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How can I prove that the space of all complex structures $Icolon Vto V$ is homotopy equivalent to the space of all non-degenerate skew-symmetric $2$-forms $omega in Lambda^2V$?

Skew-symmetry is seems to be related with the property $I^2=-operatorname{id}_V$, but I can't formalize it to complete proof.

First of all, recall that a vector space admits a complex structure if and only if it is even-dimensional, which also means it admits symplectic forms.

The proof outlined by Olivier Bégassat in the comments is correct. In this answer, I give some more details but fall short of providing a full proof. As Olivier argued, the idea is to show that the space $mathcal J(V)times_text{comp}Omega (V)$ of pairs $(J,omega)$, where $J$ is $omega$-compatible (meaning $omega(-,J-)$ is a scalar product) is a fibration over $mathcal J(V)$ and $Omega(V)$, the spaces of complex structures and symplectic forms on $V$, respectively, with contractible fibers. This should yield a homotopy equivalence $mathcal J(V)simeq Omega(V)$ after applying the long exact sequence associated to a fibration. In this post, I will focus on the contractibility of the fibers. I invite anyone willing to complete the proof, in which case I will turn this answer into a community wiki.

First, we consider the map $mathcal J(V)times_text{comp} Omega(V)to mathcal J(V)$. The fiber over $Jin mathcal J(V)$ is the space of symplectic forms that $J$ is compatible with, $Omega(V,J)$. Note that the fiber over any $Jin mathcal J(V)$ is not the empty set, since any $Jin mathcal J(V)$ is compatible with some symplectic form: Using induction, it is not hard to show that every $V^{2n}$ equipped with a complex structure $J$ admits a basis of the form ${x_1,J x_1,dots, x_n,J x_n}$. Declaring this basis to be orthonormal, we obtain a scalar product $g(-,-)$ on $V$ such that $J$ is orthogonal with respect to it. Then $omega(-,-)=g(J-,-)$ is a symplectic form on $V$ and $J$ is compatible with $omega$.

It is easy to see that each fiber is a convex space: If $J$ is compatible with $omega$ and $omega'$, then for $tin[0,1]$ the combination $(1-t)omega(-,J-)+tomega'(-,J-)$ is an inner product. Thus, the fiber over $J$ is contractible.

Now, consider the fiber of the map $mathcal J(V)times_text{comp} Omega(V)to Omega(V)$, i.e. the space of $omega$-compatible complex structures $mathcal J(V,omega)$. Proving contractibility of this space is more cumbersome, and I will not prove all the details. It is shown in e.g. proposition 2.50 of McDuff and Salamon's book Introduction to Symplectic Geometry. The idea is to show that $mathcal J(V,omega)$ is homeomorphic to the space of symmetric, positive-definite symplectic matrices. After picking a standard basis such that $omega$ is the standard symplectic form on $Bbb R^{2n}$, this map is given by $Jmapsto J_0J$ where
$$ J_0=begin{pmatrix}0 & -operatorname{id}_n \ operatorname{id}_n & 0 end{pmatrix}$$ represents multiplication by $i$. The inverse map sends $Pmapsto -J_0 P$. To prove contractibility, use the fact that if $P$ is a symmetric, positive-definite and symplectic matrix, then $P^alpha$ for any $alphain Bbb R_+$ is also a symplectic matrix, and hence this space is contractible by letting $alpha$ run from $1$ to $0$, mapping $P$ to the identity matrix.

Now, once we establish that the maps from $mathcal J(V)times_text{comp} Omega(V)$ are indeed fibrations you obtain the result we were after.