How the Yoneda Lemma Illuminates the Riesz Representation Theorem

An enriched Yoneda interpretation of the Riesz representation theorem

Let $\mathsf{Hil}$$\mathsf{Hil}$ denote the category of complex Hilbert spaces, with bounded linear maps as morphisms. For any $X,Y\in \mathsf{Hil}$$X,Y\in\mathsf{Hil}$, the operator space $\mathsf{Hil}(X,Y)$$\mathsf{Hil}(X,Y)$ carries the operator norm, hence is a Banach space. Therefore $\mathsf{Hil}$$\mathsf{Hil}$ may naturally be regarded as a $\mathsf{Ban}$$\mathsf{Ban}$-enriched category.

Now consider the following two $\mathsf{Ban}$$\mathsf{Ban}$-valued contravariant functors:

• the dual space functor

$$\mathsf{Hil}(-,\mathbb{C}):{\mathsf{Hil}}^{op}\to \mathsf{Ban},$$

which sends $X$$X$ to the Banach space of continuous linear functionals $\mathsf{Hil}(X,\mathbb{C})$$\mathsf{Hil}(X,\mathbb C)$;

• the duality functor

$$†:{\mathsf{Hil}}^{op}\to \mathsf{Ban},$$

where $†(X)$$\dagger(X)$ denotes the conjugate dual of $X$$X$.

By the enriched Yoneda lemma, there is an isomorphism of internal hom-objects

$$[{\mathsf{Hil}}^{op},\mathsf{Ban}](\mathsf{Hil}(-,\mathbb{C}),†)\cong †(\mathbb{C}).$$

Since $\mathbb{C}$$\mathbb C$ is a one-dimensional Hilbert space, we have $†(\mathbb{C})\cong \mathbb{C}$$\dagger(\mathbb C)\cong \mathbb C$. Hence

$$[{\mathsf{Hil}}^{op},\mathsf{Ban}](\mathsf{Hil}(-,\mathbb{C}),†)\cong \mathbb{C}.$$

Therefore, each scalar $c\in \mathbb{C}$$c\in\mathbb C$ determines an enriched natural transformation

$${\eta }^c:\mathsf{Hil}(-,\mathbb{C})\Rightarrow †.$$

Its component at an object $X$$X$ is given by the Yoneda formula

$${\eta }_X^c(f)=f^{†}(c),f:X\to \mathbb{C}.$$

In particular, if we choose $c=1\in \mathbb{C}$$c=1\in\mathbb C$, we obtain

$${\eta }_X(f)=f^{†}(1).$$

Obviously the inverse map is just $v⟼⟨-,v⟩$$v\longmapsto\langle -,v\rangle$.

This is precisely the standard correspondence in the Riesz representation theorem. If we adopt the convention that the inner product is linear in the first variable, then for every $x\in X$$x\in X$ we have

$$⟨x,{\eta }_X(f)⟩=⟨x,f^{†}(1)⟩=⟨f(x),1⟩=f(x).$$

Therefore

$$f(x)=⟨x,{\eta }_X(f)⟩,$$

so ${\eta }_X(f)$$\eta_X(f)$ is exactly the Riesz representing vector of $f$$f$. Thus $\eta$$\eta$ yields a natural isomorphism

$$\mathsf{Hil}(X,\mathbb{C})\cong †(X).$$

In other words, the standard natural isomorphism in the Riesz representation theorem is exactly the one generated by $1\in \mathbb{C}$$1\in\mathbb C$ via the enriched Yoneda lemma.

More generally, since $\mathbb{C}$$\mathbb C$ is one-dimensional, for every $c\in \mathbb{C}$$c\in\mathbb C$ one has

$${\eta }^c=c{\eta }^1$$

(up to a complex conjugation factor if one uses the opposite inner product convention). Hence:

• if $c\ne 0$$c\neq 0$, then ${\eta }^c$$\eta^c$ is still a natural isomorphism;

• if $c=0$$c=0$, then ${\eta }^c$$\eta^c$ is the zero transformation and is not an isomorphism.

Thus all nonzero natural isomorphisms arising in this way are parametrized by ${\mathbb{C}}^{\times }$$\mathbb C^\times$.

To understand which of these are isometric isomorphisms, one must use the fact that the Hilbert norm is induced by the inner product:

$$\parallel v{\parallel }^2=⟨v,v⟩.$$

$$⟨f(v),f(v)⟩=⟨f^{†}(v),f^{†}(v)⟩$$

This is the essential extra structure beyond Yoneda. Since the adjoint is compatible with the inner product, and since $\mathbb{C}$$\mathbb C$ is a one-dimensional Hilbert space, $f^{†}(c)$$f^\dagger(c)$ is simply a scalar multiple of $f^{†}(1)$$f^\dagger(1)$:

$$f^{†}(c)=c{f}^{†}(1)$$

(or $\bar{c}{f}^{†}(1)$$\overline{c}\,f^\dagger(1)$, depending on convention). Consequently,

$$\parallel f^{†}(c){\parallel }^2=⟨f^{†}(c),f^{†}(c)⟩=|c{|}^2⟨f^{†}(1),f^{†}(1)⟩=|c{|}^2\parallel f^{†}(1){\parallel }^2,$$

and hence

$$\parallel f^{†}(c)\parallel =|c|\parallel f^{†}(1)\parallel .$$

Since ${\eta }^1$$\eta^1$ is the standard Riesz map and is isometric, it follows that

$$\parallel {\eta }_X^c(f)\parallel =|c|\parallel f\parallel .$$

Therefore:

• if $|c|=1$$|c|=1$, then ${\eta }^c$$\eta^c$ is an isometric isomorphism;

• if $|c|\ne 1$$|c|\neq 1$ and $c\ne 0$$c\neq 0$, then ${\eta }^c$$\eta^c$ is still an isomorphism, but not an isometry.

It follows that all isometric natural isomorphisms are parametrized exactly by the unit circle

$$S^1=\{c\in \mathbb{C}:|c|=1\}.$$

That is, every element of $S^1$$S^1$ gives another isometric version of the Riesz isomorphism; these differ from one another only by a phase factor.

Nevertheless, among all these isometric choices, one usually selects $1\in S^1$$1\in S^1$ as the standard normalization. The reason is that $1$$1$ is both the multiplicative unit and a unit vector, so it introduces neither an additional scaling factor nor an additional phase twist. Thus the "standard" Riesz isomorphism is the one corresponding to $1$$1$.

Well, if we are working on $\mathbb{R}$$\mathbb R$, then we only have two ways to define the isometric natural isomorphism.