.

Any enlightenment will be greatly appreciated.

• The answer to your main question is always, provided you include the hypothesis that $F:\xi \to \xi$ has constant rank. – Andrew Sep 20 '17 at 22:43
• – Mr. Chip Sep 20 '17 at 23:03

For the special case of $T^{1,0}M$ and $T^{0,1}M$, there is a simple direct argument: Given a local frame $\{v_j\}$ for $TM$, the sections $v_j-iJ(v_j)$ form a local frame for $T^{1,0}M$ and the sections $v_j+iJ(v_j)$ form a local frame for $T^{0,1}M$. So you get local trivializations for the bundles.
As stated in the comment by @Andrew, in the general setting (which even works for bundle maps between different bundles over the same manifold which have the identity as their base map), the only condition for $\text{ker}(F)$ being a smooth subbundle is that the rank is constant. To prove this, it is easier to first show that $\text{im}(F)$ is a smooth subbundle. Given $x\in M$, choose elements $v_1(x),\dots,v_k(x)\in E_x$ for which the images under $F$ form a basis for $F(E_x)$. Extending them arbitrarily to local sections defined on a neighbourhood of $x$, the images under $F$ are linearly independent in a (possibly smaller) neighbourhood of $x$. Since the rank is constant, they have to form a local frame on this neighbourhood.
Having this at hand, you can choose a basis $\{\tilde v_{k+1}(x),\dots,\tilde v_n(x)\}$ of $\text{ker}(F_x)$ and again extend to local sections $\tilde v_j$ arbitrarily. By assumption, $F(\tilde v_j)$ is a smooth linear combination of $F(v_1),\dots,F(v_k)$ for each $j=k+1,\dots,n$. If you subtract the corresponding multiples of $v_1,\dots,v_k$ from $\tilde v_j$ you get a smooth section $v_j$ of $\text{ker}(F)$. Since the resulting sections are linearly independent in $x$, they are linearly independent on a neighbourhood of $x$ and thus define a local frame.