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Periodicity modulo 8 appears in the classification of real Clifford algebras $C\ell_{p,q}(\mathbb{R})$ (usualy refered to as the "Clifford Clock"), in real Bott periodicity and in the definition of a real structure of KO-dimension on a spectral triple. The latter concept can be found in Connes-Marcolli book http://alainconnes.org/docs/bookwebfinal.pdf, for instance.

Spectral triples are a generalization of spin$^c$ manifolds and real spectral triples of spin manifolds. In fact, every (real) spectral triple over a commutative $*$-algebra is a spin manifold, by certain reconstruction theorems proven by Connes and, independently and under other conditions, by A.Rennie and J.Várilly. The KO-dimension $N\in\mathbb{Z_8}$ of a real spectral triple is enterly determined by knowing whether certain operators on a Hilbert space $H$ commute or anticommute. $H$ generalizes the square-integrable spinors Hilbert space.

Being alien to K-theory, I suspect that the definition of KO-dim is motivated (as many concepts in noncommutative geometry are) by what happens in the "commutative case" (spin geometry). I want to know where do such commutation and anticommutation relations appear in KO-theory. Otherwise put, what is the motivation for the definition of KO-dim, from the point of view of K-theory? can this periodicity be related to real Bott periodicity or the periodicity of the Clifford clock?

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What's before the AND is covered by the classic text Cliford modules by Atiyah-Bott-Shapiro. What comes afterwards is probably explained by those who introduced those concepts... –  t.b. Jun 7 '12 at 0:35
Not being an expert in the subject, my impression is that this question is too broad. «I want to understand...» is rarely a question: do you have something concrete that you want to ask? Also, giving details about your background will surely be useful to anyone answering this. –  Mariano Suárez-Alvarez Jun 7 '12 at 3:08
(a reference to some work where this KO-dimension is introduced and/or discussed would not hurt, either! :) ) –  Mariano Suárez-Alvarez Jun 7 '12 at 3:10
I can't really answer your question, but here's how I think this is supposed to work. Remember that in order to extract a spectral triple from a manifold you need to choose a $spin^c$ structure AND a Riemannian metric. To get a real spectral triple you need to choose a spin structure and a metric tensor, but you might not want the tensor to be positive definite (you still get a Dirac operator either way). I believe the KO-dimension in this case is just the signature of the metric mod 8 (we reduce mod 8 because this and the dimension is all the Clifford algebra cares about) –  Paul Siegel Jun 8 '12 at 16:25
That said, I don't remember exactly how you recover the KO-dimension from taking commutators. My guess is that you look at the commutators between the Dirac operator and various other stuff that tell you the dimension and judiciously insert the operator $J$ which gives your spectral triple a real structure. –  Paul Siegel Jun 8 '12 at 16:28

1 Answer 1

up vote 5 down vote accepted

I'm afraid I'm rather late to the party, but let me throw out a few thoughts, in the hope that something will be of use to someone. You probably know everything under 1. and 2., so if you want the punchline, do forgive the tl;dr and just skip ahead to 3.

  1. To be absolutely clear about the state of the art, Connes's theorem actually tells you the following:

    • A unital Frechet pre-$C^\ast$-algebra $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact orientable $p$-manifold if and only if there exists a $\ast$-representation of $A$ on a Hilbert space $H$ and a self-adjoint unbounded operator $D$ on $H$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$.
    • In particular, $A$ is isomorphic to $C^\infty(X)$ for $X$ a compact spin$^{\mathbb{C}}$ $p$-manifold if and only if there exist $H$ and $D$ such that $(A,H,D)$ is a commutative spectral triple of metric dimension $p$ and $A^{\prime\prime}$ acts on $H$ with multiplicity $2^{\lfloor p/2\rfloor}$.

    Once you know that $A \cong C^\infty(X)$, you can then apply the much earlier "baby reconstruction theorem" (for lack of a better phrase) announced by Connes and proved in detail by Gracia-Bondia--Varilly--Figueroa to conclude that:

    • In the general case, $(A,H,D) \cong (C^\infty(X),L^2(X,E),D)$ where $E \to X$ is a Hermitian vector bundle and $D$ can be interpreted as an essentially self-adjoint elliptic first-order differential operator on $E$.
    • In the case where $A^{\prime\prime}$ acts with multiplicity $2^{\lfloor p/2 \rfloor}$, $E \to X$ is in fact a spinor bundle (i.e., irreducible Clifford module bundle) and $D$ is Dirac-type (viz, a perturbation of a spin$^{\mathbb{C}}$ Dirac operator by a symmetric bundle endomorphism of $E$).

    So, whilst you can refine the reconstruction theorem to a characterisation of compact spin$^{\mathbb{C}}$ manifolds with spinor bundle and essentially self-adjoint Dirac-type operator, the general result is really just a statement about compact orientable manifolds. Indeed, one can even refine the reconstruction theorem to a characterisation of compact oriented Riemannian manifolds with self-adjoint Clifford module and essentially self-adjoint Dirac-type operator.

  2. After that detour, let's get down to brass tacks---everthing here is basically taken from Varilly's excellent lecture notes. It is well known in NCG-land that a compact oriented manifold $X$ is spin$^{\mathbb{C}}$ if and only if it admits an irreducible Clifford module (i.e., spinor bundle) $S \to X$, in which case the Picard group of line bundles (up to isomorphism) acts freely and transitively on the spinor bundles by $([L],[S]) \mapsto [L \otimes S]$.

    Now, with a little bit of care, if $S \to X$ is a spinor bundle, then you can make the dual bundle $S^\ast \to X$ into a spinor bundle as well, so that $S^\ast \cong L \otimes S$ for some line bundle $S$. It is then a famous (in NCG-land) theorem of Plymen's that $X$ is actually spin if and only if there exists a spinor bundle $S$ with $S^\ast \cong S$ as Clifford modules, in which case $S$ is the spinor bundle for the underlying spin structure. By the Riesz representation theorem (for Hermitian vector bundles) together with a little bit of care, the existence of this isomorphism of Clifford modules is equivalent to the existence of the famed charge conjugation operator $J$, whose commutation or anticommutation with the Dirac operator and chirality element is, ultimately, forced by the algebraic structure of $\mathrm{Cl}(\mathbb{R}^{\dim X})$---see Landsman's excellent but seemingly little-known lecture notes for details. Hence, by Bott periodicity for real Clifford algebras, these relations only depend on $\dim X \bmod 8$, yielding Connes's famous table---for subtleties, including why Connes's table doesn't (explicitly) include all $8$ possibilities for the three signs, see Landsman's notes.

  3. So, what about $KO$-theory? Here's what I can piece together as a relative layperson from the only source that goes into any detail, Gracia-Bondia--Varilly--Figueroa. So, by Section 9.5 of GBVF, there's a nice, concrete (indeed, basically algebraic) one-to-one correspondence between real spectral triples of $KO$-dimension $j \bmod 8$ $(A,H,D,J)$, aka reduced $KR^j$-cycles $(A \otimes A^o,H,D,J)$, and so-called unreduced $KR^j$-cycles $(A \otimes A^o,H,D,J,\rho)$, which are (roughly speaking) real spectral triples of $KO$-dimension $j \bmod 8$ endowed with a compatible action of $\mathrm{Cl}(\mathbb{R}^j)$. Under this correspondence (roughly speaking!), the Dirac operator of a compact spin manifold $X$, which can be viewed as living in the $K$-homology $K_0(X)$ of $X$, should correspond to a certain $\mathrm{Cl}(\mathbb{R}^j)$-linear (twisted) Dirac operator on $X$ (see Lawson--Michelson, S II.7), which can be viewed as living in the $KO$-homology $KO_j(X)$ of $X$, where Real Bott periodicity tells you that there are only the eight distinct $KO$-homology groups. So, to cut a long story short, a real spectral triple of $KO$-dimension $j \bmod 8$ is, well, said to have $KO$-dimension $j \bmod 8$ because it lives (morally) in the relevant $j$-th $KO$-homology group. This is probably what you actually wanted, so I hope it makes some sense!

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Very rich answer indeed! I appreciate specially the reference to Landsman's notes, which I didn't knew. –  c.p. Jan 1 '13 at 17:09

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