Betti numbers of quotients of spheres by circle actions Suppose $X=\mathbb{S}^{n+1}/\mathbb{S}^1$, the action of $\mathbb{S}^1$ being isometric and with closed orbits, and $n=\dim X$ (as an Alexandrov space) being even. My question is:

Is is true that all odd Betti numbers of $X$ vanish?

I know this is true when the action is almost free because in this case $X$ is a weighted projective space, by the classification of riemannian $1$-foliations on spheres, but i think there must be some simpler way to show this in general.
 A: Almost certainly yes. Here is a partial answer, though very algebraic and probably unnecessarily complicated. I only need to assume smoothness for this.
1) Smith theory and fixed points. A version of Smith theory (best known in the case of involutions as saying $\text{rk } H_*(M^{fix}; \Bbb F_2) \leq \text{rk } H_*(M;\Bbb F_2)$; here rank means total rank) for $S^1$ actions says that $\text{rk } H_*(M^{fix}; \Bbb Q) \leq \text{rk } H_*(M; \Bbb Q)$. For $M = S^{2n+1}$, we see that the fixed point set has the homology of a sphere of lower dimension. Indeed it must be the homology of a sphere of odd dimension (this comes from following along my favorite proof of Smith theory, which requires spectral sequences and localization theorems for equivariant homology.)
2) Quotients. Associated to an $S^1$ space $Y$ are both its homology $H_*(Y)$ and its equivariant homology $H_*^{S^1}(Y)$. These graded groups have a Gysin sequence $$H_*(Y) \to H_*^{S^1}(Y) \to H_{*-2}^{S^1}(Y) \to H_{*-1}(Y) \to \cdots$$ 
When the action of $S^1$ on $Y$ is semi-free, $H_*^{S^1}(Y;\Bbb Q) = H_*(Y/S^1;\Bbb Q)$.
3) Complements. Consider $Y^{S^1}$, the fixed point set of your action. Because it has the homology of a sphere $S^{2k+1}$, Alexander duality says that its complement has the homology of $S^{2n-2k-1}$. Its complement, in addition, has a semi-free $S^1$ action. 
4) Finally, whenever $Y$ has the homology of an odd-dimensional sphere, we can calculate what $H_*^{S^1}(Y)$ could possibly be. $H_*(Y)$ is supported in a single positive degree $d$; then we have two cases, depending on whether or not the map $H_d(Y) \to H_d^{S^1}(Y)$ is an isomorphism. If it is, then the Gysin sequence implies $H_*^{S^1}(Y) \cong H_*(\Sigma^d \Bbb{CP}^\infty)$ (that is, there's a $\Bbb Q$ in degree $d, d+2, d+4, ...$). Otherwise, the periodicity in the Gysin sequence has the opposite effect, and we see that it must have the same homology as $\Bbb{CP}^{d/2}$.
5) $Y \setminus Y^{S^1}$ has the homology of $S^{2n-2k-1}$ and is semi-free, so $H_*((Y \setminus Y^{S^1})/S^1) = H_*^{S^1}(Y \setminus Y^{S^1})$; because $(Y \setminus Y^{S^1})/S^1$ has the homotopy type of a finite CW complex, our equivariant homology calculation above must take the second form (we can't have homology in arbitrarily high degrees!) So the homology of the quotient of the complement of the fixed set is that of $\Bbb{CP}^{n-k-1}$.
Now I would need to run a Mayer-Vietoris sequence, but I got bored while calculating $(\partial N)/S^1$, $N$ a tubular neighborhood of the fixed set. 
