Regularity of spectral triples

The arguments of the previous section are not applicable to determine whether $[\vert D\vert, a]$ is bounded, in the case $r = 1$. This must be formulated as an assumption. In fact, we shall ask for much more: we want each element $a \in \sA$, and each bounded operator $[D, a]$ too, to lie in the smooth domain of the following derivation.

The regularity condition does not depend on the invertibility of $D$ (that is, the condition $\ker D = \{0\}$) which we have been assuming, to simplify certain calculations. One can always replace $\vert D\vert$ by $\langle D\rangle := (D^2 + 1)^{1/2}$ in the definition, since $f(D) := \langle D\rangle - \vert D\vert$ is bounded. If $\dl'$ denotes the derivation $\dl'(T) := [\langle D\rangle, T] = \dl(T) + [f(D), T]$, then clearly $\Dom\dl' = \Dom\dl$, and it is easy to show by induction that $\Dom\dl'^k = \Dom\dl^k$ for each $k \in \bN$, so one may instead define regularity using $\dl'$. This is the approach taken in the work of Carey et al [CPRS], who use the term ``$Q\Coo$'' instead of ``regular'' for this class of spectral triples.

Since $a \in \sA$ implies $a \in \Dom\dl$, we see that $a(\sH^1) \subseteq \sH^1$, and then we can write $a(\vert D\vert\xi) = \vert D\vert(a\xi) - [\vert D\vert,a] \xi$ for $\xi \in \sH^1$. Also,
$$\begin{align*} \Vert a\xi\Vert _1^2 &= \Vert a\xi\Vert^2 + \Vert \vert D\vert a\... ...t^2 + 2\Vert\dl(a)\Vert^2, 2\Vert a\Vert^2\} \Vert\xi\Vert _1^2, \end{align*}$$
where we have used the parallelogram law $\Vert\xi + \eta\Vert^2 + \Vert\xi - \eta\Vert^2 = 2\Vert\xi\Vert^2 + 2\Vert\eta\Vert^2$. Therefore, $a$ extends to a bounded operator on $\sH^1$. If $(\sA, \sH, D)$ is regular, then by induction we find that $a(\sH^k) \subset \sH^k$ continuously for each $k$, so that $a(\sH^{\infty}) \subset \sH^{\infty}$ continuously, too.

Suppose $(\sA, \sH, D)$ is regular. Then $\sA \subset \Op_D^0$ and $[D,\sA] := \set{[D,a] : a \in \sA} \subset \Op_D^0$, too. Moreover, if $a \in \sA$, then
$$\begin{align*}[D^2, a]&= [\vert D\vert^2, a] = \vert D\vert [\vert D\vert, a] +... ...(a) - [\vert D\vert, \dl(a)] \\ &= 2\vert D\vert\dl(a) - \dl^2(a), \end{align*}$$
so that $[D^2, a] \in \Op_D^1$. Also $[D^2, [D, a]] \in \Op_D^1$ in the same way.

If $b$ lies the subalgebra of $\sL(\sH)$ generated by $\sA$ and $[D,\sA]$, we introduce
$$\begin{align} L(b) &:= \vert D\vert^{-1} [D^2, b] = 2\dl(b) - \vert D\vert^{-1}... ...[D^2, b] \vert D\vert^{-1} = 2\dl(b) + \dl^2(b) \vert D\vert^{-1}. \end{align}$$
If $b \in \bigcap_{k\geq 0}\Dom \dl^k$, then $L(b)$ and $R(b)$ lie in $\Op_D^0$. The operations $L$ and $R$ commute: indeed,

$$L(R(b)) = \vert D\vert^{-1} [D^2, [D^2, b] \vert D\vert^{-... ...vert D\vert^{-1} [D^2, [D^2, b] \vert D\vert^{-1} = R(L(b)).$$

Note also that $L^2(b) = \vert D\vert^{-2} [D^2, [D^2, b]]$.

This example also shows why regularity is defined using the derivation $\dl = [\vert D\vert,\cdot ]$ instead of the apparently simpler derivation $[D,\cdot ]$. Indeed, we have just seen that for $a \in \Coo(M)$, the operator $[\vert\Dslash\vert, [\Dslash, a]]$ has order zero (and therefore, it lies in $\Op_\Dslash^0$. On the other hand, $[\Dslash, [\Dslash, a]]$ is in general a $\Psi$DO of order $1$ (and so it lies in $\Op_\Dslash^1$). Indeed, the first-order terms in its symbol are

$$[\sg^{\Dslash}, \sg^{[\Dslash,a]}](x, \xi) = [c^j\xi_j, -i c^k \del_k a(x)] = -i [c^j,c^k] \xi_j \del_k a(x)$$

which need not vanish since $c^j$, $c^k$ do not commute. In contrast, the principal symbol of $\vert\Dslash\vert$ is a scalar matrix, which commutes with that of $[\Dslash, a]$, and the order of the commutator drops to zero.