Riesz representation theorem for a Hilbert space built not on a field but on a division ring

Question

Dual to a Hilbert space $H$ is a space of continuous functionals $\Phi:\;H\longrightarrow \mathbb K$ mapping a vector $v\in H$ to an element of the underlying field $\mathbb K$.

The Rietsz representation theorem states that for each continuous linear functional $\Phi$ there is a unique vector $\varphi_{\Phi}\in H$ such that $\Phi(v)=(v,\,\varphi_{\Phi})$ for $\forall v\in H\,$.

QUESTION 1.
Does the theorem stay valid for an arbitrary field $\mathbb K$ (not necessarily $\mathbb R$ or $\mathbb C$) ?

QUESTION2.
Does the theorem stay for a division ring (say, the ring $\mathbb H$ of quaternions) ?

Answer 1

This is a long response to your question in the comments. You cannot even state the definition of a Hilbert space over a general field or division ring. Let's go over the definition carefully: a Hilbert space over $K = \mathbb{R}, \mathbb{C}$ is

1. a vector space $H$
2. equipped with an inner product $\langle \cdot, \cdot \rangle : H \times H \to K$, namely a map which is conjugate-linear in the first variable, linear in the second variable, satisfies $\langle x, y \rangle = \overline{ \langle y, x \rangle }$, and satisfies $\langle x, x \rangle \ge 0$ with equality iff $x = 0$ (positive-definite),
3. such that $H$ is complete with respect to the metric $d(x, y) = \| x - y \| = \sqrt{ \langle x - y, x - y \rangle }$.

If we want to state this definition over a ring $K$, then

1. $K$ must be a division ring (or we don't get a vector space; weakening this requirement might be a thing you want to do for various reasons, but I think it would be misleading to call the result a Hilbert space, and they aren't),
2. equipped with an involution $(-)^{\ast} : K \to K$ (which is needed to say what "conjugate-linear" means),
3. such that the self-adjoint elements $\{ r \in K : r^{\ast} = r \}$ have an order (which is needed to say what "positive-definite" means), and
4. such that $d(x, y) = \sqrt{ \langle x - y, x - y \rangle }$ is a real number (which is needed to state completeness).

If you really wanted to you might be able to find clever ways around these issues. What I was saying in the comments is just that if you want to state the definition of a Hilbert space in the most straightforward way, without doing anything clever, then already $K$ can't just be an arbitrary division ring but must have an involution such that the self-adjoint elements can be identified with $\mathbb{R}$. The most straightforward way this happens is if $K$ is a division algebra over $\mathbb{R}$.

Now: surely $K$ itself must be a Hilbert space over $K$, in fact the $1$-dimensional Hilbert space, when equipped with the inner product $\langle x, y \rangle = x^{\ast} y$, yes? So $K$ must be a division $^{\ast}$-algebra over $\mathbb{R}$ which is complete with respect to the norm $\| x \| = \sqrt{ x^{\ast} x }$. Positive-definiteness now means that $K$ is a real $C^{\ast}$-algebra, and:

Real Gelfand-Mazur theorem: The only real $C^{\ast}$-algebras which are also division algebras are $\mathbb{R}, \mathbb{C}$, and $\mathbb{H}$.

So, again, if you want to state the Hilbert space axioms in a straightforward way without doing anything clever, $K = \mathbb{R}, \mathbb{C}$, and $\mathbb{H}$ are your only options.

This is a different and easier result than Solèr's theorem which weakens the requirements by not explicitly imposing either positive-definiteness or completeness. But since Solèr's theorem implies that even with these weaker requirements $\mathbb{R}, \mathbb{C}$, and $\mathbb{H}$ are your only options, the above result should not be too surprising.

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That means this question is about quaternionic Hilbert spaces only. Googling "quaternionic Riesz representation theorem" turns up Chi-Keung Ng's On quaternionic functional analysis. The Riesz representation theorem is Theorem 3.12.

Answer 2

Just some thoughts, too long for a comment.

Let's take the example of $\mathbb{H}^n$ for some finite $n$. Take the product

$$\langle v, w\rangle = \sum_{\ell=1}^n \bar v_k w_k$$

It is clearly biadditive, and moreover

$$\langle v a, w b\rangle = \bar a \langle v, w\rangle b$$

So you should look at $\mathbb{H}^n$ as a right vector space over $\mathbb{H}$.

Question: is every linear map $\mathbb{H}^n\to \mathbb{H}$ of the form $w \mapsto \langle v, w \rangle$ ? ( This is an algebra question, no analysis). You can check that the answer is yes.

Now consider something like $\ell^2(\mathbb{N})$ but with quaternionic entries. Define what the space is, the inner product, the topology, etc. Can you formulate the statement? Does the result hold?

Can you define a Hilbert space over $\mathbb{H}$?

Try to prove the Riesz theorem in the case of the field $\mathbb{R}$ or $\mathbb{C}$. Do you need commutativity of the coefficients?

An observation: topological vector spaces over topological fields or topological division rings are probably well studied, but perhaps it's interesting to see what results hold when $\mathbb{K}$ is not $\mathbb{R}$ or $\mathbb{C}$. Good luck!

$\bf{Added:}$ Let $\mathcal{H}$ be a right vector space over $\mathbb{H}$. Let $\langle \cdot , \cdot \rangle \colon \mathcal{H}\times \mathcal{H} \to \mathbb{H}$, $\langle w, v \rangle = \overline{\langle v, w \rangle}$, $\langle v, w_1+w_2 \rangle=\langle v, w_1 \rangle+\langle v, w_2 \rangle$, $\langle v, w a \rangle =\langle v, w \rangle a$ and $\langle v, v \rangle \ge 0$, with equality if and

Can you get any of the usual statements for (real or complex) spaces right away? Yes. For consider $\mathcal{H}$ as a vector space over $\mathbb{R}$ with the inner product

$$\langle v, w\rangle_1 \colon = \operatorname{Re}( \langle v, w\rangle)$$

Now let's prove C-B-S inequality. Take $a$ a quaternion of unit norm such that $\langle v,w a\rangle \in \mathbb{R}$, etc.

Now the fact that we have a norm follows from the real case.

Let's take $\mathcal{K}$ an $\mathbb{H}$ subspace. Check that the orthogonal of $\mathcal{K}$ for the quaternionic inner product coincides with the orthogonal wr to the real product. This allows to prove all of the desired properties by passing to the real.

One could ask why bother with quaternionic spaces if they "are" like real ones. Well, they do have extra structure that we could use.