Calculation of the top Chern class of spinor bundle over $S^{2n}$ It's well known that for a complex vector bundle $E$, we have
$$c_n(E)=e_n(E_\mathbb{R}) $$
But I'm very curious about the relationship between the top Chern class of spinor bundle and the Euler class of tangent bundle over the same even-dimensional sphere $S^{2n}$. Why is the top Chern class of spinor bundle over $S^{2n}$ is the non-zero multiple of Euler number of $S^{2n}$?
Could you give me some help with the details? Thanks!
 A: Let $\Sigma \to S^{2n}$ be the spinor bundle, then one has \begin{align*}
e(TS^{2n}) &= \frac{(-1)^{n}}{(n-1)!} \left( c_n(\Sigma^+)-c_n(\Sigma^-) \right),\\
0 &= c_n(\Sigma).
\end{align*}
Proof: The Atiyah-Singer index theorem for spin manifolds $M^{2n}$ says
\begin{align*}
 \chi(M) &= (-1)^{n}\int_M \hat{A}(TM)ch(\Sigma^+ - \Sigma^-)=\int_M e(TM) \\
 sign(M) &= \int_M \hat{A}(TM)ch(\Sigma) = \int_M L(TM)
\end{align*}
Since $M=S^{2n}$ has stably trivial tangent bundle one has $\hat{A}(TM) = 1 = L(TM)$. The only thing contributing to the first integral is some rational multiple of $c_n(\Sigma^+) - c_n(\Sigma^-)$.
Now I claim for any complex vector bundle $E \to S^{2n}$ one has
$$ch(E) = \frac{(-1)^{n-1}}{(n-1)!} c_n(E).$$
To see this, let $x_1,...,x_n$ denote chern roots for $E$. We have by definition $$ch(E)[S^{2n}]= \left( \sum\limits_{i=1}^{n} e^{x_i}\right) [S^{2n}] = \frac{1}{n!}\left(\sum\limits_{i=1}^n x_i^n\right)[S^{2n}].$$
Now we have to search for the coefficient of $c_n$ in the (unique) polynomial $P \in \mathbb{Z}[c_1,...,c_n]$ with $$P(c_1,\ldots,c_n)= x_1^n + \ldots + x_n^n$$ (where $c_i$ is the $i$-th elemtary symmetric polynomial in $n$ variables). Specializing to negatives of roots of unity namely
$$T^n - 1 = (T + x_1)...(T+x_n) = T^n + c_{1} T^{n-1} + ... + c_{n-1} T +c_n$$ we can see that this coefficient is $- (-1)^n n$.
EDIT: Alternative Without the Index theorem one can show (using the splitting principle) that for any real spin vector bundle $E$ of even rank $2r$
$$ ch(\Sigma^+(E) - \Sigma^-(E)) = \prod\limits_{i=1}^{r} (e^{-x_i/2} - e^{x_i/2}) $$
where $x_i$ are the chern roots of $E \otimes \mathbb{C}$. The euler class $e(E) = x_1 ... x_n$ divides this expression and one has (by definition)
$$ch(\Sigma^+(E)- \Sigma^-(E)) = (-1)^r e(E) \hat{A}(E)^{-1}$$
If $E$ is stabily trivial one has $\hat{A}(E) = 1$ and we are again left with determining $ch(\Sigma^+(E) - \Sigma^-(E))$ in degree $2r$.
Analogously $$ ch(\Sigma(E)^+ + \Sigma^-(E)) = \prod(e^{-x_i/2} + e^{x_i/2}) = \prod (2 + \frac{1}{4}x_i^2 + ... )$$
can be expressed only in Pontrjagin classes of $E$, so in the stably trivial case every chern class of the  spinor bundle has to vanish.
As a reference: Chapter III, § 11 & 12 in Spin Geometry by Lawson and Michelsohn.
