Let V be a symplectic vector space. Let $L_0$ be a Lagrangian subspace. Show that the space of all Lagrangian subspaces of V with transverse intersection with $L_0$ is contractible.
This is actually part of a bigger problem. Prior to this I have shown that there is a bijection between $J$ and $L(V,L_0)$ x $G(L_0)$. $J$ = space of all almost complex structures on V compatible with the symplectic structure. $L(V,L_0)$ is the space of all Lagrangian subspaces of V with transverse intersection with $L_0$ and $G(L_0)$ is the space of all inner-products on $L_0$
The above mentioned bijection is J going to $(JL_0,G_J(restricted L_0))$
The hint given is to show that $L(V,L_0)$ is identified with the space of all symmetric matrices. How do I do that?
First, let's try and make this problem simpler: whenever $L$ is Lagrangian, we can identify $V$ with $L\oplus L^{\ast}$ where $L^{\ast}$ is the dual vector space to $L$, with symplectic form given by $$\omega((v,v^{\ast}), (w,w^{\ast})) = w^{\ast}(v) - v^{\ast}(w)$$ Now here is a hint: you should show that any half-dimensional subspace $L^{\prime}$ of $V$ that is transverse to $L$ can be written as the graph of a linear map $A:L^{\ast} \to L$, in the sense that $L^{\prime} = \{(A(x), x): x \in L^{\ast}\}$. Then you should think about what condition on $A$ makes the graph of $A$ a Lagrangian subspace: you must have $$ \omega( (A(x),x), (A(y), y)) = y(A(x)) - x(A(y)) = 0$$ for all $x,y \in L^{\ast}$. How does this imply that $A$ gives a symmetric bilinear form on $L$? Lastly, you should show that the space of symmetric bilinear forms on a vector space is contractible by giving an explicit homotopy to the zero matrix.