Math 202A: Lecture 8

While arguing our case that the category \mathbf{FHil} of finite-dimensional Hilbert spaces is better than the category \mathbf{FSet} of finite sets, we found ourselves obliged to address the question: is \mathbf{FHil} enriched over itself?

We did not answer this question in Lecture 7, but we did show that \mathbf{FHil} is enriched over the category \mathbf{FBan} of finite-dimensional Banach spaces. More precisely, we proved that every linear transformation A \in \mathrm{Hom}(V,W) is a Lipschitz function with respect to the Hilbert norms on V and W, and we defined the operator norm on \mathrm{Hom}(V,W) to be the Lipschitz constant \|A\| of A \in \mathrm{Hom}(V,W).

The fact that \mathrm{Hom}(V,W) is a Banach space under operator norm is important and we will return to this often. However, we posed a problem in Lecture 7 whose solution rules out the possibility that the operator norm on \mathrm{Hom}(V,W) is induced by a scalar product. Thus, we cannot hope to promote \mathrm{Hom}(V,W) from a Banach space to a Hilbert space by means of polarization.

On the other hand, the main tool we used in Lecture 7, namely the existence of an explicit basis \{E_{yx} \colon x \in X,\ y \in Y\} of \mathrm{Hom}(V,W) associated to any given pair of orthonormal bases X \subset V and Y \subset W, can be utilized to define a scalar product \langle \cdot,\cdot\rangle_{XY} on \mathrm{Hom}(V,W) simply by declaring \{E_{yx} \colon x \in X,\ y \in Y\} of \mathrm{Hom}(V,W) to be orthonormal. This is a general fact: any vector space basis can be used to define a scalar product in which it is orthonormal. Translating this into an explicit formula in the present case of interest, we get

\langle A,B \rangle_{XY} = \sum\limits_{x \in X}\sum\limits_{y\in Y} \overline{\langle y,Ax\rangle}\langle y,Bx\rangle.

The corresponding norm is

\|A\|_{XY} = \sqrt{\sum\limits_{x \in X} |\langle y,Ax \rangle|^2}.

So, we have constructed a scalar product on \mathrm{Hom}(V,W).

Definition 8.1. For any two orthonormal bases X \subset V and Y \subset W, the corresponding scalar product \langle \cdot,\cdot\rangle_{XY} is called the Frobenius scalar product on \mathrm{Hom}(V,W).

According to Definition 8.1, there are many Frobenius scalar products on \mathrm{Hom}(V,W) — one for every choice of Cartesian coordinate systems in the source and target spaces. Our objective is to show that all of these are in fact the same sesquilinear form on \mathrm{Hom}(V,W). This means that we may speak of the Frobenius scalar product on \mathrm{Hom}(V,W), and that this scalar product is intrinsically geometric.

It is in fact very easy to see that \langle \cdot,\cdot \rangle_{XY} has no dependence on Y.

Proposition 8.2. For any A,B \in \mathrm{Hom}(V,W) we have

\langle A,B \rangle = \sum\limits_{x \in X} \langle Ax,Bx\rangle.

Proof: We calculate

\sum\limits_{x \in X} \langle Ax,Bx\rangle = \sum\limits_{x \in X} \left\langle \sum\limits_{y \in Y} \langle y,Ax\rangle y,\sum\limits_{y \in Y} \langle y,Bx\rangle y\right\rangle=\sum\limits_{x \in X}\sum\limits_{y\in Y} \overline{\langle y,Ax\rangle}\langle y,Bx\rangle,

and this is \langle A,B\rangle_{XY}. \square

Showing that the Frobenius scalar product \langle \cdot,\cdot \rangle_{XY} on \mathrm{Hom}(V,W) has no dependence on the Cartesian coordinate system X \subset V is more challenging. To do so, we will use an argument which is inductive in the dimension of W. The base case, where \dim W=1, is essentially the finite-dimensional Riesz representation theorem. The induction step motivates the introduction of orthogonal decompositions and direct sums of Hilbert spaces.

For the rest of this Lecture, let V and W be a pair of Hilbert spaces with fixed orthonormal bases X \subset V and Y \subset W, and denote by \langle \cdot,\cdot \rangle_{XY} the corresponding Frobenius scalar product on \mathrm{Hom}(V,W).

For the rest of this lecture we focus on the case where W is one-dimensional. This means that our specified orthonormal basis of W is a singleton set Y=\{y\} containing a unit vector y \in W. Thus, W=\mathbb{C}y = \{\alpha y \colon y \in Y\} and \mathrm{Hom}(V,W) = \mathrm{Hom}(V,\mathbb{C}y). Because the target space W = \mathbb{C}y is one-dimensional, there are only |X|=\dim V elementary transformations in \mathrm{Hom}(V,\mathbb{C}y), namely

E_{yx}(\cdot) = y \langle x,\cdot \rangle, \quad x \in X.

Let us generalize the elementary transformations to a broader class of linear transformations in \mathrm{Hom}(V,\mathbb{C}y) defined as follows.

Definition 8.2. For every vector v \in V, the corresponding covector L_{yv} \in \mathrm{Hom}(V,\mathbb{C}y) is the linear transformation defined by

L_{yv}(\cdot) = y \langle v,\cdot \rangle.

It is useful to view the assignment v \mapsto L_{yv} as defining a function

\Phi_y \colon V \longrightarrow \mathrm{Hom}(V,\mathbb{C}y).

This function is called the Riesz mapping, and it injects V into \mathrm{Hom}(V,\mathbb{C}y). That the Riesz mapping is injective follows from the fact that \langle v_1-v_2,v_1-v_2\rangle=0 forces v_1=v_2. The Riesz mapping is not linear but antilinear,

L_{y(\alpha_1v_1+\alpha_2v_2)} = \overline{\alpha}_1L_{yv_1} + \overline{\alpha}_2L_{v_2}.

Like linearity, antilinearity implies that the image of V in \mathrm{Hom}(V,\mathbb{C}y) is a vector subspace – the space of covectors.

We claim that that the space of covectors is all of \mathrm{Hom}(V,\mathbb{C}y), i.e. that the Riesz mapping is surjective. Indeed, let A \in \mathrm{Hom}(V,\mathbb{C}y) be an arbitrary transformation. Then, for each x \in X, we have Ax = \alpha_x y for some \alpha_x \in \mathbb{C}. Thus, for any \tilde{v} \in V, we have

A \tilde{v} = \sum\limits_{x \in X} \langle x,\tilde{v}\rangle \alpha_x y.

In order to show that A \in \mathrm{Hom}(V,\mathbb{C}y) is in fact a covector, consider the vector v \in V defined by

v = \sum\limits_{x \in X} \overline{\alpha_x} x.

Then, we have

L_{yv}(\tilde{v}) = y \langle v,\tilde{v}\rangle = y \sum\limits_{x \in X} \alpha_x \langle x,\tilde{v}\rangle = A\tilde{v}.

This completes the proof of the finite-dimensional Riesz representation theorem.

Theorem 8.3. The Riesz mapping v \mapsto L_{yv} is an antilinear bijection V \to \mathrm{Hom}(V,\mathbb{C}y).

Note that the injectivity of the Riesz mapping was proved without any reference to the finite-dimensionality of V. However, our surjectivity argument did make use of the finite-dimensionality of V. Let us also remark that, typically, Definition 8.2 and Theorem 8.3. are stated as theorems about the relationship between V and its dual space V^*=\mathrm{Hom}(V,\mathbb{C}). Note that \mathbb{C}=\mathbb{C}y where y =e^{i\theta} may be any complex number of modulus one: the orthonormal bases of \mathbb{C} are exactly singleton sets containing a point on the unit circle. The standard definition of covectors is made assuming that we take y=1 as the basis of \mathbb{C}.

According to Theorem 8.3, the vector space \mathrm{Hom}(V,\mathbb{C}y) is precisely the space of covectors L_{yv}, v \in V. Thus, we may define a scalar product on \mathrm{Hom}(V,\mathbb{C}y) by

\langle L_{yv_1},L_{yv_2} \rangle := \langle v_1,v_2\rangle, \quad v_1,v_2 \in V.

Let us call this the Riesz scalar product on \mathrm{Hom}(V,\mathbb{C}y). It is defined solely in terms of the scalar product in V, and therefore is canonical and geometric, independent of a choice of basis in V. On the other hand, since L_{yx} = E_{yx}, the Riesz scalar product \langle \cdot,\cdot\rangle on \mathrm{Hom}(V,\mathbb{C}y) coincides with the Frobenius scalar product \langle \cdot,\cdot \rangle_{XY} on \mathrm{Hom}(V,\mathbb{C}y). This proves that the Frobenius scalar product is independent of coordinates in the case of a one-dimensional target space.

Published by Jonathan Novak

Mathematician at UC San Diego.

Leave a Reply