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The Schrödinger-Lichnerowicz formula

If $E \to M$ is any smooth vector bundle with connection $\nb^E$ on $\sE = \Ga(M, E)$ , we can consider not only $\nb^E\: \sE \to \sA^1(M) \ox_{\sA} \sE$ , but also the connection $\nb^{E'} := \nb \ox 1 + 1 \ox \nb^E$ on the tensor product bundle $\sE' = \sA^1(M) \ox_{\sA} \sE$ ; here $\nb$ is once again the Levi-Civita connection on $\sA^1(M)$ . Their composition is an operator $\nb^{E'} \circ \nb^E$ from $\sE$ to $\sA^1(M) \ox_{\sA} \sA^1(M) \ox_{\sA} \sE$ ; using the metric on $\sA^1(M)$ we can take the trace over the first two factors, ending up with a Laplacian:

$\begin{displaymath} \Dl^E := -\Tr_g(\nb^{E'} \circ \nb^E) : \sE \to \sE. \end{displaymath}$

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The minus sign is a convention to yield a positive operator (instead of a negative one) [BGV]. Locally, this means:

$\begin{displaymath} \Dl^E = - g^{ij}(\nb^E_{\del_i} \nb^E_{\del_j} - \Ga_{ij}^k \nb^E_{\del_k}). \end{displaymath}$

$\begin{defn} In particular, when $E = M \x \bC$ is the trivial line bundle, we ... ...n $E = S$, we get the \textbf{spinor Laplacian} for a spin manifold. \end{defn}$

Before examining the relation between the Dirac operator and the spinor Laplacian, we collect a few well-known formulas for the Riemann curvature tensor, . These can be found in many places, for instance [BGV]; perhaps the best reference is Milnor's little book [Mil].

The square of the Levi-Civita connection on $\gX(M)$ is $\Coo(M)$ -linear, so it is given by $\nb^2 X = R(X)$ , where $R \in \sA^2(M, \End TM)$ . In local coordinates, its components are $R_{ijkl} := g(\del_i, R(\del_k,\del_l) \del_j)$ .

Taking a trace over the first and third indices, we get the Ricci tensor, whose components are $R_{jl} := g^{ik} R_{ijkl}$ . The trace of the Ricci tensor is the scalar curvature (or ``curvature scalar'') $s:= g^{jl} R_{jl} = g^{jl} g^{ik} R_{ijkl} \in \Coo(M)$ . Under exchange of indices, has the following skewsymmetry and symmetry relations:

$\begin{displaymath} R_{ijkl} = - R_{jikl} = - R_{ijlk}; \qquad R_{ijkl} = R_{klij}. \end{displaymath}$

The (first) Bianchi identity says that the cyclic sum over three indices vanishes:

$\begin{displaymath} R_{ijkl} + R_{iljk} + R_{iklj} = 0. \end{displaymath}$

Moreover, the Ricci tensor is symmetric: $R_{jl} = R_{lj}$ .

The formula in the next Proposition is generally attributed to Lichnerowicz [Lich, 1963], but was anticipated by Schrödinger in a little-known paper [Sch1, 1932].

$\begin{prop} Let $(M,g)$ be a compact Riemannian spin manifold with spinor modu... ...Dslash^2 = \Dl^S + \quarter s \end{equation}as an operator on $\sS$, \end{prop}$

$\begin{proof} It is enough to prove the equality when applied to spinors $\psi$ ... ...\quarter R_{jl} g^{jl} = \quarter s. \tag*{\qed} \end{align}\hideqed \end{proof}$

One consequence of the formula () is a famous ``vanishing theorem'' of Lichnerowicz.

$\begin{cor} If $s(x) \geq 0$ for all $x \in M$, and $s(x_0) > 0$ at some point $x_0 \in M$, then $\ker\Dslash = \{0\}$. \end{cor}$

$\begin{proof} Suppose that $\psi \in \sS$ satisfies $\Dslash\psi = 0$. Then \be... ...$0 = k \int_M s \nu_g$, which entails $k = 0$ and then $\psi = 0$. \end{proof}$

We saw by example (Appendix A.2) that on $\bS^2$ , the Dirac operator for the round metric has spectrum $\spec(\Dslash) = \bN \setminus \{0\}$ : here $s \equiv 2$ and $\ker\Dslash = \{0\}$ . Thus there are no ``harmonic spinors'' on $\bS^2$ .

Next: The spectral growth of Up: Dirac operators Previous: Selfadjointness of the Dirac Contents

Pawel Witkowski 2006-03-14