Question about Clifford multiplication

Question

Let $X$ be a smooth vector field on the even dimensional sphere $S^n$. Let $S(TS^n)=S^+(TS^n)\oplus S^-(TS^n)$ be the spinor bundle over $S^n$ equipped with a bundle metric that is compatible with the clifford multiplication, which is trivial for all $n$ if the choose the trivial spin structure in the situation $n=1$. Let $v=c(X):S^+(TS^n)\to S^-(TS^n)$ be the Clifford multiplication of $X$ and $v^*:S^-(TS^n)\to S^+(TS^n)$ be the adjoint of $v$ with respect to Hermitian metrics on $S^\pm (TS^n)$. Let

$$V=v+v^*:S(TS^n)\to S(TS^n)$$

My question is how to prove that: $$V^2=|X|^2.$$

I guess that it needs some properties about Clifford multiplication, but I have no idea to prove the identity.

Could you give me some help with the details? Thanks!

Answer 1

Your statement is a pointwise statement. It essentially follows directly from the definition of a "Clifford multiplication". To understand it, you first have to be aware of the fact that there are two conventions for the Clifford relations in the literature: 1) $XY+YX+2<X,Y>=0$ and 2) $XY+YX-2<X,Y>=0$. It seems that you use convention 2), most references I know well, use convention 1). The next question is what kind of spinor bundle you are using -- I guess you use spinors defined by complex representations. In that case one can pass from convention 1) to 2) and vice versa by replacing $V$ by $iV$.

From my point of view you should ask differently. The trouble with the question you asked is that your is an immediate consequence of $XY+YX-2<X,Y>=0$ if you put $X=Y$. But what you probably want to know is: How does one define $S(TS^n)$ and its splitting in odd and even part and the Clifford multiplication such that you have the property above. The references that I know well are

• Chapter 1 in Lawson, Michelsohn, Spin geometry, beware that the proof of the injectivity $W\to Cl(W)$ has to be repaired.
• Chapter 1 of Bourguignon, Hijazi, Milhorat, Moroianu, Moroianu, A spinorial approach to Riemann and Conformal geometry

In these books it will be evident from the definition of a Clifford action, that $X\cdot X\cdot \phi=-|X|^2 \phi$. Note that these books will not distinguish in notation between the vector $X$ and its action on the spinor bundle, called $V$ in your question. So in the language you introduced above, this gives $V\cdot V\cdot \phi=-|X|^2 \phi$

Now define $V*\phi:= iV\cdot \phi$. Then $*$ is a Clifford multiplication in your sense, and it satisfies $V* V* \phi=|X|^2 \phi$.