Let $M$ be a smooth manifold of even dimension $2n$. I would like to understand the equivalence of the following two standard definitions for an almost complex structure on $M$.

Denote by $TM$ the tangent bundle of $M$. Then the first definition I wish to consider is the following.

Definition 1 An almost complex structure on $M$ is a vector bundle morphism $J:TM\rightarrow TM$ satisfying $J^2=-id_{TM}$.

Now $TM$ is a real vector bundle of rank $2n$ over $M$ and is thus classified up to bundle isomorphism by the homotopy class of a map $\tau_M:M\rightarrow BGl(\mathbb{R}^{2n})$, where $BGl(\mathbb{R}^{2n})$ is the classifying space of the Lie group $Gl(\mathbb{R}^{2n})$. The complex general linear group $Gl(\mathbb{C}^n)$ also has a classifying space $BGl(\mathbb{C}^n)$, and the map $\mathbb{C}^n\rightarrow \mathbb{R}^{2n}$, $\underline z=\underline x+i\underline y\mapsto (\underline x,\underline y)$, corresponds to a homomorphism $r:Gl(\mathbb{C}^n)\rightarrow Gl(\mathbb{R}^{2n})$ given by

$$r(X)=r(A+iB)=\begin{pmatrix}A & -B\\B & A\end{pmatrix}$$

where $A$, $B$ are real matrices. This homomorphism subsequently induces a map of classifying spaces $Br:Gl(\mathbb{C}^n)\rightarrow BGl(\mathbb{R}^{2n})$ with homotopy fibre $Gl(\mathbb{R}^{2n})/Gl(\mathbb{C}^n)$.

Definition 2 An almost complex structure on $M$ is a lift through $Br$, up to homotopy, of the classifying map $\tau_M$. $$\begin{array}{ccccccccc} & & BGl(\mathbb{C}^n) \\ & \nearrow & \downarrow{Br}\\ M & \xrightarrow{\tau_M} & BGl(\mathbb{R}^{2n}). \end{array}$$

Remark that $BGl(\mathbb{R}^{2n})$ is modelled by the Grassmannian $Gr_{2n}(\mathbb{R}^{\infty})$ of $2n$-planes in $\mathbb{R}^\infty$, and a corresponding statement holds for $BGl(\mathbb{C}^{n})$. Since $M$ is finite dimensional the map $\tau_M$ will factor through the inclusion $Gr_{2n}(\mathbb{R}^{2N})\hookrightarrow Gr_{2n}(\mathbb{R}^{\infty})$ for some $N\geq n$, and $Gr_{2n}(\mathbb{R}^{2N})$ may be given the structure of a smooth manifold and $\tau_M$ represented by a smooth map. Similarly $Gr_{N}(\mathbb{C}^{N})$ carries the structure of a complex manifold and $Br|$ may be chosen to be a smooth map between these spaces.

Now my question: How are these two definition equivalent?


2 Answers 2


The key point is to understand how the classifying map works. Here's one way to:

One can produce a bundle-embedding $TM \hookrightarrow M \times \bf R^\infty$ (which can moreover factor through some trivial bundle with finite-dimensional fibers), in which case there is a Gauss map $\mathscr{G} : M \to \text{Gr}_{2n}(\mathbf{R}^\infty)$ defined by sending each point $p \in M$ to the tangent space $T_p M \subset \mathbf{R}^\infty$. Modelling $B\text{GL}_{2n}(\mathbf{R})$ on $\text{Gr}_{2n}(\mathbf{R}^\infty)$, this map $\mathscr{G}$ is exactly the classifying map of the tangent bundle of $M$. This map does not depend on the bundle-embedding upto homotopy and one can recover $TM$ as $\mathscr{G}^* \gamma_{2n}$ where $\gamma_k \to \text{Gr}_k(\mathbf{R}^\infty)$ is the tautological $k$-plane bundle.

If $M$ admits a complex structure, then the fibers of $TM$ admit a (fiberwise continuously varying) complex structure. This enables one to construct a bundle embedding $TM \hookrightarrow M \times \bf R^\infty$ such that if $\mathbf{R}^\infty$ is realized as a "$\infty/2$"-dimensional complex vector space $\mathbf{C}^\infty$, then the bundle-embedding respects the fiberwise complex structure, i.e., $T_p M \subset \mathbf{C}^\infty$ is a $n$-dimensional complex subspace. Once that is done, let $\mathscr{G}$ be the corresponding Gauss map.

Then $\mathscr{G}$ has a natural lift to a map $M \to \text{Gr}_n(\mathbf{C}^\infty)$, factoring through the map of Grassmannians $\text{Gr}_n(\mathbf{C}^\infty) \to \text{Gr}_{2n}(\mathbf{R}^\infty)$ sending a complex $n$-subspace of $\mathbf{C}^\infty = \mathbf{R}^{2\infty}$ to it's underlying real $2n$-subspace. Modelling $B\text{GL}_n(\mathbf{C})$ as $\text{Gr}_n(\mathbf{C}^\infty)$ once again, this is the same as Definition 2. If the classifying space were modeled otherwise, one would have to do the checking that the map $Br : B\text{GL}_n(\mathbf{C}) \to B\text{GL}_{2n}(\mathbf{R})$ you wrote down is naturally isomorphic to the Grassmannian-level map I wrote down, by a commutative diagram.

This shows Definition 1 implies Definition 2. Can you try the converse?

  • $\begingroup$ Thanks for the answer. Actually the main thing I had trouble with was with constructing the correct embedding $TM\hookrightarrow M\times\mathbb{C}^\infty\subseteq M\times\mathbb{R}^\infty$. Could you elaborate on that at all, please? $\endgroup$
    – Tyrone
    Jul 8, 2018 at 16:23
  • $\begingroup$ @Tyrone I have written up here how to embed vector bundles in trivial Euclidean bundles. If the fibers admit complex structure, the exact same construction should produce a fiberwise complex structure preserving embedding to some $M \times \Bbb C^k$. $\endgroup$ Jul 8, 2018 at 16:31
  • $\begingroup$ Ok, I think I have sorted it. Thanks! $\endgroup$
    – Tyrone
    Jul 8, 2018 at 17:10

1 implies 2. Suppose defined an almost complex structure on $M$. Consider an atlas $U_i$ of $M$ where $U_i$ is contractible and diffeomorphic to an open subset of $ \mathbb {R} ^n$. The restriction $TM_i$ of the tangent bundle $TM$ to $U_i$ is trivial. Let $J_i$ be the restriction of $J$ to $U_i$, there exists an isomorphism of pseudocomplex bundles $g_{ij} $ between the restriction of $(TM_j, J_j)$ and $(TM_x, J_x)$ to $U_i\cap U_j$. The family of morphisms $g_{ij} $ is a trivialisation of $TM$. Remark that $g_{ij}(x) $ is a complex automorphism of $(TM_x, J_x)$. Thus it is an element of $Gl(n, \mathbb{C})$ and $TM$ has a $Gl(n, \mathbb{C})$ reduction.

2 implies 1. The existence of $\tau_M$ is equivalent to saying that $TM$ has a $Gl(n, \mathbb{C})$ reduction, which is equivalent to saying that we can suppose that $g_{ij} (x) \in Gl(n, \mathbb{C})$. Denote by $J_i$ the standard complex structure on $U_i$ and $f_i$ a partition of unity associated to $U_i$, write $J=\sum f_iJ_i$.

  • $\begingroup$ I'm sorry but I don't see how $1)\Rightarrow 2)$ from this. If $g_{if}(x)$ are supposed to be the transition functions coming from an arbitrary real atlas on $M$ then I don't see how they will commute with $J$ without further assumption. I also don't understand how to generate a homotopy class of map $M\rightarrow BGl(\mathbb{C}^n)$ lifting $\tau_M$ from this data. $\endgroup$
    – Tyrone
    Jul 8, 2018 at 16:15
  • $\begingroup$ For $2)\Rightarrow 1)$, I would like to understand the equivalence rather than assume it. What is the "standard complex structure on $U_i$"? $\endgroup$
    – Tyrone
    Jul 8, 2018 at 16:16
  • $\begingroup$ The atlas is not chosen arbitrary, they may exist different pseudo complex structures, but it one exists such an atlas exists. Consider the restriction of the pseudo complex structures on charts and write the condition they coincide on the intersection. $\endgroup$ Jul 8, 2018 at 16:19
  • $\begingroup$ $U_i$ is an open subset of $\mathbb{R} ^{2n}$. $\endgroup$ Jul 8, 2018 at 16:20
  • $\begingroup$ Then I think you are suggesting to restrict to such an atlas. How do you get a homotopy class $M\rightarrow BGl(\mathbb{C}^n)$ from this data? $\endgroup$
    – Tyrone
    Jul 8, 2018 at 17:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .