1b. Quantum spaces

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In order to talk about noncommutative geometry, the idea will be that of defining our quantum spaces [math]X[/math] as being abstract manifolds, whose coordinates [math]x_1,\ldots,x_N[/math] do not necessarily commute. Thus, we are in need of some good algebraic geometry correspondence, between such abstract spaces [math]X[/math], and the corresponding algebras of coordinates [math]A[/math]. Following Heisenberg, von Neumann and many others, we will use here the correspondence [math]A=C(X)[/math] coming from operator algebra theory. Let us start with:

Definition

A Hilbert space is a complex vector space [math]H[/math], given with a scalar product [math] \lt x,y \gt [/math], satisfying the following conditions:

• [math] \lt x,y \gt [/math] is linear in [math]x[/math], and antilinear in [math]y[/math].
• [math]\overline{ \lt x,y \gt }= \lt y,x \gt [/math], for any [math]x,y[/math].
• [math] \lt x,x \gt \gt 0[/math], for any [math]x\neq0[/math].
• [math]H[/math] is complete with respect to the norm [math]||x||=\sqrt{ \lt x,x \gt }[/math].

Observe that we are using here mathematicians' convention for linearity, as opposed to Dirac's convention in , used by physicists. Ironically, this change came from Dirac himself, who advised his students and mankind to “shut up and compute”, in anything related to quantum mechanics. Many mathematicians, including myself, followed his advice, shut up and computed, and concluded that linearity at left is better.

Back to mathematics now, in the above definition, the fact that [math]||x||=\sqrt{ \lt x,x \gt }[/math] is indeed a norm comes from the Cauchy-Schwarz inequality, [math]| \lt x,y \gt |\leq||x||\cdot||y||[/math], which itself comes from the fact that the following degree 2 polynomial, with [math]t\in\mathbb R[/math] and [math]w\in\mathbb T[/math], being positive, its discriminant must be negative:

[[math]] f(t)=||x+wty||^2 [[/math]]

In finite dimensions, any algebraic basis [math]\{f_1,\ldots,f_N\}[/math] can be turned into an orthonormal basis [math]\{e_1,\ldots,e_N\}[/math], by using the Gram-Schmidt procedure. Thus, we have [math]H\simeq\mathbb C^N[/math], with this latter space being endowed with its usual scalar product, namely:

[[math]] \lt x,y \gt =\sum_ix_i\bar{y}_i [[/math]]

The same happens in infinite dimensions, once again by Gram-Schmidt, coupled if needed with the Zorn lemma, in case our space is really very big. In other words, any Hilbert space has an orthonormal basis [math]\{e_i\}_{i\in I}[/math], and we have:

[[math]] H\simeq l^2(I) [[/math]]

Of particular interest is the “separable” case, where [math]I[/math] is countable. According to the above, there is up to isomorphism only one Hilbert space here, namely:

[[math]] H=l^2(\mathbb N) [[/math]]

All this is, however, quite tricky, and can be a bit misleading. Consider for instance the space [math]H=L^2[0,1][/math] of square-summable functions [math]f:[0,1]\to\mathbb C[/math], with:

[[math]] \lt f,g \gt =\int_0^1f(x)\overline{g(x)}dx [[/math]]

This space is of course separable, because we can use the basis [math]f_n=x^n[/math] with [math]n\in\mathbb N[/math], orthogonalized by Gram-Schmidt. However, the orthogonalization procedure is something non-trivial, and so the isomorphism [math]H\simeq l^2(\mathbb N)[/math] that we obtain is something non-trivial as well. Doing some computations here is actually an excellent exercise.

In what follows we will be interested in the linear operators [math]T:H\to H[/math] which are bounded. Regarding such operators, we have the following result:

Theorem

Given a Hilbert space [math]H[/math], the linear operators [math]T:H\to H[/math] which are bounded, in the sense that

[[math]] ||T||=\sup_{||x||\leq1}||Tx|| [[/math]]

is finite, form a complex algebra [math]B(H)[/math], having the following properties:

• [math]B(H)[/math] is complete with respect to [math]||.||[/math], so we have a Banach algebra.
• [math]B(H)[/math] has an involution [math]T\to T^*[/math], given by [math] \lt Tx,y \gt = \lt x,T^*y \gt [/math].

In addition, the norm and involution are related by the formula [math]||TT^*||=||T||^2[/math].

Show Proof

The fact that we have indeed an algebra follows from:

[[math]] ||S+T||\leq||S||+||T|| [[/math]]

[[math]] ||\lambda T||=|\lambda|\cdot||T|| [[/math]]

[[math]] ||ST||\leq||S||\cdot||T|| [[/math]]

(1) Assuming that [math]\{T_n\}\subset B(H)[/math] is Cauchy then [math]\{T_nx\}[/math] is Cauchy for any [math]x\in H[/math], so we can define indeed the limit [math]T=\lim_{n\to\infty}T_n[/math] by setting:

[[math]] Tx=\lim_{n\to\infty}T_nx [[/math]]

(2) Here the existence of [math]T^*[/math] comes from the fact that [math]\varphi(x)= \lt Tx,y \gt [/math] being a linear form [math]H\to\mathbb C[/math], we must have [math]\varphi(x)= \lt x,T^*y \gt [/math], for a certain vector [math]T^*y\in H[/math]. Moreover, since this vector is unique, [math]T^*[/math] is unique too, and we have as well:

[[math]] (S+T)^*=S^*+T^*\quad,\quad (\lambda T)^*=\bar{\lambda}T^* [[/math]]

[[math]] (ST)^*=T^*S^*\quad,\quad (T^*)^*=T [[/math]]

Observe also that we have indeed [math]T^*\in B(H)[/math], because:

[[math]] \begin{eqnarray*} ||T|| &=&\sup_{||x||=1}\sup_{||y||=1} \lt Tx,y \gt \\ &=&\sup_{||y||=1}\sup_{||x||=1} \lt x,T^*y \gt \\ &=&||T^*|| \end{eqnarray*} [[/math]]

Regarding now the last assertion, we have the following estimate:

[[math]] ||TT^*|| \leq||T||\cdot||T^*|| =||T||^2 [[/math]]

On the other hand, we have as well the following estimate:

[[math]] \begin{eqnarray*} ||T||^2 &=&\sup_{||x||=1}| \lt Tx,Tx \gt |\\ &=&\sup_{||x||=1}| \lt x,T^*Tx \gt |\\ &\leq&||T^*T|| \end{eqnarray*} [[/math]]

By replacing [math]T\to T^*[/math] we obtain from this [math]||T||^2\leq||TT^*||[/math], and we are done.

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Observe that when [math]H[/math] comes with an orthonormal basis [math]\{e_i\}_{i\in I}[/math], the linear map [math]T\to M[/math] given by [math]M_{ij}= \lt Te_j,e_i \gt [/math] produces an embedding as follows:

[[math]] B(H)\subset M_I(\mathbb C) [[/math]]

Moreover, in this picture the operation [math]T\to T^*[/math] takes a very simple form, namely:

[[math]] (M^*)_{ij}=\overline{M}_{ji} [[/math]]

However, as explained before Theorem 1.9, it is better in general not to use bases, and this because very simple spaces like [math]L^2[0,1][/math] do not have simple bases.

The conditions found in Theorem 1.9 suggest the following definition:

Definition

A [math]C^*[/math]-algebra is a complex algebra [math]A[/math], having:

• A norm [math]a\to||a||[/math], making it a Banach algebra.
• An involution [math]a\to a^*[/math], satisfying [math]||aa^*||=||a||^2[/math].

Generally speaking, the elements [math]a\in A[/math] are best thought of as being some kind of “generalized operators”, on some Hilbert space which is not present. By using this idea, one can emulate spectral theory in this setting, as follows:

Proposition

Given [math]a\in A[/math], define its spectrum as being the set

[[math]] \sigma(a)=\left\{\lambda\in\mathbb C\Big|a-\lambda\not\in A^{-1}\right\} [[/math]]

and its spectral radius [math]\rho(a)[/math] as the radius of the smallest centered disk containing [math]\sigma(a)[/math].

• The spectrum of a norm one element is in the unit disk.
• The spectrum of a unitary element [math](a^*=a^{-1}[/math]) is on the unit circle.
• The spectrum of a self-adjoint element ([math]a=a^*[/math]) consists of real numbers.
• The spectral radius of a normal element ([math]aa^*=a^*a[/math]) is equal to its norm.

Show Proof

Our first claim is that for any polynomial [math]f\in\mathbb C[X][/math], and more generally for any rational function [math]f\in\mathbb C(X)[/math] having poles outside [math]\sigma(a)[/math], we have:

[[math]] \sigma(f(a))=f(\sigma(a)) [[/math]]

This indeed something well-known for the usual matrices. In the general case, assume first that we have a polynomial, [math]f\in\mathbb C[X][/math]. If we pick an arbitrary number [math]\lambda\in\mathbb C[/math], and write [math]f(X)-\lambda=c(X-r_1)\ldots(X-r_k)[/math], we have then, as desired:

[[math]] \begin{eqnarray*} \lambda\notin\sigma(f(a)) &\iff&f(a)-\lambda\in A^{-1}\\ &\iff&c(a-r_1)\ldots(a-r_k)\in A^{-1}\\ &\iff&a-r_1,\ldots,a-r_k\in A^{-1}\\ &\iff&r_1,\ldots,r_k\notin\sigma(a)\\ &\iff&\lambda\notin f(\sigma(a)) \end{eqnarray*} [[/math]]

Assume now that we are in the general case, [math]f\in\mathbb C(X)[/math]. We pick [math]\lambda\in\mathbb C[/math], we write [math]f=P/Q[/math], and we set [math]F=P-\lambda Q[/math]. By using the above finding, we obtain, as desired:

[[math]] \begin{eqnarray*} \lambda\in\sigma(f(a)) &\iff&F(a)\notin A^{-1}\\ &\iff&0\in\sigma(F(a))\\ &\iff&0\in F(\sigma(a))\\ &\iff&\exists\mu\in\sigma(a),F(\mu)=0\\ &\iff&\lambda\in f(\sigma(a)) \end{eqnarray*} [[/math]]

Regarding now the assertions in the statement, these basically follows from this:

(1) This comes from the following formula, valid when [math]||a|| \lt 1[/math]:

[[math]] \frac{1}{1-a}=1+a+a^2+\ldots [[/math]]

(2) Assuming [math]a^*=a^{-1}[/math], we have the following norm computations:

[[math]] ||a||=\sqrt{||aa^*||}=\sqrt{1}=1 [[/math]]

[[math]] ||a^{-1}||=||a^*||=||a||=1 [[/math]]

If we denote by [math]D[/math] the unit disk, we obtain from this, by using (1):

[[math]] ||a||=1\implies\sigma(a)\subset D [[/math]]

[[math]] ||a^{-1}||=1\implies\sigma(a^{-1})\subset D [[/math]]

On the other hand, by using the rational function [math]f(z)=z^{-1}[/math], we have:

[[math]] \sigma(a^{-1})\subset D\implies \sigma(a)\subset D^{-1} [[/math]]

Now by putting everything together we obtain, as desired:

[[math]] \sigma(a)\subset D\cap D^{-1}=\mathbb T [[/math]]

(3) This follows from (2). Indeed, for [math]t \gt \gt 0[/math] we have:

[[math]] \left(\frac{a+it}{a-it}\right)^* =\frac{a-it}{a+it} =\left(\frac{a+it}{a-it}\right)^{-1} [[/math]]

Thus the element [math]f(a)[/math] is a unitary, and by using (2) its spectrum is contained in [math]\mathbb T[/math]. We conclude from this that we have:

[[math]] f(\sigma(a))=\sigma(f(a))\subset\mathbb T [[/math]]

But this shows that we have [math]\sigma(a)\subset f^{-1}(\mathbb T)=\mathbb R[/math], as desired.

(4) We already know that we have [math]\rho(a)\leq ||a||[/math], for any [math]a\in A[/math]. For the reverse inequality, when [math]a[/math] is normal, we fix a number [math]\rho \gt \rho(a)[/math]. We have then:

[[math]] \begin{eqnarray*} \int_{|z|=\rho}\frac{z^n}{z -a}\,dz &=&\int_{|z|=\rho}\sum_{k=0}^\infty z^{n-k-1}a^k\,dz\\ &=&\sum_{k=0}^\infty\left(\int_{|z|=\rho}z^{n-k-1}dz\right)a^k\\ &=&a^{n-1} \end{eqnarray*} [[/math]]

By applying the norm and taking [math]n[/math]-th roots we obtain from this formula, modulo some elementary manipulations, the following estimate:

[[math]] \rho\geq\lim_{n\to\infty}||a^n||^{1/n} [[/math]]

Now recall that [math]\rho[/math] was by definition an arbitrary number satisfying [math]\rho \gt \rho(a)[/math]. Thus, we have obtained the following estimate, valid for any [math]a\in A[/math]:

[[math]] \rho(a)\geq\lim_{n\to\infty}||a^n||^{1/n} [[/math]]

In order to finish, we must prove that when [math]a[/math] is normal, this estimate implies the missing estimate, namely [math]\rho(a)\geq||a||[/math]. We can proceed in two steps, as follows:

\underline{Step 1}. In the case [math]a=a^*[/math] we have [math]||a^n||=||a||^n[/math] for any exponent of the form [math]n=2^k[/math], by using the [math]C^*[/math]-algebra condition [math]||aa^*||=||a||^2[/math], and by taking [math]n[/math]-th roots we get:

[[math]] \rho(a)\geq||a|| [[/math]]

Thus, we are done with the self-adjoint case, with the result [math]\rho(a)=||a||[/math].

\underline{Step 2}. In the general normal case [math]aa^*=a^*a[/math] we have [math]a^n(a^n)^*=(aa^*)^n[/math], and by using this, along with the result from Step 1, applied to [math]aa^*[/math], we obtain:

[[math]] \begin{eqnarray*} \rho(a) &\geq&\lim_{n\to\infty}||a^n||^{1/n} =\sqrt{\lim_{n\to\infty}||a^n(a^n)^*||^{1/n}}\\ &=&\sqrt{\lim_{n\to\infty}||(aa^*)^n||^{1/n}} =\sqrt{\rho(aa^*)}\\ &=&\sqrt{||a||^2} =||a|| \end{eqnarray*} [[/math]]

Thus, we are led to the conclusion in the statement.

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We can now formulate a key theorem, as follows:

Theorem (Gelfand)

If [math]X[/math] is a compact space, the algebra [math]C(X)[/math] of continuous functions [math]f:X\to\mathbb C[/math] is a commutative [math]C^*[/math]-algebra, with structure as follows:

• The norm is the usual sup norm, [math]||f||=\sup_{x\in X}|f(x)|[/math].
• The involution is the usual involution, [math]f^*(x)=\overline{f(x)}[/math].

Conversely, any commutative [math]C^*[/math]-algebra is of the form [math]C(X)[/math], with its “spectrum” [math]X=Spec(A)[/math] appearing as the space of characters [math]\chi :A\to\mathbb C[/math].

Show Proof

In what regards the first assertion, everything here is trivial. Conversely, given a commutative [math]C^*[/math]-algebra [math]A[/math], we can define [math]X[/math] to be the set of characters [math]\chi :A\to\mathbb C[/math], with the topology making continuous all the evaluation maps [math]ev_a:\chi\to\chi(a)[/math]. Then [math]X[/math] is a compact space, and [math]a\to ev_a[/math] is a morphism of algebras:

[[math]] ev:A\to C(X) [[/math]]

Our first claim is that [math]ev[/math] is involutive. Indeed, we can use the following formula:

[[math]] a=\frac{a+a^*}{2}-i\cdot\frac{i(a-a^*)}{2} [[/math]]

Thus it is enough to prove the equality [math]ev_{a^*}=ev_a^*[/math] for self-adjoint elements [math]a[/math]. But this is the same as proving that [math]a=a^*[/math] implies that [math]ev_a[/math] is a real function, which is in turn true, because [math]ev_a(\chi)=\chi(a)[/math] is an element of [math]\sigma(a)[/math], contained in [math]\mathbb R[/math]. Thus, claim proved. Finally, since [math]A[/math] is commutative, each element is normal, so [math]ev[/math] is isometric:

[[math]] ||ev_a||=\rho(a)=||a|| [[/math]]

It remains to prove that [math]ev[/math] is surjective. But this follows from the Stone-Weierstrass theorem, because [math]ev(A)[/math] is a closed subalgebra of [math]C(X)[/math], which separates the points.

■

The Gelfand theorem suggests formulating the following definition:

Definition

Given a [math]C^*[/math]-algebra [math]A[/math], not necessarily commutative, we write

[[math]] A=C(X) [[/math]]

and call the abstract object [math]X[/math] a “compact quantum space”.

In other words, the category of compact quantum spaces is by definition the category of [math]C^*[/math]-algebras, with the arrows reversed. We will be back to this, with examples, and with some technical comments as well, including a modification, the idea being that the above definition is in fact quite naive, and needs a fix. More on this later.

Let us discuss now the other basic result regarding the [math]C^*[/math]-algebras, namely the GNS representation theorem. We will need some more spectral theory, as follows:

Proposition

For a normal element [math]a\in A[/math], the following are equivalent:

• [math]a[/math] is positive, in the sense that [math]\sigma(a)\subset[0,\infty)[/math].
• [math]a=b^2[/math], for some [math]b\in A[/math] satisfying [math]b=b^*[/math].
• [math]a=cc^*[/math], for some [math]c\in A[/math].

Show Proof

This is something very standard, as follows:

[math](1)\implies(2)[/math] Since our element [math]a[/math] is normal the algebra [math] \lt a \gt [/math] that is generates is commutative, and by using the Gelfand theorem, we can set [math]b=\sqrt{a}[/math].

[math](2)\implies(3)[/math] This is trivial, because we can set [math]c=b[/math].

[math](3)\implies(1)[/math] We proceed by contradiction. By multiplying [math]c[/math] by a suitable element of [math] \lt cc^* \gt [/math], we are led to the existence of an element [math]d\neq0[/math] satisfying [math]-dd^*\geq0[/math]. By writing now [math]d=x+iy[/math] with [math]x=x^*,y=y^*[/math] we have:

[[math]] dd^*+d^*d=2(x^2+y^2)\geq0 [[/math]]

Thus [math]d^*d\geq0[/math]. But this contradicts the elementary fact that [math]\sigma(dd^*),\sigma(d^*d)[/math] must coincide outside [math]\{0\}[/math], which can be checked by explicit inversion.

■

Here is now the GNS representation theorem, along with the idea of the proof:

Theorem (GNS theorem)

Let [math]A[/math] be a [math]C^*[/math]-algebra.

• [math]A[/math] appears as a closed [math]*[/math]-subalgebra [math]A\subset B(H)[/math], for some Hilbert space [math]H[/math].
• When [math]A[/math] is separable (usually the case), [math]H[/math] can be chosen to be separable.
• When [math]A[/math] is finite dimensional, [math]H[/math] can be chosen to be finite dimensional.

Show Proof

Let us first discuss the commutative case, [math]A=C(X)[/math]. Our claim here is that if we pick a probability measure on [math]X[/math], we have an embedding as follows:

[[math]] C(X)\subset B(L^2(X))\quad,\quad f\to(g\to fg) [[/math]]

Indeed, given a function [math]f\in C(X)[/math], consider the operator [math]T_f(g)=fg[/math], acting on [math]H=L^2(X)[/math]. Observe that [math]T_f[/math] is indeed well-defined, and bounded as well, because:

[[math]] ||fg||_2 =\sqrt{\int_X|f(x)|^2|g(x)|^2dx} \leq||f||_\infty||g||_2 [[/math]]

Thus, [math]f\to T_f[/math] provides us with a [math]C^*[/math]-algebra embedding [math]C(X)\subset B(H)[/math], as claimed. In general now, assuming that a linear form [math]\varphi:A\to\mathbb C[/math] has some suitable positivity properties, making it analogous to the integration functionals [math]\int_X:A\to\mathbb C[/math] from the commutative case, we can define a scalar product on [math]A[/math], by the following formula:

[[math]] \lt a,b \gt =\varphi(ab^*) [[/math]]

By completing we obtain a Hilbert space [math]H[/math], and we have an embedding as follows:

[[math]] A\subset B(H)\quad,\quad a\to(b\to ab) [[/math]]

Thus we obtain the assertion (1), and a careful examination of the construction [math]A\to H[/math], outlined above, shows that the assertions (2,3) are in fact proved as well.

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So long for operator theory and operator algebras. Obviously, some non-trivial things going on here, and although the above is basically all that we need, in what follows, more familiarity with all this would be desirable. The learning here starts with Rudin , perfectly mastered, and then with some basic functional analysis, basic operator theory, and basic operator algebras, say from Lax . For more, a good, useful, and especially modern book is Blackadar . And for even more, go with Connes .

Be said in passing, speaking Connes, with our [math](S,T,U,K)[/math] philosophy we are already a bit away from what he does, because in his vision, the noncommutative Riemannian manifolds [math]X[/math] do not need coordinates, while in our vision, based on Nash , they do. But this is only a slight difference, everything here being heavily inspired by .

General references

Banica, Teo (2024). "Affine noncommutative geometry". arXiv:2012.10973 [math.QA].