# classification of Hilbert spaces

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 $\ell^{2}(X)$ space (http://planetmath.org/EllpXSpace), where $X$ has the cardinality of any orthonormal basis of the Hilbert space in consideration.

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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 $\ell^{2}(I)$.

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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 $U:H\longrightarrow\ell^{2}(I)$ be defined by

 $Ux\,(i):=\langle x,e_{i}\rangle$

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 $\ell^{2}(I)$ it follows that

 $\|x\|^{2}=\sum_{i\in I}|\langle x,e_{i}\rangle|^{2}=\sum_{i\in I}|Ux\,(i)|^{2}% =\|Ux\|^{2}_{\ell^{2}(I)}$

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

Let $f\in\ell^{2}(I)$. By definition of the space $\ell^{2}(I)$ we must have $\displaystyle\sum_{i\in I}|f(i)|^{2}<\infty$, 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

 $Ux_{0}\,(j)=\langle x_{0},e_{j}\rangle=\sum_{i\in I}f(i)\langle e_{i},e_{j}% \rangle=f(j)$

or in other , $Ux_{0}=f$. Hence, $U$ is surjective. $\square$

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Proof of the classification theorem :

• $(\Longrightarrow)$ 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 $\{Ue_{i}\}_{i\in I}$ is an orthonormal basis for $H_{2}$. Hence, $H_{1}$ and $H_{2}$ have the same dimension.

• $(\Longleftarrow)$ 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 $\ell^{2}(I)$. Hence $H_{1}$ and $H_{2}$ are isometrically isomorphic. $\square$

 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 $\ell^{2}(X)$ space