5d. Quantum spaces

[math] \newcommand{\mathds}{\mathbb}[/math]

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Let us end this preliminary chapter on operator algebras with some philosophy, a bit a la Heisenberg. In relation with general “quantum space” goals, Theorem 5.18 is something very interesting, philosophically speaking, suggesting us to formulate:

Definition

Given a von Neumann algebra [math]A\subset B(H)[/math], we write

[[math]] A=L^\infty(X) [[/math]]

and call [math]X[/math] a quantum measured space.

As an example here, for the simplest noncommutative von Neumann algebra that we know, namely the usual matrix algebra [math]A=M_N(\mathbb C)[/math], the formula that we want to write is as follows, with [math]M_N[/math] being a certain mysterious quantum space:

[[math]] M_N(\mathbb C)=L^\infty(M_N) [[/math]]

So, what can we say about this space [math]M_N[/math]? As a first observation, this is a finite space, with its cardinality being defined and computed as follows:

[[math]] |M_N| =\dim_\mathbb CM_N(\mathbb C) =N^2 [[/math]]

Now since this is the same as the cardinality of the set [math]\{1,\ldots,N^2\}[/math], we are led to the conclusion that we should have a twisting result as follows, with the twisting operation [math]X\to X^\sigma[/math] being something that destroys the points, but keeps the cardinality:

[[math]] M_N=\{1,\ldots,N^2\}^\sigma [[/math]]

From an analytic viewpoint now, we would like to understand what is the integration over [math]M_N[/math], giving rise to the corresponding [math]L^\infty[/math] functions. And here, we can set:

[[math]] \int_{M_N}A=tr(A) [[/math]]

To be more precise, on the left we have the integral of an arbitrary function on [math]M_N[/math], which according to our conventions, should be a usual matrix:

[[math]] A\in L^\infty(M_N)=M_N(\mathbb C) [[/math]]

As for the quantity on the right, the outcome of the computation, this can only be the trace of [math]A[/math]. In addition, it is better to choose this trace to be normalized, by [math]tr(1)=1[/math], and this in order for our measure on [math]M_N[/math] to have mass 1, as it is ideal:

[[math]] tr(A)=\frac{1}{N}\,Tr(A) [[/math]]

We can say even more about this. Indeed, since the traces of positive matrices are positive, we are led to the following formula, to be taken with the above conventions, which shows that the measure on [math]M_N[/math] that we constructed is a probability measure:

[[math]] A \gt 0\implies \int_{M_N}A \gt 0 [[/math]]

Before going further, let us record what we found, for future reference:

Theorem

The quantum measured space [math]M_N[/math] formally given by

[[math]] M_N(\mathbb C)=L^\infty(M_N) [[/math]]

has cardinality [math]N^2[/math], appears as a twist, in a purely algebraic sense,

[[math]] M_N=\{1,\ldots,N^2\}^\sigma [[/math]]

and is a probability space, its uniform integration being given by

[[math]] \int_{M_N}A=tr(A) [[/math]]

where at right we have the normalized trace of matrices, [math]tr=Tr/N[/math].

Show Proof

This is something half-informal, mostly for fun, which basically follows from the above discussion, the details and missing details being as follows:

(1) In what regards the formula [math]|M_N|=N^2[/math], coming by computing the complex vector space dimension, as explained above, this is obviously something rock-solid.

(2) Regarding twisting, we would like to have a formula as follows, with the operation [math]A\to A^\sigma[/math] being something that destroys the commutativity of the multiplication:

[[math]] L^\infty(M_N)=L^\infty(1,\ldots,N^2)^\sigma [[/math]]

In more familiar terms, with usual complex matrices on the left, and with a better-looking product of sets being used on the right, this formula reads:

[[math]] M_N(\mathbb C)=L^\infty\Big(\{1,\ldots,N\}\times\{1,\ldots,N\}\Big)^\sigma [[/math]]

In order to establish this formula, consider the algebra on the right. As a complex vector space, this algebra has the standard basis [math]\{f_{ij}\}[/math] formed by the Dirac masses at the points [math](i,j)[/math], and the multiplicative structure of this algebra is given by:

[[math]] f_{ij}f_{kl}=\delta_{ij,kl} [[/math]]

Now let us twist this multiplication, according to the formula [math]e_{ij}e_{kl}=\delta_{jk}e_{il}[/math]. We obtain in this way the usual combination formulae for the standard matrix units [math]e_{ij}:e_j\to e_i[/math] of the algebra [math]M_N(\mathbb C)[/math], and so we have our twisting result, as claimed.

(3) In what regards the integration formula in the statement, with the conclusion that the underlying measure on [math]M_N[/math] is a probability one, this is something that we fully explained before, and as for the result (1) above, it is something rock-solid.

(4) As a last technical comment, observe that the twisting operation performed in (2) destroys both the involution, and the trace of the algebra. This is something quite interesting, which cannot be fixed, and we will back to it, later on.

■

In order to advance now, based on the above result, the key point there is the construction and interpretation of the trace [math]tr:M_N(\mathbb C)\to\mathbb C[/math], as an integration functional. But this leads us into the following natural, and quite puzzling question: \begin{question} In the general context of Definition 5.27, where we formally wrote [math]A=L^\infty(X)[/math], what is the underlying integration functional [math]tr:A\to\mathbb C[/math]? \end{question} This is a quite subtle question, and there are several possible answers here. For instance, we would like the integration functional to have the following property:

[[math]] tr(ab)=tr(ba) [[/math]]

And the problem is that certain von Neumann algebras do not possess such traces. This is actually something quite advanced, that we do not know yet, but by anticipating a bit, we are in trouble, and we must modify Definition 5.27, as follows:

Definition (update)

Given a von Neumann algebra [math]A\subset B(H)[/math], coming with a faithful positive unital trace [math]tr:A\to\mathbb C[/math], we write

[[math]] A=L^\infty(X) [[/math]]

and call [math]X[/math] a quantum probability space. We also write the trace as [math]tr=\int_X[/math], and call it integration with respect to the uniform measure on [math]X[/math].

At the level of examples, passed the classical probability spaces [math]X[/math], we know from Theorem 5.28 that the quantum space [math]M_N[/math] is a finite quantum probability space. But this raises the question of understanding what the finite quantum probability spaces are, in general. For this purpose, we need to examine the finite dimensional von Neumann algebras. And the result here, extending Theorem 5.13, is as follows:

Theorem

The finite dimensional von Neumann algebras [math]A\subset B(H)[/math] over an arbitrary Hilbert space [math]H[/math] are exactly the direct sums of matrix algebras,

[[math]] A=M_{n_1}(\mathbb C)\oplus\ldots\oplus M_{n_k}(\mathbb C) [[/math]]

embedded into [math]B(H)[/math] by using a partition of unity of [math]B(H)[/math] with rank [math]1[/math] projections

[[math]] 1=P_1+\ldots+P_k [[/math]]

with the “factors” [math]M_{n_i}(\mathbb C)[/math] being each embedded into the algebra [math]P_iB(H)P_i[/math].

Show Proof

This is standard, as in the case [math]A\subset M_N(\mathbb C)[/math]. Consider the center of [math]A[/math], which is a finite dimensional commutative von Neumann algebra, of the following form:

[[math]] Z(A)=\mathbb C^k [[/math]]

Now let [math]P_i[/math] be the Dirac mass at [math]i\in\{1,\ldots,k\}[/math]. Then [math]P_i\in B(H)[/math] is an orthogonal projection, and these projections form a partition of unity, as follows:

[[math]] 1=P_1+\ldots+P_k [[/math]]

With [math]A_i=P_iAP_i[/math], we have then a non-unital [math]*[/math]-algebra decomposition, as follows:

[[math]] A=A_1\oplus\ldots\oplus A_k [[/math]]

On the other hand, it follows from the minimality of each of the projections [math]P_i\in Z(A)[/math] that we have unital [math]*[/math]-algebra isomorphisms [math]A_i\simeq M_{n_i}(\mathbb C)[/math], and this gives the result.

■

We can now deduce what the finite quantum measured spaces are, in the sense of the old Definition 5.27. Indeed, we must solve here the following equation:

[[math]] L^\infty(X)=M_{n_1}(\mathbb C)\oplus\ldots\oplus M_{n_k}(\mathbb C) [[/math]]

Now since the direct unions of sets correspond to direct sums at the level of the associated algebras of functions, in the classical case, we can take the following formula as a definition for a direct union of sets, in the general, noncommutative case:

[[math]] L^\infty(X_1\sqcup\ldots\sqcup X_k)=L^\infty(X_1)\oplus\ldots\oplus L^\infty(X_k) [[/math]]

With this, and by remembering the definition of [math]M_N[/math], we are led to the conclusion that the solution to our quantum measured space equation above is as follows:

[[math]] X=M_{n_1}\sqcup\ldots\sqcup M_{n_k} [[/math]]

For fully solving our problem, in the spirit of the new Definition 5.30, we still have to discuss the traces on [math]L^\infty(X)[/math]. We are led in this way to the following statement:

Theorem

The finite quantum measured spaces are the spaces

[[math]] X=M_{n_1}\sqcup\ldots\sqcup M_{n_k} [[/math]]

according to the following formula, for the associated algebras of functions:

[[math]] L^\infty(X)=M_{n_1}(\mathbb C)\oplus\ldots\oplus M_{n_k}(\mathbb C) [[/math]]

The cardinality [math]|X|[/math] of such a space is the following number,

[[math]] N=n_1^2+\ldots+n_k^2 [[/math]]

and the possible traces are as follows, with [math]\lambda_i \gt 0[/math] summing up to [math]1[/math]:

[[math]] tr=\lambda_1tr_1\oplus\ldots\oplus\lambda_ktr_k [[/math]]

Among these traces, we have the canonical trace, appearing as

[[math]] tr:L^\infty(X)\subset\mathcal L(L^\infty(X))\to\mathbb C [[/math]]

via the left regular representation, having weights [math]\lambda_i=n_i^2/N[/math].

Show Proof

We have many assertions here, basically coming from the above discussion, with only the last one needing some explanations. Consider the left regular representation of our algebra [math]A=L^\infty(X)[/math], which is given by the following formula:

[[math]] \pi:A\subset\mathcal L(A)\quad,\quad \pi(a):b\to ab [[/math]]

We know that the algebra [math]\mathcal L(A)[/math] of linear operators [math]T:A\to A[/math] is isomorphic to a matrix algebra, and more specifically to [math]M_N(\mathbb C)[/math], with [math]N=|X|[/math] being as before:

[[math]] \mathcal L(A)\simeq M_N(\mathbb C) [[/math]]

Thus, this algebra has a trace [math]tr:\mathcal L(A)\to\mathbb C[/math], and by composing this trace with the representation [math]\pi[/math], we obtain a certain trace [math]tr:A\to\mathbb C[/math], that we can call “canonical”:

[[math]] tr:A\subset\mathcal L(A)\to\mathbb C [[/math]]

We can compute the weights of this trace by using a multimatrix basis of [math]A[/math], formed by matrix units [math]e_{ab}^i[/math], with [math]i\in\{1,\ldots,k\}[/math] and with [math]a,b\in\{1,\ldots,n_i\}[/math], and we obtain:

[[math]] \lambda_i=\frac{n_i^2}{N} [[/math]]

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

■

We will be back to quantum spaces on several occasions, in what follows. In fact, the present book is as much on operator algebras as it is on quantum spaces, and this because these two points of view are both useful, and complementary to each other.

General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].