# Is there a concept of a “free Hilbert space on a set”?

I am looking for a "good" definition of a Hilbert space with a distinct orthonormal basis (in the Hilbert space sense) such that each basis element corresponds to an element of a given set $X$. Before I explain my attempt for a definition of this, let me talk about something analogous.

Analogy: There exists something analogous for vector spaces, namely the free vector space on the set $X$. This is a vector space $V(X)$ with a distinct basis such that every basis element corresponds to an element of $X$. More precisely, a free vector space on $X$ is a vector space $V(X)$, together with a map $i: X \rightarrow V(X)$ such that the following universal property is satisfied: For every vector space $W$ (over the same field) and every map $\phi: X \rightarrow W$, there is a unique linear map $\psi: V(X) \rightarrow W$ such that $\phi = \psi \circ i$. For such a free vector space over $X$, the set $i(X) = \{i(x) \mid x \in X\}$ is a basis of $V(X)$ such that each element of the basis corresponds to an element of $X$. I heard that one calls $V(X)$ the free vector space because $V(X)$ free object on $X$, but I don't fully understand this concept.

Let's restrict to complex vector spaces. For a given set $X$, one can construct a free vector space on $X$ as follows: One takes the set $V(X)$ of functions $f: X \rightarrow \mathbb{C}$ with finite support, endowed with pointwise addition and scalar multiplication. This is a vector space with the Kronecker delta functions $\delta_x$ (which evaluate to 1 on $x$ and to zero elsewhere) as a basis. Then $V(X)$, together with the map $i: X \rightarrow V(X), \ x \mapsto \delta_x$ is a free vector space on $X$.

My attempt: Inspired by the above construction of a free vector space on a set $X$, I want to define a free Hilbert space on $X$. Again, let's restrict to complex Hilbert spaces. For a given set $X$, let $\mathcal{H}(X)$ be the set of functions $f: X \rightarrow \mathbb{C}$ with countable support such that $\sum_{x \in \text{supp}(f)} \vert f(x) \vert^2 < \infty$, endowed with pointwise addition and scalar multiplication and the inner product $\langle f, g \rangle = \sum_{x \in \text{supp}(f) \cap \text{supp}(g)} f(x) \overline{g(x)}$. In other words, set $\mathcal{H}(X) := \ell^2(X)$. This is a Hilbert space where the Kronecker delta functions $\delta_x$ for $x \in X$ form an orthonormal basis. Let $i: X \rightarrow \mathcal{H}(X)$ be the map $x \mapsto \delta_x$.

My question: Is this a "good" definition of a free Hilbert space on a set $X$? Does it satisfy a universal property analogous to the one for free vector spaces? Is this a "Hilbert space with a distinct orthonormal basis such that each basis element corresponds to an element of $X$"?

What makes me skeptical is the fact that I found a document on the functor $\ell^2$ in which it is said that "The important $\ell^2$–construction is in many ways the closest thing there is to a free Hilbert space" (page 1) but also "Lemma 4.8 showed that $\ell^2(X)$ is not the free Hilbert space on X, at least not in the categorically accepted meaning." What is this "categorically accepted meaning" and how does it relate to the universal property of the free vector space i mentioned above?

• As described in the paper, you do not need to highlight the countable support of the functions, if $f:X\to\Bbb C$ satisfies $\sum_{x\in X}|f(x)|^2<\infty$, then there are always only finitely many points satisfying $|f(x)|>2^{-n}$ for any $n\in\Bbb N$, so that the support is automatically countable. – Dr. Lutz Lehmann May 29 '14 at 9:33
• Thanks. I was aware of that, but I thought it's easier to define it that way so that I do not need to (at least a priori) define the sum as a sum over a net rather than over a sequence (since I would sum over possibly uncountably many numbers). – Corsin Pfister May 29 '14 at 9:40

Consier the category $\mathsf{Hilb}$ of Hilbert spaces and continuous linear maps between them. The free Hilbert space $H(X)$ on a set $X$ (in the sense of category theory) would have to satisfy the adjunction $$\hom_\mathsf{Hilb}(H(X),K) \cong \hom_{\mathsf{Set}}(X,|K|),$$ where $K$ is a Hilbert space with underlying set $|K|$. Notice that $H(X)=\ell^2(X)$ for finite sets $X$. More precisely, we have for arbitrary sets $X$ $$\hom_\mathsf{Hilb}(\ell^2(X),K) = \{f \in \hom_{\mathsf{Set}}(X,|K|) : \sum_{x \in X} ||f(x)||^2 < \infty\}.$$ Now let $X$ be any infinite set. Assume that $H(X)$ exists. Let $e : X \to |H(X)|$ be the unit. For every map $f : X \to |K|$ with $K \in \mathsf{Hilb}$ there is a unique $\tilde{f} : H(X) \to K$ such that $\tilde{f}(e(x))=f(x)$ for all $x \in X$. If $C:=||\tilde{f}|| \in \mathbb{R}_{\geq 0}$, it follows $||f(x)|| \leq C ||e(x)||$. Certainly we may choose $f(x) \neq 0$, so that also $e(x) \neq 0$. Hence, if $g : X \to |K|$ is any map, then $g$ is bounded (consider $f(x):=g(x) \cdot ||e(x)||$). This is a contradiction (assume $\mathbb{N} \subseteq X$ and define $g(n):= n \cdot u$ for some unit vector $u$).

• I thought categories were written with $\textbf{Boldface}$ font, not with $\textsf{Sans Serif}$ font! – Asaf Karagila May 30 '14 at 0:16
• @AsafKaragila, as long as they aren't written $\mathcal{Set},$ I'm happy. – goblin May 30 '14 at 1:21
• You have to be careful here I think. $g$ is map in $Set$ so it makes no sense to say it's bounded. Do you mean the restriction of $\tilde{g}$ to $X$ which embeds in $H(X)$ via $e : X \to |H(X)|$? And how can $Hilb$ even be a category? The only morphisms that preserve the inner product are the isometries. – Matematleta Jun 12 '14 at 3:01

Here is the answer to the analogous question about Banach spaces. The same kind of argument as in Martin Brandenburg's answer shows that there is no free Banach space on a set if you think the morphisms in the category of Banach spaces are the continuous linear maps.

However, there is a better choice of morphisms available: you can instead let the morphisms be the continuous linear maps of norm at most $1$. (See this blog post for a defense of this choice.) One of the many nice things about this choice of morphism is that an isomorphism in this category is an isometric isomorphism; in other words, this category really remembers the norm on a Banach space and not just the norm up to Lipschitz equivalence. It is also categorically extremely well-behaved: the corresponding category is complete, cocomplete, and has a symmetric monoidal structure, the Banach space tensor product, with respect to which it is closed monoidal.

In this category there is such a thing as the free Banach space on a set $S$, and it turns out to be precisely $\ell^1(S)$ provided that you also modify the forgetful functor: the new forgetful functor sends a Banach space not to its underlying set but to the underlying set of its unit ball. (This is also $\text{Hom}(\mathbb{C}, -)$ where $\mathbb{C}$ is the monoidal unit for the Banach space tensor product.)

Unfortunately, I don't think this argument can be adapted to the case of Hilbert spaces. There are a couple of different choices of morphism and forgetful functor you can try and I think none of them work. Potentially the real problem with Hilbert spaces is that they do not really form a category: they form a dagger category with involution given by the adjoint, and that structure really needs to be taken into account when thinking categorically about Hilbert spaces.

$\ell^2(S)$ can be thought of as satisfying a universal property, but it's not free on a collection of vectors: instead, it's free on a collection of orthonormal vectors.