Real spectral triples

Recall that a spin structure on an oriented compact manifold $(M, \eps)$ is represented by a pair $(\sS,C)$, where $\sS$ is a $\sB$-$\sA$-bimodule and, according to Proposition , $C\: \sS \to \sS$ is an antilinear map such that $C(\psi a) = C(\psi) \bar a$ for $a \in \sA$; $C(b \psi) = \chi(\bar b) C(\psi)$ for $b \in \sB$; and, by choosing a metric $g$ on $M$, which determines a Hermitian pairing on $\sS$, we can also require that $\pairing{C\phi}{C\psi} = \pairing{\psi}{\phi} \in \sA$ for $\phi,\psi \in \sS$. $\sS$ may be completed to a Hilbert space $\sH = L^2(M,S)$, with scalar product $\braket{\phi}{\psi} = \int_M\pairing{\phi}{\psi} \nu_g$. It is clear that $C$ extends to a bounded antilinear operator on $\sH$ such that $\braket{C\phi}{C\psi} = \braket{\psi}{\phi}$ by integration with respect to $\nu_g$, so that (the extended version of) $C$ is antiunitary on $\sH$. Moreover, the Dirac operator is $\Dslash = -i\hat{c} \circ \nabla^S$, where by construction the spin connection $\nb^S$ commutes with $C$: that is, $\nb^S_X$ commutes with $C$, for each $X \in \gX(M)$.

The property $C(\psi a) = C(\psi) \bar{a}$ shows that, for each $x \in X$, $\psi(x) \mapsto C(\psi)(x)$ is an antilinear operator $C_x$ on the fibre $S_x$ of the spinor bundle, which is a Fock space with $\dim_{\bC} S_x = 2^m$. Thus, to determine whether $C$ commutes with $\Dslash$ or not, we can work with the local representation $\Dslash = -i\ga^\al \nabla_{E_\al}^S$. Here $\ga^\al = c(\th^\al)$, for $\al = 1,\dots,n$, is a local section of the Clifford algebra bundle $\bCl(T^*M) \to M$, and the property $C(b\psi) = \chi(\bar{b}) C(\psi)$ says that $C(\ga^\al\psi) = - \ga^\al C(\psi)$ whenever $\psi$ is supported on a local chart domain.

However, replacing $b$ by $\ga^\al \in \Ga(U, \bCl^1(T^*M))$ is only allowed when the dimension $n$ is even. In the odd case, $\sB$ consists of sections of the bundle $\bCl^0(T^*M)$, and we can only write relations like $C(\ga^\al\ga^\bt\psi) = \ga^\al\ga^\bt C(\psi)$ for $\psi \in \Ga(U, S)$. But since $C$ is antilinear, in the even case we get

$$C\Dslash \psi = C(-i\ga^\al \nb_{E_\al}^S \psi) = iC(\ga^\a... ...}^S \psi) = -i\ga^\al C(\nb_{E_\al}^S \psi) = \Dslash C \psi.$$

Thus $[\Dslash, C] = 0$ on $\sH$, when $n = 2m$ is even.

¿What happens in the odd-dimensional case? Consider what happens on a single fibre $S_x$, which carries a selfadjoint representation of $B_x = \bCl^0(T_x^*M)$. Recall that we use the convention that $c(\om) := c(\om\ga)$ to extend the action of $\sB$ to all of $\Ga(M,\bCl(T^*M))$, where $\ga = (-i)^m \th^1\dots\th^{2m+1}$ is the chirality element. For $\om = \th^\al$, then gives $C c(\om) C^{-1} = C c(\om\ga) C^{-1} = c(\chi(\Bar{\om\ga})) = c(\Bar{\om\ga})$ since $\om\ga$ is even for $\om$ odd, and $\Bar{\om\ga} = i^m \th^\al \th^1 \dots \th^{2m+1} = (-1)^m\om\ga$. We conclude that $C c(\om) C^{-1} = (-1)^m c(\om)$ for $\om \in \sA^1(M)$ real, and therefore $C\Dslash = (-1)^{m+1} \Dslash C$ by antilinearity of $C$. We sum up:

$$C\Dslash = \begin{cases} +\Dslash C, &\text{if } n \not\equi... ...d 4, \\ -\Dslash C, &\text{if } n \equiv 1 \mod 4. \end{cases}$$

In the even case, $\sB = \Ga(M, \bCl(T^*M))$ contains the operator $\Ga = c(\ga)$ which extends to a selfadjoint unitary operator on $\sH$. Recall from Definition  that $c_J(\ga)$ is the $\bZ_2$-grading operator on the Fock space $\La^\8 W_J$, the model for $S_x$. If $\sH^\pm = L^2(M, S^\pm)$ denotes the completion of $\sS^\pm$ in the norm of $\sH$, then $\sH = \sH^+ \oplus \sH^-$, with $\Ga$ being the $\bZ_2$-grading operator. Now $\ga$ is even and $\bar\ga = (-1)^m \ga$ as before, so that $C\Ga = (-1)^m\Ga C$ whenever $n = 2m$.

When $M$ is a connected manifold, there is a third sign associated with $C$, since we know that $C^2 = \pm 1$. Once more, the sign can be found by examining the case of a single fibre $S_x$, so we ask whether an irreducible representation $S$ of $\bCl(V)$ admits an antiunitary conjugation $C \: S \to S$ such that $C c_J(v) C^{-1} = \pm c_J(v)$ for $v\in V$ (plus sign if $\dim V = 1 \mod 4$) and either $C^2 = +1$ or $C^2 = -1$. By periodicity of the Clifford algebras, the sign depends only on $n \bmod 8$, where $n = \dim V$.

Note that if $\{\ga^1,\dots,\ga^n\}$ generate $\Cl_{n,0}$, then $\{-i\ga^1,\dots,-i\ga^n\}$ generate $\Cl_{0,n} = \Cl(\bR^n, g)$ with $g$ negative-definite. Thus one can equally well work with $\Cl_{0,q}$, for $q = 0,1,\dots, 7$. Since $\Cl_{p,0} \ox_\bR M_N(\bR) \isom \Cl_{0,8-p} \ox_\bR M_{N'}(\bR)$ for $p = 0,1,\dots,7$ and suitable matrix sizes $N,N'$, we get, from our classification () of the Clifford algebras $\Cl_{p,0}$:

• for $q \equiv 0,6,7 \bmod 8$, $\Cl_{0,q}$ is an algebra over $\bR$,
• for $q \equiv 1,5 \bmod 8$, $\Cl_{0,q}$ is an algebra over $\bC$,
• for $q \equiv 2,3,4 \bmod 8$, $\Cl_{0,q}$ is an algebra over $\bH$.

On a case-by-case basis, using this classification, one finds that $C^2 = -1$ if and only if $n = 2,3,4,5 \bmod 8$.

Summary: There are two tables of signs

$$\begin{array}[t]{\vert c\vert cccc\vert} \hline n \bmod 8 & ... ...C & - & + & - & + \rule[-5pt]{0pt}{17pt} \\ \hline \end{array}$$

There is a deeper reason why only these signs can occur, and why they depend on $n \bmod 8$: the data set $(\sA, \sH, \Dslash, C, \Ga)$ determines a class in the ``Real'' KR-homology $\mathrm{KR}^\8(\sA)$, and $\mathrm{KR}^{j+8}(\sA) \isom \mathrm{KR}^j(\sA)$ by Bott periodicity. We leave this story for Prof. Brodzki's course. (But see [GVF, Sec. 9.5] for a pedestrian approach.)

``Real'' KR-homology is a theory for algebras with involution: in the commutative case, we may just take $a\mapsto a^*$, and we ask that $C a C^{-1} = a^*$ i.e., that $C$ implement the involution. This is trivial for the manifold case, since $C(\psi a) = C(\psi)\bar{a} =: a^*C(\psi)$, the $a^*$ here being multiplication by $\bar a$.

In the noncommutative case, the operator $Ca^*C^{-1}$ would generate a second representation of $\sA$, in fact an antirepresentation (that is, a representation of the opposite algebra $\sA^\opp$) and we should require that this commute with the original representation of $\sA$.

We have seen that in the standard commutative example, the even case arises when the auxiliary algebra $\sB$ contains a natural $\bZ_2$-grading operator, and this happens exactly when the manifold dimension is even. Now, the manifold dimension is determined by the spectral growth of the Dirac operator, and this spectral version of dimension may be used for noncommutative spectral triples, too. To make this more precise, we must look more closely at spectral growth.