# Equivalent definitions of almost complex structures

The definition I have seen for an almost complex structure is the following

$$J:TM\to TM$$ which is linear fibre by fibre, such that $J^2 = -\text{Id}$, and such that $\pi(J(X_x)) = x$ where $\pi$ is the projection of $TM$ on $M$.

I see that this implies a map $$\tilde{J}:\mathfrak{X}(M)\to\mathfrak{X}(M)$$ such that $(\tilde{J})^2 = -\text{Id}$ and given by $\tilde{J}(X)(x) = J(X_x)$

My question is:

Given a linear map $\tilde{J}:\mathfrak{X}(M)\to\mathfrak{X}(M)$ such that $(\tilde{J})^2 = -\text{Id}$, can we reconstruct an almost complex structure on $M$?

• What is the definition of $J(X_x)$ in $J(X)(x) = J(X_x)$? – user99914 Jun 8 '18 at 10:30
• Okay, I'll just. it isn't very well written – tomak Jun 8 '18 at 10:32
• Are there more properties assumed? For now $\tilde J$ might not preserve the fiber, and might not be linear on the fiber. – user99914 Jun 8 '18 at 10:36
• The problem is that I don't know. I can go from $J$ to $\tilde{J}$, but I don't know the properties that I would need to be able to go the other way. So if there are properties that could be added to $\tilde{J}$ (that $\tilde{J}$ does indeed have if it comes from some $J$) we can add them. – tomak Jun 8 '18 at 11:55
• for instance, it would be a good idea to add linearity to $\tilde{J}$ – tomak Jun 8 '18 at 12:00

Let $V$ be an infinite-dimensional vector space. By the axiom of choice, there is a basis $\{v_i\}_{i\in I}$ for $V$. As $I$ is infinite, we can find $A, B \subset I$ with $A\cap B = \emptyset$, $A\cup B = I$, and $|A| = |B|$. Let $\sigma : A \to B$ be a bijection. Then we can define a linear map $\tilde{J} : V \to V$ by setting $\tilde{J}(v_a) = v_{\sigma(a)}$ and $\tilde{J}(v_b) = -v_{\sigma^{-1}(b)}$; note that $\tilde{J}^2 = -\operatorname{id}_V$.
Now note that $\mathfrak{X}(M)$ is an infinite-dimensional vector space if $\dim M > 0$. If we could construct a $J$ from $\tilde{J}$, we would see that every positive-dimensional manifold admits an almost complex structure which is absurd.