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.
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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 ${\mathrm{\ell}}^{2}(I)$.
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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.
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Proof of Theorem 1: Let ${\{{e}_{i}\}}_{i\in I}$ an orthonormal basis indexed by the set $I$. Let $U:H\u27f6{\mathrm{\ell}}^{2}(I)$ be defined by
$$Ux(i):=\u27e8x,{e}_{i}\u27e9$$ 
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
$${\parallel x\parallel}^{2}=\sum _{i\in I}{\u27e8x,{e}_{i}\u27e9}^{2}=\sum _{i\in I}{Ux(i)}^{2}={\parallel Ux\parallel}_{{\mathrm{\ell}}^{2}(I)}^{2}$$ 
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 $$, and therefore, using the RieszFischer theorem, the series $\sum _{i\in I}}f(i){e}_{i$ converges to an element ${x}_{0}\in H$. We now see that
$$U{x}_{0}(j)=\u27e8{x}_{0},{e}_{j}\u27e9=\sum _{i\in I}f(i)\u27e8{e}_{i},{e}_{j}\u27e9=f(j)$$ 
or in other , $U{x}_{0}=f$. Hence, $U$ is surjective. $\mathrm{\square}$
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Proof of the classification theorem :

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$(\u27f9)$ 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.

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$(\u27f8)$ 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  20130322 17:56:18 
Last modified on  20130322 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 