# In Verse: Inverse Functions

Functions compose,
let us suppose,
as f of g of x.
When one undoes the other
now’s the case we shall inspect:
Inputs, outputs, switch their place,
then double back with grace
Domain and range,
these sets exchange,
the relation thus bijects
upon return,
as we will learn;
At first, to keep things terse,
when f of g
is an identity,
then f is g’s inverse.

But wait, there’s more!
I must implore,
please stay and hear my pitch.
it’s not without a hitch:
This minus one
is sure to stun
an unsuspecting schmuck,
for though it hints
at exponents,
you’ll find you’re out of luck
compose at all
with the reciprocal
most likely, you’ll get stuck.

Next let us graph,
observe as half
the plane swaps with the other.
Our mirror is
y equals x
as you will soon discover.
Once vertical, now horizontal,
lines can now define
when functions are invertible
so we’re not steering blind

That is all I have to say,
For questions on the homework task,

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# Commutative Geometry

Spin Manifolds and Spin Connections

Definition 1.  Let $M$ be a Riemannian manifold.  The Clifford bundle over $M$ is the Clifford algebra $Cl(T^*M)=\bigoplus_{p\in M}Cl(T_p^*M)$ generated with respect to the inverse Riemannian metric.  A left Clifford module for a compact Riemannian manifold is the finitely generated projective $C(M)$-module $S=\Gamma(E)$ corresponding to a complex vector bundle $E$ over $M$ together wth a $C(M)$-homomorphism $c:\Gamma(Cl(M))\to\Gamma(End(S)).$  The Clifford module is self-adjoint if $c(\sigma)^*=c(\sigma^*).$

Thus a Clifford module $S=\Gamma(E)$ is a $\Gamma(Cl(M))-C(M)$ bimodule where the left scalar multiplication is given by

$\displaystyle (\sigma s)_x=c(\sigma)_x(s_x)$

for $\sigma\in\Gamma(Cl(M))$ and $s\in S.$  Hence $Cl_x(M)=End(S_x).$  The chirality element $\gamma\in\Gamma(Cl(M))$ is given locally by $\gamma_x=(-i)^m\,dx_1\wedge\cdots\wedge dx_n$ with $\dim(M)\in\{2m,2m+1\}.$  For an orientable manifold, this induces a $\mathbb{Z}_2$-grading on $Cl(M).$  $S$ decomposes into $(\pm 1)$-eigenspaces of $c(\gamma),$ which in turn yields a decomposition of $Cl(M).$

Recall that an elementary C*-algebra is one that is isomorphic to a C*-algebra of compact operators on some Hilbert space $H,$ and the algebra is either finite dimensional, or infinite dimensional and separable.  Then for $M$ even (or odd) dimensional, $Cl(M)$ (or $Cl^+(M)$) is in fact a continuous field of locally trivial elementary C*-algebras.  Depending on the dimension of $M,$ we’ll use $Cl^{(+)}(M)$ for the variable Clifford bundle.

The forthcoming machinery is a bit cumbersome, some I will omit proofs.

Proposition 2.  A locally trivial continuous field of elementary C*-algebras $\mathcal{A}$ determines a Cech cohomology class $\delta(\mathcal{A})\in H^3(M,\mathbb{Z}),$ called a Dixmier-Douady class, such that $\delta(\mathcal{A})=\delta(\mathcal{A}')$ iff $\mathcal{A}=\mathcal{A'}.$

Theorem 3 (Plymen). If M is a compact Riemannian manifold, then $C(M)$ and $\Gamma(Cl^{(+)}(M))$ are Morita equivalent iff $\delta(Cl^{(+)}(M))=0.$

Recall the Picard group $Pic(A)$ of a C*-algebra $A$ is defined as the group $Mrt(A,A)$ under the operation

$[E][E']=[E\otimes_A E'].$

Proposition 4.  Let $M$ be a compact Riemannian manifold, $A=C(M),$ and $B=\Gamma(Cl^{(+)}(M)).$  Then $Pic(A)$ acts freely and transitively (on the right) on $Mrt(B,A).$

Definition 5.  If $M$ is a compact orientable manifold with $\Gamma(Cl(M))$ and $C(M)$ Morita equivalent (i.e. $\delta(Cl^{(+)}(M))=0$), then a pair $(v,S)$ with $v$ an orientation of $M$ and $S\in Mrt(\Gamma(Cl^{(+)}(M)),C(M))$ (hence in particular, it is a Clifford module) is called a spinc-structure on $M$, and $(M,v,S)$ is called a spinc-manifold.

Proposition 6.  Any self-adjoint Clifford module over a compact spin^c manifold $(M,v,S)$ is of the form $E=S\otimes_{C(M)}F$ for a finitely generated projective $C(M)$-module $F.$

If $\mathcal{A}$ is a continuous field of elementary C*-algebras and $\delta{\mathcal{A}}=0,$ it turns out that they can be classified by Cech classes $\kappa(\mathcal{A})\in H^2(M,\mathbb{Z}_2).$  And just as the vanishing of $\delta(\Gamma(Cl^{(+)}(M)))\in H^3(M,\mathbb{Z})$ was equivalent to Morita equivalence of $\delta(\Gamma(Cl^{(+)}(M)))$ and $C(M),$ the vanishing of $\kappa(\Gamma(Cl^{(+)}(M)))\in H^2(M,\mathbb{Z}_2)$ has an equivalence as well.

Theorem 7.  Let $(M,v,S)$ be a spin^c manifold with $A=C(M)$ and $B=\Gamma(Cl^{(+)}(M)).$  Then $\kappa(B)=0$ iff at least one Morita equivalence $B-A$ bimodule $S$ admits a bijective antilinear map $C:S\to S$ satisfying

$\begin{array}{rcl}C(sa)&=&(Cs)a^*\\C(bs)&=&\chi(b^*)(Cs)\\\langle Cr,Cs\rangle&=&\langle r,s\rangle.\end{array}$

Proposition 8.  Any map $C$ satisfying the above equations further satisfies $C^2=\pm 1.$

Definition 9.  A spin structure is a triple $(v,S,C)$ with $(v,S)$ a spin^c structure on $M$ and $C$ a conjugation map satisfying properties of the above theorem.  $(M,v,S,C)$ is called a spin manifold.  The bundle $E\to M$ for which $S=\Gamma(E)$ will be called the spinor bundle.

Proposition 10.  In fact, if $\kappa(B)=0,$ the spin structures of $M$ are classified by $H^1(M,\mathbb{Z}_2).$

Hermitian connection $\nabla:E\to E\otimes\Omega^1(A)$ on an $A$-inner product space is one that satisfies

$\langle\nabla x,y\rangle+\langle x,\nabla y\rangle=d\langle x,y\rangle$

with $d$ being the universal derivation and $\langle x\otimes a\,db,y\rangle=\langle x,y\rangle\,a^*\,db^*.$

Theorem 11.  Let $M$ be a spin manifold with spin structure $(v,S,C).$  Then there exists a unique Hermitian connection $\nabla^S:S\to S\otimes\Omega^1(M)$ satisfying

$\displaystyle\nabla^S(c(\sigma)s)=c(\nabla^g\sigma)s+c(\sigma)\nabla^Ss$

for $\sigma\in\Gamma^\infty(Cl(M))$ and $s\in S.$ (Note $\nabla^g:\Gamma^\infty(M)\to\Gamma^\infty(M)\otimes\Omega^1(M)$ extends to $\nabla^g:\Gamma^\infty(Cl(M))\to\Gamma^\infty(Cl(M))\otimes\Omega^1(M)$ by duality).

The Hermitian connection $\nabla^S$ is called the spin connection of the spin manifold $(M,v,S,C).$  The spin connection may be generalized to a Clifford connection of any self-adjoint Clifford module.

Proposition 12.  If $E=S\otimes F$ for a given spin structure $(v,S,C)$ and $\nabla^F$ is a Hermitian connection on $F,$ then $\nabla^S\otimes 1_F+1_S\otimes\nabla^F$ is a Clifford connection on $E.$  Conversely, if $E=S\otimes F$ is a self-adjoint Clifford module over a compact spin manifold $(M,v,S,C)$ and $\nabla^E$ is a Clifford connection on $E,$ then there is a unique Hermitian connection $\nabla^F$ on $F$ such that $\nabla^E=\nabla^S\otimes 1_F+1_S\otimes\nabla^F.$

Dirac Operators

Definition 13.  Let $E$ be a self-adjoint Clifford module over a compact Riemannian manifold $E$ with the the induced action $\hat{c}:\Gamma(Cl(M))\otimes E\to E$ defined by $\hat{c}(\sigma\otimes x)=c(\sigma)x.$  If $\nabla$ be a Clifford connection, the generalized Dirac operator associated to $\nabla$ is a map $D:E\to E$ defined by

$\displaystyle D=-i(\hat{c}\circ\nabla).$

After applying $\nabla,$ one embeds $\Omega^1(M)$ into $\Gamma(Cl(M)).$  In the case of a compact spin manifold and spin connection $\nabla^S,$ the Dirac operator is defined as

$\displaystyle D=-i(\hat{c}\circ\nabla^S).$

Proposition 14.  If $D$ is a generalized Dirac operator on a self-adjoint Clifford module $E$ and $a\in C^\infty(M),$ then

$\displaystyle [D,a]=-ic(da).$

Proof.  Let $s\in E.$

$\begin{array}{rcl}i[D,a](s)&=&\hat{c}(\nabla(as))-a\hat{c}(\nabla s)\\&=&\hat{c}(\nabla(as)-a\nabla s)\\&=&\hat{c}(da\otimes s+a\nabla s-a\nabla s)\\&=&\hat{c}(da\otimes a)\\&=&c(da)s.\end{array}$

So the Dirac operator on a spin manifold is ultimately defined by the Riemannian metric of the manifold.  A key starting point for noncommutative geometry is the ability to recover the metric from the Dirac operator alone.

First note that the spinor module $S$ is an inner product space with the inner product

$\displaystyle\langle r,s\rangle=\int_M\langle r,s\rangle_S\,|v_g|$

where $\langle\cdot,\cdot\rangle_S$ is $\Gamma^\infty(Cl(M))$-valued.  It’s completion is the Hilbert space $L^2(M,S),$ called the space of L^2 spinors.  Now suppose that $\gamma:[0,1]\to M$ is a shortest path between points $x$ and $y$ in $M.$  That is, the length

$\displaystyle\ell(\gamma)=\int_0^1\|\gamma'(t)\|\,dt=\int_0^1\sqrt{g(\gamma'(t),\gamma'(t))}\,dt$

is minimized over all $\gamma$ with $\gamma(0)=x$ and $\gamma(1)=y.$  Then for $a\in C^\infty(M)$ we have

$\begin{array}{rcl}\displaystyle\left|a(y)-a(x)\right|&=&\displaystyle\left|a(\gamma(1))-a(\gamma(0))\right|\\&=&\displaystyle\left|\int_0^1\frac{d}{dt}a(\gamma(t))\,dt\right|\\&=&\displaystyle\left|\int_0^1da_{\gamma(t)}(\gamma'(t))\,dt\right|\\&=&\displaystyle\left|\int_0^1g_{\gamma(t)}(\text{grad}_{\gamma(t)}a,\gamma'(t))\,dt\right|\\&\leq&\displaystyle\int_0^1\left\|g_{\gamma(t)}(\text{grad}_{\gamma(t)}a,\gamma'(t))\right\|\,dt\\&\leq&\displaystyle\int_0^1\|\text{grad}_{\gamma(t)}a\|\|\gamma'(t)\|\,dt\\&\leq&\displaystyle\|\text{grad}\,a\|_\infty\int_0^1\|\gamma'(t)\|\,dt\\&=&\displaystyle\|\text{grad}\,a\|_\infty\,\ell(\gamma).\end{array}$

Hence if $\|\text{grad}\,a\|_\infty\leq 1,$ then $|a(x)-a(y)|\leq\ell(\gamma).$  Thus we’d have $|a(x)-a(y)|\leq d(x,y).$  That is,

$\displaystyle\sup\{|a(x)-a(y)|:a\in C^\infty(M)\mbox{~and~}\|\text{grad}\,a\|_\infty\leq 1\}\leq d(x,y).$

The inequality is in fact an equality.  And we have the following.

Proposition 15.  Let $x,y\in M$ with $M$ a compact spin manifold.  Then

$\displaystyle d(x,y)=\sup\{|a(x)-a(y)|:a\in C(M)\mbox{~and~}\|[D,a]\|\leq 1\}.$

Proof.  We need only to show that the condition $\|\text{grad}\,a\|_\infty\leq 1$ is equivalent to the condition $\|[D,a]\|\leq 1.$  By proposition 14 above, we have $[D,a]=-ic(da)$ and hence

$\begin{array}{rcl}\displaystyle\|[D,a]\|^2&=&\displaystyle\|c(da)\|^2\\&=&\sup_{x\in M}\|c(da)(x)\|^2\\&=&\displaystyle\sup_{x\in M}g_x^{-1}(d\bar{a}(x),da(x))\\&=&\displaystyle\sup_{x\in M}g_x(\text{grad}_x\bar{a},\text{grad}_xa)\\&=&\displaystyle\|\text{grad}\,a\|_\infty^2,\end{array}$

which yields the result.

A key fact regarding the Dirac operator involves its behavior on the space of L^2 spinors.  Recall an operator $D$ on an inner product space is self-adjoint if

$\langle Dr,s\rangle=\langle r,Ds\rangle.$

An densely-defined operator (defined on a dense subset) is essentially self-adjoint if its closure is self-adjoint.

Theorem 16.  The Dirac operator on a compact spin manifold is essentially self-adjoint on the spinor space $H=L^2(M,S).$

The $\mathbb{Z}_2$-grading of $S$ then extends to a $\mathbb{Z}_2$-grading of $H.$   We have $H^{\pm}=L^2(M,S^{\pm}).$  The chirality operator $c(\gamma)$ anticommutes with the Dirac operator on $S$, and this extends to an operator $\chi$ on $H$ where we also have $\{\chi,D\}=0.$

An important result is the Lichnerowicz theorem that says on a compact spin manifold with $D$ the Dirac operator, $\Delta^S$ the spinor Laplacian (induced by spin connection), and $s$ the scalar curvature, we have

$\displaystyle D^2=\Delta^S+\frac{1}{4}s.$

We also end up obtaining that

$\displaystyle\mbox{Wres}\,(a|D|^{-n})=2^m\mbox{Wres}\,(a\Delta^{-n/2}),$

or equivalently

$\displaystyle\mbox{tr}\,(a|D|^{-n})=2^m\mbox{tr}\,(a\Delta^{-n/2}).$

This allows us to define the integral of an element $a\in C^\infty(M)$ via the Dirac operator instead of the Laplacian, obtaining

$\displaystyle \int_M a\,|v_g|=\frac{n(2\pi)^n}{2^m\Omega_n}\text{tr}\,(a|D|^{-n}).$

This, in turn, gives a way of defining a noncommutative integral on elements of a generic algebra once a suitable Dirac operator is chosen.

[1]  Varilly, Joeseph et al.  Elements of Noncommutative Geometry.

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# Noncommutative Differentiation #2

In the previous post, we saw that the Chern character gave a correspondence between the $K_0$-group of a commutative Riemannian manifold and its de Rham cohomology.  We now wish to extend the Chern character to the noncommutative manifold setting.

Hochschild (Co)Homology

Definition 1.  Let $A$ be an algebra and $E$ be a left $A$-bimodule.  Define $C_n(E,A)=E\otimes A^{\otimes n}$ and $\partial:C_*(E,A)\to C_* (E,A)$ linearly by

$\begin{array}{rcl}\partial_0(e\otimes a_1\otimes\cdots\otimes a_n)&=&ea_1\otimes a_2\otimes\cdots\otimes a_n\\\partial_i(e\otimes a_1\otimes\cdots\otimes a_n)&=&e\otimes a_1\otimes\cdots\otimes a_ia_{i+1}\otimes\cdots\otimes a_n\\\partial_n(e\otimes a_1\otimes\cdots\otimes a_n)&=&a_ne\otimes a_1\otimes\cdots\otimes a_{n-1}\end{array}$

and then

$\displaystyle\partial(e\otimes a_1\otimes\cdots\otimes a_n)=\sum_{i=0}^n(-1)^i\partial_i(e\otimes a_1\otimes\cdots\otimes a_n).$

This is a differential which yields the Hochschild chain complex with Hochschild homology $HH_n(E,A)=\ker\partial_{n-1}/\text{im}\,\partial_n.$  (note the abuse of notation I use here with the subscripts;  we will not use the original $\partial_i$‘s anymore, so $\partial_n$ will denote the differential from $C_n(E,A)$ to $C_{n-1}(E,A)$).  This induces a canonical Hochschild cochain complex where $C^n(E,A)=Hom(E\otimes A^{\otimes n},R)$ and $d_n:C^n(E,A)\to C^{n+1}(E,A)$ with

$d_n(f)(e\otimes a_1\otimes\cdots\otimes a_{n+1})=f(\partial_n(e\otimes a_1\otimes\cdots\otimes a_{n+1}))$

and corresponding Hochschild cohomology $HH^*(E,A).$

The functors are (co)homological functors.  Also note that a $0$-cocycle is a trace on $A$ when $E=A,$ for if $f\in C^0(A,A)=Hom(A,R)$ is a cocycle, we have

$\displaystyle 0=df(a_0\otimes a_1)=f(\partial(a_0\otimes a_1))=f(a_0a_1-a_1a_0)=f(a_0a_1)-f(a_1a_0).$

Definition 2.  An $n$-dimensional cycle over $A$ is a graded algebra $\Omega=\oplus_{k=0}^n\Omega^kA$ (with $d\omega=0$ for an $n$-form $\omega$) together with an integral $\int:\Omega^*\to\mathbb{C}$ (linear) such that $\int\omega=0$ for any $k$-form unless $\omega$ is a nonboundary $n$-form.  Furthermore we want

$\displaystyle\int\omega_k\omega_l=(-1)^{kl}\int\omega_l\omega_k$

for a $k$-form $\omega_k$ and $l$-form $\omega_l.$

The most obvious example of an $n$-cycle is the de Rham complex over the exterior algebra of an $n$-dimensional compact manifold with boundary.

Definition 3.  Let $\int$ be an $n$-cycle over $A.$  Then the Chern character is a map $\text{ch}\,:A^{\otimes (n+1)}\to\mathbb{C}$ defined linearly by

$\displaystyle\text{ch}\,(a\otimes a_1\otimes\cdots\otimes a_n)=\int a\,da_1\cdots da_n.$

Definition 4.  If $M$ is a compact differentiable $n$-dimensional manifold with boundary, a de Rham k-current $C$ is a continuous linear functional on $\Omega^k(M).$  The boundary of the current is denoted $\partial C,$ and integration of $k-1$ forms with respect to it is defined by

$\displaystyle\int_{\partial C}\omega=\int_Cd\omega$

for $\omega\in\Omega^{k-1}(A).$

The main point is to establish two isomorphisms for such manifolds:

$\displaystyle HH_k(C^\infty(M))=\Omega^k(M)$

and

$\displaystyle HH^k(C^\infty(M))=D_k(M)$

where $D_k(M)$ is the space of de Rham $k$-currents on $M.$  This is done by first defining an abstract $k$-current $C_\varphi$ for a $k$-cocycle $\varphi\in C^k(A,A)$ by

$\displaystyle\int_{C_\varphi}a\,da_1\wedge\cdots\wedge da_k=\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\text{sgn}\,(\pi)}\varphi(a\otimes a_{\pi(1)}\otimes\cdots\otimes a_{\pi(n)}).$

When $A$ is commutative, one can verify that the right-hand side takes cocycles to cocycles and kills coboundaries.  Thus it gives a homomorphism (in fact an isomorphism) between the de Rham currents of $M$ and $HH^*(C^\infty(M)).$  The first isomorphism then follows.

[1]  Varilly, Josephy et al.  Elements of Noncommutative Geometry.

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# Noncommutative Differentiation #1

First begin with a unital algebra $A$ and an $A$-bimodule $E.$  (One may easily extend what follows to rings).  Recall a derivation $d$ on $A$ is a linear map that satisfies $d(ab)=d(a)b+ad(b).$  We can easily extend this to a derivation $d:A\to E$ with the same requirement.

Definition 1.  The universal derivation on $A$ is a derivation $d:A\to A\otimes A$ defined by

$\displaystyle d(a)=1\otimes a-a\otimes 1.$

Further define $\Omega^0(A)=A,$ $\Omega^1(A)$ as the subbimodule of $A\otimes A$ generated by all elements of the form $a\,dx$ for $a,x\in A$ (here the left multiplication by $a$ is defined by multiplication in the first component).  Then define  $\Omega^n(A)=\bigotimes_{i=1}^n\Omega^1(A).$  The derivation extends to $d:\Omega^n(A)\to\Omega^{n+1}(A)$ by defined

$\displaystyle d(a\,dx_1\otimes\cdots\otimes dx_n)=da\otimes dx_1\otimes\cdots\otimes dx_n.$

Proposition 2.  The universal derivation of $A$ is both universal, and a derivation.  That is, for any derivation $D:A\to E,$ there exists a unique homomorphism $h:\Omega^1(A)\to E$ such that $h\circ d=D.$

Proof.  We prove it is a derivation; universality is left to the reader.

$\begin{array}{rcl}d(ab)&=&1\otimes ab-ab\otimes 1\\&=&1\otimes ab-ab\otimes 1+a\otimes b-a\otimes b\\&=&a(1\otimes b-b\otimes 1)+(a\otimes 1-1\otimes a)b\\&=&a\,db+(da)\,b.\end{array}$

Definition 3.  A connection on a right $A$-module $E$ is a map $\nabla:E\to E\otimes\Omega^1(A)$ satisfying

$\displaystyle\nabla(sa)=\nabla(s)a+s\otimes da$

for $s\in E$ and $a\in A.$  This easily extends inductively to $n$-forms: $\nabla:E\otimes\Omega^n(A)\to E\otimes\Omega^{n+1}(A)$ by

$\displaystyle\nabla(s\otimes\omega)=\nabla s\otimes\omega+s\otimes d\omega.$

We may also view $E\otimes_A\Omega^*(A)$ as a right $\Omega^*(A)$-module.  In this case we have

$\displaystyle\nabla(\sigma\omega)=(\nabla\sigma)\omega+(-1)^k\sigma\,d\omega$

where $\sigma\in E\otimes_A\Omega^k(A)$ and $\omega\in\Omega^*(A).$

Proposition 4.  If a right $A$-module $E$ admits a universal connection, then it is projective.

Example 5.  Let $e\in A$ be an idempotent so that $eA^n$ is a finitely generated projective $A$-module.  Then we have a composition

$\displaystyle eA^n\stackrel{i}{\longrightarrow}A^n\stackrel{d}{\longrightarrow}A^n\otimes_A\Omega^1A\stackrel{e}{\longrightarrow}eA^n\otimes_A\Omega^1(A).$

Defining $\nabla=e\circ d\circ i$ gives a connection on $eA^n$ called the Levi-Cevita connection.  We write $\nabla s=(e\circ d)(s)=e\,ds.$

Note that the square connection $\nabla^2:E\otimes_A\Omega^n(A)\to E\otimes_A\Omega^{n+2}(A)$ is also an $A$-module homomorphism:

$\begin{array}{rcl}\nabla^2(s\otimes\omega)&=&\nabla(\nabla s\otimes\omega+s\otimes d\omega)\\&=&\nabla^2s\otimes\omega-\nabla s\otimes d\omega+\nabla s\otimes d\omega+s\otimes d^2\omega\\&=&\nabla^2s\otimes\omega.\end{array}$

Definition 6.  $\nabla^2$ is called the curvature of the connection.

Riemannian Geometry as a Special Case

In Riemannian geometry, we may view the metric as a mapping

$g:\Gamma^\infty(M,TM)\times\Gamma^\infty(M,TM)\to C^\infty(M)$

which is $C^\infty(M)$-bilinear and satisfies $g(X,X)>0.$  This may be viewed as an inner product on the tangent spaces: $g(X_p,Y_p)=g(X,Y)(p).$  We’ll then define the $C^\infty(M)$-module $\Omega^k(M)=\bigwedge^k TM^*$ (with $\Omega^0(M)=C^\infty(M)$), and let $E$ in this case be the $C^\infty(M)$-module $\Gamma^\infty(M,TM)=\Gamma^\infty(M).$  One may easily see that the exterior derivative

$\displaystyle df=\sum_{i=0}^n\frac{\partial f}{\partial x_i}\,dx_i$

is a map $d:\Omega^0(M)\to\Omega^1(M)$ as desired.  This easily extends to $d:\Omega^k(M)\to\Omega^{k+1}(M).$  We’d then want a connection on the Riemannian manifold to begin as a map $\nabla:\Gamma^\infty(M)\to\Gamma^\infty(M)\otimes_{C^\infty(M)}\Omega^1(M)$ that satisfies the Leibniz rule for Lie brackets, and this is precisely how it is defined.  As above, it extends to a map $\nabla:\Gamma^\infty(M)\otimes\Omega^k(M)\to\Gamma^\infty(M)\otimes\Omega^{k+1}(M)$ (requiring the extended Leibniz rule as above).

Definition 7.  For any vector field $X$ on $M,$ there is a degree $-1$ operation on $\Gamma^\infty(M)\otimes\Omega^k(M),$ called the contraction by $X,$ given by

$\displaystyle\iota_X(Y\otimes\omega)=Y\otimes\iota_X\omega$

where

$\displaystyle (\iota_X\omega)(Y_1,...,Y_{k-1})=\omega(X,Y_1,...,Y_{k-1}).$

The Lie derivative with respect to the vector field $X$ is the map $\mathcal{L}_X:\Omega^k(M)\to\Omega^k(M)$ defined by

$\displaystyle\mathcal{L}_X=\iota_X\circ d+d\circ\iota_X.$

The Lie derivative can be extended to $\mathcal{L}_X:\Gamma^\infty(M)\otimes\Omega^k(M)\to\Gamma^\infty(M)\otimes\Omega^k(M)$ by defining $\mathcal{L}_XY=[X,Y]$ and thus

$\displaystyle\mathcal{L}_X(Y\otimes\omega)=[X,Y]\otimes\omega+Y\otimes\mathcal{L}_X\omega.$

Even more generally, for any connection $\nabla$ we may define the directional connection

$\displaystyle\nabla_X=\iota_X\circ\nabla+\nabla\circ\iota_X$

so that $\nabla:\Gamma^\infty(M)\otimes\Omega^k(M)\to\Gamma^\infty(M)\otimes\Omega^k(M)$ is defined by

$\displaystyle\nabla_X(Y\otimes\omega)=\nabla_XY\otimes\omega+Y\otimes\mathcal{L}_X\omega.$

And, even more generally, we can extend to connection to tensor fields $\nabla:T_r^s(M)\to T_r^{s+1}(M).$

Definition 8.  If $\nabla$ is a connection on $M$ and $X$ and $Y$ are vector fields, we define the torsion of $X$ and $Y$ as a map $\theta:\Gamma^\infty(M)\otimes\Gamma^\infty(M)\to\Gamma^\infty(M)$ defined by

$\displaystyle\theta(X,Y)=\nabla_XY-\nabla_YX-[X,Y].$

The Riemannian connection is then the unique connection which is torsion-free and satisfies

$\displaystyle g(\nabla_ZX,Y)+g(X,\nabla_ZY)=Z(g(X,Y))$

for any vector field $Z.$  The Riemannian curvature with respect to the vector fields $X$ and $Y$ is a map $R:\Gamma^\infty(M)\to\Gamma^\infty(M)$ defined by

$\displaystyle R(X,Y)(Z)=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.$

We may view this more generally as a $(1,3)$-tensor $R(X,Y,Z),$ or even as a $(0,4)$-tensor:  $R(X,Y,Z,W)=g(R(X,Y,Z),W)$

Chern Character

We return to differential algebra.  So the Riemannian curvature was a connection unique (or intrinsic) to the manifold.  Similarly, when we viewed $eA^n$ as a projective $A$-module, we had an intrinsic connection defined by $\nabla s=e\,ds$ for $s\in eA^n$ (or $\nabla=e\,d$).  Thus its curvature is given by

$\displaystyle Rs=\nabla^2s=(ed)(ed)s=e\,de\,ds.$

Since $e^2=e,$ we have $e\,de+de\,e=de,$ so $e\,de=de\,(1-e)$ and $de\,e=(1-e)\,de.$  Multiplying either one of these by $e$ gives $e\,de\,e=0.$  Further, we may write $s=es$ since $s\in eA^n.$  This gives us $ds=de\,s+e\,ds$ and hence $de\,s=(1-e)\,ds.$  Hence we have

$\displaystyle Rs=e\,de\,ds=de\,(1-e)\,ds=de\,de\,s=de\,de\,es.$

So, finally, we may write

$R=de\,de\,e=de\,(1-e)\,de=e\,de\,de.$

Thus in the algebraic setting, we have $R\in eA^n\otimes\Omega^2(A).$  But note in this setting that we have fixed the module as $eA^n.$  Similarly, if we fix a vector field $Z$ over a Riemannian manifold $M,$ then the curvature gives us an intrinsic $(1,2)$-form as well:

$\displaystyle K_Z(f,X,Y)=R(X,Y,Z)(f).$

Definition 9.  The Chern character of $p\in Proj(M_k(A))$ is defined as

$\displaystyle\text{ch}\, p=\text{tr}\,\exp(R)=\sum_{n=0}^\infty\frac{\text{tr}\,\left(p\,dp\,dp\right)^n}{n!}=\sum_{n=0}^\infty\frac{\text{tr}\,p\,(dp)^{2n}}{n!}$

Now we have some main facts:

Proposition 10.  If $A=C^\infty(M),$ then the Chern character defines de Rham cohomology classes (namely each term $\text{ch}_{2k}p$ in the sum).

Proposition 11.  The de Rham classes $[\text{ch}_{2k}p]$ depend only on the algebraic equivalence class $[p]\in K_0(A).$

Theorem 12.  $\text{ch}\,:K_0^{oa}(C^\infty(M))=K_0^{top}(M)\to H_{dR}^{\text{even}}(M)$ is a ring homomorphism.

Theorem 13.  $\text{ch}$ gives isomorphisms $K_0(C^\infty(M))\otimes_{\mathbb{Z}}\mathbb{Q}=H^{even}(M)$ and $K_0^{top}(M)\otimes_{\mathbb{Z}}\mathbb{C}=H_{dR}^{even}(M).$

These can all be found in [1].

[1]  Varilly, Joeseph et al.  Elements of Noncommutative Geometry.
[2]  Petersen, Peter.  Riemannian Geometry.  Second Edition.

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# Noncommutative Integration

The goal here is to interpret elements of (noncommutative) algebras as smooth functions on (noncommutative) manifolds, say by sending $a\mapsto \tilde{a}$.  When restricting to commutative algebras, say $C^\infty(M),$ the interpretation $f\mapsto\tilde{f}$ should be an identity.  That is, we should have $\tilde{f}(x)=f(x)$ for all $x\in M.$

Dixmier Trace

Let $H$ be a Hilbert space and $L^{1+}(H)$ be the space of compact linear operators on $H$ satisfying

$\displaystyle\|T\|_{1+}=\sup_{n}\frac{\sum_{i=1}^n\mu_i(T)}{\ln n}<\infty$

where $\mu_i(T)$ are the eigenvalues of $T,$ decreasing in order.  If we let $a_n=\frac{\sum_i\mu_i(T)}{\ln n},$ then the Dixmier trace is defined as

$\displaystyle\text{tr}\,(T)=\lim_{n\to\omega}a_n$

where the limit is taken over all bounded sequences.  More generally we could extend this to elements of, say, to endomorphisms of vector spaces.

Now let $M$ be a compact Riemannian manifold of dimension $n$ and $f\in C^\infty(M).$  We have the main theorem:

Theorem 1.  Let $\Omega_n=\frac{2\pi^{n/2}}{\Gamma(n/2)}.$  For $f\in C^\infty(M),$

$\displaystyle\int_M f(x)\,dx=\frac{n(2\pi)^n}{\Omega_n}\text{tr}(f\Delta^{-n/2}).$

Here we’re using the generalized differential operator defined by

$\displaystyle\Delta^{z/2}f(x)=\left((i|\xi|)^z\hat{f}(\xi)\right)^\vee(x)$

and the unitary Fourier transform.  This suggests a method of defining $\int_M a(x)\,dx$ for an element $a\in A$ of some noncommutative algebra $A.$  The precise class of algebras for which we define integration over $M$ will be addressed when we reach the fundamental theorem of noncommutative geometry later.

We now paint a picture of the proof of this theorem.

Pseudodifferential Operators and the Wodzicki Residue

Let $P(D)$ be a polynomial differential operator on $C^\infty(\mathbb{R}^n)$.  That is,

$\displaystyle P(D)=\sum_{\alpha}a_\alpha D^\alpha$

where $D^{\alpha}=(-i\partial_1)^{\alpha_1}\cdots(-i\partial_n)^{\alpha_n}.$  Then for $f\in C^\infty(\mathbb{R}^n)$ we can write

$\displaystyle P(D)f(x)=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}P(\xi)f(y)\,dy\,d\xi$

via the Fourier inversion formula.  A pseudodifferential operator will be an operator whose form is a bit more general than that of the above one.

Definition 2.  A function $p\in C^\infty(U\times\mathbb{R}^n)$ is a symbol of order $d,$ denoted $p\in S^d(U),$ iff for any compact $K\subset U$ and any $\alpha,\beta\in\mathbb{N}^n,$ there exists a constant $C_{K,\alpha,\beta}$ such that

$\displaystyle |D_x^\beta D_\xi^\alpha p(x,\xi)|\leq C_{K,\alpha,\beta}\left(1+|\xi|^2\right)^{(d-|\alpha|)/2}$

for all $x\in K$ and $\xi\in\mathbb{R}^n.$  We will also define a function $a\in C^\infty(U\times U\times\mathbb{R}^n)$ to be an amplitude of order $d,$ denoted $a\in A^d(U),$ iff for any compact $K\subset U$ and $\alpha,\beta,\gamma\in\mathbb{N}^n,$ there exists a constant $C_{K,\alpha,\beta,\gamma}$ such that

$\displaystyle |D_x^\gamma D_y^\beta D_\xi^\alpha a(x,y,\xi)|\leq C_{K,\alpha,\beta,\gamma}\left(1+|\xi|^2\right)^{(d-|\alpha|)/2}$

for all $x,y\in K$ and $\xi\in\mathbb{R}^n.$  A pseudodifferential operator of order $d$ is an operator $P(x,D)$ on $C^\infty(U)$ such that for $a\in C^\infty(U)$ we have

$\displaystyle P(x,D)f(x)=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot \xi}p(x,\xi)f(y)\,dy\,d\xi$

where either $p\in S^d(U)$ or $a(x,y,\xi)=p(x,\xi)f(y)\in A^d(U).$

It turns out that the two conditions (on $p$ or $a$) in the definition are equivalent, but that the latter will give more technical flexibility.  So in general we may simply write $P\in\Psi^d(U)$ for a pseudodifferential operator $P.$

Now if $M$ is a differentiable manifold, we can extend this definition.

Definition 3.  An operator $P:C_c^\infty(M)\to C^\infty(M)$ is a pseudodifferential operator of order $d$ iff the operator $P(f)\circ\varphi^{-1}\in\Psi^d(\varphi(U))$ and $\ker p$ is smooth off the diagonal of $M\times M$ (i.e. in the integrand).  Or more generally, we may extend it to operators between smooth vector fields.

Recall the dual of $C_c^\infty(U)$ is the space of distributions on $U,$ denoted $D(U).$  If $a\in A^d(U),$ then its kernel is the distribution $k_a\in D(U\times U)$ given by

$\displaystyle k_a(x,y)=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}a(x,y,\xi)\,d\xi.$

An amplitude $a\in A^d(U)$ is smoothing iff its kernel is in $C^\infty(U\times U).$  A pseudodifferential operator $p$ whose amplitude $a(x,y,\xi)$ is smoothing is said to be in $\Psi^{-\infty}(U).$

Definition 4.  Let $M$ be a compact smooth manifold.  The quotient algebra $P(M)=\Psi_{cl}^\infty(M)/\Psi^{-\infty}(M)$ is called the algebra of classical symbols on $M.$  A classical symbol in here has an asymptotic expansion

$\displaystyle a(x,\xi)\sim\sum_{i=0}^\infty a_{d-i}(x,\xi)$

where $a_r(x,\xi)$ satisfies $a_r(x,t\xi)=t^ra_r(x,\xi).$

Definition 5.  If $V$ is an $n$-dimensional vector space, an $\alpha$-density is a continuous mapping $d:V^n\to\mathbb{R}$ such that

$\displaystyle d(Av_1,...,Av_n)=|\det A|^\alpha$

for all $v_1,...,v_n\in V$ and $A\in End(V).$

Theorem 6. (Wodzicki)  If $A\in\Psi_{cl}^\infty(M)$ is a classical symbol on a smooth compact $n$-manifold $M,$ then there exists a $1$-density on TM whose local expression is given by

$\displaystyle\text{wres}_xA=\int_{|\xi|=1}a_{-n}(x,\xi)|\sigma_\xi||dx_1\wedge\cdots\wedge dx_n|.$

The Wodzicki residue of $A$ is the integral of this density over $M$

$\displaystyle\text{Wres}\,A=\int_M\text{wres}_xA.$

When dropped to the quotient algebra $P(M),$ the Wodzicki residue is a trace.  We’re assuming the manifold’s volume has been normalized to $1$ as well.  The following then lead to the result.

Proposition 7.

$\displaystyle\text{Wres}\,\Delta^{-n/2}=\Omega_n.$

Theorem 8 (Connes).  If $H$ is an elliptic pseudodifferential operator of order $-n$ on a complex vector bundle $E$ of a compact Riemannian manifold $M,$ then

$H\in L^{1+}=\left\{T\in K(End(E)):\|T\|_{1+}\sup_{n}\frac{\mu_n(T)}{\ln n}<\infty\right\}$

and

$\displaystyle\frac{\text{Wres}\,H}{n(2\pi)^n}=\text{tr}\,H.$

Corollary 9.

$\displaystyle\frac{\text{Wres}\,\Delta^{-n/2}}{\text{tr}\,\Delta^{-n/2}}=n(2\pi)^n.$

Then, armed with all of these, we return to Theorem 1.  One can show $f\Delta^{-n/2}$ is a pseudodifferential operator of order $-n$ where $f$ acts as a bounded multiplication operator.  The local expression of the Wodzicki residue has the form

$\displaystyle\text{wres}_x(f\Delta^{-n/2})=\Omega_nf(x)\rho(x)|dx|$

where $\rho(x)|dx|=\sqrt{\det g}|dx_1\wedge\cdots\wedge dx_n|.$  Integrating both sides and applying Corollary 9 then gives

$\displaystyle\Omega_n\int_Mf(x)\,dx=\text{Wres}\,(f\Delta^{-n/2})=n(2\pi)^n\text{tr}\,(f\Delta^{-n/2}),$

which gives the result.

[1]  Varilly, Joseph et al.  Elements of Noncommutative Geometry.
[2]  http://en.wikipedia.org/wiki/Pseudo-differential_operator
[3]  Grafakos, Loukas.  Modern Fourier Analysis.

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# Clifford Algebras and Spinors

Clifford Algebras

Recall an $F$-valued quadratic form of two variables on an $F$-vector space $V$ is a map $Q:V\times V\to F$ satisfying

$Q(u,v)=aQ(u,u)+bQ(u,v)+cQ(v,v)$

for $a,b,c\in F.$ Assuming linearity allows us to extend this to a map $Q:V\otimes V\to F$ (also denoted by $Q$).

Definition 1.  If $V$ is a finite dimensional $F$-vector space with quadratic form $Q,$ then the Clifford algebra of $V$ with respect to $Q$ is defined as $Cl(V,Q)=T(V)/I$ where $T(V)$ is the tensor algebra of $V$ and $I$ is the two-sided ideal generated by all elements of the form $v\otimes v-Q(v,v).$

For example, if $Q(v,v)=0,$ then $CL(V,Q)=\bigwedge(V).$  Observe that if the characteristic of $F$ is not $2,$ then we have the following identity in the Clifford algebra:

$\begin{array}{lcl}u\otimes v+v\otimes u&=&u\otimes v+v\otimes u+u\otimes u+v\otimes v-u\otimes u-v\otimes v\\&=&(u+v)\otimes(u+v)-u\otimes u-v\otimes v\\&=&Q(u+v,u+v)-Q(u,u)-Q(v,v)\\&=&2\langle u,v\rangle\end{array}$

where $\langle u,v\rangle=(Q(u+v,u+v)-Q(u,u)-Q(v,v))/2$ is a symmetric bilinear form.  Moreover, in the exterior algebra, one easily sees $\langle u,v\rangle=0.$  Another point on notation, while we will denote the Lie bracket (or commutator) of two vectors as $[u,v]=u\otimes v-v\otimes u,$ we will define the Poisson bracket as $\{u,v\}=u\otimes v+v\otimes u.$  This gives us

$\{u,v\}=2\langle u,v\rangle.$

In this context, it also suffices for our quadratic form to be defined only on the diagonal (i.e. $Q(u,u)$).  We can extend it as $Q(u,v)=\langle u,v\rangle.$  Indeed this is an extension for

$Q(u,u)=\langle u,u\rangle=\frac{1}{2}(u\otimes u+u\otimes u)=u\otimes u,$

which is true in the Clifford algebra (again assuming $char(F)\neq 2$).  Note it follows that the quadratic form is symmetric: $Q(u,v)=Q(v,u).$

Example 2.  Consider the following quadratic form on $\mathbb{R}^{p+q}.$

$\displaystyle Q(x,y)=\sum_{i=1}^px_iy_i-\sum_{i=p+1}^qx_iy_i.$

Then we denote the corresponding Clifford algebra as $Cl_{p,q}=Cl(\mathbb{R}^{p+q},Q).$  The corresponding quadratic form on $\mathbb{R}^{3+1}$ is used in Minkowski space.  It turns out that we get $Cl_{1,0}=\mathbb{R}\oplus\mathbb{R},$ $Cl_{0,1}=\mathbb{C},$ $Cl_{2,0}=M_2(\mathbb{R}),$ $Cl_{1,1}=M_2(\mathbb{R}),$ $Cl_{0,2}=\mathbb{H},$ $Cl_{3,0}=M_2(\mathbb{C}).$  In fact

Proposition 3.  For $p,q\in\mathbb{N},$ we have

$\begin{array}{lcl}C_{p+1,p+1}&=&Cl_{p,q}\otimes_{\mathbb{R}}M_2(\mathbb{R})\\Cl_{p+4,q}&=&Cl_{p,q}\otimes_{\mathbb{R}}M_2(\mathbb{H})\\&=&Cl_{p,q+4}\end{array}.$

It turns out all finite dimensional real Clifford algebras (up to isomorphism) have the above quadratic form.  That is, they are isomorphic to some $Cl_{p,q}.$  So what is $C_{p,q}$ for any $p,q$?  First consider the spinorial clock:

Source [1]

Theorem 4 (Classification of Real Clifford Algebras).  For $p,q\in\mathbb{N},$ $Cl_{p,q}=M_n(B)$ where $m:A\to B$ in the spinorial clock with $m=p-q\bmod{8},$ and $n$ is the integer for which $\dim_\mathbb{R}M_n(B)=2^{p+q}.$  Furthermore, each Clifford algebra has a $\mathbb{Z}_2$-grading with even part $Cl_{p,q}^+=M_n(A)$ if $n$ is odd or $M_{n/2}(A)$ for even $n.$

The $\mathbb{Z}_2$-grading $Cl(V,Q)$ is determined by the extension of the anti-identity map on $V.$  Call this map $\chi,$ which respects multiplication: for example $\chi(e_1\otimes\cdots\otimes e_k)=(-1)^ke_1\otimes\cdots\otimes e_k.$  The fixed subspace is the even part, while the unfixed is the odd part.  $\chi^2=1,$ so the grading is by $\mathbb{Z}_2.$

Spin Groups

Let us now assume $V$ is a vector space over $\mathbb{C}$ with quadratic form $Q$ (and that the corresponding Clifford algebras are complex).  A unitary vector $u\in Cl(V,Q)$ satisfies $u\otimes u^*=u^*\otimes u=1$ for some $u^*.$  We’ll now suppress tensor notation and $Q$.  For every invertible $x\in Cl(V),$ we define the twisted conjugation map $\phi_x:Cl(V)\to Cl(V)$ by $\phi_x(y)=\chi(x)yx^{-1}.$

Definition 5.  The orthogonal group $O(V)$ is the group generated by all unitary twisted conjugations $\phi_u.$  This group has two connected components, and the connected component of the identity is denoted $SO(V).$  The spin^c group of $V,$ denoted $spin^c(V),$ is the subgroup of $Cl^+(V)$ generated by an even number of unitaries $u_1\cdots u_{2n}.$

For a unitary $u\in spin^c(V),$ the map $\phi_u(x)\in SO(V).$  That is, $\phi:spin^c(V)\to SO(V),$  and $\ker\phi=\mathbb{T}.$  Thus we have an exact sequence of groups

$\displaystyle 1\longrightarrow\mathbb{T}\to spin^c(V)\stackrel{\phi}{\longrightarrow}SO(V)\longrightarrow 1.$

If $u\in spin^c(V)$ has the decomposition $u_1\cdots u_{2n},$ define $u^!=u_{2n}\cdots u_1.$  Then there is a homomorphism $\psi:spin^c(V)\to\mathbb{T}$ defined by $\psi(u)=u^!u=\lambda_1\cdots\lambda_{2n}$ where $u_i^2=\lambda_i\in\mathbb{T}.$  Thus we have a map $\Phi:spin^c(V)\to SO(V)\times\mathbb{T}$ defined by

$\displaystyle\Phi(u)=(\phi_u,\psi(u))=(\phi_u,u^!u).$

Its kernel is $\{\pm 1\}.$  The spin group of $V$ is defined as $spin(V)=\ker\psi.$  Thus we have another short exact sequence

$\displaystyle 1\longrightarrow\{\pm 1\}\longrightarrow spin(V)\stackrel{\Phi}{\longrightarrow} SO(V)\longrightarrow 1.$

Fock Space Representations

For now, assume $\dim_\mathbb{R}V=2n$ and that $V$ has a quadratic form $Q.$  We will turn $V$ into a complex Hilbert space if it possesses the following structure.  An orthogonal complex structure on $V$ is an operator $J\in End_\mathbb{R}(V)$ that satisfies $J^2=-1$ and

$\displaystyle Q(u,v)=Q(Ju,Jv).$

In this case, we define an inner product on $V$ by

$\displaystyle\langle u,v\rangle_J=Q(u,v)+iQ(Ju,v).$

The completed Hilbert algebra will be denoted $V_J.$  Fock space is then defined as the Hilbert algebra induced by exterior algebra $\bigwedge V_J$ under the inner product

$\displaystyle\langle u_1\wedge\cdots\wedge u_n,v_1\wedge\cdots\wedge v_m\rangle=\delta_{mn}\det[\langle u_i,v_j\rangle_J].$

We’ll denote Fock space by $F_J(V),$ and its elements will be called spinors.  The goal is to define a representation of $Cl(V)$ on Fock space.  We do this by defining a representation of $V$ on Fock space (which then trivially extends to $Cl(V)$).  We begin by defining a codifferential and differential representation on $\bigwedge V_J$ (which, in quantum field theory, are respectively called creation and annihilation operators).  Define the codifferential representation $d:V\to End(\bigwedge V_J)$ by

$\displaystyle d(v)(u_1\wedge\cdots\wedge u_k)=v\wedge u_1\wedge\cdots\wedge u_k.$

Define the differential representation $\partial:V\to End(\bigwedge V_J)$ by

$\displaystyle \partial(v)(u_1\wedge\cdots\wedge u_k)=\sum_{i=1}^k(-1)^{i-1}\langle v,u_i\rangle_J\,u_1\wedge\cdots\wedge\hat{u_i}\wedge\cdots\wedge u_k.$

The representation we seek is then the morphism $D:V\to End(\bigwedge V_J)$ defined by

$\displaystyle D(v)=\partial(v)+d(v).$

This representation trivially extends to one $D:Cl(V)\to End(\bigwedge V_J)$ (also denoted $D$), which is irreducible.

Now if $\dim_\mathbb{R}V=2n+1,$ we can define Fock space the same, but we’ll have two choices of irreducible representations.  If $\{e_1,...,e_{2n+1}\}$ is a basis for $V,$ then let $U$ be the subspace generated by the vectors $\{e_1,...,e_{2n}\}.$  Then we’ll have an action as before $D:Cl(U)\to End(\bigwedge U_J).$  We extend this to an action $D:Cl(V)\to End(\bigwedge V_J)$ such that

$\displaystyle D((-1)^ne_1\otimes\cdots\otimes e_{2n+1})=\pm 1.$

The restriction of $D$ to $spin^c(V)$ is called the spin^c representation of $spin^c(V)$ on $\bigwedge V_J.$

Infinite Dimensional Clifford Algebras and Fock Space

We begin with a triple as before: $(V,Q,J)$ where $V$ is a complex vector space, $Q$ is a symmetric positive-definite bilinear form on $V$, and $J$ is a complex orthogonal structure on $V.$  We can define $Cl(V)$, however this would have tensors of infinite degree, which we want to exclude.  So instead we take a limit.

For any finite dimensional subspaces $V_1,V_2$ of $V$ with $V_1\leq V_2,$ there is a canonical inclusion $\iota:V_1\to V_2$ which extends to an inclusion $\iota_*:Cl(V_1)\to Cl(V_2).$  Then we can define $Cl_{fin}(V)=\lim_{\longrightarrow} Cl(U)$ over all finite dimensional subspaces $U$ of $V.$  Then our limit Clifford algebra consists of all finite degree tensors of $Cl(V).$  Endow it with the Hilbert algebra structure induced by the same inner product as before.  It is also a C*-Hilbert algebra with involution given as follows:  for  $u\in Cl_{fin}(V),$ let $u^\circ$ be the corresponding tensor in the opposite Clifford algebra $Cl_{fin}(V)^\circ.$  Then we define

$\displaystyle a^*=(a^\circ)^!$

for $a\in Cl_{fin}(V).$  As a C*-algebra, $Cl_{fin}(V)$ can be realized as a C*-algebra of bounded operators on a Hilbert space via the GNS construction.  Since we won’t be needing the formal Clifford algebra $Cl(V)$ with infinite degree tensors, we’ll just write $Cl(V)=Cl_{fin}(V).$

We’ll define Fock space in the infinite dimensional setting as before: $F_J(V)=\bigwedge V_J$ completed under the inner product as before.  $D:V\to End(F_J(V))$ extends to a irreducible Fock representation $D:Cl(V)\to End(F_J(V)).$

[1] Varilly, Joeseph et al.  Elements of Noncommutative Geometry.

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# Noncommutative Mathematics and Virtual Objects

Noncommutative Topology

For a commutative C*-algebra $A,$ the spectrum (or dual space) is denoted $\hat{A}$ and consists of all nonzero *-homomorphisms from $A$ to $\mathbb{C}.$  The primitive spectrum of $A,$ denoted $Prim(A),$ is the set of maximal ideals of $A.$  The Gelfand map $\Gamma:A\to C(\hat{A})$ is defined by $\Gamma(x)(f)=f(x).$

Theorem 1 (Gelfand-Naimark)  If $A$ is a commutative C*-algebra, then the Gelfand map is an isometric *-isomorphism of $A$ and $C_0(\hat{A}).$

One can impose a Jacobson topology on $Prim(A)$ as follows.  If $\mathcal{U}$ is a set of ideals in $Prim(A)$, its closure is defined as

$\displaystyle\overline{\mathcal{U}}=\left\{\mathfrak{p}\in Prim(A):\bigcap_{\mathfrak{a}\in\mathcal{U}}\mathfrak{a}\subseteq\mathfrak{p}\right\}.$

Then we have a map $\Omega:\hat{A}\to Prim(A)$ defined by $\Omega(f)=\ker f.$  It is clearly injective, moreover if $I\in Prim(A),$ then necessarily $A/I=\mathbb{C},$ so $I=\ker\pi$ and $\pi\in\hat{A}.$  This induces a topology on $\hat{A}$ whose open sets are preimages of open sets–thereby making $\Omega$ continuous (and in fact a homeomorphism).

Theorem 2  There is a categorical equivalence between the category of compact Hausdorff spaces and commutative unital C*-algebras.

Proof Idea.  For a compact Hausdorff space $X,$ one simply associates the unital C*-algebra $C(X),$ which is easily seen to be a contravariant functor: if $f:X\to Y,$ then the induced morphism $f':C(Y)\to C(X)$ is defined by $f'(g)(x)=(g\circ f)(x).$  Conversely if $A$ is a commutative unital C*-algebra, then the Gelfand-Naimark theorem says $A=C(\hat{A}),$ so $\hat{A}$ is the corresponding compact Hausdorff space.

This theorem in fact generalizes to

Theorem 3  There is a categorical equivalence between the category of locally compact Hausdorff spaces and commutative C*-algebras.

The axiom of noncommutative topology is the extension of this functor to all C*-algebras. When $A$ is noncommutative, the corresponding topological space is a virtual noncommutative space.  The motivation for this can come from the motivation of wanting all real numbers to be a square.  So we introduce a virtual number $i$ satisfying $i^2=-1$ so that the bijection $\sqrt{-}$ between nonnegative reals extends to negative reals.  Similarly the virtual noncommutative spaces allow an extended bijection with all C*-algebras.

Or more elegantly consider the group isomorphism $\exp:(\mathbb{R},+)\to(\mathbb{R}_{>0},\cdot)$—or even a categorical equivalence if you wish.  Now what if we’d like extend this isomorphism so that the range becomes all of $\mathbb{R}-\{0\}$?  What would need to change in the domain to keep this isomorphism? Consider changing the domain to

$\displaystyle\mathbb{R}^s=\{r+si:r\in\mathbb{R},s\in\{0,\pi\}\}$

with addition defined by $(r+si)+(r'+s'i)=r+r'+(s+s')i$ where we define $\pi+\pi=0.$  Then this does the trick via the exponential map. (The $s$ here alludes to spin).

Proposition 4  $\exp:\mathbb{R}^s\to\mathbb{R}-\{0\}$ is a group isomorphism.

Proof.  The bijection is obvious, and we already have the isomorphism between $\mathbb{R}_{+}^s$ and $\mathbb{R}_{>0}.$  Now we have

$\begin{array}{rcl}\exp((r+\pi i)+(r'+\pi i))&=&\exp(r+r')\\&=&e^re^{r'}\\&=&(-e^r)(-e^{r'})\\&=&\exp(r+\pi i)\exp(r'+\pi i)\end{array}$

and similarly for the mixed spin cases.

Noncommutative Measure Theory

Theorem 5  There is a categorical equivalence between $\sigma$-finite measure spaces and commutative von Neumann algebras.

Proof Idea.  Let $(X,\mathcal{A},\mu)$ be $\sigma$-finite and $A=\{T_f:f\in L^\infty(X)\}$ be operators on $L^2(X)$ defined by left multiplication.  This is well-defined since for $g\in L^2(X)$ we have

$\displaystyle\|T_fg\|_2^2=\|fg\|_2^2=\||f|^2|g|^2\|_1\leq\|f\|_\infty^2\|g\|_2^2<\infty.$

by the general Holder inequality.  $A$ can then be seen as a commutative *-subalgebra of $B(L^2(X))$ with $A'=A$ (and thus $A''=A'=A$), so $A$ is von Neumann.  In fact, there is an isometric *-isomorphism between $A$ and $L^\infty(X).$

Conversely if $A$ is a commutative von Neumann algebra, then if $f\in C_0(\hat{A}),$ there is a positive form on $C_0(\hat{A})$ given by $\phi(f)=\langle T_fx,x\rangle$ with $T_f=\Omega(f)\in A$ and $x$ a cyclic vector with respect to the irreducible representations of $\hat{A}.$  So by the Riesz representation theorem we have

$\displaystyle\phi(f)=\int_{\hat{A}}f\,d\mu_{x,x}$

for a measure $\mu_{x,x}$ on the Borel $\sigma$-algebra of $\hat{A}.$  Furthermore we have an isometric *-isomorphism $A=L^\infty(\hat{A},\mu_{x,x}).$

Thus the study of noncommutative von Neumann algebras can be equated to that of virtual noncommutative measure spaces.

Along the same lines, a pair $(A,\phi)$ of a von Neumann algebra and a state on it generalize to the notion of a noncommutative probability space.

Noncommutative Geometry

The equivalence becomes a bit more complicated here.  On one side we’ll want a connected compact orientable Riemannian manifold, and on the other we’ll have an irreducible spin geometry.

Definition 6  A spectral triple is a triple $(A,H,D)$ where $A$ is an algebra with a bounded representation $\rho$ on a Hilbert space $H,$ together with $D$ is a fixed self-adjoint operator on $H$ with compact resolvent (complement of spectrum) such that $\|[D,a]\|=\|D\rho(a)-\rho(a)D\|<\infty$ for all $a\in A.$  The spectral triple is called real if there is also an antiunitary operator $C$ such that $[a,Cb^*C^{-1}]=0$ for all $a,b\in A.$

Theorem 7  It turns out that if $(A,H,D)$ is a spectral triple of dimension $n,$ then there exists a compact orientable $n$-manifold such that $A=C^\infty(M).$

A (noncommutativespin geometry is a quintuple $(A,H,D,C,\chi)$ where $(A,H,D)$ is a real spectral triple satisfying 7 additional technical conditions (see [1]).  The geometry is irreducible iff no nontrivial projective operator on $H$ commutes with the action of $A,$ $D,C,$ or $\chi.$  Part of the main result is

Theorem 7  If $(A,H,D,C,\chi)$ is an irreducible spin geometry with $A=C^\infty(M)$ for a compact orientable connected manifold $M$ without boundary and of dimension $n,$ then there is a unique Riemannian metric on $M$ whose induced metric is given by

$\displaystyle d(x,y)=\sup_{a\in A}\{|a(x)-a(y)|:\|[D,a]\|\leq 1\}.$

It turns out that the commutativity of $C^\infty(M)$ is not used in the proof of this theorem, so one can say that commutative Riemannian manifolds are the Riemannian manifolds obtained by invoking this theorem for a commutative algebra of smooth functions defined on $M,$ and the noncommutative Riemannian manifolds are those obtained from starting with a noncommutative algebra of “function elements” defined on $M.$

The converse question asks “what are necessary conditions on a spectral triple ensuring it comes from a compact orientable Riemannian manifold?”.  Connes addressed this in [3].  And a recent paper [2] by Cacic builds on this.

[1]  Varilly, Joseph et al.  Elements of Noncommutative Geometry
[2]  Cacic, Branimir.  A Reconstruction Theorem For Almost-Commutative Spectral Triples.  2012.  http://arxiv.org/abs/1101.5908v4
[3]  Connes, Alain.  On the Spectral Characterization of Manifolds.  2008.  http://arxiv.org/abs/0810.2088v1

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