M721: Index Theory

Ruminations of a Graduate Class

Hodge star operator and Signature operator

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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 pforms,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 charF\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)

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Written by hailhu

October 21, 2012 at 4:13 pm

Posted in Uncategorized

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