2
$\begingroup$

I'm trying to understand how to form the ultrapower of a normed space.

I read how to construct the underlying vector space here: http://en.wikipedia.org/wiki/Ultraproduct

But I don't see an obvious norm which can be defined on the resulting equivalence classes.

Is there a standard one which always works?

$\endgroup$
  • $\begingroup$ Taking the ultraproducts of the norms gives an "ultranorm" (it takes values in the nonstandard reals) and then maybe you can take standard parts? $\endgroup$ – Qiaochu Yuan Jun 22 '13 at 19:29
  • $\begingroup$ First you're going to have to describe how you are treating vector spaces as first-order structures. What is the language you are using? Is it a two-sort language or a one-sort language with just a lot of function symbols? $\endgroup$ – Asaf Karagila Jun 22 '13 at 19:29
  • $\begingroup$ @Qiaochu: Not every non-standard real has a standard part. Consider transfinite ones, what would be their standard part? $\endgroup$ – Asaf Karagila Jun 22 '13 at 19:30
  • 1
    $\begingroup$ @Asaf: it seems one should take the ultraproduct in an "internal" fashion; in particular, when taking the product one should restrict as GEdgar does. $\endgroup$ – Qiaochu Yuan Jun 22 '13 at 19:43
2
$\begingroup$

Let $I$ be an index set, let $\mathcal U$ be an ultrafilter on $I$. For each $i \in I$ let $(X_i, \|\cdot\|_i)$ be a normed space. Let $X$ be the $l^\infty$-product of the $X_i$, consisting in all families $(x_i)_{i \in I}$ such that $x_i \in X_i$ for all $i$ and $\sup_i \|x_i\|_i < \infty$. The ultraproduct is the quotient $X/N$ where $N$ consists of all $(x_i)$ with $\lim_{i,\mathcal U} \|x_i\|_i = 0$. (That's the limit along an ultrafilter.) In space $X/N$ use norm $\lim_{i,\mathcal U}\|x_i\|_i$.

$\endgroup$
  • $\begingroup$ Ahh so you are indeed restricting the Cartesian product to elements upon which the norm makes sense. $\endgroup$ – roo Jun 22 '13 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.