# M721: Index Theory

## The Dirac operator

with one comment

We begin by endowing a vector bundle with a Clifford module structure. It is with additional structure that we may define a Dirac operator.

Let $(X^n, g)$ be an $n$-dimensional Riemannian manifold with covariant derivative $\nabla^X$ (on $TX$) , and let $\mathbb{R}^N\to S\to X$ be a vector bundle.

Clifford Module Bundles and a Dirac “Type” Operator

Definition (Clifford module)

A Clifford module $T$ for a real inner product space $(V,\beta)$ is a left module over $Cl(V,\beta)$. Equivalently, there is a -algebra homomorphism $c: Cl(V,\beta) \to \text{Hom(T,T)}$ given by $c(v) = (t\mapsto v\cdot t)$. Since $v^2:=v\cdot v = -||v||^2$ for any $v\in Cl(V,\beta)$ (see the Glossary below), one has that $c$ satisfies $c(v)^2 = - ||v||^2\text{Id}$.

Definition (Bundle of Clifford Modules)

A bundle $S$ (as above) is a bundle of Clifford modules if there is a map of bundles of $\mathbb{R}$-algebras $C:Cl(TX)\to \text{Hom}(S,S)$ such that $c(v)^2s = -||v||^2 s$ for any section $s\in C^{\infty}(S)$. In other words, for each $x\in X$, $\text{Hom}(S_x, S_x)$ is a Clifford module for $(T_x X, g_x)$.

Definition (Dirac type operator)

Let $S$ be a Clifford module bundle equipped with a covariant derivative $\nabla^S$. Let $D:C^\infty(S)\to C^\infty(S)$ be the map defined by the composition

$C^\infty(S)=\Omega^0(S) \xrightarrow{\nabla^S} \Omega^1(S) = C^\infty(\text{Hom}(TX,S)) \xrightarrow{H} C^\infty(TX\otimes S)\xrightarrow{c} \Omega^0(S)$

where $H$ is the inverse of the bundle isomorphism $C^\infty(TX\otimes S) \to C^\infty(\text{Hom}(TX,S)), v\otimes s \mapsto g(v,.)\otimes s$, and where $c(v\otimes s) = v\cdot s$ is Clifford multiplication. We call such a map (which depends on the Clifford module bundle $S$, $\nabla^S$ and $g$) a Dirac type operator.

If we a fix an orthonormal frame $\{e_1,\ldots, e_n\}$ for $TX$ over some neighborhood $U\subset X$ and, using the metric $g$, let $\{e^i\}$ be the corresponding frame for $T^\ast X|_U$, we may write this composition locally as

$s\overset{\nabla^S}{\mapsto} \sum_{i=1}^n e^i\otimes \nabla^S_{e_i} s \overset{H}{\mapsto} \sum_{i=1}^n e_i\otimes \nabla^S_{e_i} s \overset{c}{\mapsto} \sum_{i=1}^n e_i\cdot \nabla^S_{e_i} s$

That is, a Dirac type operator is locally of the form

$Ds = \sum_{i=1}^n e_i\cdot \nabla^S_{e_i} s$.

Proposition

A Dirac type operator is a first order differential operator.

Proof

Let $s\in C^\infty(S)$ and $f\in C^{\infty}(X)$. Using the local description above, we compute:

$[D, f]s = D(fs) - f(Ds)$

$= \sum_{i=1}^n \bigl[ e_i\cdot \nabla^S_{e_i} (fs) - f e_i\cdot \nabla^S_{e_i} s\bigr]$

$= \sum_{i=1}^n \bigl[ e_i\cdot (e_i(f)s + f \nabla^S_{e_i} s) - f e_i\cdot \nabla^S_{e_i} s\bigr]$

$= \sum_{i=1}^n \bigl[ e_i\cdot e_i(f)s + e_i\cdot f \nabla^S_{e_i} s - f e_i\cdot \nabla^S_{e_i} s \bigr]$

$= \sum_{i=1}^n \bigl[ e_i\cdot e_i(f)s + f e_i\cdot \nabla^S_{e_i} s - f e_i\cdot \nabla^S_{e_i} s \bigr]$

$= \sum_{i=1}^n e_i\cdot e_i(f)s$.

In particular, $[D,f](gs) = g[D,f](s)$ for $g\in C^{\infty}(X)$, so $[D,f]$ is $C^\infty(X)$-linear and hence in $DO_0(S,S)$. Thus $D\in DO_1(S,S)$, and as $D$ itself is not $C^\infty(X)$-linear, $D$ is of order 1. $\square$

Remark

The above proof extends (by incorporating induction) to show that the composition of a $k$– and an $m-$ order differential operator is a differential operator of order $\leq k+m$. Here, $\nabla^S$ and $c$ are differential operators of order 1 and 0, respectively.

We next show that $D$ is elliptic.

Lemma (Symbol of a Dirac type operator)

Let $D$ be a Dirac type operator and let $\xi^\ast\in T^\ast X$. Then the symbol $\sigma_D(\xi^\ast)$ of $D$ at $\xi^\ast$ is given by

$\sigma_D(\xi^\ast)(s) = \frac{1}{i} \xi\cdot s$,

where $\xi\in TX$ is the dual to $\xi^\ast$ determined by the metric, i.e., such that $\xi^\ast = g(\xi, \cdot)$.

Proof

Fix $x\in X$, and let $U$ be an open neighborhood of $x$ in $X$. Using $g$ choose an orthonormal frame $\{e_i\}$ of $TX|_U$ with dual frame $\{e^i\}$ for $T^\ast X|_U$. Being a bundle homomorphism $\pi^\ast(S)\to \pi^\ast(S)$ (over $\mathbb{R}$), $\sigma_D$ is $\mathbb{R}$-linear in the $T^\ast X$-ordinate. Thus it suffices to verify the proposition for $e^j$; that is, we wish to show that $\sigma_D(e^j)(s) = \frac{1}{i} e_j\cdot s$ for $s\in C^{\infty}(S|_U)$.

Choose a local chart $\varphi: U\to \mathbb{R}^n$ such that $\varphi(x)=0$; let $x_i=\text{pr}_i\circ \varphi: U\to \mathbb{R}$. Note that $\{e_i(x)\}$ is an orthonormal basis for $T_xX$. Let $\frac{\partial}{\partial x_i} = d\varphi_x(e_i(x))$; then ${dx_j}_x(e_i(x)) = {d\text{pr}_j}|_0\circ d\varphi_x(e_i(x)) = {d\text{pr}_j}|_0\bigl(\frac{\partial}{\partial x_i}\bigr) = \delta_{ji}$, so ${dx_j}_x = e^j(x)$.

Thus (see the Glossary below), as $D$ is order $1$, we have

$\sigma_D(e^j)(s(x)) = \frac{1}{i} D(x_j s)(x)$

$= \frac{1}{i} (\sum_{i=1}^n e_i\cdot \nabla_{e_i}(x_j s))(x)$

$= \frac{1}{i} \sum_{i=1}^n e_i(x) \cdot \bigl[ {dx_j}_x(e_i(x)) s(x) + \underbrace{x_j(x)}_{=0}\nabla_{e_i}s(x)\bigr]$

$= \frac{1}{i} \sum_{i=1}^n e_i(x) \cdot \bigl[ \delta_{ji} s(x) + 0]$

$= \frac{1}{i} e_j(x)\cdot s(x)$,

as required. $\square$

Corollary

A Dirac type operator is elliptic.

Proof

If $\xi^\ast\in TX^\ast-X$ then $\sigma_D(\xi^\ast)$ has inverse $\sigma_D^{-1}(\xi^\ast)(s) = i \frac{1}{\xi^\ast} \cdot s$. $\square$

Remark

Noting that $\sigma_{D^2}(\xi) = \sigma_D(\xi)\sigma_D(\xi)$ (the product of linear maps), we observe that $\sigma_{D^2}(\xi)s = -\xi\cdot \xi \cdot s = ||\xi||^2 s$. Taking $S=Cl(TX)$, this — apparently — implies that a Dirac type operator is, at the symbol level, the square root of the Laplacian.

$\mathbb{Z}_2$-grading and “a” Dirac Operator

A Dirac type operator is formally self-adjoint (so that its index is $0$) if we impose the following further restrictions on the Clifford-module bundle.

Definition (Clifford-Compatible)

Let $S\to X$ be a bundle of Clifford modules. We say that $S$ is Clifford compatible if it is equipped with a metric $\langle, \rangle$ and a covariant derivative $\nabla^S$ such that

(1) $\nabla^S$ is Riemannian, i.e., for all sections $s, t\in C^\infty(S)$:

$d\langle s, t\rangle = \langle \nabla^S s, t\rangle + \langle s, \nabla^S t \rangle$, and

(2) for all vector fields $V, W\in C^{\infty}(TX)\subset C^{\infty}(Cl(TX))$ and for any section $s\in C^\infty(S)$:

$\nabla^S_V(W\cdot s) = \nabla^X_V W \cdot s + W\cdot \nabla^S_V s$.

Definition (A Dirac operator)

A differential operator $D: C^\infty(S)\to C^\infty(S)$ is a Dirac operator if

(1) $D$ is a Dirac type operator, and

(2) $S$ is Clifford compatible.

Lemma

If $X$ is oriented and $D$ is a Dirac operator, then $D$ is formally self-adjoint. That is,
$\langle\langle Ds, t\rangle\rangle = \langle\langle s, Dt\rangle\rangle$,
where $\langle\langle s,t\rangle\rangle = \int_{X} \langle s, t \rangle$ (and the integration is with respect to the volume form on $X$).

Proof

Omitted 😦 $\square$

Consequently we have $\text{index}(D)=0$ (see Ning’s blog). To make use of the index, then, we introduce a $Z_2$-grading on Clifford module bundles.

Definition

Recall that a Clifford algebra is $\mathbb{Z}$-filtered — $Cl_n=\cup_{i} {Cl_n^i}$ with $Cl_n^i\cdot Cl_n^j\subset Cl_n^{i+j}$ — and $\mathbb{Z}_2$-graded — $Cl_n={Cl_n}^{+}\oplus {Cl_n}^-$ (even and odd products).

A Clifford module bundle $S$ is $\mathbf{Z}_2$-graded if it decomposes into a direct sum $S=S^+\oplus S^-$ of vector bundles such that for each $x\in X$ and $v\in T_xX$, one has $v\cdot S_x^{\pm} \subset S_x^{\mp}$.

Such a $\mathbb{Z}_2$-graded bundle is compatible if this decomposition is both orthogonal with respect to $\, \langle, \rangle$ and parallel with respect to the covariant derivative $\nabla^S$, ie. $\nabla^S S^{\pm} \subset S^{\pm}$.

Example

If $X$ is oriented, the Clifford bundle $Cl(TX) = Cl^+(TX) \oplus Cl^-(TX)$ is a $\mathbb{Z}_2$-graded compatible Clifford bundle. Some words which may be connected to verify this: Levi Civita connection, induced connection on $F(TX)$, lift to principal spin bundle, induced covariant derivative on associated vector bundle, compatibility with the metric.

Examples of Dirac Operators

We now look at four examples of Dirac operators. The first two are familiar; here we reinterpret them in terms of Clifford modules.

Example 1: The De Rham Operator

Recall that the filtered algebra $Cl(V^n,q) = \cup Cl^i(V,q)$ has associated graded algebra $Gr(Cl(V,q))\cong \Lambda^\ast V$. (See the Glossary below.)

Lemma

Let $f:\Lambda^\ast(\mathbb{R})\to Cl_n(\mathbb{R})$ be the map defined by

$v_1\wedge \ldots \wedge v_r\mapsto \frac{1}{r!} \sum_{\sigma\in \Sigma_r} \text{sign}(\sigma) v_{\sigma(1)}\cdot v_{\sigma(2)}\cdots v_{\sigma(r)}$.

Then $f$ is

(1) an isomorphism of vector spaces

(2) filtration preserving, i.e., $f(\oplus_{i\leq r} \Lambda^i(\mathbb{R}))\subset {Cl_n}^r(\mathbb{R})$, and

(3) $O(n)$-equivariant, i.e., $f(A\omega) = A f(\omega)$ for $\omega\in \Lambda^\ast(\mathbb{R})$ and $A\in O(n)$.

Proof

(1) and (2). Let $e_1, \ldots, e_n$ is an orthonormal basis for $\mathbb{R}^n$. Since (by an equivalent definition of the exterior algebra) $v_1\wedge \ldots \wedge v_r = \frac{1}{r!} \sum_{\sigma\in \Sigma_r} \text{sign}(\sigma) v_{\sigma(1)}\otimes v_{\sigma(2)}\otimes\ldots \otimes v_{\sigma(r)}$, we see that $f$ is induced by the map $T(\mathbb{R}^n) \to Cl_n(\mathbb{R})$ taking $e_{i_1}\otimes \cdots\otimes e_{i_r}$ to $e_{i_1}\ldots e_{e_r}$ and descending to $\Lambda^\ast(V)$; that is, $f(e_{i_1}\wedge\ldots \wedge e_{i_r}) = e_{i_1}\cdots e_{i_r}$. It follows that $f$ is an isomorphism and preserves the grading.

(3) Using that $Ae_1\wedge \ldots Ae_n = (\text{det} A) (e_1\wedge \ldots e_n)$, I feel like we need to be working with $SO(n)$ here. Please comment! $\square$

Corollary

The exterior algebra bundle over a(n oriented?) manifold $\Lambda^\ast(T^\ast X)$ and the clifford bundles $Cl(TX)$ are isomorphic as vector bundles.

Proof

Let $F(TX)$ denote the principal $O(n)$-bundle associated to $TX$. By parts (1) and (3) of the lemma, the map $f$ above induces a vector bundle isomorphism $\Lambda^\ast(TX) = F(TX)\times_{O(n)} \Lambda^\ast(\mathbb{R}^n) \to F(TX)\times _{O(n)} Cl_n(\mathbb{R}) = Cl(TX)$. $\square$

Theorem 2.5.12 (Lawson, Michelsohn)

Under this bundle isomorphism $\Lambda^\ast(T^\ast X) \cong Cl(TX)$, the de Rham operator $d+d^\ast$ corresponds to the Dirac operator $D$.

Corollary

Since we have already established (see Hailiang’s(?) blog post) that the Euler characteristic $\chi(X)$ of $X$ is equal to $\text{Index}(d+d^\ast|_{\Omega^{\text{even}}}:\Omega^{\text{even}}\to \Omega^{\text{odd}})$, the theorem (along with the grading-preserving property of $f$) implies we may also compute it as $\text{Index}(D|_{C^\infty(Cl^+(TX))}: C^\infty(Cl^+(TX))\to C^\infty(Cl^-(TX)))$.

Example 2: The Signature Operator

We now look to reinterpret the signature operator in terms of Clifford bundles.

Recall (see Hailiang’s blog) in the case that $n=2p$ and $p$ is even, the Hodge star operator $\ast: \Omega^p\to \Omega^{n-p}$ is an involution so we can decompose $\Omega^p = \Omega^p_+\oplus \Omega^p_-$ into the $+1$ and $-1$ eigenspaces of $\ast$. We defined the signature operator $D = d+d^\ast$. Since $D\ast = -\ast D$, we saw that $D$ took $\Omega^p_{\pm}$ to $\Omega^p_{\mp}$ and letting $D_+=D|_{\Omega^p_+}$, we found that $\text{Index}(D_+) = \text{Sign}(X)$. The signature $\text{sign}(X)$ of $X$ was defined to be the signature of the quadratic form on $\mathcal{H}^p$ given by $(\alpha,\beta)\mapsto \int_X \alpha\wedge \beta$.

In the case that $p$ is odd, we had to modify the construction. We complexified, taking $\Omega_\mathbb{C}^p = C^\infty(\Lambda^p(T^\ast X\otimes \mathbb{C}))$, and defined $\tau = i\ast$. Then the above paragraph went through with $\tau$ replacing $\ast$ and $\Omega_\mathbb{C}$ replacing $\Omega$.

Let $\{e_1, \ldots, e_n\}$ be an oriented orthonormal basis for $\mathbb{R}^n$. Let $\omega = e_1\cdots e_n$ in $Cl_n$. Then by the lemma above and the corresponding properties of the volume form $e_1\wedge\ldots \wedge e_n$ in $\Lambda^\ast(\mathbb{R}^n)$, we obtain that $\omega$ is a basis-independent section of $Cl_n(\mathbb{R})$.

Lemma

(1) We have $\omega^2 = \begin{cases} 1 \qquad n\equiv 0, \, 3 \mod 4\\ -1 \qquad n\equiv 1, \, 2 \mod 4 \end{cases}$

(2) If $n$ is even and $v\in \mathbb{R}^n$, then $\omega \cdot v = -v \cdot \omega$.

Proof

(1) We compute $(e_1\cdots e_n)(e_1\cdots e_n) = (-1)^n(-1)^{n-1}\cdots (-1)^2 (-1)^1 = (-1)^{n(n+1)/2}$. Writing $n=4k+l$ for $0\leq l\leq 3$, one finds that $n(n+1)/2$ is even if and only if $l=0$ or $3$.

(2) It suffices to verify for $v=e_i$. We have $(e_1\cdots e_n)\cdot e_i = (-1)^{n-i} e_1\cdots \hat{e}_i\cdots e_n$ and $e_i \cdot (e_1\cdots e_n) = (-1)^i e_1\cdots \hat{e}_i\cdots e_n$. (Here, a hat $\hat{}$ indicates that the element be omitted from the product.) Since $n$ is even, $(-1)^{n-i}=(-1)^i$. $\square$

Now, $\omega$ acts on any Clifford module via $v\mapsto w\cdot v$, and by part (1) of the lemma this defines an involution in the case $n\equiv 0, 3\mod 4$; in the case $n\equiv 1, 2\mod 4$, $i\omega$ defines an involution. Compare with the Hodge star operator recalled above. So define

$\Gamma = \begin{cases} \omega \qquad n\equiv 0, \, 3 \mod 4,\\ i\omega \qquad n\equiv 1, \, 2 \mod 4; \end{cases}$

then $\Gamma^2=\text{Id}$. Thus if $X$ is an oriented manifold and $S$ is a Clifford module bundle of $X$, putting $S^{\pm} = \{v\in S: \Gamma v = \pm v\}$, we have

if $n\equiv 0, 3 \mod 4$: $S = S^+\oplus S^-$, or

if $n\equiv 1,2 \mod 4$: $S_\mathbb{C} = S_\mathbb{C}^+\oplus S_\mathbb{C}^-$

Corollary

If $n$ is even then $S$ is $\mathbb{Z}_2$-graded, i.e., for $x\in X$ and $v\in T_xX$, one has $v\cdot S_x^{\pm} \subset S_x^{\mp}$.

Proof

By part (2) of the lemma, for $v\in T_xX$ we have $\Gamma v = -\Gamma v$.
Thus if $e\in S_x^{\pm}$ (i.e. $\Gamma e = \pm e$) then $v\cdot e = v\cdot (\pm \Gamma(e)) = \mp \Gamma(v\cdot e)$, so $v\cdot e\in S^{\mp}$. $\square$

Proposition

If $n$ is even and $S$ is $\mathbb{Z}_2$-graded compatible, then the associated Dirac operator splits as
$D = \left( \begin{array}{cc}0 & D^-\\ D^+ & 0\end{array}\right): C^\infty S^+\oplus C^\infty S^-\to C^\infty S^+\oplus C^\infty S^-$.
In particular, if $S=Cl(TX)$, by Theorem 2.5.12 we have $\text{Index}(D^+) = \text{Sign}(X)$.

Proof

Since $S$ is $\mathbb{Z}_2$-graded and is compatible (so in particular, the covariant derivative preserves $S^{\pm})$, the Dirac operator
$Ds = \sum e_i\cdot \nabla_{e_i}^S s$
takes $C^\infty(S^{\pm})$ to $C^\infty(S^{\mp})$. $\square$

Example 3: twisted Dirac Operators

Preliminary: If $S$ and $E$ are vector bundles over $X$ with covariant derivatives $\nabla^S$ and $\nabla^E$, respectively, then the tensor product bundle $S\otimes E\to X$ has covariant derivative
$\nabla^S\otimes \nabla^E := \text{Id}_S\otimes \nabla^E + \nabla^S\otimes \text{Id}_E$.

Fact: If $S$ is a compatible $\mathbb{Z}_2$-graded Clifford module bundle and $(E, g, \nabla^E)$ is a Riemannian bundle (see Property (1) of a compatible Clifford Bundle above for the definition), then $S\otimes E$ is a compatible $\mathbb{Z}_2$-graded Clifford module bundle (with Clifford multiplication $v\cdot(s\otimes e) = (v\cdot s)\otimes e$ for $v\in Cl_n$, $s\in S$, $e\in E$). In the case that $S=Cl(TX_{\mathbb{C}})$, we call the Dirac operator on $S\otimes E$ a twisted Dirac operator.

Fact: (Apparently from topological K-theory) If the Index theorem holds for any twisted Signature operator then it holds for all elliptic differential operators.

Example 4: Spin Manifolds and The Atiyah-Singer Dirac Operator

Recall (see Prasit’s blog) that there is an isomorphism $\varphi: Cl_{2l}(\mathbb{C}) \xrightarrow{\cong} M_{2^l}(\mathbb{C})$. So since
$\mathbb{C}^n$ is an $M_n(\mathbb{C})$ module, one has that $\mathbb{C}^{2^l}$ is an $Cl_{2l}(\mathbb{C})$-module via $a\cdot v = \varphi(a)v$, for $a\in Cl_{2l}(\mathbb{C})$ and $v\in \mathbb{C}^{2^l}$. To avoid confusion, let us call $\mathbb{C}^{2^l}$ with this module structure $C$.

Now, any $M_{2^l}(\mathbb{C})$-module is isomorphic to $\oplus^{2^l} \mathbb{C}^{2^l}$, so it follows that any $Cl_{2l}(\mathbb{C})$-module is isomorphic to $\oplus^{2^l} C$.

Let $S$ be a Clifford module bundle. From the above paragraph we see that each fiber (a $Cl_{2l}(\mathbb{C})$-module) is isomorphic (via $\varphi_x$, say) to $\oplus^{2^l} C_x$, where $C_x$ is a copy of $C$. We may then
ask if this splitting extends over the whole bundle; that is, is there a Clifford module bundle $S\to X$ and a bundle isomorphism $\varphi:S\to \oplus^{2^l} C$ which restricts fiberwise to an isomorphism $S_x\cong \oplus^{2^l} C$.

In turns out the answer is a resounding “Yes” if $X$ is a spin manifold.

Definition (Spin Structure)

Let $E\to X$ is an $N$-dimensional vector bundle. A spin structure on $E$ is a principal $\text{Spin}(N)$-bundle $P\to X$ together with a bundle isomorphism $P\times_{\text{Spin}(N)} \mathbb{R}^N\to E$. (Then $E$ is the associated vector bundle for $P$). Using classifying space theory, we may reinterpret this
to say that a spin structure on $E$ is a lift of the classifying map $\varphi_E: X\to BGL_N(\mathbb{R})$ to $B\text{Spin}(N)$.

We may break up the existence of a spin structure into pieces as follows.

After choosing a metric on $E$, we may first reduce the structure group of $E$ to $O(n)$. (The only obstruction to doing so is the paracompactness of $X$.) So we’re left to lift a map $\varphi_E: X\to BO(n)$ to $B\text{Spin}(N)$.

Since $B\text{Spin}(N)$ is the universal cover of $BSO(N)$, we may first try to lift $\varphi_E$ to $BSO(n)$.

The short exact sequence of groups $1\to SO(N)\to O(N)\xrightarrow{\text{det}} \mathbb{Z}_2\to 1$ induces a fibration of classifying spaces
$X\xrightarrow{\varphi_E} BO(N)\to B\mathbb{Z}_2=K(\mathbb{Z}_2,1)$.

It turns out that the map $X\to BO(N)$ lifts to $BSO(N)$ if and only if the composite $X\xrightarrow{\varphi_E} BO(N)\to B\mathbb{Z}_2=K(\mathbb{Z}_2,1)$ is nullhomotopic. Since $[X, K(\mathbb{Z}_2,1)]\cong H^1(X;\mathbb{Z}_2)$, there is an element $\omega_1(E)\in H^1(X;\mathbb{Z}_2)$ that vanishes if and only if $\varphi_E$ lifts. We call $\omega_1(E)$ the first Stiefel Whitney class of $E$.

Similarly, the map $X\to BSO(N)$ lifts to $B\text{Spin}(N)$ if and only if the composition $X\to BSO(N)\to B K(\mathbb{Z}_2,1)=K(\mathbb{Z}_2,2)$ is nullhomotopic; we call the corresponding element in $H^2(X;\mathbb{Z}_2)\cong [X, K(\mathbb{Z}_2,2)]$ (that vanishes iff $X\to BSO(N)$ lifts) the second Stiefel Whitney class $\omega_2(E)$ of $E$.

Definition (Spin manifold)

We will call an oriented manifold $X$ (so $\omega_1(TX)=0$) a spin manifold if its tangent bundle $TX$ admits a spin structure (i.e., $\omega_2(TX)=0$). It can be shown that this is equivalent to the existence of a trivialization of $TX$ over the $2$-skeleton of $X$. (Compare with the fact that $X$ is orientable if and only if $TX$ is trivializable over the $1$-skeleton.)

Definition (The Atiyah-Singer Dirac Operator)

Suppose $X^n$ has a spin structure with principal $\text{Spin}(n)$-bundle $P$ (associated to $TX$). Since $Cl_n(\mathbb{C})$ acts on $C$ on the left and $\text{Spin}(n)={Cl_{n}}^+(\mathbb{C})\cap \langle S^{n-1}\rangle$ (where $\langle S^{n-1}\rangle$ has general element $x_1\cdot x_2\cdots x_k$ with $x_i\in \mathbb{C}^n$, $x_i^2=1$) is a subgroup of the group of units $Cl_n^\times$, we may define
the $C$-bundle associated to $P$ by $\mathbf{C} = P\times_{\text{Spin}(n)} C\to X$. Since $C$ is a $Cl_n(\mathbb{C})$ module, $\mathbf{C}$ is a Clifford module bundle.

Some words: By lifting the Levi-Civita connection on $P_{SO(n)}$ one obtains a connection on $P_{\text{Spin}(n)}$, and hence (see who’s blog?) a covariant derivative $\nabla^\mathbf{C}$ on $\mathbf{C}$ which makes it Clifford compatible
as a $\mathbb{Z}_2$ graded Clifford module bundle.

We may then define the Atiyah-Singer Dirac Operator by

$Ds = \sum e_i\cdot \nabla_{e_i}^\mathbf{C} s$.

Some more words:

If $Ds=0$ then $s$ is called a harmonic spinor.

If $X$ has positive scalar curvature, then $D$ is injective. So if we have ways to compute the index (using the ASHI theorem, for example), we may be able to deduce that $X$ does not admit a metric of positive scalar curvature.

Glossary

(to include links to other blog posts)

Differential Operator (global definition)

If $E$ and $F$ are vector bundles over $X$ (of the same dimension), we define the family of differential operators of order $\leq m$ from $E$ to $F$ by

$DO_m(E,F) = \{D\in \text{Hom}_{\mathbb{R}}(C^\infty E, C^\infty F):\, [D,f]\in DO_{m-1}(E,F) \, \forall f\in C^{\infty}(X)\}$

with $DO_{-1}(E,F)=\{0\}$. In particular, $DO_{0}(E,F) = \{C^{\infty}(X)\text{-linear bundle maps } C^\infty(E)\to C^\infty(F)\}$.

Symbol of a differential operator

(cf. Juanita’s blog) We recall the definition of the symbol $\sigma_D$ of an order-$m$ differential operator $D:C^\infty(E)\to C^\infty(F)$. Denote $T^\ast X\xrightarrow{\pi} X$, $E\xrightarrow{p} X$, so $\pi^\ast(E)=\{(\xi^\ast,e)\in T^\ast X\times E:\, \pi(\xi^\ast)=x=p(e)\}$. Let $\xi^\ast\in T^\ast X$. Let $x\in X$. The symbol $\sigma_D(\xi^\ast)$ of $D$ at $x$ is the homomorphism $E_x\to F_x$ defined by

$\sigma_D(\xi^\ast)(s(x)) = \frac{1}{i} \frac{1}{m!} D(f^m s)(x)$

where $s\in C^\infty(E)$, and $f\in C^\infty(X)$ is such that $df_x = \xi^\ast(x)$.

Covariant Derivative

A covariant derivative $\nabla$ on a vector bundle $E\to X$ is a map $\nabla: C^{\infty}(TX)\times C^{\infty}(E)\to C^{\infty}(E)$, $(v,s)\mapsto \nabla^S_v s$ where $\nabla_v^S$ is an $\mathbb{R}$-linear map $C^\infty(E)\to C^{\infty}(E)$ satisfying the Leibnitz rule $\nabla^S_v (fs) = v(f)s + f\nabla^S_v(s)$ for $s\in C^\infty(E)$ and $f\in C^\infty(X)$.

(cf Prasit’s blog) A $k$-algebra $A$ is $\mathbb{Z}$-graded if $A=\bigoplus_i A_i$ such that $A_i A_j\subset A_{i+j}$. A $k$-algebra is filtered over $\mathbb{Z}$
if $A=\cup A^i$ such that $A^i\subset A^{i+1}$ and $A^i A^j \subset A^{i+j}$. A graded algebra $A=\bigoplus_i A_i$ defines a filtered algebra by taking $A^i=A_0\oplus\ldots A_i$, and conversely a filtered algebra $A=\cup_i A^i$ defines a graded algebra by taking $A_i=A^i/A^{i-1}$ (with $A^{-1}=0$).

Tensor Algebra and the Clifford Algebra

If $V$ is a vector space, the tensor algebra $TV = \mathbb{R}\oplus V\oplus V\otimes V\oplus\ldots \oplus V\oplus\ldots \oplus V$ has multiplication defined by concatenation, i.e., $(x_1\otimes \ldots \otimes x_j)(y_1\otimes \ldots \otimes y_j) = x_1\otimes \ldots \otimes x_j \otimes y_1\otimes \ldots \otimes y_j$. Thus $TV$ is a graded algebra with $(TV)_i = \underbrace{V\otimes \ldots \otimes V}_{\text{\emph{i} times}}$, and filtered with $(TV)^i=\mathbb{R}\oplus \cdots \oplus V\otimes\cdots \otimes V$.

Recall that $Cl(V,q)=\cup_i Cl^i(V,q)$ with $Cl^i(V,q)=(TV)^i/{\langle v\otimes v = -q(v)\rangle}$. (Here, $q$ is a quadratic form; sometimes it is convenient to refer instead to the associated symmetric bilinear form $\beta$.) For any $v\in V$, $q(v)\in \mathbb{R}=Cl^0(V,q)$ so $q(v)$ maps to $0$ under the quotient $Gr_i(Cl(V,q))= Cl^i(V,q)/Cl^{i-1}(V,q)$. This sets up a natural identification between $Gr_i(Cl(V,q))$ and $(\Lambda^\ast V)_i = \frac{(TV)_i}{\langle v\otimes v=0\rangle}$. Whence the associated graded algebra for $Cl(V,q)$ is isomorphic to $\Lambda^\ast V$. In particular, they are isomorphic as vector spaces, with dimension $\sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}=2^n$.

Clifford bundle

The Clifford bundle $Cl(TX)\xrightarrow{\pi} X$ has fibers $Cl(T_xX, g_x) \cong Cl_n(\mathbb{R})$ ($x\in X$), where $g$ is a Riemannian metric on $X$. (The latter isomorphism is given by identifying an orthonormal (with respect to $g_x$) basis for $T_xX$ with the standard generators $\{e_i\}$.) Just as $TX$ is the $\mathbb{R}^n$-bundle associated to the orthogonal frame bundle (of the tangent bundle) $F(TX)$ over $X$, $Cl(TX)$ is the associated $Cl_n(\mathbb{R})$-bundle to $F(TX)$.

Written by aclightf

November 21, 2012 at 7:45 pm

Posted in Uncategorized