Isospectral deformations of commutative spectral triples

To some extent, one can recover the sphere $\bS^2$ from spectral triple data alone. Thus, if $\sA$ is a $*$-subalgebra of some $C^*$-algebra containing elements $x, y, z$, and if the matrix

$$p = \frac{1}{2} \twobytwo{1 + z}{x + iy}{x - iy}{1 - z} \in M_2(\sA)$$

is a projector, i.e., $p = p^* = p^2$, then by Exercise , the elements $x, y, z$ commute, they are selfadjoint, and they satisfy $x^2 + y^2 + z^2 = 1$. Thus, the commutative $C^*$-algebra $A$ generated by $x, y, z$ is of the form $C(X)$, where $X \subseteq \bS^2$ is a closed subset. If $\sA$ is now a pre-$C^*$-subalgebra of $A$ containing $x, y, z$, and is the algebra of some spectral triple $(\sA, \sH, D)$, then the extra condition $\pi_D(\cc) = \Ga$ can only hold if $X$ is the support of the measure $\nu$. This means that $X = \bS^2$.

A similar argument can be tried, to obtain an ``algebraic'' description of $\bS^4$. What follows is a heuristic motivation, following [CL]. One looks for a projector $p \in M_4(\sA)$, of the form

$$p = \begin{pmatrix} 1 + z & 0 & a & b \\ 0 & 1 + z & -b^* &... ...ord{where} q = \begin{pmatrix}a & b \ -b^* & a \end{pmatrix}.$$

Then $p = p^*$ only if $z = z^*$, and then $p^2 = p$ implies that $-1 \leq z \leq 1$ in the $C^*$-completion $A$ of $\sA$, $\bigl[\bigl( \begin{smallmatrix}z & 0 \ 0 & z \end{smallmatrix}\bigr), q \bigr] = 0$, and $qq^* = q^*q = \twobytwo{1 - z^2}{0}{0}{1 - z^2}$. From that, one finds that $a, a^*, b, b^*, z$ commute, and $aa^* + bb^* = 1 - z^2$. Thus $A = C(X)$ with $X \subseteq \bS^4$. Again, it can be shown that the equality $X = \bS^4$ is reached by some extra conditions, namely,
$$\begin{align*} \tr(p - \half) &= 0, \\ \tr((p - \half) dp dp) &= 0 \word{in}... ... \\ \pi_D((p - \half) dp dp dp dp) &= \Ga \word{in} \sL(\sH). \end{align*}$$

However, if one takes instead $q := \twobytwo{a}{b}{-\bar\la b^*}{a^*}$ with $\la = e^{2\pi i\th}$, then there is another, noncommutative, solution [CL]: now $A$ is the $C^*$-algebra generated by $a$, $b$ and $z = z^*$, where $z$ is central, and the other relations are
$$\begin{gather} ab = e^{-2\pi i\th} ba, \qquad a^*b = e^{2\pi i\th} ba^*, \notag \\ aa^* = a^*a, \qquad bb^* = b^*b, \qquad aa^* + bb^* = 1 - z^2. \end{gather}$$
To find a solution to these relations, where the central element $z$ is taken to be a scalar multiple of $1$, we substitute
$$\begin{align*} a &= u \sin\psi \cos\phi \\ b &= v \sin\psi \cos\phi \\ z &= (\cos\psi) 1 \end{align*}$$
with $-\pi \leq \psi \leq \pi$ and $-\pi < \phi \leq \pi$, say. In this way, the commutation relations () reduce to

$$uu^* = u^*u = 1, \qquad vv^* = v^*v = 1, \qquad vu = e^{2\pi i\th} uv.$$

These are the relations for the unitary generators of a noncommutative $2$-torus: see Section . Thus, by fixing values of $\phi,\psi$ with $\psi \neq \pm\pi$ and $\phi \notin \frac{\pi}{2}\bZ$, we get a homomorphism from $A$ to $C(\bT^2_\th)$, the $C^*$-algebra of the noncommutative $2$-torus with parameters $\Th = \twobytwo{0}{\th}{-\th}{0} \in M_2(A)$.

We look for a suitable algebra $\sA$, generated by elements satisfying the above relations, by examining a Moyal deformation of $\Coo(\bS^4)$. One should first note that $\bS^4 \subset \bR^5 = \bC \x \bC \x \bR$ carries an obvious action of $\bT^2$, namely,

$$(t_1, t_2) \cdot (\al, \bt, z) := (t_1\al, t_2\bt, z), \word{for} \vert t_1\vert = \vert t_2\vert = 1,$$

which preserves the defining relation $\al \bar\al + \bt \bar\bt + z^2 = 1$ of $\bS^4$. The action is not free: there are two fixed points $(0,0,\pm 1)$, and for each $t$ with $-1 < t < 1$ there are two circular orbits, namely $\set{(\al, 0, t) : \al\bar\al = 1 - t^2}$ and $\set{(\bt, 0, t) : \bt\bar\bt = 1 - t^2}$. The remaining orbits are copies of $\bT^2$. The construction which follows will produce a ``noncommutative space'' $\bS^4_\th$ that can be thought of as the sphere $\bS^4$ with each principal orbit $\bT^2$ replaced by a noncommutative torus $\bT^2_\th$, while the $\bS^1$-orbits and the two fixed points remain unchanged.

In quantum mechanics, the Moyal product of two functions $f,h \in \sS(\bR^n)$ is defined as an (oscillatory) integral of the form

$$(f \star h)(x) := (\pi\th)^{-n} \int_{\bR^n} \int_{\bR^n} f(x + s) h(x + t) e^{-2is(\Th^{-1}t)} ds dt.$$ (28)

Here $n = 2m$ is even, $\Th = - \Th^t\in M_n(\bR)$ is an invertible skewsymmetric matrix, and $\det\Th = \th^n$ with $\th > 0$. In the next section, we shall interpret this formula in a precise manner (see Definition  below), and show that $f \star h$ lies in $\sS(\bR^n)$ also. Formally, at any rate, one can rewrite it as an ordinary Fourier integral:

$$(f \star h)(x) := (2\pi)^{-n} \int_{\bR^n} \int_{\bR^n} f(x - \half\Th u) h(x + t) e^{-iut} du dt,$$

with the advantage that now $\Th$ need not be invertible (so that $n$ need no longer be even). It was noticed by Rieffel [Rie] that one can replace the translation action of $\bR^n$ on $f, h$ by any (strongly continuous) action $\al$ of some $\bR^l$ on a $C^*$-algebra $A$. Then, given $\Th = -\Th^t \in M_l(\bR)$, one can define

$$a \star b := \int_{\bR^l} \int_{\bR^l} \al_{\shalf\Th u}(a) \al_{-t}(b) e^{2\pi i ut} du dt,$$

provided that the integral makes sense. In particular, if $\al$ is periodic action of $\bR^l$, i.e., $\al_{t+r} = \al_t$ for $r \in \bZ^n$, so that $\al$ is effectively an action of $\bT^n = \bR^n/\bZ^n$, then one can describe the Moyal deformation as follows.

Let $\sA:=\set{a \in A : t \mapsto \al_t(a) \text{ is smooth}}$ be the ``smooth subalgebra'' for the action of $\bT^l$. It can be shown that $\sA$ is a Fréchet pre-$C^*$-algebra, and each $a \in \sA$ can be written as a convergent series $a = \sum_{r\in\bZ^l} a_r$, where $a_r \in A_{(r)}$ and $\Vert a_r\Vert \to 0$ rapidly as $\vert r\vert \to \infty$.

For actions of $\bT^l$, Rieffel [Rie] showed that the integral formula and the series formula for $a \star b$ are equivalent, when $a,b$ belong to the smooth subalgebra $\sA$.

To deform the spectral triple $(\Coo(M), L^2(M,S), \Dslash)$, we need a further step. Since each $\sg_t$ is an isometry of $M$, it defines an automorphism of the tangent bundle $TM$ (with $T_xM \to T_{\sg_t(x)}M$), and of the cotangent bundle $T^*M$ (with $T_{\sg_t(x)}^*M \to T_x^*M$), preserving the orientation and the metric on each bundle. But the group $\SO(T_x^*M, g_x)$ does not act directly on the fibre $S_x$ of the spinor bundle. Instead, the action of the Clifford algebra $\sB$ on $\sH = L^2(M,S)$ yields a homomorphism $\Spin(T_x^*M, g_x) \to \End(S_x)$ for each $x \in M$, and we know that there is a double covering $\Ad_x \: \Spin(T_x^*M, g_x) \to \SO(T_x^*M, g_x)$ by conjugation.

It turns out [CDV] that one can lift the isometric action $\al \: \bT^l \to SO(T^*M)$ to an action of another torus $\tau \: \Tilde\bT^l \to \Aut(S)$, where there is a covering map $\pi\: \Tilde\bT^l \to \bT^l$ such that $\pi(\pm 1) = 1$, making the following diagram commute:

We can regard $\bT^l$ as $\bR^l/(\bZ^l + \hat{\bZ}^l)$, where $\hat{\bZ}^l = \bZ^l + (\half, \half, \hdots, \half)$. With this convention, one can show that the set of commuting selfadjoint operators $P_1, \hdots, P_l$ on $\sH$ which generate the unitary representation of $\Tilde\bT^l$, i.e.,

$$\tau_{\tilde{t}} =: \exp(it_1 P_1 +\cdots+ it_n P_n),$$

have half-integer spectra: $\spec(P_j) \subseteq \half\bZ$.

Now define a family of unitary operators $\set{\sg(r,P) : r\in \bZ^l}$ by

$$\sg(r,P) := \exp\Bigl( -\pi i \tsum_{j,k} r_j \th_{jk} P_k \Bigr),$$

that is, we substitute $s_k$ by $P_k$ in the cocycle formula $\sg(r,s)$ of ().

It follows that each $\sg(r,P)$ commutes with $\Dslash$, too. However, the operators $\sg(r,P)$ need not commute with the multiplication operators $\psi \mapsto f \psi$, for $f \in \Coo(M)$. Indeed, () implies that $\tau_{\tilde{t}} f \tau_{-\tilde{t}} = \al_t(f)$ for each $\tilde{t} \in \Tilde\bT^l$.

We are now ready to exhibit the isospectral deformation of the standard commutative example for a spin manifold $M$ carrying an isometric action of $\bT^l$, with respect to a fixed matrix $\Th$ of deformation parameters. The deformation is called isospectral for the simple reason that the operator $D$ of the deformed spectral triple is the same Dirac operator of the undeformed case, so it is no surprise that its spectrum does not change. What does change is the algebra: in fact, the underlying vector space of $\sA$ is unchanged, but the product operation is deformed, and consequently its representation on $\sH$ changes, too.

When $\dim M$ is even, and $\Ga$ is the $\bZ_2$-grading operator $\Ga$ on the spinor space $\sH$, we should note that the orientation condition $\pi_\Dslash(\cc) = \Ga$ says, among other things, that $\Ga$ appears in the algebra generated by the operators $f$ and $[\Dslash, f]$, for $f \in \sA$. The representation $L$ of $\sA_\Th$ extends to this algebra of operators by using () as a definition of $L([\Dslash,f])$. In the formula () for $\pi_\Dslash(\cc)$, if we replace all terms $a_j^r$ by $L(a_j^r)$, then we obtain $L(\pi_\Dslash(\cc)) = L(\Ga) = \Ga$. Thus $\cc$ may also be regarded as a Hochschild $n$-cycle over $\sA_\Th$, and the orientation condition $\pi_\Dslash(\cc) = \Ga$ is unchanged by the deformation. In odd dimensions, the same is true, with $\Ga$ replaced by $1$.

In conclusion: the isospectral deformation procedure of Connes and Landi yields a family of noncommutative spectral triples that satisfy all of our stated conditions for a noncommutative spin geometry. (Moreover [CL], Poincaré duality in $K$-theory is stable under deformation, too.)