# Are norms intrinsically $\mathbb{R}$-valued?

Another way of phrasing this: are there any viable definitions of something which is norm-like but whose range is in a linearly ordered rig (for example) rather than $\mathbb{R}$?

I have searched a fair bit (including in fairly encyclopedic textbooks), but have come up empty handed as to why everyone just uses $\mathbb{R}$ and does not consider generalizations.

The underlying motivation comes from looking at various theories of mathematics from a minimalistic, "universal algebra" perspective. From that way of looking at things, as opposed to a more semantic perspective which focuses on applications of norms, it seems difficult to justify why norms must range over $\mathbb{R}$. But perhaps it really is important that the range be Dedekind complete -- which would justify this choice. But this is currently not apparent to me.

• Valuations on fields can be considered as taking values on arbitrary ordered abelian groups. May 27, 2014 at 18:40
• @Dirk: a ring without negatives. May 27, 2014 at 18:51
• relevant: mathoverflow.net/questions/122915/… May 27, 2014 at 18:58
• A user attempted three times to post an answer pointing out the above mentioned valuations on fields (in the equivalent form of valuation rings), and all three answers were killed by moderators. While the answers were quite poorly worded and not very informative, I fail to see what is the justification for deleting them as “not an answer”. May 28, 2014 at 10:17
• Along the lines of what @MichałKukieła said, non-Archimedean real closed fields can serve as the range of a norm, culminating in the surreal numbers. The problem is that we have a very good understanding of the second order theory of $\mathbb{R}$ (analysis, topology, etc), but a very poor understanding of the second order theory of these larger real closed fields -- by definition or by theorem (Surreals) they have the same first order theory as the reals. A rich and well understood second order theory for the range of a norm allows for many more robust choices of norms, such as integrals etc. Dec 20, 2017 at 0:44

You might be interested in the whole (mostly Russian) literature on "Banach-Kantorovich" or "lattice-normed" spaces, which are:

"a triplet $(\mathcal U,E,\lambda)$ consisting of a vector space $\mathcal U$, a Dedekind complete vector lattice $E$ and a map $\lambda:\mathcal U\to E_+$ satisfying some natural conditions that allow one to consider $\lambda$ as a vector-valued analogue of the classical notion of a norm (...) Vector spaces equipped with a norm taking values in an Archimedean vector lattice were introduced by L. V. Kantorovich in 1935."

The concept of valuation rings of arbitrary ranks exists. As a special case you get non-Archimedean valuations of higher ranks, whose corresponding norms are non-real.

Maybe you will find what you want from the following pages:

http://en.wikipedia.org/wiki/Valuation_ring (look for the notion of rank),

• Maybe it's noteworthy that Huber's Adic Spaces use crucially the idea to look at "all" valuations (taking values in totally ordered abelian groups) of appropriate rings, in contrast to the Berkovich space which only sees $\mathbb{R}$-valued ones. See example 2.20 of Scholze's Perfectoid Spaces for a nice explanation. May 29, 2014 at 10:02

Not a norm based answer, but perhaps you may still find the work on "cone metric spaces" relevant---this dates back to 1934 by D. Kurepa (a student of M. Fréchet), who considered "metrics" that may take on a value in an ordered vector space. The paper linked to seems to present an updated view.

There is a very abstract generalization of norm and the general idea is as follows:

Valued field: consider a field $K$ and a valuation $|\cdot|:K\mapsto G\cup\{0\}$ satisfying

1. $|x|\geq0$,
2. $|x|=0$ iff $x=0$,
3. $|x+y|\leq max\{|x|,|y|\}$,
4. $|xy|=|x||y|$.

where $G$ is an arbitrary multiplicative ordered group and $0$ is an element such that $0<g$ for all $g\in G$.

$G$-module: Let $G$ be a linearly ordered group. A linearly ordered set $X$ is called a $G$-module if there exists a map $G\times X\to X$, written $(g,x)\mapsto gx$, such that for all $g,g_1,g_2\in G$ and all $x,x_1,x_2\in X$ we have:

1. $g_1(g_2x)=(g_1g_2)x$
2. $1x=x$
3. $g_1\geq g_2\Rightarrow g_1x\geq g_2x$
4. $x_1\geq x_2\Rightarrow gx_1\geq gx_2$
5. $Gx$ is coinitial in $X$
6. $X$ has no smallest element.

Norm: Let $E$ be a vector space over $(K,|\cdot|)$ and let $X$ be a $G$-module. An $X$-norm on $E$ is a map $||\cdot||:E\to X\cup\{0\}$ such that for all $x,y\in E$, $\lambda\in K$:

1. $||x||=0\Leftrightarrow x=0$
2. $||\lambda x||=|\lambda|||x||$
3. $||x+y||\leq\max\{||x||,||y||\}.$

For an introduction in this area I recommend the paper:

Banach spaces over fields with a infinite rank valuation - [H.Ochsenius A., W.H.Schikhof] - 1999