I'm confused with working out the Chern character in the following special setting.

Let $E$ be a spinor bundle

$$S=P_{Spin(2n)}(S^{2n})\times_\rho \mathbb{C}^{2n}$$

over sphere $S^{2n}$, where $\rho$ is the natural left action. We all know that there is a splitting

$$S=S^+\oplus S^-$$

by the volume element.

Now I want to compute the Chern character $\rm {ch}(S^+)$ of $S^+$ via Euler class $e(TS^{2n})$ of $S^{2n}$.

My thought is as follow: By splitting principle, we may assume

$$TX=V_1\oplus \cdots\oplus V_n , \qquad V_j=L_j\oplus \overline{L}_j $$

here $V_j$ is a 2-plane bundle over $S^{2n}$ and $L_j$ is a line bundle and $\overline{L}_j$ is its conjugate line bundle.

Then the spinor bundle has the splitting

$$S(TS^n)=S(V_1)\oplus S(V_2)\oplus \cdots \oplus S(V_n),$$

So the half spinor bundle has the splitting

$$S^+(TS^n)=S^+(V_1)\oplus S^+(V_2)\oplus \cdots \oplus S^+(V_n)$$

And it's not difficult that we have

$$L_j=S^+(V_j)\otimes S^+(V_j), \quad \overline{L}_j=S^-(V_j)\otimes S^-(V_j)$$

Therefore

$$\rm{ch}(S^+(V_j))=\left(\rm {ch}(L_j)\right)^{1/2}=e^{x_j/2},\quad \text{here the Chern root $x_j=c_1(L_j)$} $$

$$\rm {ch}(S^+(TS^{2n}))=\sum_{j=1}^n \rm{ch} (S^+(V_j))=\sum_{j=1}^n e^{x_j/2}=\sum_{j=1}^n\sum_{k=0}^{\infty}\dfrac{1}{2^kk!}x_j^k\in H^{*}(S^{2n},\mathbb{Z})$$

and the sphere only has $0$-th cohomology and the top cohomology, so the above equation just remains that

$$\sum_{j=1}^n\left(1+\dfrac{1}{2^n n!}x_1\cdots x_n\right)=n+\dfrac{1}{2^n (n-1)!}x_1\cdots x_n$$

In additionally, We have easily that

$$e(TS^{2n})=\rm x_1\cdots x_n$$

Hence

$$ \rm {ch}(S^+(TS^{2n}))=n+\dfrac{e(TS^{2n})}{2^n (n-1)!} $$

And it also tell us that

$$c_n(S^+(TS^{2n}))=\dfrac{\chi(S^{2n})}{2^n}$$

I'm not sure whether my calculation is right. If not, what's the expression of $\rm {ch}(S^+(TS^{2n}))$ via $e(TS^{2n})$? Could you please tell me this stuff? Thanks in advance

I'm confused with working out the Chern character in the following special setting.

Let $E$ be a spinor bundle

$$S=P_{Spin(2n)}(S^{2n})\times_\rho \mathbb{C}^{2n}$$

over sphere $S^{2n}$, where $\rho$ is the natural left action. We all know that there is a splitting

$$S=S^+\oplus S^-$$

by the volume element.

Now I want to compute the Chern character $\rm {ch}(S^+)$ of $S^+$ via Euler class $e(TS^{2n})$ of $S^{2n}$.

My thought is as follow: By splitting principle, we may assume

$$TX=V_1\oplus \cdots\oplus V_n , \qquad V_j=L_j\oplus \overline{L}_j $$

here $V_j$ is a 2-plane bundle over $S^{2n}$ and $L_j$ is a line bundle and $\overline{L}_j$ is its conjugate line bundle.

Then the spinor bundle has the splitting

$$S(TS^n)=S(V_1)\oplus S(V_2)\oplus \cdots \oplus S(V_n),$$

So the half spinor bundle has the splitting

$$S^+(TS^n)=S^+(V_1)\oplus S^+(V_2)\oplus \cdots \oplus S^+(V_n)$$

And it's not difficult that we have

$$L_j=S^+(V_j)\otimes S^+(V_j), \quad \overline{L}_j=S^-(V_j)\otimes S^-(V_j)$$

Therefore

$$\rm{ch}(S^+(V_j))=\left(\rm {ch}(L_j)\right)^{1/2}=e^{x_j/2},\quad \text{here the Chern root $x_j=c_1(L_j)$} $$

$$\rm {ch}(S^+(TS^{2n}))=\sum_{j=1}^n \rm{ch} (S^+(V_j))=\sum_{j=1}^n e^{x_j/2}=\sum_{j=1}^n\sum_{k=0}^{\infty}\dfrac{1}{2^kk!}x_j^k\in H^{*}(S^{2n},\mathbb{Z})$$

and the sphere only has $0$-th cohomology and the top cohomology, so the above equation just remains that

$$\sum_{j=1}^n\left(1+\dfrac{1}{2^n n!}x_1\cdots x_n\right)=n+\dfrac{1}{2^n (n-1)!}x_1\cdots x_n$$

In additionally, We have easily that

$$e(TS^{2n})=\rm x_1\cdots x_n$$

Hence

$$ \rm {ch}(S^+(TS^{2n}))=n+\dfrac{e(TS^{2n})}{2^n (n-1)!} $$

And it also tells us that

$$c_n(S^+(TS^{2n}))=\dfrac{\chi(S^{2n})}{2^n}$$

I'm not sure whether my calculation is right. If not, what's the expression of $\rm {ch}(S^+(TS^{2n}))$ via $e(TS^{2n})$? Could you please tell me this stuff? Thanks in advance