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The Dixmier trace revisited

Let $(\sA, \sH, D)$ be a spectral triple, whose algebra $\sA$ is unital. We continue to assume, for convenience, that $\ker D = \{0\}$ , so that $D^{-1}$ is a compact operator on $\sH$ . Suppose now that $\vert D\vert^{-p} \in \sL^{1+}$ , for some $p \geq 1$ . Then the functional on $\sA$ given by $a \mapsto \Trw(a \vert D\vert^{-p})$ , for some particular $\om$ , is our candidate for a ``noncommutative integral''. To see why that should be so, we first examine the commutative case.

$\begin{prop} If $M$ is a compact boundaryless $n$-dimensional spin manifold, wi... ...m+1)!! \pi^{m+1}} &\word{if} n = 2m + 1. \end{cases}\end{displaymath}\end{prop}$

$\begin{proof} % latex2html id marker 9832We know that $\vert\Dslash\vert^{-n}$... ...n(2\pi)^n} \int_M a(x) \nu_g. \tag*{\qed} % \end{align*}\hideqed % \end{proof}$

Therefore, the functional $a \mapsto \Tr^+(a \vert\Dslash\vert^{-n})$ is just the usual integral with respect to the Riemannian volume form, expect for the normalization constant. Therefore, it can be adapted to more general spectral triples as a ``noncommutative integral''.

However, in the noncommutative case, it is not obvious that $a \mapsto \Tr^+(a \vert\Dslash\vert^{-n})$ will be itself a trace. ¿Why should $\Tr^+(ab \vert\Dslash\vert^{-n})$ be equal to $\Tr^+(ba \vert\Dslash\vert^{-n}) = \Tr^+(a \vert\Dslash\vert^{-n} b)$ ? To check this tracial property of the noncommutative integral, we need the Hölder inequality for Dixmier traces.

$\begin{fact}[Horn's inequality] If $T, S\in\sK$ and $n\in\bN$, then \begin{equation} \sg_n(TS) \leq \sum_{k=0}^{n-1} s_k(T) s_k(s). \end{equation}\end{fact}$

$\begin{prop} \textup{(a)}\quad If $T \in \sL^{1+}$ and $S$ is a bounded operat... ...{1/p} (\Trw \vert T\vert^q)^{1/q}. \end{equation}\end{subequations}\end{prop}$

$\begin{proof} \textrm{Ad (a):}\quad By the minimax formula \eqref{eq:sing-val} f... ...displaymath}and the result \eqref{eq:Hoelder-Dix-b} follows at once. \end{proof}$

$\begin{prop} Let $(\sA,\sH,D)$ be any spectral triple whose operator $D$ is i... ...t^r, a]$ is a bounded operator for each $r$ such that $0 < r < 1$. \end{prop}$

We postpone the proof of this Proposition until later. It is a crucial property of spectral triple that this bounded commutator property is not automatic for the case , that is, the commutators $[\vert D\vert, a]$ need not be bounded in general.

$\begin{thm} If $(\sA, \sH, D)$ is a spectral triple such that $\vert D\vert^{-... ...{-p}) = \Trw(Ta \vert D\vert^{-p}) \words{for all} \om. \end{equation}\end{thm}$

$\begin{proof} Note that $aT \vert D\vert^{-p}$ and $Ta \vert D\vert^{-p}$ li... ...> p. \end{equation}This is a consequence of the next lemma. \hideqed \end{proof}$

$\begin{lem} If $A \in \sL^{1+}(\sH)$ and $A \geq 0$, then $A^s \in \sL^1(\sH)$ for $s > 1$. \end{lem}$

$\begin{proof} We need the following result on sequence spaces. If $E$ is a Bana... ...fty, \end{displaymath}since $A \in \sL^{1+}$ implies $A \in \sL^s$. \end{proof}$

This establishes () and concludes the proof of Theorem .

$\begin{cor} If $A \geq 0$ is in $\sL^{1+}$, and $\Tr^+ A > 0$, then $\Tr^+ A^s = 0$ for $s > 1$. \end{cor}$

To establish Proposition , we use the following commutator estimate, due to Helton and Howe [HH].

$\begin{lem} Let $D$ be a selfadjoint operator on $\sH$, and let $a \in \sL(\sH)... ...D,a]\Vert \int_{\bR} \vert t \hat{g}(t)\vert dt. \end{displaymath} \end{lem}$

$\begin{proof} We may define \begin{displaymath} g(D) := \inv{2\pi}\int_{\bR} \h... ...a]$ extends to a bounded operator, and the required estimate holds. \end{proof}$

$\begin{proof} % latex2html id marker 10098 [Proof of Proposition \ref{pr:nice-co... ...\Vert[h(D), a]\Vert$ is finite since $h(D)$ is a bounded operator. \end{proof}$

Next: Regularity of spectral triples Up: Spectral Triples: General Theory Previous: Spectral Triples: General Theory Contents

Pawel Witkowski 2006-03-14