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Geometric conditions on spectral triples

We begin by listing a set of requirements on a spectral triple $(\sA, \sH, D)$, whose algebra $\sA$ is unital but not necessarily commutative, such that $(\sA, \sH, D)$ provides a ``spin geometry'' generalization of our ``standard commutative example'' $(\Coo(M), L^2(M,S), \Dslash)$. Again we shall assume, for convenience, that $D$ is invertible.


\begin{cond}[Spectral dimension]
There is an \emph{integer} $n \in \{1,2,\dots\}...
... satisfying
$a\Ga = \Ga a$ for all $a \in \sA$, and $D\Ga = -\Ga D$.
\end{cond}


\begin{remark}
% latex2html id marker 13115It is useful to allow the case $n =...
...
traces $\Trw$ by the ordinary matrix trace $\tr$.
\end{itemize}
\end{remark}


\begin{cond}[Regularity]
For each $a \in \sA$, the bounded operators $a$ and $[...
...a])\Vert$. This ensures that $\sA$ is a
Fr\'echet pre-$C^*$-algebra.
\end{cond}


\begin{cond}[Finiteness]
The subspace of smooth vectors
$\sH^\infty := \bigcap_{...
...\sA)$ such that
$\sH^\infty \isom \sA^N p$ as left $\sA$-modules.
\end{cond}


\begin{cond}[Real structure]
There is an antiunitary operator $J : \sH \to \sH$\...
...th}[a, J b^* J^{-1}]= 0, \word{for all} a,b \in \sA.
\end{displaymath}\end{cond}


\begin{cond}[First order]
For each $a,b \in \sA$, the following relation holds:
...
...}this is equivalent to the condition that $[a, [D, Jb^*J^{-1}]] = 0$.
\end{cond}


\begin{cond}[Orientation]
There is a Hochschild $n$-cycle
\begin{displaymath}
\c...
...$ is even}, \\
1, &\text{if $n$ is odd}. \end{cases}\end{equation}\end{cond}

In many examples, including the noncommutative examples we shall meet in the next two sections, one can often take $b_j^0 = 1$, so that $\cc$ may be replaced, for convenience, by the cycle $\sum_j a_j^0 \ox a_j^1 \oxyox a_j^n \in Z_n(\sA, \sA)$. In the commutative case, where $\sA^\opp = \sA$, this identification may be justified: the product map $m\: \sA \ox \sA \to \sA$ is a homomorphism.

The data set $(\sA, \sH, D; \Ga \text{ or } 1, J, \cc)$ satisfying these six conditions constitute a ``noncommutative spin geometry''. In the fundamental paper where these conditions were first laid out [Con2], Connes added one more nondegeneracy condition (Poincaré duality in $K$-theory) as a requirement. We shall not go into this matter here.



To understand the orientation condition in the standard commutative example, we show that $\cc$ arises from a volume form on the oriented compact manifold $M$. Choose a metric $g$ on $M$ and let $\nu_g$ be the corresponding Riemannian volume form. Furthermore, let $\{(U_j,a_j)\}$ be a finite atlas of charts on $M$, where $a_j\: U_j \to \bR^n$, and let $\{f_j\}$ be a partition of unity subordinate to the open cover $\{U_j\}$; then for $r = 1,\dots,n$, each $f_j a_j^r$ lies in $\Coo(M)$ with $\supp(f_j a_j^r) \subset U_j$. Over each $U_j$, let $\{\th_j^1,\dots,\th_j^n\}$ be a local orthonormal basis of $1$-forms (with respect to the metric $g$). Then

\begin{displaymath}
\nu_g \bigr\vert _{U_j} = \th_j^1 \wyw \th_j^n = h_j  da_j^1 \wyw da_j^n,
\end{displaymath}

for some smooth functions $h_j \: U_j \to \bC$. We write $a_j^0 := (-i)^m  f_j h_j \in \Coo(M)$, where as usual, $n = 2m$ or $n = 2m + 1$. With that notation, we get

\begin{displaymath}
(-i)^m \nu_g = (-i)^m \tsum_j f_j \bigl( \nu_g \bigr\vert _{U_j} \bigr)
= \tsum_j a_j^0  da_j^1 \wyw da_j^n.
\end{displaymath}

Now we define
\begin{displaymath}
\cc := \inv{n!} \sum_{\sg\in S_n} (-1)^\sg
\sum_j a_j^0 \ox a_j^{\sg(1)} \oxyox a_j^{\sg(n)}.
\end{displaymath} (26)


\begin{exer}
Show that the Hochschild boundary $b\cc$ of the chain
\eqref{eq:Hoch-cycle} is zero because $\sA$ is commutative.
\end{exer}

Therefore, $\cc$ is a Hochschild $n$-cycle in $Z_n(\sA,\sA)$, for $\sA = \Coo(M)$. Its representative as a bounded operator on $\sH$ is
\begin{align*}
\inv{n!} \sum_{\sg\in S_n} (-1)^\sg \sum_j
a_j^0  [\Dslash, a_j^...
...^m  c(\th_j^1) \dots c(\th_j^n)
\\
&= c(\ga) = \Ga \text{ or } 1,
\end{align*}
since $c(\ga) = \Ga$ for $n = 2m$, and $c(\ga) = 1$ for $n = 2m + 1$.

This calculation shows that the elements $a_j^1,\dots,a_j^n$ occurring in the cycle $c$ are local coordinate functions for $M$. An alternative approach would be to embed $M$ in some $\bR^N$ and take the $a_j^r$ to be some of the cartesian coordinates of $\bR^N$, regarded as functions on $M$. This is illustrated in the following example.


\begin{example}
% latex2html id marker 13193By regarding the sphere $\bS^2$ a...
...s the completion of the spinor
module $\sS = \sS^+ \oplus \sS^-$.
\end{example}

Consider the following element of $M_2(\sA)$, with $\sA = \Coo(\bS^2)$:

\begin{displaymath}
p := \frac{1}{2} \twobytwo{1 + z}{x + iy}{x - iy}{1 - z}.
\end{displaymath} (27)

Note that $\tr(p - \half) = 0$, where $\tr(a) := a_{11} + a_{22}$ means the matrix trace $\tr \: M_2(\sA) \to \sA$.


\begin{exer}
Show that, if $\sA$ is any $*$-algebra and $p \in M_2(\sA)$ is gi...
...
\\
[x,y] = [x,z] = [y,z] = 0,
\\
x^2 + y^2 + z^2 = 1.
\end{gather*}\end{exer}


\begin{exer}
Check that $\tr(p  dp \w dp) = - \frac{i}{2} \nu$.
\end{exer}

If we replace $- \frac{i}{2} \nu$ by the Hochschild $2$-cycle $\cc$, the same calculation that solves the previous exercise also shows that $\pi_D(\cc) = \Ga$.

This computation has a deeper significance. One can show that the left $\sA$-module $M_2(\sA) p$ is isomorphic to $\sE_1$ in our classification of $\sA$-modules of sections of line bundles over $\bS^2$; and we have seen in Section [*] that $\sE_1 \isom \Ga(\bS^2,L)$ where $L \to \bS^2$ is the tautological line bundle. The first Chern class $c_1(L)$ equals (a standard multiple of) $[\nu] \in H_\dR^2(\bS^2)$. One can trace a parallel relation between spin$^c$ structures on $\bS^2$ defined, via the principal $U(1)$-bundle $SU(2) \to \bS^2$, on associated line bundles, and the Chern classes of each such line bundle. For that, we refer to [BHMS].


next up previous contents
Next: Isospectral deformations of commutative Up: Spectral Triples: Examples Previous: Spectral Triples: Examples   Contents
Pawel Witkowski 2006-03-14