nLab
isomorphism classes of Banach spaces

This page is inspired by the following question, which appeared on MathOverflow.

Let $p,q\in (1,\mathrm{\infty })$ with $p\ne q$. Are the Banach spaces $L^p(ℝ)$, $L^q(ℝ)$ isomorphic?

More generally, one can ask:

Given two Banach spaces, $X$ and $Y$, when are they isomorphic?

The following started out as an adapted version of Bill Johnson’s answer to the MathOverflow question.

One way to prove that a Banach space $X$ is not isomorphic to a Banach space $Y$ is to exhibit a property which is preserved under isomorphisms that $X$ has but $Y$ does not. For example, among the spaces $L_p(ℝ)$ for $p\in [1,\mathrm{\infty }]$, $L_{\mathrm{\infty }}$ is the only nonseparable space, and $L_1$ is the only separable space with a nonseparable dual. Thus $L_1$ and $L_{\mathrm{\infty }}$ are not isomorphic to each other or to any $L_p$ with $p\in (1,\mathrm{\infty })$.

To distinguish among the $L_p$ with $p\in (1,\mathrm{\infty })$ finer properties are needed. Type and cotype are examples of such properties. The (best) type and cotype of $L_p$ are standard calculations: if $p\in [1,2]$ then $L_p$ has type $p$ and cotype $2$ (and no better), and if $p\in [2,\mathrm{\infty })$ then $L_p$ has type $2$ and cotype $p$ (and no better). See for example in Theorem 6.2.14 of AK06. From that, one can see that if $p\ne q$, then $L_p$ and $L_q$ either have different (best) type or different (best) cotype.

Type and cotype depend only on the collection of finite dimensional subspaces of a space (we call such a property a local property?). So neither can be used to prove, e.g., that for $p\ne 2$, $L_p$ is not isomorphic to ${\ell }_p$. One way of proving this is to show that for $p\ne 2$, ${\ell }_2$ embeds isomorphically into $L_p$ but not into ${\ell }_p$ (see also AK).

References

• AK06 Albiac, Fernando and Kalton, Nigel. Topics in Banach space theory. Graduate Texts in Mathematics, 233. Springer, New York, 2006. xii+373 pp. ISBN: 978-0387-28141-4; MR2192298