1a. Operator algebras

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

The quantum groups are not groups in the usual sense, but rather abstract generalizations of the groups, motivated by quantum mechanics. To be more precise, a quantum group [math]G[/math] is something similar to a classical group, except for the fact that the functions on it [math]f:G\to\mathbb C[/math] do not necessarily commute, [math]fg\neq gf[/math]. And due to this, [math]G[/math] is not exactly a set of points, or transformations, but rather an abstract object, described by the algebra [math]A[/math] of functions on it [math]f:G\to\mathbb C[/math], which can be noncommutative.

In order to introduce the quantum groups, we need some sort of algebraic geometry correspondence, between “quantum spaces” and noncommutative algebras. Which is not exactly a trivial business, because such correspondences are not really needed in the context of usual mathematics, or of usual physics, such as classical mechanics.

However, we can use some inspiration from quantum mechanics. Phenomena of type [math]fg\neq gf[/math] are commonplace there, known since the early 1920s, and the work of Heisenberg. And it is also known from there that the good framework for understanding such phenomena is the mathematics of the infinite dimensional complex Hilbert spaces [math]H[/math], with as main example the Schrödinger space [math]H=L^2(\mathbb R^3)[/math] of wave functions of the electron.

So, problem solved, and in order to start our hunt for quantum groups, we just need to understand the Hilbert space basics. And do not worry, we will see later, with some further inspiration from quantum mechanics helping, coming this time from the work of von Neumann and others, that this will naturally lead us into the correspondence between “quantum spaces” and noncommutative algebras that we are looking for.

As a starting point, we have the following basic definition:

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].

Here the fact that [math]||.||[/math] is indeed a norm comes from the Cauchy-Schwarz inequality, which states that if (1,2,3) above are satisfied, then we have:

[[math]] | \lt x,y \gt |\leq||x||\cdot||y|| [[/math]]

Indeed, this inequality 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+twy||^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:

[[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 so the Hilbert space itself is:

[[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 separable, because we can use the basis [math]\{x^n\}_{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.

Again inspired by physics, and more specifically by the “matrix mechanics” of Heisenberg, we will be interested in the linear operators over a Hilbert space. We have:

Proposition

Let [math]H[/math] be a Hilbert space, with orthonormal basis [math]\{e_i\}_{i\in I}[/math]. The algebra [math]\mathcal L(H)[/math] of linear operators [math]T:H\to H[/math] embeds then into the matrix algebra [math]M_I(\mathbb C)[/math], with [math]T[/math] corresponding to the matrix [math]M_{ij}= \lt Te_j,e_i \gt [/math]. In particular:

In the finite dimensional case, where [math]\dim(H)=N \lt \infty[/math], we obtain in this way a usual matrix algebra, [math]\mathcal L(H)\simeq M_N(\mathbb C)[/math].
In the separable infinite dimensional case, where [math]I\simeq\mathbb N[/math], we obtain in this way a subalgebra of the infinite matrices, [math]\mathcal L(H)\subset M_\infty(\mathbb C)[/math].

Show Proof

The correspondence [math]T\to M[/math] in the statement is indeed linear, and its kernel is [math]\{0\}[/math]. As for the last two assertions, these are clear as well.

■

The above result is something quite theoretical, because for basic spaces like [math]L^2[0,1][/math], which do not have a simple orthonormal basis, the embedding [math]\mathcal L(H)\subset M_\infty(\mathbb C)[/math] that we obtain is not very useful. Thus, while the operators [math]T:H\to H[/math] are basically some infinite matrices, it is better to think of these operators as being objects on their own.

Normally quantum mechanics, or at least basic quantum mechanics, is about operators [math]T:H\to H[/math] which are densely defined, also called “unbounded”. In what concerns us, the correspondence between “quantum spaces” and noncommutative algebras that we want to establish rather belongs to advanced quantum mechanics, where the operators [math]T:H\to H[/math] are usually everywhere defined, as in Proposition 1.2, and bounded.

So, departing now from the quantum mechanics of the 1920s, and its difficulties, and turning instead to pure mathematics for inspiration, a natural question is that of understanding what the correct infinite dimensional analogue of the matrix algebra [math]M_N(\mathbb C)[/math] is. And the answer here is provided by 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 with unit, denoted [math]B(H)[/math]. This algebra has the following properties:

[math]B(H)[/math] is complete with respect to [math]||.||[/math], and 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 the involution are related by the formula [math]||TT^*||=||T||^2[/math].

Show Proof

The fact that the bounded operators form indeed an algebra, as stated, follows from the following estimates, which are all clear from definitions:

[[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, the sequence [math]\{T_nx\}[/math] is Cauchy for any [math]x\in H[/math], so we can define the limit [math]T=\lim_{n\to\infty}T_n[/math] by setting:

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

It is routine to check that this formula defines indeed a bounded operator [math]T\in B(H)[/math], and that we have [math]T_n\to T[/math] in norm, and this gives the result.

(2) Here the existence of [math]T^*[/math] comes from the fact that [math]\varphi(x)= \lt Tx,y \gt [/math] being a linear map [math]H\to\mathbb C[/math], we must have a formula as follows, for a certain vector [math]T^*y\in H[/math]:

[[math]] \varphi(x)= \lt x,T^*y \gt [[/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]]

(3) Regarding now the last assertion, we have:

[[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 in this formula [math]T\to T^*[/math] we obtain [math]||T||^2\leq||TT^*||[/math]. Thus, we have proved both the needed inequalities, and we are done.

■

Observe that, in view of Proposition 1.2, we embeddings of [math]*[/math]-algebras, as follows:

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

In this picture the adjoint operation [math]T\to T^*[/math] constructed above takes a very simple form, namely [math](M^*)_{ij}=\overline{M}_{ji}[/math], at the level of the associated matrices.

We are now getting close to the point where we wanted to get, namely algebras of operators, corresponding to “quantum spaces”. But one more mathematical trick, in order to get to this. Instead at looking at algebras of operators [math]A\subset B(H)[/math], depending on a Hilbert space [math]H[/math], that is after all something quite cumbersome, that we don't want to have in our theory, it is more convenient to simply take some inspiration from Theorem 1.3, and based on that, axiomatize a nice class of complex algebras, as follows:

Definition

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

A norm [math]a\to||a||[/math], making it a Banach algebra (the Cauchy sequences converge).
An involution [math]a\to a^*[/math], which satisfies [math]||aa^*||=||a||^2[/math], for any [math]a\in A[/math].

We know from Theorem 1.3 that the full operator algebra [math]B(H)[/math] is a [math]C^*[/math]-algebra, for any Hilbert space [math]H[/math]. More generally, any closed [math]*[/math]-subalgebra [math]A\subset B(H)[/math] is a [math]C^*[/math]-algebra. The celebrated Gelfand-Naimark-Segal (GNS) theorem states that any [math]C^*[/math]-algebra appears in fact in this way. This is something non-trivial, and we will be back to it later on.

For the moment, we are interested in developing the theory of [math]C^*[/math]-algebras, without reference to operators, or Hilbert spaces. Our goal, that we can now state quite precisely, is to show that any [math]C^*[/math]-algebra appears as follows, with [math]X[/math] being a “quantum space”:

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

In order to do this, let us first look at the commutative [math]C^*[/math]-algebras [math]A[/math], which should normally correspond to usual spaces [math]X[/math]. As a first observation, we have:

Proposition

If [math]X[/math] is an abstract compact space, the algebra [math]C(X)[/math] of continuous functions [math]f:X\to\mathbb C[/math] is a [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].

This algebra is commutative, in the sense that [math]fg=gf[/math], for any [math]f,g\in C(X)[/math].

Show Proof

Almost everything here is trivial. Observe also that we have indeed:

[[math]] \begin{eqnarray*} ||ff^*|| &=&\sup_{x\in X}|f(x)\overline{f(x)}|\\ &=&\sup_{x\in X}|f(x)|^2\\ &=&||f||^2 \end{eqnarray*} [[/math]]

Finally, we have [math]fg=gf[/math], since [math]f(x)g(x)=g(x)f(x)[/math] for any [math]x\in X[/math].

■

Our claim now is that any commutative [math]C^*[/math]-algebra appears in this way. This is a non-trivial result, which requires a number of preliminaries. Let us begin with:

Definition

The spectrum of an element [math]a\in A[/math] is the set

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

where [math]A^{-1}\subset A[/math] is the set of invertible elements.

As a basic example, the spectrum of a usual matrix [math]M\in M_N(\mathbb C)[/math] is the collection of its eigenvalues. Also, the spectrum of a continuous function [math]f\in C(X)[/math] is its image. In the case of the trivial algebra [math]A=\mathbb C[/math], the spectrum of an element is the element itself.

As a first, basic result regarding spectra, we have:

Proposition

We have the following formula, valid for any [math]a,b\in A[/math]:

[[math]] \sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\} [[/math]]

Moreover, there are examples where [math]\sigma(ab)\neq\sigma(ba)[/math].

Show Proof

We first prove that we have the following implication:

[[math]] 1\notin\sigma(ab)\implies1\notin\sigma(ba) [[/math]]

Assume indeed that [math]1-ab[/math] is invertible, with inverse [math]c=(1-ab)^{-1}[/math]. We have then [math]abc=cab=c-1[/math], and by using these identities, we obtain:

[[math]] \begin{eqnarray*} (1+bca)(1-ba) &=&1+bca-ba-bcaba\\ &=&1+bca-ba-bca+ba\\ &=&1 \end{eqnarray*} [[/math]]

A similar computation shows that we have as well [math](1-ba)(1+bca)=1[/math]. We conclude that [math]1-ba[/math] is invertible, with inverse [math]1+bca[/math], which proves our claim. By multiplying by scalars, we deduce from this that we have, for any [math]\lambda\in\mathbb C-\{0\}[/math], as desired:

[[math]] \lambda\notin\sigma(ab)\implies\lambda\notin\sigma(ba) [[/math]]

Regarding now the last claim, let us first recall that for usual matrices [math]a,b\in M_N(\mathbb C)[/math] we have [math]0\in\sigma(ab)\iff 0\in \sigma(ba)[/math], because [math]ab[/math] is invertible if any only if [math]ba[/math] is. However, this latter fact fails for general operators on Hilbert spaces. As a basic example, we can take [math]a,b[/math] to be the shift [math]S(e_i)=e_{i+1}[/math] on the space [math]l^2(\mathbb N)[/math], and its adjoint. Indeed, we have [math]S^*S=1[/math], and [math]SS^*[/math] being the projection onto [math]e_0^\perp[/math], it is not invertible.

■

Given an element [math]a\in A[/math], and a rational function [math]f=P/Q[/math] having poles outside [math]\sigma(a)[/math], we can construct the element [math]f(a)=P(a)Q(a)^{-1}[/math]. For simplicity, we write:

[[math]] f(a)=\frac{P(a)}{Q(a)} [[/math]]

With this convention, we have the following result:

Theorem

We have the “rational functional calculus” formula

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

valid for any rational function [math]f\in\mathbb C(X)[/math] having poles outside [math]\sigma(a)[/math].

Show Proof

In order to prove this result, we can proceed in two steps, as follows:

(1) Assume first that we are in the polynomial case, [math]f\in\mathbb C[X][/math]. We pick [math]\lambda\in\mathbb C[/math], and we write [math]f(X)-\lambda=c(X-r_1)\ldots(X-r_n)[/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_n)\in A^{-1}\\ &\iff&a-r_1,\ldots,a-r_n\in A^{-1}\\ &\iff&r_1,\ldots,r_n\notin\sigma(a)\\ &\iff&\lambda\notin f(\sigma(a)) \end{eqnarray*} [[/math]]

(2) 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 (1), we obtain:

[[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]]

Thus, we have obtained the formula in the statement.

■

Given an element [math]a\in A[/math], its spectral radius [math]\rho (a)[/math] is the radius of the smallest disk centered at [math]0[/math] containing [math]\sigma(a)[/math]. We have the following key result:

Theorem

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

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

We use the various results established above.

(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 by using (2), and the rational function [math]f(z)=(z+it)/(z-it)[/math], with [math]t\in\mathbb R[/math]. Indeed, for [math]t \gt \gt 0[/math] the element [math]f(a)[/math] is well-defined, and 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 [math]f(a)[/math] is a unitary, and by (2) its spectrum is contained in [math]\mathbb T[/math]. We conclude that we have [math]f(\sigma(a))=\sigma(f(a))\subset\mathbb T[/math], and so [math]\sigma(a)\subset f^{-1}(\mathbb T)=\mathbb R[/math], as desired.

(4) We have [math]\rho(a)\leq ||a||[/math] from (1). Conversely, given [math]\rho \gt \rho(a)[/math], we have:

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

By applying the norm and taking [math]n[/math]-th roots we obtain:

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

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], and by taking [math]n[/math]-th roots we get [math]\rho\geq ||{a}||[/math]. This gives the missing inequality, namely:

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

In the general case, [math]aa^*=a^*a[/math], we have [math]a^n(a^n)^*=(aa^*)^n[/math]. We obtain from this [math]\rho(a)^2=\rho(aa^*)[/math], and since [math]aa^*[/math] is self-adjoint, we get [math]\rho(aa^*)=||a||^2[/math], and we are done.

■

Summarizing, we have so far a collection of useful results regarding the spectra of the elements in [math]C^*[/math]-algebras, which are quite similar to the results regarding the eigenvalues of the usual matrices. We will heavily use these results, in what follows.

General references

Banica, Teo (2024). "Introduction to quantum groups". arXiv:1909.08152 [math.CO].