Please forgive me if the question is too elementary, but however I was unable to manage by myself. The question comes from J.Varilly, H.Figueroa and J. Gracia-Bondia book "Elements of noncommutative geometry". The framework is as follows: on a spin (Riemanian, without boundary, compact) manifold one constructs the (unique) Dirac operator $D$. One then shows that $[D,a]=-ic(da)$ where $a \in C^{\infty}(M)$ is regarded as a multiplication operator on $L^2(M,S)$---$L^2$ sections of spinor bundle $S$. This space can be regarded as a completion of the space of smooth sections of spinor bundle $\mathcal{S}:=\Gamma^{\infty}(M,S)$ with respect to the norm obtained from the scalar product $\langle \varphi,\psi \rangle:=\int_{M}(\varphi|\psi)d\nu_g$. Here $(\cdot|-)$ denotes the hermitian pairing on $\mathcal{S}$ defined by $(\varphi | \psi)(x):=\langle \varphi_x,\psi_x \rangle_x$ whete $\langle \cdot, - \rangle_x$ is the scalar product on the fiber $S_x$ of the spinor bundle $S$. $\nu_g=\sqrt{det[g_{ij}]}dx^1 \wedge ... \wedge dx^n$ is the Riemannian volume form.
Further, $c$ is the action of Clifford bundle $\Gamma^{\infty}(M,\mathbb{C}l(T^*M))$ on $\mathcal{S}$: therefore $da$ is regarded as an element of $\Gamma^{\infty}(M,\mathbb{C}l(T^*M))$ and $c(da)$ is an element of $\Gamma^{\infty}(M,End(S))$. Therefore $c(da)$ becomes an (bounded) operator on $L^2(M,S)$. This space can be viewed as a direct integral of Hilbert spaces $S_x$. Therefore the operator norm $\|c(da)\|=\sup_{x \in M} \|c(da)(x)\|$ where $c(da)(x):S_x \to S_x$. The question is the following:
Why this norm is equal to $g^{-1}_x(d\bar{a}(x),da(x))$?
$\begingroup$
$\endgroup$
Add a comment
|