Homology groups of unit tangent bundle Let $M$ be the closed connected orientable $n$-dimensional manifold and let $S(M)$ be the unit tangent bundle i.e. a set $\lbrace (x,v)| x\in M,\ v\in \mathrm{T}M,\ |v|=1\rbrace$. I want to compute homology groups $H_{*}(S(M))$ in terms of $H_{*}(M)$. How can I do it?
My attempt: There is a fiber bundle $\mathbb{S}^{n-1} \hookrightarrow S(M) \rightarrow M$. Second page of Serre spectral sequence is determined by formula $E^{2}_{p,q} = H_{p}(M; H_{q}(\mathbb{S}^{n-1})) = H_{p}(M) \otimes H_{q}(\mathbb{S}^{n-1})$. By dimension reasoning, $E^{2} = E^{3} = \dots = E^{n}$ and $E^{n+1} = E^{n+2} = \dots = E^{\infty}$. At $E^{n}$ we have only one nontrivial differential $\partial: H_{n}(M) \rightarrow H_{0}(M)$. So I can deduce that
$\begin{cases}
H_{k}(S(M)) = H_{k}(M),\ \text{if}\ k < n-1\\
H_{n+k-1}(S(M)) = H_{k}(M),\ \text{if}\ k > 1\\
H_{n-1}(S(M)) = H_{n-1}(M) \oplus \mathbb{Z}/\mathrm{Im}\partial\\
H_{n}(S(M)) = H_{1}(M) \oplus \mathrm{Ker}\partial
\end{cases}$
It is well-known that $H_{0}(M) = H_{n}(M) = \mathbb{Z}$ so $\partial$ is just multiplication by some number $d$. How can one calculate this number?
Hypothesis: $d=\chi(M)$ where $\chi(M)$ is Euler characteristic of manifold $M$. I know that it is true for genus $g$ surfaces. Also if $\chi(M) = 0$ then $d=0$. It can be derived from the Gysin exact sequence. More precisely, $\chi(M) = 0$ is equivalent to existence of nonvanishing vector field i.e. section $i$ of fiber bundle $p: S(M) \rightarrow M$. Gysin sequence at term $m=n$ looks like
$H_{n}(S(M)) \stackrel{p_{*}}{\rightarrow}H_{n}(M) \stackrel{\partial}{\rightarrow}H_{0}(M) $
Since $p$ have section $i$ then $p_{*}$ have section $i_{*}$, hence $p_{*}$ is surjection. Then $\mathrm{Ker}\partial = \mathrm{Im}p_{*} = H_{n}(M)\Rightarrow d = 0$. Is there any way to calculate number $d$ without assumption that $\chi(M) = 0$?
 A: Here's a proof that $d = \pm \chi(M)$.
Consider the universal orientable $n$-plane bundle $ESO(n)\rightarrow BSO(n)$.  It turns out that the unit sphere bundle in $ESO(n)$ is homotopy equivalent to $BSO(n-1)$ with projection $BSO(n-1)\rightarrow BSO(n)$ induced from the natural inclusion $SO(n-1)\rightarrow SO(n)$.  (See, e.g., Corollary 2.28 in Ralph Cohen's notes.)
So we have a bundle $S^{n-1}\rightarrow BSO(n-1)\rightarrow BSO(n)$.
Using the Gysin sequence, we see that the Euler class $e\in H^n(BSO(n))$ is characterized as being a generator of the cyclic group $\ker (H^n(BO(n))\rightarrow H^n(BO(n-1)))$.  Note that this cyclic group is either $\mathbb{Z}/2\mathbb{Z}$ (if $n$ is odd) or $\mathbb{Z}$ (if $n$ is even), so $e$ is either uniquely determined, or determined up to sign.
On the other hand, in the spectral sequence, the image of $\partial$ is precisely this kernel.
Now, using naturality, it follows that for any (linear) sphere bundle $S^{n-1}\rightarrow E\rightarrow M$, that the image of $\partial$ in the spectral sequence is, up to sign, the Euler class of the bundle.  For the tangent bundle, the Euler class $e\in H^n(M)\cong \mathbb{Z}$ is $\pm \chi(M)$, which gives the result.
