7
$\begingroup$

Let $\Bbbk$ be a field; I am interested in the following ring (which I suspect is a field). Its elements are formal expressions that look like

$$ \sum_{n=0}^{\infty} a_n x^{b_n} $$

where $a_n\in \Bbbk$ and $b_n\in \mathbb{R}$, with $b_n$ strictly increasing, and $\lim_{n\to\infty} b_n = \infty$. (Technically, we should quotient these by some relation allowing us to insert and remove terms where $a_n=0$. Or we could require all the $a_n$'s to be nonzero, but then we'd have to include finite sums as well. Or we could represent them by functions $a : \mathbb{R} \to \Bbbk$ assigning a coefficient to each exponent, whose support is left-finite.) We can add and multiply these expressions in fairly evident ways; the condition on $b_n$ ensures that multiplication works (i.e. the resulting set of exponents can again be enumerated with order type $\omega$ and limit $\infty$).

This ring of power-series-like-objects is closely related to some others. Specifically, if $\Bbbk=\mathbb{R}$ then it contains the Levi-Civita field as the elements for which each $b_n\in\mathbb{Q}$, while it is contained in the Hahn series field $\Bbbk[[x^{\mathbb{R}}]]$. Note that the set of all Hahn series with order type $\omega$ is not closed under multiplication; this is a natural subset thereof that is. I believe that it is also the set of Hahn series with order type $\omega$ that converge to themselves in the valuation topology of the Hahn series field, and also that it is the closure of the field $\Bbbk(x^{\mathbb{R}})$ of generalized rational functions inside the Hahn series field, and the Cauchy completion of $\Bbbk(x^{\mathbb{R}})$ in its valuation uniformity.

Does this field have a standard name and/or a notation? Is it an instance of some more general construction (e.g. replacing $\mathbb{R}$ by something more general, which would presumably then also include the Levi-Civita field as the case of $\mathbb{Q}$)?

$\endgroup$
  • 3
    $\begingroup$ This is the completion of $\mathbb{k}(x^{\mathbb{R}})$ as a valued field, i.e. the unique up to isomorphism dense valued field extension without proper dense valued field extension. $\endgroup$ – nombre Jul 6 at 10:29
  • 1
    $\begingroup$ It is a field because the multiplication is well-defined commutative and the geometric series implies $1+\sum_{n=1}^\infty a_n x^{b_n}, 0 < b_n <b_{n+1} \to \infty$ has an inverse. $\endgroup$ – reuns Jul 6 at 22:05
10
$\begingroup$

I think your ring looks similar to the Novikov ring (see topology papers).

$\endgroup$
  • $\begingroup$ Following Google to en.wikipedia.org/wiki/Novikov_ring and thence by an external link back to MO, I found mathoverflow.net/a/13226/49 where my ring is called the "universal Novikov field" (hence I guess it is a field!). Going back to Google, I see that this terminology seems to be common. Unfortunately the most common notation for it is $\Lambda$ or $\Lambda(\Bbbk)$, which works in the context of a particular paper but can't really be a global notation (like $\Bbbk[[x]]$) -- for one thing, it clashes with the standard notation for exterior algebras. But a name helps! $\endgroup$ – Mike Shulman Jul 6 at 15:48
  • 2
    $\begingroup$ @MikeShulman: Funny: 4 reviewers out of 4 voted to delete this post. It was only you accepting it as an answer that stopped the review process and saved the post. $\endgroup$ – Alex M. Jul 6 at 17:38
  • 3
    $\begingroup$ @AlexM. Yes, that is funny. I would argue that those reviewers were being too trigger-happy. Along the lines of meta.stackexchange.com/q/225370, I think this is an answer, and it did contain enough information to lead me to what I wanted to know, even if it would have been nice for it to contain more information to save me the trouble of Googling and following links. $\endgroup$ – Mike Shulman Jul 6 at 18:00

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.