# Calculation of $H^{2}(S^{2k-1}/G,\mathbb{Z})$

Hi!

I'm sure it is well known but i don't know enough algebraic topology... Let $k\geq 2$, $$S^{2k-1}\hookrightarrow \mathbb{C}^{k}$$ be the unit sphere and $G$ be a finite subgroup of $U(k)$ acting linearly (the action is induced by the one on $\mathbb{C}^{k}$) and freely on $S^{2k-1}$ so the quotient $S^{2k-1}/G$ is a manifold. Is there a quick and dirty method to calculate $H^{2}(S^{2k-1}/G,\mathbb{Z})$? I could be happy with $H^{2}(S^{2k-1}/G,\mathbb{R})$ but it would be better $H^{2}(S^{2k-1}/G,\mathbb{Z})$.

• Under these conditions, the cohomology of the quotient (with real coefficients) is the space of $G$ invariants in the cohomology of the total space $S^{2k-1}$ and hence $H^2$ of the quotient , with real coeffts., vanishes. – Venkataramana Apr 11 '13 at 13:46
This spectral sequence in your setup has $$E_2^{p,q} = H^p(G;H^q(S^{2k-1};\mathbb{Z}))$$ and converges to a graded group associated to $H^\ast(S^{2k-1}/G;\mathbb{Z})$. Since you have kindly specified that $2k-1\ge 3$, this makes it easy to see that $$H^2(S^{2k-1}/G;\mathbb{Z})\cong H^2(G;H^0(S^{2k-1};\mathbb{Z}))\cong H^2(G;\mathbb{Z}).$$ So it's the second cohomology group of $G$, which classifies complex line bundles on $BG$ (if you like topology) or classifies central extensions of $G$ by $\mathbb{Z}$ (if you like group theory). Perhaps you already know what this group is, or can look it up.