Is there a notion of point in noncommutative geometry?

Question

It is not clear to me whether there is a general notion of point in NCG. I have heard (more through physics) that the notion of a point becomes meaningless or ill-defined in noncommutative spaces, but I have also learned that a point in a noncommutative algebra $A$ is a pure state on $A$. What are the obstacles or difficulties in defining such a notion? Can anyone enlighten me? Any insight is welcome

Answer 1

A caveat: everything correct and/or good in this is stolen. I should have probably written this more carefully but given some comments above I felt too much time may have been a bad idea. I pause for a quote here from William Thurston:

This question brings to the fore something that is fundamental and pervasive: that what we [mathematicians] are doing is finding ways for people to understand and think about mathematics.

Essentially the below gives an interpretation, and it is one that has served me in my area and has helped me understand and think about mathematics. However, the question is, if nobody else likes or share such an interpretation, does it comprise mathematics?

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I will work in the setting of a unital $\mathrm{C}^*$-algebra $\mathcal{A}$. If $\mathcal{A}$ is commutative, then we know that $\mathcal{A}=C(X)$ is the algebra of continuous functions on a compact Hausdorff space $X$. This is a categorical equivalence given by Gelfand's Theorem: $$\text{compact Hausdorff spaces}\simeq (\text{unital commutative $\mathrm{C}^*$-algebras})^{\text{op}}.$$ Starting with a compact Hausdorff space $X$, the algebra of continuous functions on $X$, $C(X)$, is a unital commutative $\mathrm{C}^*$-algebra; and starting with a unital commutative $\mathrm{C}^*$-algebra $\mathcal{A}$, its spectrum, $\Omega(\mathcal{A})$, the set of characters, non-zero *-homomorphisms $\mathcal{A}\to \mathbb{C}$, is a compact Hausdorff space such that $\mathcal{A}\cong C(\Omega(\mathcal{A}))$.

A first push in the general direction of your question is to consider a non-commutative unital $\mathrm{C}^*$-algebra as the algebra of continuous functions on a compact quantum space, and a general unital commutative $\mathrm{C}^*$-algebra can be denoted $\mathcal{A}=C(\mathbb{X})$. Inspired by this, one can define the category of `compact quantum spaces' as $$\text{compact quantum spaces}:\simeq (\text{unital $\mathrm{C}^*$-algebras})^{\text{op}}.$$

One way to interpret your question is to note that a point in the commutative setting is an element $x\in X$, so to extend to the noncommutative setting the compact quantum space $\mathbb{X}$ must be identified with a set.

The most obvious and trivial obstruction to this is that in the commutative case the product on $\mathcal{A}=C(X)$ is the point-wise multiplication: $$(fg)(x)=f(x)g(x)\qquad (f,g\in C(X),\,x\in X),$$ and the commutativity of $\mathbb{C}$ gives commutativity to this product. We cannot, in general, consider the product on noncommutative $\mathcal{A}=C(\mathbb{X})$ to be a pointwise product because the product is no longer commutative.

So, the first answer to your question is to say that there are no points, that $\mathbb{X}$ is not a set-of-points, but a virtual object. While we can use the $C(\mathbb{X})$ to help with intuition, making definitions, etc., it is futile to ask your question. We will use this notation from now on, $C(\mathbb{X})$ is a not-necessarily commutative unital $\mathrm{C}^*$-algebra. This might be called the Gelfand Philosophy.

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Now the question is: could we interpret $\mathbb{X}$ as a set...

My first attempt to do this went something like this (I won't explain fully, but at some point was inspired by the text Nik mentioned (I remember something like... a self-adjoint operator is like a labelling of eigenspaces)): suppose $C(\mathbb{X})\subseteq B(\mathsf{H})$ for a Hilbert space $\mathsf{H}$. Let self-adjoint $f\in C(\mathbb{X})$ be of finite spectrum (if you are interested in measuring something that is not of finite spectrum, partition its spectrum into finitely many Borel subsets, and use Borel functional calculus in $C(\mathbb{X})''$ to get something finite spectrum). What does $f$ 'see' when it looks at $\mathsf{H}$? It sees a finite set: the set of eigenspaces $X_f:=\{E_\lambda\,|\,\lambda\in\sigma(f)\}$, and $f$ is like a function $X_f\to \mathbb{C}$, $f(E_\lambda)=\lambda$. If you take a different finite-spectrum self-adjoint $g\in C(\mathbb{G})$, $g$ also sees a set, $X_g$, but if $f$ and $g$ don't commute, then $f$ and $g$ don't agree on the set that they see, and so we don't get a self-adjoint $fg$. If $f$ and $g$ do commute, they will agree what set they see, $X_f=X_g$, and it will make sense to consider $fg:X_f\to \mathbb{C}$. So the Hilbert space is the quantum space, and the manifestation of the quantumness is that what set it is depends on what observable is looking at it. That is a first pass. I got into all kinds of problems around unitary equivalence, etc., and abandoned this.

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My second attempt went like this. Start with $C(X)$ with $X$ a classical space.

1. Each self-adjoint $f\in C(X)$ is a real-valued function: input $x$, output $f(x)$.

2. $X$ is compact Hausdorff.

3. Each such $f$ is continuous: $\lim_\lambda f(x_\lambda)=f(\lim_\lambda x_\lambda)$.

4. If we consider $X$ as a set of "states", we can use elements of $C(X)$ to learn more and more about the state, and we can measure away without affecting the state.

Next consider the actual state space $\mathcal{S}(C(X))$, which consists via integration of regular Borel measures on $X$. Let us look naively at this, at $C(X)\sim C(\mathcal{S}(A))$:

1. Each self-adjoint $f\in C(X)$ is a real valued function: input $\nu\in \mathcal{S}(C(X))$, output $\nu(f)$. Using the isometric embedding into the bidual $\imath:C(X)\hookrightarrow C(X)^{**}$, we could write this as input $\nu$, output $\imath(f)(\nu)$... to share the "$f(x)$" form of notation.

2. $\mathcal{S}(C(X))$ is compact Hausdorff.

3. Each such $f$ is continuous with respect to the weak*-topology: $$\lim_\lambda\imath(f)(\nu_\lambda)=\imath(f)\left(\lim_\lambda \nu_\lambda\right)$$

But when we consider point 4, we get into bother, because 1 above is wrong. We are working in probability here: $\nu$ is an expectation: $$\nu(f)=\mathbb{E}_\nu[f]\qquad (f\in C(X)).$$ Next I argue that we cannot in general measure with infinite precision (relevant: Conditioning a $\mathrm{C}^*$-algebra state with infinite precision), so we must consider finite spectrum $f$ (which might place $f\in C(\mathbb{X})^{**}$ via some Borel functional calculus, in which case $p_\lambda$ will be some $\mathbf{1}_{B_\lambda}(f)$ with the $B_\lambda$ forming a partition of $\sigma(f)$ into Borel sets). So in fact, we should have something like: $$ \begin{aligned} f&=\sum_{\lambda \in \sigma(f)}\lambda\,p_\lambda \\ \implies \nu(f)&=\sum_{\lambda\in\sigma(f)}\lambda\,\nu(p_\lambda), \end{aligned} $$ and so if $\nu$ is an expectation, in fact:

$$\mathbb{E}_\nu(f)=\sum_{\lambda\in\sigma(f)}\lambda\,\nu(p_\lambda)\implies\mathbb{P}_\nu[f=\lambda]=\nu(p_\lambda).$$

This is just classical probability, $p_\lambda$ is just the indicator function $$p_\lambda=\mathbf{1}_{\{x\in X\,\colon\,f(x)=\lambda\}}.$$ but of course given a state $\nu$, when we measure a random variable $f\in C(X)$, and find $f=\lambda$ we have a conditioning, and you can see that it is given as follows:

$$ \begin{aligned} \nu&\mapsto \widetilde{p_\lambda}\nu, \qquad\text{ where} \\\widetilde{p_\lambda}\nu(f)&=\frac{\nu(p_\lambda fp_\lambda)}{\nu(p_\lambda)}\qquad (f\in C(X)) \end{aligned} $$ This is just classical probability theory, and the 1-4. get replaced by:

1. Each self-adjoint $f\in C(X)$ is a real-valued random variable function: given a state $\nu\in\mathcal{S}(C(X))$, output $f=\lambda$ occurs with probability $\nu(p_\lambda)$, and $\mathbb{E}_\nu(f)=\nu(f)$.

2. $\mathcal{S}(C(X))$ is compact Hausdorff.

3. Each such $f$ is continuous as a function $\mathcal{S}(C(X))\to \mathbb{C}$: $$\lim_\lambda f(\nu_\lambda)=f(\lim_\lambda \nu_\lambda).$$

4. If we consider $\mathcal{S}(C(X))$ as a set of "states", we can use elements of $C(X)$ to learn more and more about the state, but measurement involves state conditioning.

So I want us to make a jump here from $C(X)$ with points $x\in X$ to $C(X)$ with points $\nu\in \mathcal{S}(C(X))$. Now... why is this a cogent leap? I quote from https://www.sciencedirect.com/science/article/pii/S072308692100075X (note $M_p(G)$ via integration is the same as the state space):

This Gelfand-Birkhoff picture requires a leap to be made before ever 'going quantum'. Pick up a fresh deck of $N$ cards in some known order and "randomly" shuffle the deck. The shuffle is distributed according to some probability $\nu$ in the set of probabilities on $S_{N}$, $M_p(S_N)$. Without turning over the cards, i.e. without making measurements, it can not be said exactly what permutation acted on the deck. The leap here is to not just consider as permutations the deterministic permutations in $S_N$, but also the random permutations in $M_p(S_N)$ (which includes via the Dirac measures the deterministic permutations). Bilinearly extending the group law to $M_p(S_N)$, gives the random group law, the convolution $$(\mu\star \nu)(\{\sigma\})=\sum_{\tau\in S_N}\mu(\{\sigma\tau^{-1}\})\nu(\{\tau\})\qquad\qquad(\mu,\,\nu\in M_p(S_N)).$$ The Dirac measure concentrated at the identity is an identity for the random group law. Furthermore the inverse can be extended to a map $\operatorname{inv}:M_p(S_N)\to M_p(S_N)$, which gives the inverse of a Dirac measure for the random group law. Precisely because the map $\operatorname{inv}$ is not an inverse on the whole of $M_p(S_N)$, the set of random permutations does not form a group, but it is nonetheless a monoid whose elements can be well-interpreted, understood, and studied in their own right. Note that where $C(S_N)$ is the algebra of complex-valued functions on $S_N$, $M_p(S_N)$ is the subset of positive functionals of norm one on $C(S_N)$: the set of states of $C(S_N)$. Once this leap is made, to accepting that the elements of $M_p(S_N)$ can be studied as "permutations", it isn't so difficult to leap to quantum permutation groups $S_N^+$, where a state on an algebra $C(S_N^+)$ defining the quantum permutation group can be interpreted as a ``permutation'', a quantum permutation.

So, for a not-necessarily commutative $\mathrm{C}^*$-algebra $C(\mathbb{X})$, I choose to interpret the state-space as the set of points:

1. Each self-adjoint $f\in C(\mathbb{X})$ is a real-valued random variable: given a state $\varphi\in\mathcal{S}(C(\mathbb{X}))$, output $f=\lambda$ occurs with probability $\varphi(p_\lambda)$, and $\mathbb{E}_\varphi(f)=\varphi(f)$.

2. $\mathcal{S}(C(\mathbb{X}))$ is compact Hausdorff.

3. Each such $f$ is continuous as a function $\mathcal{S}(C(\mathbb{X}))\to \mathbb{C}$: $$\lim_\lambda f(\varphi_\lambda)=f(\lim_\lambda \varphi_\lambda).$$

4. If we consider $\mathcal{S}(C(\mathbb{X}))$ as a set of "states", we can use elements of $C(\mathbb{X})$ to learn more and more about the state, but measurement involves state conditioning... and things are very different to classical case...

I am happy to explain further in comments, but perhaps if you are interested in this approach the above link might be the way to go.

Answer 2

In the early days of noncommutative topology (the 60s and 70s) the idea that pure states are "points" was common. But I think Yemon's comments about there being too many pure states are well taken. I don't believe this idea is pursued much any more. I also think Yemon is correct in saying that one of the motivating ideas in noncommutative geometry is to deal with "spaces" that do not have points.

The following approach may or may not be helpful. In quantum mechanics the phase space of a system is modelled by a Hilbert space $H$, and the states of the system are the unit vectors modulo multiplication by scalars of modulus 1. In classical physics the phase space is just a set $X$, and we can describe a topology on $X$ by specifying a C${}^\ast$-subalgebra of $l^\infty(X)$. So we could say that we describe a "quantum topology" on $H$ by specifying a C${}^\ast$-subalgebra of $B(H)$, and the "points" of this topology are the unit vectors in $H$ modulo multiplication by scalars of modulus 1. This point of view is developed in my book Mathematical Quantization.

As I said, this may or may not be helpful for what you want to do. I personally have gotten significant mileage out of it in my work on quantum graphs. (Specifically, here and here.) A quantum graph is a weak${}^\ast$-closed operator system $A \subseteq B(H)$, and if the unit vectors in $H$ are the "vertices" of the quantum graph then we can, e.g., define the degree of the vertex $v$ to be the dimension of $Av$. So in this setting the perspective I'm offering was very helpful.