<p>Consider a variety of algebras $\mathbb{V} = \mathbb{V}(\sigma,\tau)$ which consists of the set of algebras defined over a fixed signature $\sigma$ and satisfying a set of identities $\tau$. We may then define its <em>ordered</em> version $\mathbb{V}_{or}(\sigma,\tau)$ cas the set of algebras admitting a total order $\leq$ compatible with $\sigma$. By this I mean that:</p>

<p>(*) for each $f : X^n \rightarrow X$ in $\sigma$, for each tuples $t,t' \in X^n$, if $t_i \leq t'_i$ for every $i \in [n]$ then $f(t) \leq f(t')$.</p>

<p>Given such an ordered algebra $A$, we may define its order type $ot(A)$ as the maximum order type of a well-founded order $\leq' \subseteq \leq$ without accumulation points. We may then define the <em>spectrum</em> of the variety $\mathbb{V}$ as the set of ordinals obtainable in that way.</p>

<p>Is this notion known in universal algebra, and if so, which varieties have a known spectra?</p>

<p>An easy case seems to be the variety of ordered fields: for instance, the field of reals $\mathbb{R}$ would have order type $\omega$, and the field of rational fractions $\mathbb{R}(x)$ would have order type $\omega^{\omega}$. So is the following true:</p>

<p>Conjecture: the spectrum of the ordered fields consists of the ordinals closed under addition and multiplication?</p>

<p>(NOTE: I added the requirement that $\leq'$ has no accumulation points, as otherwise any countable ordinal would be embeddable in $\mathbb{R}$. The precise meaning is relative to $\leq$: if $S$ is the set of elements of $\leq'$, we ask that for every $x \in S$, there is an interval $I$ of $\leq$ properly containing $x$ s.t. $I \cap S = \{x\}$.)</p>

CC BY-SA 3.0
2014-04-07T23:08:52.760Z
Your questions may have a higher chance of being answered at Mathoverflow, where they would be seen by a broader spectrum of mathematicians.
Sasho Nikolov
CC BY-SA 3.0
2014-04-08T05:37:30.640Z
Thanks for the advice, but I guess that when you've lost yourself, it doesn't really matter where you end...
NisaiVloot
CC BY-SA 3.0
2014-04-08T13:22:05.743Z
@HeloLobo: I agree with Sasho Nikolov. (As a side note, you might want to call you notion something other than "order type" as that term already has a specific meaning: http://en.wikipedia.org/wiki/Order_type).
Joshua Grochow
CC BY-SA 3.0
2014-04-08T16:37:33.320Z
Regarding the conjecture, Dana Scott seems to have asked a similar question: http://www.cs.nyu.edu/pipermail/fom/2003-August/007150.html
András Salamon
CC BY-SA 3.0
2014-04-11T17:59:23.240Z
@Andras: I do not think it is consistent. I would think the consistent usage would be that the order type of $A$ is the isomorphism class of the total ordering $(A, \leq)$, considered up to isomorphism of total orders (note that this notion makes sense for arbitrary total orderings, not just well-orderings). Instead, HeloLobo is considering the order type of the "largest" *well-subordering* of $(A, \leq)$.
Joshua Grochow
<algebra><order-theory>
Spectrum of a variety: a possible connection btw ordinals and structures?
NisaiVloot
# Spectrum of a variety: a possible connection btw ordinals and structures?

(Asked by: username_0 on 2014-04-07)

Consider a variety of algebras $\mathbb{V} = \mathbb{V}(\sigma,\tau)$ which consists of the set of algebras defined over a fixed signature $\sigma$ and satisfying a set of identities $\tau$. We may then define its *ordered* version $\mathbb{V}\_{or}(\sigma,\tau)$ cas the set of algebras admitting a total order $\leq$ compatible with $\sigma$. By this I mean that:

(\*) for each $f : X^n \rightarrow X$ in $\sigma$, for each tuples $t,t' \in X^n$, if $t\_i \leq t'\_i$ for every $i \in [n]$ then $f(t) \leq f(t')$.

Given such an ordered algebra $A$, we may define its order type $ot(A)$ as the maximum order type of a well-founded order $\leq' \subseteq \leq$ without accumulation points. We may then define the *spectrum* of the variety $\mathbb{V}$ as the set of ordinals obtainable in that way.

Is this notion known in universal algebra, and if so, which varieties have a known spectra?

An easy case seems to be the variety of ordered fields: for instance, the field of reals $\mathbb{R}$ would have order type $\omega$, and the field of rational fractions $\mathbb{R}(x)$ would have order type $\omega^{\omega}$. So is the following true:

Conjecture: the spectrum of the ordered fields consists of the ordinals closed under addition and multiplication?

(NOTE: I added the requirement that $\leq'$ has no accumulation points, as otherwise any countable ordinal would be embeddable in $\mathbb{R}$. The precise meaning is relative to $\leq$: if $S$ is the set of elements of $\leq'$, we ask that for every $x \in S$, there is an interval $I$ of $\leq$ properly containing $x$ s.t. $I \cap S = \{x\}$.)

**Tags:** algebra, order-theory

> **Comments:**
> **username_1**: Your questions may have a higher chance of being answered at Mathoverflow, where they would be seen by a broader spectrum of mathematicians.  
> **username_0**: Thanks for the advice, but I guess that when you've lost yourself, it doesn't really matter where you end...  
> **username_2**: @HeloLobo: I agree with username_1. (As a side note, you might want to call you notion something other than "order type" as that term already has a specific meaning: http://en.wikipedia.org/wiki/Order_type).  
> **username_3**: Regarding the conjecture, Dana Scott seems to have asked a similar question: http://www.cs.nyu.edu/pipermail/fom/2003-August/007150.html  
> **username_2**: @Andras: I do not think it is consistent. I would think the consistent usage would be that the order type of $A$ is the isomorphism class of the total ordering $(A, \leq)$, considered up to isomorphism of total orders (note that this notion makes sense for arbitrary total orderings, not just well-orderings). Instead, HeloLobo is considering the order type of the "largest" *well-subordering* of $(A, \leq)$.  

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## 0 answers

