4
$\begingroup$

Valuations and seminorms on rings play a big role in number theory and analytic geometry, with seminorms being heavily used in Berkovich geometry and valuations featuring heavily in adic geometry.

Let's quickly recall their definitions:

A valuation on a ring $A$ is a pair $(\Gamma,\nu)$ consisting of

  • The Value Group. A totally ordered abelian group $\Gamma$;
  • The Valuation. A map of of sets $$\nu\colon A\to\Gamma\mathbin{\textstyle\coprod}\{0\},$$ where $\Gamma\mathbin{\textstyle\coprod}\{0\}$ is the totally ordered abelian group with zero consisting of the disjoint union of $\Gamma$ with $\{0\}$, with multiplication extending that of $\Gamma$ by declaring $0\cdot0=0$ and $0\cdot a=0$ for all $a\in\Gamma$, and where we have $0<a$ for all $a\in\Gamma$.

satisfying the following conditions:

  1. We have $\nu(0)=0$.
  2. We have $\nu(1)=1$.
  3. For each $a,b\in A$, we have $\nu(ab)=\nu(a)\nu(b)$.
  4. For each $a,b\in A$, we have $\nu(a+b)\leq\max(\nu(a),\nu(b))$.

A seminorm on a ring $A$ is a map $|-|\colon A\to\mathbb{R}_{\geq0}$ satisfying the following conditions:

  1. We have $|0|=0$.
  2. We have $|1|=1$.
  3. For each $a,b\in A$, we have $|a+b|\leq|a|+|b|$.
  4. For each $a,b\in A$, we have $|ab|\leq|a||b|$.

Moreover, $|-|$ is a norm if $|a|=0$ implies $a=0$.

Question. Are there notions of valuations, seminorms, and norms on ring spectra extending the above notions?

$\endgroup$
8
  • $\begingroup$ (P.S. By "ring spectra" I mean either commutative algebras in $\mathrm{Sp}$ or $\mathbb{E}_k$-rings for some $1\leq k\leq\infty$, and by "extending" the classical notions I mean that a valuation/seminorm/norm on a ring spectrum $E$ should induce an ordinary valuation/seminorm/norm on $\pi_0(E)$.) $\endgroup$
    – Emily
    Commented Mar 22, 2023 at 1:47
  • 1
    $\begingroup$ It does not seem reasonable to consider $E_1$-rings, as this definition does not seem to be reasonable for classical associative rings. On the other hand, it is unclear whether one should only look at connective guys — for example, if one looks at the sphere spectrum, do we need to see the "chromatic information" (for which one should pass to nonconnective guys)? $\endgroup$
    – Z. M
    Commented Mar 22, 2023 at 5:09
  • $\begingroup$ In the definition of valuation, the disjoint union is not a group with the structure you give it. In any case, with the correct definitions, if you define valuations and semi norms on a ring spectrum $R$ as those in $\pi_0R$ then this seems to satisfy your conditions. $\endgroup$ Commented Mar 22, 2023 at 6:06
  • $\begingroup$ @Z.M I hadn't really though of this before, but apparently people do consider valuations on noncommutative rings (e.g. link), although I'm not sure how useful they are in that context. Probably we should indeed look at at least $\mathbb{E}_2$-rings. I'm also not sure about connectivity, although I think having such a theory for nonconnective ring spectra as well would certainly be nice $\endgroup$
    – Emily
    Commented Mar 22, 2023 at 10:23
  • 1
    $\begingroup$ I was not only talking about a "theory for nonconnective rings" — the concept of valuations should be equivalent to the concept of valuation rings, and the question is whether there are nonconnective valuation rings? Let's be a bit simpler — what are fields? For example, would we consider Morava K-theories (not $E_2$, but quasi-commutative) being fields? $\endgroup$
    – Z. M
    Commented Mar 22, 2023 at 12:01

0

You must log in to answer this question.