Cohomology of the Lagrangian Grassmannian I was wondering if some of you cold provide me with a reference where I can find what the cohomology ring of the Lagrangian Grassmannian of $\mathbb{C}^n$ looks like. In particular, I would need to know whether the second cohomology group $H^2(\Lambda(n); \mathbb{Z})$ is zero or not.
To give a bit of context, I have been reading the paper by Viterbo "Intersection de sous-variétés lagrangiennes, fonctionnelles d’action et indice des systèmes hamiltoniens" (one can find it here), and more specifically the proof that if the first Chern class of a symplectic vector bundle is 2-torsion, then there exists a cohomology class which induces the Maslov class on the fibres. The idea of the proof is to use the Serre spectral sequence for a quotient of the classifying bundle for $U(n)$, and then apply results which can be found in the collection "Cohomologie des espaces localement compacts d'apres J. Leray" by Armand Borel to deduce that there is such a class if it lies in the intersection of the kernels of all the differentials $d_k$ applied to elements in $E^{0, 2}_2$. One therefore needs to prove that $d_2(\mu)=\pm 2c_1$, where $\mu$ is the Maslov class $\mu\in H^1(\Lambda(n); \mathbb{Z}).$ The piece I miss to do so is exactly that $H^2(\Lambda(n); \mathbb{Z})=0$, or that the differential there is injective (which looks harder to prove).
 A: As $SO(n)$ is normal and index $2$ in $O(n)$, we may view $U(n)/O(n)$ as $\left(U(n)/SO(n)\right) / \mathbb{Z}_2$.
Now, let $f:U(n)\rightarrow SU(n)\times S^1$ be the diffeomorphism (which is NOT a group homomorphism) $f(A) = (B,\det A)$ where $B$ is obtained from $A$ by multiplying the first row of $A$ by $\det A^{-1}$.
Now, use $f$ to transport the $O(n)$ action on $U(n)$ to $SU(n)\times S^1$, i.e., we define an action of $O(n)$ on $SU(n)\times S^1$ by $C\ast(A,z) = f(C\ast f^{-1}(A,z))$.
Note that if $C\in SO(n)$, then the second coordinate of $C\ast(A,z)$ is $z$.  Further, for $g:=diag(-1,1,1,...,1)\in O(n)\setminus SO(n)$, the second coordinate of $g\ast(A,z)$ is $-z$.
So, viewing $U(n)/O(n)$ as $\left((SU(n)\times S^1)/SO(n)\right)/\mathbb{Z}_2$, we see that it's of the form $\left((SU(n)/SO(n))\times S^1\right)/\mathbb{Z}_2$.  Further, from the computation of $g$ above, we see that $g$ acts on this space diagonally, using the usual antipodal map on the $S^1$ factor.  In other words, this bundle is the associated bundle to the usual covering $\mathbb{Z}_2\rightarrow S^1\rightarrow S^1$.
Using the associated bundle construction, it follows that $U(n)/O(n)$ is a bundle over $S^1$ with fiber $SU(n)/SO(n)$.  From here, we see $\pi_1(SU(n)/SO(n)) = 0$ since $SU(n)$ is simply connected.  Thus, the LES in homotopy groups associated to $SU(n)/SO(n)\rightarrow U(n)/O(n)\rightarrow S^1$ now shows that $\pi_1(U(n)/O(n)) = \mathbb{Z}$.  Hurewicz and universal coefficients give $H^2(\Lambda(n);\mathbb{Z})$ is torsion free.
Thus, $H^2(\Lambda(n);\mathbb{Z}) = 0$ iff $H^2(\Lambda(n);\mathbb{Q}) = 0$.  Using the transfer homomorphism, we will be done showing $H^2(\Lambda(n);\mathbb{Z}) = 0$ if we show that $H^2(U(n)/SO(n);\mathbb{Q}) = 0$.  This space is diffeomorphic to $SU(n)/SO(n)\times S^1$, and $SU(n)/SO(n)$ has $\pi_1 = 0$ and $\pi_2 = \mathbb{Z}_2$.  It follows that $$H^1(SU(n)/SO(n);\mathbb{Q})\cong H^2(SU(n)/SO(n);\mathbb{Q}) = 0.$$  Kunneth now gives the final answer that $H^2(U(n)/SO(n);\mathbb{Q}) = 0$, so we are done.
