# M721: Index Theory

## Hodge star operator and Signature operator

Let $k=\mathbb{R}$ or $\mathbb{C}$ and $V$ be an inner product space with$\{e_1,\cdot\cdot\cdot,e_n\}$ a fixed orthonormal basis,$\Lambda^p V$ be the space of $p$ form

Hodge Star Operator

Lemma 1  There is a unique map $\ast:\Lambda^p V\rightarrow\Lambda^{n-p} V$ s.t. for any $\alpha,\beta\in\Lambda^p V$

$\langle\alpha,\beta\rangle vol=\alpha\wedge\ast\beta$

PROOF. (Uniqueness) suppose we have another map $\ast'$, then

$\alpha\land(\ast-\ast')\beta=\alpha\land\ast\beta-\alpha\land\ast'\beta=0$ for every $\alpha$

so $(\ast-\ast')\beta=0$ for every $\beta$,i.e.$\ast=\ast'$

(Existence) Fix an oriented orthonormal basis $\{e_1,\cdot\cdot\cdot,e_n\}$, for $\sigma\in S_n$

we define                            $\ast(e_{\sigma(1)}\land\cdot\cdot\cdot\land e_{\sigma(n)})=\rm{sign}(\sigma)e_{\sigma(p+1)}\land\cdot\cdot\cdot\land e_{\sigma(n)}$

We have      $\ast(e_1\land\cdot\cdot\cdot\land e_p)=e_{p+1}\land\cdot\cdot\cdot \land e_{n}$  and   $\ast\ast=(-1)^{p(n-1)}$

Suppose $X^n$ is an oriented closed Riemannian Manifold,$\alpha,\beta$ are $p$forms,define the $l^2$ inner product by

$\langle\alpha,\beta\rangle_{l^2}=\int_{X}\alpha\land\ast\beta$

Using integration by parts and stokes theorem,we have the following equalities:

$\langle d\alpha,\beta\rangle =\int_{X} d\alpha\land\ast\beta$

$= (-1)^p\int_{X} \alpha\land d\ast\beta$

$= (-1)^p\int_{X}\alpha\land (-1)^{(n-p-1)(p+1)}\ast\ast d\ast\beta$

$= \langle\alpha,(-1)^{n(p+1)+1}\ast d\ast\beta\rangle$

hence we yield
Lemma 2 The formal adjoint of $d$ is $\delta=(-1)^{n(p+1)+1}\ast d \ast$,i.e.$\langle d\alpha,\beta\rangle=\langle\alpha,\delta\beta\rangle$.

Exercise   Define $\Delta=d\delta+\delta d$,show that $\ast\Delta=\Delta\ast$
Corollary  $\ast:\mathcal{H}^p\xrightarrow{\cong}\mathcal{H}^{n-p}$

Harmonic Form and Signature

If $\tau:V\rightarrow V$ is an $involution$ and char$F\neq 2$,then

$V=V_+\oplus V_{-}$

where $V_{\pm}$ denote the $\pm 1$ eigenspace of $\tau$

If $n=4k$,$p=2k$,then $\ast\ast=1$

Theorem 1                                   $\mbox{sign}(X)=\dim\mathcal{H}_{+}^p-\dim\mathcal{H}_{-}^p$

PROOF using corollary,we could define the following $nondegenerate$  bilinear form:

$I:\mathcal{H}^p\times\mathcal{H}^p\rightarrow\mathbb{R}\quad(\alpha,\beta)\mapsto\int_{X}\alpha\land\beta$

Let $\mathcal{H}_{\pm 1}^p$ be the $\pm 1$ eigenspace of $\ast$. For $\alpha\in\mathcal{H}_{+}^p$,$\beta\in\mathcal{H}_{-}^p$,we have:

$I(\alpha,\alpha)=I(\alpha,\ast\alpha)=\langle\alpha,\alpha\rangle\geq0$
$I(\beta,\beta)=-I(\beta,\ast\beta)=-\langle\beta,\beta\rangle\leq 0$
$-I(\alpha,\beta)=\langle\alpha,\beta\rangle=\langle\beta,\alpha\rangle=I(\alpha,\beta)$

Hence there is a decomposation

$\mathcal{H}^p=\mathcal{H}_{+}^p\oplus\mathcal{H}_{-}^p$

and $I$ is positive definite on $\mathcal{H}_+\times\mathcal{H}_+$ and negative definite on $\mathcal{H}_-\times\mathcal{H}_-$

Using Hodge-de Rham isomorphism          $\mathcal{H}^p(X)\cong H^p(X,\mathbb{R})$

the above non degenerate bilinear form is equivalent to the intersection form:

$I':H^p(X,\mathbb{R})\times H^p(X,\mathbb{R})\rightarrow\mathbb{R},(\alpha,\beta)\mapsto\langle\alpha\cup\beta,[X]\rangle$

So we have

$\mbox{sign}(X)=\dim H_{+}^p-\dim H_{-}^p=\dim\mathcal{H}_{+}^p-\dim\mathcal{H}_{-}^p$

Signature Operator

If $n=2l$,then $\ast\ast=(-1)^{p(n-p)}=(-1)^p$

Let $\Omega^p=C^{\infty}(\Lambda^p T_{\mathbb{C}}^{\ast}X^{2l})$ be the complex-valued $p$ forms,define

$\tau=i^{p(p-1)+n/2}\ast:\Omega^p\rightarrow\Omega^{n-p}$

we have                           $\tau^2=Id$     and       $(d+\delta)\tau=-\tau(d+\delta)$

so if we write $\Omega^{\ast}=\Omega_{+}^{\ast}\oplus\Omega_{-}^{\ast}$,where $\Omega_{\pm 1}^{\ast}$ denotes the $\pm 1$-eigenspace of $\tau$,then $D=d+\delta$ interchanges $\Omega_{\pm 1}^{\ast}$, due to its anti-commutitivity with $\tau$, i.e.:

$D=d+\delta=\begin{bmatrix} 0 & D_{-} \\ D_{+} & 0 \end{bmatrix}: \Omega_{+}^{\ast}\oplus\Omega_{-}^{\ast}\rightarrow\Omega_{+}^{\ast}\oplus\Omega_{-}^{\ast}$

Definition $D_{+}:\Omega_{+}^{\ast}\rightarrow\Omega_{-}^{\ast}$ is called the  signature operator

Theorem 2                                        $\mbox{Index}(D_{+})=\mbox{sign}(X)$

PROOF  we have following facts:

• $(d+\delta)^2=\Delta$ is elliptic,hence $d+\delta$ is elliptic,so are $D_{+}$ and $D_{-}$. $\dim (\mbox{ker} D_{+})$ and $\dim (\mbox{ker} D_{-})$ are finite
• $D$ is self-adjoint,so $(D_{+})^{\ast}=D_{-}$
• $\mbox{ker} (d+\delta)$=$\mbox{ker}\Delta$.so $\mbox{ker} D_{\pm 1}$ consists of harmonic forms for the $\pm 1$ eigenvectors of $\tau$
• $\dim\mathcal{H}_{+}^p=\dim\mathcal{H}_{-}^{n-p}$ for $p\neq l$

Using these facts,we yield:     $\mbox{Index} (D_{+}) = \dim (\mbox{ker} D_{+})-\dim(\mbox{coker} D_{+})$

$=\dim(\mbox{ker}D_{+})-\dim(\mbox{ker} D_{+}^{\ast})$

$=\dim(\mbox{ker}D_{+})-\dim(\mbox{ker} D_{-})$

$=\Sigma_p\dim\mathcal{H}_{+}^p-\Sigma_p \dim\mathcal{H}_{-}^p$

$=\dim\mathcal{H}_{+}^{l}-dim\mathcal{H}_{-}^{l}$

$=\mbox{sign}(X)$