Boundedness Spectral Triple Axioms for de Rham Complex

Question

In Connes' axioms for a spectral triple $(A,H,D)$, they have a representation of an algebra $A$ in bounded operators on a Hilbert space $H$, and (unbounded) operator $D$, such that $[D,a]$ is bounded. One of the classical motivating examples is Dirac--de Rham operator D:= d + d$^*$ on de Rham complex of a Riemannian manifold. Is it clear that the two boundedness axioms discussed by me are actually satisfied?

Sorry if this is too much basic a question, this is for me quite new material.

Answer 1

Yes and yes. To fix notation, what we're dealing with is the commutative spectral triple $(C^\infty(X),L^2(X,\wedge T^\ast X),d+d^\ast)$, where $(X,g)$ is a compact oriented Riemannian manifold. Recall, in particular, that $\wedge T^\ast X$ is a Hermitian vector bundle with Hermitian metric $(\cdot,\cdot)$ induced by the Riemannian metric $g$, so that the inner product on $L^2(X,\wedge T^\ast X)$ is $$ \langle \xi,\eta \rangle := \int_X (\xi,\eta) \,\mathrm{dVol}_g. $$

1. Since the underlying manifold is compact, any continuous function on $X$ is bounded, so that for any $f \in C^\infty(X)$ and $\xi \in L^2(X,\wedge T^\ast X)$, $$ \|f \xi \|^2 = \int_X (f \xi,f\xi) \,\mathrm{dVol}_g = \int_X \lvert f\rvert^2 (\xi,\xi) \,\mathrm{dVol}_g \leq \| \lvert f \rvert^2 \|_\infty \int_X (\xi,\xi) \,\mathrm{dVol}_g = \|f\|_\infty^2 \|\xi\|^2, $$ which shows that multiplication by $f \in C^\infty(X)$ is bounded with operator norm $\|f\| \leq \|f\|_\infty$. More generally, you can show that any bundle endomorphism $E$ of $\wedge T^\ast X$ defines a bounded operator $E : L^2(X,\wedge T^\ast X) \to L^2(X,\wedge T^\ast X)$ with operator norm $$ \|E\| \leq \left\|x \mapsto \|E_x\| \right\|_\infty, $$ where $\|E_x\|$ denotes the operator norm of the linear operator $E_x$ on the finite-dimensional inner product space $\wedge T^\ast_x X$; off the top of my head, use compactness of $X$ and your favourite partition of unity to treat $E$ as just a matrix-valued function.

2. Now, let $f \in C^\infty(X)$, let $\xi$, $\eta \in C^\infty(X,\wedge T^\ast X)$. By a straightforward computation, you can show that $$ \langle \xi, [d+d^\ast,f]\eta \rangle = \langle \xi, df \wedge \eta \rangle + \langle d\overline{f} \wedge \xi, \eta \rangle. $$ From this, it follows that for all $g \in C^\infty(X)$ and $\xi$, $\eta \in C^\infty(X,\wedge T^\ast X)$, $$ \langle \xi, [d+d^\ast, f]g\eta \rangle = \langle \xi, df \wedge g\eta \rangle + \langle d\overline{f} \wedge \xi, g \eta \rangle = \langle \overline{g} \xi, df \wedge \eta \rangle + \langle d\overline{f} \wedge \overline{g}\xi,\eta \rangle\\ = \langle \overline{g}\xi,[d+d^\ast,f]\eta \rangle = \langle \xi,g[d+d^\ast,f]\eta \rangle, $$ so that $[d+d^\ast,f]$ is $C^\infty(X)$-linear on $C^\infty(X,\wedge T^\ast X)$, i.e., $[d+d^\ast,f]$ is a bundle endomorphism of $\wedge T^\ast X$. Hence, we can apply 1. to conclude that $[d+d^\ast,f]$ is bounded. It's worth noting that the fact that $[d+d^\ast,f]$ is a bundle endomorphism is just another way of saying that $d+d^\ast$ is a first-order differential operator.