The Moyal plane as a nonunital spectral triple

In order to extend the notion of spectral triple $(\sA, \sH, D)$ to include the case where the algebra $\sA$ may be nonunital, we modify Definition  as follows.

In general, $D$ may have continuous spectrum, so that the operator $(D^2 + 1)^{-1/2}$ will usually not be compact. But it is enough to ask that it become compact when mollified by any multiplication operator in $\sA$. An equivalent condition is that $a(D - \la)^{-1}$ be compact, for all $\la \notin \spec(D)$. In the nonunital case, there is no advantage in supposing that $D$ be invertible, so it is better to work directly with $(D^2 + 1)^{1/2}$ instead of $\vert D\vert$.

The simplest noncommutative, nonunital example is an isospectral deformation of this commutative case, where we use the same Dirac operator $\Dslash = -i \ga^j \del/\del x^j$ on the same spinor space $\sH = L^2(\bR^n) \ox \bC^{2^m}$, but we change the algebra by replacing the ordinary product of functions by a Moyal product.

Before giving the details, we summarize the effect of this nonunital isospectral deformation on the conditions given in Section  to define a ``noncommutative spin geometry''.

• The reality and first-order conditions are unchanged: we use the same charge conjugation operator $C$ as in the undeformed case.

• The regularity condition is essentially unchanged: all that is needed is to replace the derivation $\dl \: T \mapsto [\vert D\vert, T]$ by the derivation $\dl_1 \: T \mapsto [(\Dslash^2 + 1)^{1/2}, T]$, because $\Dom \dl_1^k = \Dom \dl^k$ for each $k \in \bN$ since $(\Dslash^2 + 1)^{1/2} - \vert D\vert$ is a bounded operator.

• For the orientation condition, the Hochschild $n$-cycle will not lie in $Z_p(\sA, \sA \ox \sA^\opp)$ but rather in $Z_p(\tsA, \tsA \ox \tsA^\opp)$, where $\tsA$ is a unitization of $\sA$, that is, a unital $*$-algebra in which $\sA$ is included as an essential ideal.

• For the finiteness condition, we ask that $\sH^\infty = \sA^N p$, for some projector $p = p^2 = p^*$ lying in $M_N(\tsA)$. Thus $\sH^\infty$ can be regarded as the pullback, via the inclusion $\sA \hookto \tsA$, of the finitely generated projective left $\tsA$-module $\tsA^N p$.

• To define the integer $n$ as the spectral dimension, we would like to be able to assert that $a (D^2 + 1)^{-1/2}$ lies in $\sL^{n+}(\sH)$ for each $a \in \sA$, and that $0 < \Trw(a (D^2 + 1)^{-n/2}) < \infty$ whenever $a$ is positive and nonzero. It turns out that we can only verify this for $a$ belonging to a certain dense subalgebra of $\sA$, in the Moyal-plane example: see below.

In what follows, we will sketch the main features of the Moyal-plane spectral triple. A complete treatment can be found in Gayral et al [GGISV], on which this outline is based. Our main concern here is to identify the ``correct'' algebra $\sA$ and its unitization $\tsA$ so that the modified spin-geometry conditions will hold.

We now recall the Moyal product over $\bR^n$, discussed in the previous Section. It depends on a real skewsymmetric matrix $\Th \in M_n(\bR)$ of ``deformation parameters''. For $n = 2$, such a matrix is of the form $\begin{pmatrix}0 & \th \ -\th & 0 \end{pmatrix}$ for some $\th \in \bR$; and for $n = 2m$ or $n = 2m + 1$, $\Th$ is similar to a direct sum of $m$ such matrices with possibly different values of $\th$ (so $\Th$ cannot be invertible if $n$ is odd). For convenience, we now take $n$ to be even, and we shall suppose that all values of $\th$ are the same. (In applications to quantum mechanics, where the Moyal product originated [Moy], $\th = \hbar$ is the Planck constant.) Thus, we choose

$$\Th := \th S \in M_{2m}(\bR), \word{with} S := \begin{pmatrix}0 & 1_m \ -1_m & 0 \end{pmatrix}, \quad \th > 0.$$ (29)

Note that $\det\Th = \th^{2m} > 0$.

The first or second integral in () can also be regarded as defining the Moyal product $f \star_\th h$, where $f$ and $h$ need not be Schwartz functions, provided that the integrals are understood in some generalized sense. Thus Rieffel [Rie], for instance, considers them as oscillatory integrals. Here we shall extend the Moyal product by duality, as follows. It is easy to see that $\Vert f \star_\th h\Vert _\infty \leq (\pi\th)^{-n} \Vert f\Vert _1 \Vert h\Vert _1$, from the first integral in (). By applying similar estimates to the functions $x^\al \del^\bt (f \star_\th h)$, for $\al,\bt \in \bN^n$, one can verify that the product $(f,h) \mapsto f \star_\th h$ is a jointly continuous bilinear map from $\sS(\bR^n) \x \sS(\bR^n)$ to $\sS(\bR^n)$.

Here are some elementary properties of the Moyal product that are easy to check formally; they can be verified rigorously by some work with oscillatory integrals: see [Rie].

1. The Moyal product is associative: $(f \star_\th g) \star_\th h = f \star_\th (g \star_\th h)$.
2. The Leibniz rule holds: $\del_j(f \star_\th h) = \del_j f \star_\th h + f \star_\th \del_j h$ for $j = 1,\dots,n$.
3. Complex conjugation is an involution: $\overline{f \star_\th h} = \bar h \star_\th \bar f$.
4. Integration over $\bR^n$ is a trace for the Moyal product:
   

   $$\int_{\bR^n} (f \star_\th h)(x) dx = \int_{\bR^n} (h \star_\th f)(x) dx = \int_{\bR^n} f(x) h(x) dx.$$ (30)

   
   

We denote $\sS_\th := (\sS(\bR^n), \star_\th)$. It is a Fréchet $*$-algebra, with the usual topology of $\sS(\bR^n)$.

The trace property gives us a (suitably normalized) bilinear pairing:

$$\dst{f}{h} := (\pi\th)^{-m} \int_{\bR^n} (f \star_\th h)(x) dx.$$

Together with associativity, this gives the relation

$$\dst{f \star_\th g}{h} = \dst{f}{g \star_\th h} = (\pi\th)^{-m} \int_{\bR^n} (f \star_\th g \star_\th h)(x) dx,$$

valid for $f,g,h \in \sS(\bR^n)$. Now, if $T \in \sS'(\bR^n)$ is a tempered distribution, and if $f \in \sS(\bR^n)$, we can define $T \star_\th f, f\star_\th T \in \sS'(\bR^n)$ by the continuity of the Moyal product:

$$\dst{T \star_\th f}{h} := \dst{T}{f \star_\th h}, \qquad \dst{f \star_\th T}{h} := \dst{T}{h \star_\th f}.$$

In this way, $\sS'(\bR^n)$ becomes a bimodule over $\sS_\th$. Inside this bimodule, we can identify a multiplier algebra in the obvious way.

This Moyal algebra is very large: for instance, it contains all polynomials on $\bR^n$. However, because it contains many unbounded elements, it cannot serve as a coordinate algebra for a spectral triple. Even so, it is a starting point for a second approach, developed in [GV]. Consider the quadratic polynomials $H_r := \half(x_r^2 + x_{m+r}^2)$ for $r = 1,\ldots,m$. In the quantum-mechanical interpretation, these are Hamiltonians for a set of $m$ independent harmonic oscillators; but for now, it is enough to know that they belong to $\sM_\th$. It turns out that the left and right Moyal multiplications by these $H_r$ have a set of joint eigenfunctions $\set{f_{kl} : k,l \in \bN^m}$ belonging to the Schwartz space $\sS(\bR^n)$, with the following properties:

• The eigenvalues are half-integer multiples of $\th$, namely,
  
  $$H_r \star_\th f_{kl} = \th(k_r + \half) f_{kl}, \qquad f_{kl} \star_\th H_r = \th(l_r + \half) f_{kl}.$$
  
  

• The eigenfunctions form a set of matrix units for the Moyal product: $f_{kl} \star_\th f_{rs} = \dl_{lr} f_{ks}$ and $\bar f_{kl} = f_{lk}$ for all $k,l,r,s \in \bN^m$.
• Any $f \in \sS(\bR^n)$ is given by a series $f = (2\pi\th)^{-m/2} \sum_{kl} \al_{kl} f_{kl}$, converging in the topology of $\sS(\bR^n)$, such that $\al_{kl} \to 0$ rapidly.
• The subset $\set{(2\pi\th)^{-m/2} f_{kl} : k,l \in \bN^m}$ of $\sS(\bR^n)$ is an orthonormal basis for $L^2(\bR^n)$.

For example, when $n = 2$ and $k = l \in \bN$, $f_{kk}$ is given by

$$f_{kk}(x_1,x_2) := 2 (-1)^k e^{-(x_1^2 + x_2^2)/\th} L_k^0(\tfrac{2}{\th}(x_1^2 + x_2^2)),$$

where $L_k^0$ is the Laguerre polynomial of order $k$.

Because of these properties, we can extend the Moyal product to pairs of functions in $L^2(\bR^n)$. If $f = (2\pi\th)^{-m/2} \sum_{k,l} \al_{kl} f_{kl}$ and $h = (2\pi\th)^{-m/2} \sum_{k,l} \bt_{kl} f_{kl}$, we define

$$f \star_\th h := (2\pi\th)^{-m} \sum_{k,r,l} \al_{kr}\bt_{rl} f_{kl}.$$ (31)

The Schwarz inequality for sequences shows that
$$\begin{align} \Vert f \star_\th h\Vert _2^2 &= (2\pi\th)^{-2m} \biggl\Vert \sum_... ...{rl}\vert^2 = (2\pi\th)^{-m} \Vert f\Vert _2^2 \Vert h\Vert _2^2. \end{align}$$
This calculation guarantees that the series () converges whenever $f,h \in L^2(\bR^n)$; and that the operator $L(f) \: h \mapsto f \star_\th h$ extends to a bounded operator in $\sL(L^2(\bR^n))$ whenever $f \in L^2(\bR^n)$. Moreover, it gives a bound on the operator norm:

$$\Vert L(f)\Vert \leq (2\pi\th)^{-m/2} \Vert f\Vert _2.$$

Now the Schwartz-multiplier algebra $\sM_\th$ can be replaced by an $L^2$-multiplier algebra. By duality in sequence spaces, any $T \in \sS'(\bR^n)$ can be given an expansion in terms of the $\{f_{kl}\}$ basis, and in this way one can define an algebra

$$A_\th := \set{R \in \sS'(\bR^n) : R \star_\th f \in L^2(\bR^n) \text{ for all } f \in L^2(\bR^n)}.$$

This is actually a $C^*$-algebra, with operator norm $\Vert L(R)\Vert := \sup\set{\Vert R \star_\th f\Vert _2/\Vert f\Vert _2 : f \neq 0}$.

There is a unitary isomorphism $W \: L^2(\bR^n) \to L^2(\bR^m) \ox L^2(\bR^m)$ (tensor product of Hilbert spaces), such that $W L(f) W^{-1} = \sg(f) \ox 1$, where $\sg$ is the (irreducible) Schrödinger representation; that is to say, $f \mapsto L(f)$ is equivalent to the Schrödinger representation with infinite multiplicity. One can show that $A_\th = W^{-1} \sL(L^2(\bR^m)) W$, whereas the norm closure of the $*$-algebra $(\sS(\bR^n),\star_\th)$ is $W^{-1} \sK(L^2(\bR^m)) W$. For the details, consult [VG] and [GGISV].

The analogue of Lemma  holds, too: $\sS_\th$ is a nonunital pre-$C^*$-algebra. As in the proof of Lemma , if $f \in \sS_\th$, suppose the equation $(1 + f) \star_\th (1 + g) = 1$ has a solution $g$ in the unital $C^*$-algebra $A_\th$. We may also write

$$f + g + f \star_\th g = 0 \words{and} f + g + g \star_\th f = 0,$$ (32)

and we wish to show that $g \in \sS_\th$. Since $g = - f - f \star_\th g$, it is enough to show that $f \star_\th g \in \sS_\th$. Now, left-multiplying the second equation in () by $f$ gives $f \star_\th f + f \star_\th g + f \star_\th g \star_\th f = 0$, so it is enough to check that $f \star_\th g \star_\th f \in \sS(\bR^n)$ whenever $f \in \sS(\bR^n)$ and $g \in A_\th$. This turns out to be true: the necessary norm estimates are given in [VG].

However, the algebra $\sS_\th$ is not the best candidate for the coordinate algebra of the Moyal spectral triple. We now introduce a better algebra.

It does not matter whether these derivatives $\del^\al f$ are taken to be distributional derivatives only, since arguments based on Sobolev's Lemma show that if $f$ and all its distributional derivatives are square-integrable, then $f$ is actually a smooth function.

The algebra $\sA_\th$ is nonunital. Next, we introduce the preferred unitization of $\sA_\th$.

It is proved in Schwartz' book that $\sD_{L^2}(\bR^n) \subset \sB(\bR^n)$, and that the inclusion is continuous for the given topologies. (This is not as obvious as it seems, because in general square-integrable functions on $\bR^n$ need not be bounded.) Combining this with knowledge of the Moyal multiplier algebras, we end up with the following inclusions [GGISV]:

$$\sS_\th \subset \sA_\th \subset \tsA_\th \subset A_\th \cap \sM_\th.$$

The inclusion $\tsA_\th \subset A_\th$ is a consequence of the Calderón-Vaillancourt theorem, which says that a pseudodifferential operator of order zero on $\bR^n$, whose symbol is differentiable to a high enough order, gives a bounded operator on $L^2(\bR^n)$; we may notice that the second integral in () says that $L(f)$ is pseudodifferential, with symbol $p(x,\xi) = f(x - \half\th S\xi)$.

Rieffel, in [Rie], showed that $\tsA_\th$ is the space of smooth vectors for the action of $\bR^n$ (by translations) on its $C^*$-completion; this entails that $\tsA_\th$ is a pre-$C^*$-algebra.

Now, the inclusion $\tsA_\th \subset A_\th$ means that $\Vert\del^\al f \star_\th \del^\bt h\Vert _2$ is finite, whenever $f \in \tsA_\th$ and $h \in \sA_\th$; therefore, $f \star_\th h$ lies in $\sA_\th$ also. A similar argument shows that $h \star_\th f$ lies in $\sA_\th$. Thus, $\sA_\th$ is an ideal in $\tsA_\th$. (In fact, it is an essential ideal; that is to say, if $f \star_\th h = 0$ for all $h \in \sA_\th$, then $f = 0$; this can be seen by taking $h = f_{kl}$ for any $k,l \in \bN^n$ and checking that $f$ must vanish.)

An important family of elements in $\tsA_\th$ that do not belong to $\sA_\th$ are the plane waves:

$$u_k(x) := e^{2\pi ikx}, \word{for each} k \in \bR^n.$$

It is immediate from the formulas () that

$$u_k \star_\th u_l = e^{-\pi i\th k(Sl)} u_{k+l}, \word{for all} k,l \in \bR^n.$$

In particular, by taking $k,l \in \bZ^n$ to be integral vectors, we get an inclusion $\Coo(\bT^n_{\th S}) \hookto \tsA_\th$: the smooth algebra of the noncommutative $n$-torus, for $\Th = \th S$, can be identified with a subalgebra of periodic functions in $\tsA_\th$.

In particular, the Hochschild $n$-cycle $\cc$ representing the orientation of this noncommutative torus can also be regarded as an $n$-cycle over $\tsA_\th$. We can write $u_k = v_1^{k_1} \star_\th \cdots \star_\th v_n^{k_n}$ where $v_j = u_{e_j}$ for the standard orthonormal basis $\{e_1, \dots, e_n\}$ of $\bR^n$. The expression for $\cc$ is

$$\cc = \inv{n! (2\pi i)^n} \sum_{\sg\in S_n} (-1)^\sg (v_{\s... ...\sg(n)})^{-1} \ox v_{\sg(1)} \ox v_{\sg(2)} \oxyox v_{\sg(n)}.$$

(When $\th = 0$, we can write $v_j = e^{2\pi it_j}$, and the right hand side reduces to $dt_1 \wyw dt_n$, the usual volume form for either $\bR^n$ or the flat torus $\bT^n = \bR^n/\bZ^n$.)

We refer to [GGISV] for the discussion of the spectral dimension properties of the triple $(\sA_\th, L^2(\bR^n)\ox\bC^{2^m}, \Dslash)$. Briefly, the facts are these. If $\pi(f) := L(f) \ox 1_{2^m}$ denotes the representation of $\sA_\th$ on the spinor space $\sH$ by componentwise left Moyal multiplication, then one can show that, for any $f \in \sA_\th$, we get

$$\pi(f) (\Dslash^2 + 1)^{-1/2} \in \sL^p(\sH), \word{for all} p > n.$$

In particular, these operators are compact, so this triple is indeed a nonunital spectral triple. However, this is not quite enough to guarantee that

$$\pi(f) (\Dslash^2 + 1)^{-1/2} \in \sL^{n+}(\sH),$$ (33)

for every $f \in \sA_\th$. Instead, what is found in [GGISV] is that () holds for $f$ lying in the (dense) subalgebra $\sS_\th$. The key lemma which makes the proof work is a ``strong factorization'' property of $\sS_\th$, proved in [GV]: namely, that any $f \in \sS_\th$ can be expressed (without taking finite sums) as a product $f = g \star_\th h$, with $g,h \in \sS_\th$. This factorization property fails for the full algebra $\sA_\th$.

Once () has been established, one can proceed to compute its Dixmier trace. It turns out that $\Trw(\pi(f) (\Dslash^2 + 1)^{-1/2})$ is unchanged from its value when $\th = 0$, namely $(2^m \Omega_n/n (2\pi)^n) \int_{\bR^n} f(x) dx$. The end result is that the spectral dimension condition for nonunital spectral triples is the expected one, but that Dixmier-traceability as in () should only be required for a dense subalgebra of the original algebra.