Being nonseparable Banach space is in fact nothing special: one meets the first
examples in the standard functional analysis course, when one learns about
$ell^p$ or $L^p[0,1]$ spaces-these spaces are nonseparable when $p=infty$.
However I spoke to one person which said to me that nonseparable Hilbert
spaces are rather exotic and in "real life" one rarely come across them. Here I list some examples when I came across nonseparable Hilbert spaces—but to be honest, none of them have truly convinced me about the importance of those spaces:
When $A$ is a $C^*$-algebra, then if $A$ is separable then $A$ embeds into $B(H)$ where $H$ is separable—nevertheless sometimes one is interested in so called universal representation $pi_u$ of $A$ (one feature of this representation is that any state of $pi_u(A)$ is in fact a vector state). The Hilbert space of this representation is hardly ever separable.
When $Q$ denotes the Calkin algebra (the quotient of $B(H)$ by compact operators) one can show that it cannot be represented on separable Hilbert space (this is very elegant argument: there is an uncountable family of infinite subsets of $mathbb{N}$ such that the intersection of any two members of this family is finite—this gives an uncountable family of mutually disjoint projections in the quotient—in fact this argument applies also to $ell^{infty}/c_0$).
There is a notion of almost periodic function on $mathbb{R}$ (this can be generalized to other groups but let us stick to this particular case) and the space of almost periodic functions contains all functions of the type $e_s(t)=e^{ist}$. One can introduce the scalar product on the linear span of all $e_s$-s and those function turns out to be mutually orthogonal. Completing one obtains a nonseparable Hilbert spaces.
There is a notion of tensor product of Hilbert spaces in particular the so called complete tensor product which is due to von Neumann. This construction yields a nonseparable Hilbert space when the tensored family is infinite. But as far as I know, in most application one restricts to the preferred so called $C_0$-sequence thus obtaining the separable space (this is some sort of choosing a ,,stabilization”).
That’s all-I don’t know any other examples of situations when one is faced with nonseparable Hilbert spaces-so ,,huge” for the first sight spaces as Fock spaces used in second quantization or $L^2$ over some infinite dimensional spaces with Gaussian measures are all separable.
So I would like to ask:
What are some interesting examples of nonseparable Hilbert spaces which you have ever met?
Forgive me if my question is too vaque but I tried to give some motivation around it.
MathOverflow Asked by truebaran on January 23, 2021
6 AnswersSome simple arguments (which, to be honest, I'm also not completely convinced of) about the non-necessity of non-separable Hilbert spaces. If "in real life" means working into a given Hilbert space, and not meditating over the category of all Hilbert spaces, then we should admit that there is a lot of not used room, thus maybe unnecessary, in a non-separable Hilbert space $H$. Even if a Hilbert basis has been fixed, and one identifies the space with $ell_2(S)$ for a set $S$, any element of $H$, any sequence of elements, any separable subspace, are already included in some $ell_2(D)$ for a countable subset $Dsubset S$. Any closed subspace $V$ of $H$ splits into an orthogonal sum of subspaces $V_jsubset ell_2(D_j)$, for a partition of $S$ into countable subsets ${D_j}_{jin I}$. Any bounded operator $T$ on $H$ splits into such an orthogonal sum of invariant subspaces, and $T$ in a sense is just a family of independent operators on $ell_2:=ell_2(mathbb{N})$. So it seems there is nothing new in the structure of a non-separable Hilbert space, that cannot be described by means of the already rich structure of $ell_2$ (which on the contrary is really a new object wrto the finite dimensional $ell_2^n$, or at least a new way of speaking about them).
Answered by Pietro Majer on January 23, 2021
Example 3 of almost-periodic functions. It is more natural than the description given in the OP.
Harald Bohr defined almost periodic functions in order to study Dirichlet series as applicable in analytic number theory.
Bohr's definition. A bounded measurable function $f : mathbb R to mathbb R$ is almost periodic iff the set of translates ${T_s f : s in mathbb R}$ is relatively compact in $L^infty$, where $$ (T_s f)(x) = f(x-s) $$
Equivalently: for every $epsilon > 0$ there exists $s in mathbb R$ so that for all $x in mathbb R$ $$ |f(x)-f(x-s)| le epsilon . $$
The almost periodic functions naturally come with an "invariant mean" $$ M(f) = lim_{T to infty}frac{1}{2T}int_{-T}^T f(x);dx tag1$$ and from that we get an inner product $$ langle f,g rangle = lim_{T to infty}frac{1}{2T}int_{-T}^T f(x)overline{g(x)};dx tag2$$ and a seminorm $$ |f| = lim_{T to infty}frac{1}{2T}int_{-T}^T |f(x)|^2;dx tag3$$
The space of almost perioduc function is non-separable according to seminorm (3). So of course its completion is a nonseparable Hilbert space.
In fact, Bohr's definition may be used on general locally compact groups. Indeed, Bohr's invariant mean on such groups was developed before Haar's invariant integral. The invariant mean $M(f)$ is obtained as follows. If $f$ is a bounded almost periodic function on a group $G$, then the closed (according to uniform convergence) convex hull of the set of all translates $T_s f$ contains exactly one constant. Define $M(f)$ to be that constant.
Answered by Gerald Edgar on January 23, 2021
Your example #4 might be of profound relevance to the understanding of the difference be quantum physics and classical physics. In quantum physics, the state of a system is described by a vector in a Hilbert space (its space of states). When two systems are associated, the space of states of the resulting system is the von Neumann tensor product of the spaces of states of the associated two systems. Simple quantum systems have separable space of states.
Now one big question in physics is why elementary particles such as electrons behave according to the laws of quantum physics while macroscopic objects which are built out of quantum objects behave differently, i.e. according to classical physics. The question is all the more important for measurement devices that interact with quantum objects while behaving classically. They operate projective measurements on the quantum systems thus breaking the unitary evolution of quantum systems.
The answer might lie in the difference in properties of spaces of states of the microscopic and macroscopic systems : the very large (infinite for all practical purposes) number of tensor products of elementary particle spaces of states needed to build the space of states of macroscopic objects such as measurement devices could be non separable for the reason you mention. This non-separability could open the door to different properties including non unitary evolution.
Your question thus triggers another one: what are the properties of non-separable Hilbert spaces that are built out of infinite von Neumann tensor products of separable Hilbert space?
Answered by Mathias on January 23, 2021
@Yemon Choi: the answer to your comment is too long for a reply so putting it in another answer to the original question – hope this is not off-topic for the original post.
The space ${mathbb{R}}^{[0,1]}$ and the $L^2$ functions on it arise not as a model of how the brain actually works, but as a counterfactual, pathological example.
In neuroscience, a sensory stimulus (for example an image seen by the eye) causes a pattern of activity in a set of neurons, known as a neural code for the stimulus. We can model this activity using a real-valued function $f(n,s)$, where $n$ represents the neuron, and $s$ represents the stimulus. (The same holds for artificial neural networks used in machine learning.) The actual brain of course contains a finite number of neurons, but to make mathematical models it is useful to consider them as drawn randomly from a continuous probability distribution, and the same for the stimuli.
A classical hypothesis of neuroscience, originally inspired by information theory, holds that the brain would operate most efficiently if neural responses were independent. However if you require the responses of all neurons be independent whatever the stimulus, you soon end up considering examples like neurons as random functions drawn from ${mathbb{R}}^{[0,1]}$ with the product Gaussian measure, and $f(n,s)$ as evaluation at the point $sin[0,1]$. This is a pathological model, as such functions are almost surely discontinuous, and it is also not consistent with experimental brain recordings. You are correct that stochastic processes are a much better model for how the brain actually works, and brain recordings actually let us further constrain the smoothness of these stochastic processes. Nevertheless it is still useful to consider counterfactual models such as the above. The paper I cited deals with the smoothness constraints, and does not yet mention the counterfactual, but a forthcoming revision will.
Answered by Neuromath on January 23, 2021
Let A = $mathbb{R}^{[0,1]}$, with a product Gaussian measure - in other words, consider an independent Gaussian-distributed random variable for each real number in $[0,1]$. The space $L^2[A]$ is non separable.
Believe it or not, this example has come up in neuroscience.
Answered by Neuromath on January 23, 2021
To the contrary, my feeling is that nonseparable Hilbert spaces are in some sense artifacts and can almost always be avoided. And more generally, to your comment about nonseparable Banach spaces being "nothing special", most of the nonseparable Banach spaces one meets in practice are duals of separable Banach spaces, and therefore are weak* separable. Algebras of almost periodic functions, and their noncommutative analogs, the CCR algebras, are rare examples of interesting nonseparable Banach spaces which are not dual spaces, but even here one is mainly interested in regular representations, which are determined by their behavior on separable subalgebras (the span of the functions $e_s$ with $s$ rational, say).
Answered by Nik Weaver on January 23, 2021
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