most recent 30 from http://mathoverflow.net 2013-05-25T03:56:15Z http://mathoverflow.net/feeds/question/87602 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87602/topological-vector-spaces-that-are-isomorphic-to-their-duals Oliver 2012-02-05T17:58:54Z 2012-02-06T04:08:54Z

<p>After reviewing the (locally convex) topological vector spaces that I know, the only examples I could find where there is an isomorphism from the space to its (anti)dual, are Hilbert spaces. So my question is :</p> <p>Are there topological vector spaces $V$ such that the topology does not come from a Hilbert structure, and such that there exists an isomorphism $\chi : V \to V'$, where $V'$ denotes the antidual of $V$ (continuous antilinear forms on $V$) ? </p>

http://mathoverflow.net/questions/87602/topological-vector-spaces-that-are-isomorphic-to-their-duals/87603#87603 Todd Trimble 2012-02-05T18:05:52Z 2012-02-05T18:05:52Z

<p>Take a reflexive TVS $V$, and consider $V \times V^\ast$. </p>

http://mathoverflow.net/questions/87602/topological-vector-spaces-that-are-isomorphic-to-their-duals/87604#87604 Gerald Edgar 2012-02-05T18:06:38Z 2012-02-05T18:06:38Z

<p><code>If $X$ is any reflexive space, then</code>$X \oplus X^{*}$<code>is isomorphic to its dual $X^* \oplus X^{**}$</code>.</p>

http://mathoverflow.net/questions/87602/topological-vector-spaces-that-are-isomorphic-to-their-duals/87632#87632 S. Carnahan 2012-02-06T04:08:54Z 2012-02-06T04:08:54Z

<p>An interesting family of examples comes from number theory (or algebraic geometry, depending on who you ask): If you have a field $k$, the Laurent power series field $k((t))$ has an ultrametric topology where <code>$\{ t^n k[[t]] \}_{n \in \mathbb{Z}}$</code> form a neighborhood basis of zero. This space is isomorphic to its topological dual: by adjoining $(dt)^{1/2}$, one obtains the perfect residue pairing $$\langle f(t) (dt)^{1/2}, g(t) (dt)^{1/2} \rangle = \operatorname{Res} f(t)g(t) dt.$$ You can do a similar trick with finite dimensional vector spaces over $k((t))$.</p> <p>If you want nontrivial antilinearity, you may choose $k$ to be a separable quadratic extension of some underfield $F$, and change the residue pairing to be sesquilinear: $$\langle f(t) (dt)^{1/2}, g(t) (dt)^{1/2} \rangle = \operatorname{Res} f(t) \bar{g}(t) dt.$$</p>