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!
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.
I will present an explicit calculation using Chern-Weil theory, which makes an amusing use of Legendre's duplication formula for the gamma function.
The Chern character form of a vector bundle $E$ with connection $\nabla$ and curvature tensor $R^\nabla$ is $$ \mathrm{ch}(E) = \sum_{k=0}^\infty \frac{1}{k!} \mathrm{tr}(R^\nabla)^k. $$ On the $2n$-sphere, only the $n$-th summand survives.
On a general Riemannian manifold, the curvature of the spinor bundle is $$R^\Sigma(X, Y) \psi = \frac{1}{4}\sum_{ij=1}^m \langle R(X, Y) e_i, e_j\rangle e_i \cdot e_j,$$ where $R$ is the Riemann tensor (of the tangent bundle), $X$, $Y$ are vector fields and $e_1, \dots, e_m$ is a local ON-frame. For the round sphere $S^m$, we have $R(X, Y) Z = \langle Y, Z\rangle X - \langle X, Z\rangle Y$, hence $$ \sum_{ij=1}^{m}\langle R(X, Y) e_i, e_j\rangle e_i \cdot e_j = \sum_{ij=1}^m\bigl(\langle Y, e_i\rangle \langle X e_j\rangle - \langle X, e_i\rangle \langle Y e_j\rangle\bigr)e_i \cdot e_j = X \cdot Y - Y \cdot X. $$ We now use the general formula $$ (\theta_1 \wedge \cdots \wedge \theta_k)(X_1, \dots, X_{2k}) = \frac{1}{2^k}\sum_{\sigma \in S_{2k}} \mathrm{sgn}(\sigma) \prod_{i=1}^k \theta_i(X_{\sigma_{2i-1}}, X_{\sigma_{2i}}) $$ for the wedge product of $2$-forms $\theta_1, \dots, \theta_k$, which can be easily shown using induction. We then get on $S^m = S^{2n}$ that $$ (R^\Sigma)^{2n}(e_1, \dots, e_{2n}) = \frac{1}{4^n \cdot 2^n} \sum_{\sigma \in S_{2n}} \mathrm{sgn}(\sigma) \prod_{i=1}^{n} (e_{\sigma_{2i-1}} \cdot e_{\sigma_{2i}} - e_{\sigma_{2i}} \cdot e_{\sigma_{2i-1}})\\ =\frac{1}{4^{n}} \sum_{\sigma \in S_{2n}}\mathrm{sgn}(\sigma) e_{\sigma_1} \cdots e_{\sigma_{2n}} = \frac{(2n)!}{4^{n}} e_1 \cdots e_{2n}. $$ Here $e_1, \dots, e_{2n}$ is a positively oriented ON-basis of the tangent space. Because $i^{n} e_1\cdots e_{2n}$ is the grading operator of $\Sigma S^{2n}$ and the even and odd chirality spinor spaces have equal dimension, we see that it has zero trace over the whole spinor bundle. However, the grading operator is $\pm 1$ on $\Sigma^\pm S^{2n}$, we obtain \begin{equation*} \mathrm{tr}_{\Sigma^\pm}(i^{n}e_1 \cdots e_{2n}) = \pm 2^{n-1}, \end{equation*} as $\dim(\Sigma^\pm S^{2n}) = 2^{n-1}$. We therefore get $$ \mathrm{tr}_{\Sigma^\pm}\big((iR^\Sigma)^{n}(e_1, \dots, e_{2n})\big) = \pm \frac{(2n)!}{4^{n}} \cdot 2^{n-1} = \pm \frac{(2n)!}{2^{n+1}}, $$ which implies that $$ \mathrm{tr}_{\Sigma^\pm}(iR^\Sigma)^{n} = \pm \frac{(2n)!}{2^{n+1}} \cdot \mathrm{vol} = \pm \frac{2n \cdot \Gamma(2n)}{2^{n+1}} \cdot \mathrm{vol} , $$ where $\mathrm{vol} = e_1 \wedge \cdots \wedge e_{2n}$ is the volume form. Therefore, the $2n$-form part of the Chern character form is $$ \mathrm{ch}_{2n}(R^{\Sigma^\pm}) = \frac{1}{n!} \mathrm{tr}_{\Sigma^\pm } \left(\frac{i R^\nabla}{2\pi}\right)^{n} = \pm \frac{1}{n\cdot \Gamma(n)} \frac{1}{(2\pi)^{n}}\frac{2n \cdot \Gamma(2n)}{2^{n+1}} \cdot \mathrm{vol} \\ = \pm \frac{\Gamma(2n)}{ \Gamma(n)} \frac{1}{2^{2n} \pi^{n}} \cdot \mathrm{vol}. $$ Integrating over the sphere, we get $$ \int_{S^n} \mathrm{ch}(R^{\Sigma^\pm}) = \pm \frac{\Gamma(2n)}{ \Gamma(n)} \frac{1}{2^{2n} \pi^{n}} \cdot \underbrace{\frac{2 \pi^{(2n+1)/2}}{\Gamma\left(\frac{2n+1}{2}\right)}}_{=\mathrm{vol}(S^{2n})} = \pm \frac{\Gamma(2n)}{\Gamma(n)\Gamma(n+1/2)} \cdot \frac{\sqrt{\pi}}{2^{2n-1}}. $$ At this point, the right hand side looks rather hopeless, but then Legendre's duplication formula saves the day and states that the whole mess is just $\pm 1$.
Another explicit calculation can be found in these notes by Baum and Erp, see section 2.11. They realize the spinor bundle as a subbundle of a trivial bundle using a certain projection-valued function, but ultimately, the calculations are very similar and also use the duplication formula in the very last step of the proof (see the proof of Prop. 6 on p.10).
