classification of Hilbert spaces

Hilbert spaces can be classified, up to isometric isomorphism, according to their dimension. Recall that an isometric isomorphism of Hilbert spaces is an unitary transformation, therefore it preserves the vector space structure along with the inner product structure (hence, preserving also the topological structure). Recall also that the dimension of a Hilbert space is a well defined concept, i.e. all orthonormal bases of an Hilbert space share the same cardinality.

The classification theorem we describe here states that two Hilbert spaces $H_1$ and $H_2$ are isometrically isomorphic if and only if they have the same dimension, i.e. if and only if an orthonormal basis of $H_1$ has the same cardinality of an orthonormal basis of $H_2$.

This will be achieved by proving that every Hilbert space is isometrically isomorphic to an ${\mathrm{\ell }}^2(X)$ space (http://planetmath.org/EllpXSpace), where $X$ has the cardinality of any orthonormal basis of the Hilbert space in consideration.

Theorem 1 - Suppose $H$ is an Hilbert space and let $I$ be a set that indexes one (and hence, any) orthonormal basis of $H$. Then, $H$ is isometrically isomorphic to ${\mathrm{\ell }}^2(I)$.

Theorem [Classification of Hilbert spaces] - Two Hilbert spaces $H_1$ and $H_2$ are isometrically isomorphic if and only if they have the same dimension.

Proof of Theorem 1: Let ${\{e_i\}}_{i\in I}$ an orthonormal basis indexed by the set $I$. Let $U:H⟶{\mathrm{\ell }}^2(I)$ be defined by

$$Ux(i):=⟨x,e_i⟩$$

We claim that $U$ is an isometric isomorphism. It is clear that $U$ is linear. Using Parseval’s equality and the definition of norm in ${\mathrm{\ell }}^2(I)$ it follows that

$${\parallel x\parallel }^2=\sum _{i\in I}{|⟨x,e_i⟩|}^2=\sum _{i\in I}{|Ux(i)|}^2={\parallel Ux\parallel }_{{\mathrm{\ell }}^2(I)}^2$$

We conclude that $U$ is isometric. It remains to see that it is surjective (since injectivity follows from the isometric condition).

Let $f\in {\mathrm{\ell }}^2(I)$. By definition of the space ${\mathrm{\ell }}^2(I)$ we must have , and therefore, using the Riesz-Fischer theorem, the series ${\displaystyle \sum _{i\in I}}f(i)e_i$ converges to an element $x_0\in H$. We now see that

$$U{x}_0(j)=⟨x_0,e_j⟩=\sum _{i\in I}f(i)⟨e_i,e_j⟩=f(j)$$

or in other , $U{x}_0=f$. Hence, $U$ is surjective. $\mathrm{\square }$

Proof of the classification theorem :

• $(⟹)$ Of course, if the Hilbert spaces $H_1$ and $H_2$ are isometrically isomorphic, with isometric isomorphism $U$, then if ${\{e_i\}}_{i\in I}$ is an orthonormal basis for $H_1$ than ${\{U{e}_i\}}_{i\in I}$ is an orthonormal basis for $H_2$. Hence, $H_1$ and $H_2$ have the same dimension.

• $(⟸)$ If the Hilbert spaces $H_1$ and $H_2$ have the same dimension, then we can index any orthonormal basis of $H_1$ and any orthonormal basis of $H_2$ by the same set $I$. Using Theorem 1 we see that $H_1$ and $H_2$ are both isometrically isomorphic to ${\mathrm{\ell }}^2(I)$. Hence $H_1$ and $H_2$ are isometrically isomorphic. $\mathrm{\square }$

Title	classification of Hilbert spaces
Canonical name	ClassificationOfHilbertSpaces
Date of creation	2013-03-22 17:56:18
Last modified on	2013-03-22 17:56:18
Owner	asteroid (17536)
Last modified by	asteroid (17536)
Numerical id	10
Author	asteroid (17536)
Entry type	Theorem
Classification	msc 46C15
Classification	msc 46C05
Synonym	Hilbert spaces of the same dimension are isometrically isomorphic
Related topic	EllpXSpace
Related topic	OrthonormalBasis
Related topic	ClassificationOfSeparableHilbertSpaces
Related topic	CategoryOfHilbertSpaces
Related topic	RieszFischerTheorem
Related topic	QuantumGroupsAndVonNeumannAlgebras
Defines	every Hilbert space is isometrically isomorphic to a ${\mathrm{\ell }}^2(X)$ space