# Extending homomorphisms between ordered abelian groups

Let $\Omega$ be a linearly (i.e. fully) ordered set, and let $\Lambda_{\Omega}$ be the ordered abelian group consisting of those $(\lambda_\omega)_{\omega\in\Omega}\in\mathbb{R}^{\Omega}$ with well-ordered support. Note that $\mathbb{R}^{\Omega}$ itself has incomparable elements, but restricting to elements of $\Lambda_{\Omega}$ ensures that all pairs of elements are comparable: $(\lambda_\omega)_{\omega\in\Omega}<(\lambda'_\omega)_{\omega\in\Omega}$ if $\lambda_{\omega_0}<\lambda'_{\omega_0}$ where $\omega_0$ is the least element of $\Omega$ for which $\lambda_{\omega_0}\neq\lambda'_{\omega_0}$. A group of the form $\Lambda_{\Omega}$ is called a Hahn group.

For $\lambda,\lambda'\in\Lambda_{\Omega}$ with $\lambda'>0$, write $\lambda\ll\lambda'$ if $\lambda=0$ or the first non-zero entry of $\lambda'$ precedes that of $\lambda$. (This means that $\lambda$ is infinitely smaller than $\lambda'$: that is, $n\lambda<\lambda'$ for all $n\in\mathbb{Z}$.)

Let $\Lambda_0$ be a subgroup of $\Lambda_{\Omega}$ and $\zeta:\Lambda_0\to\Lambda_{\Omega}$ a (not necessarily order-preserving) homomorphism for which $\zeta(\lambda)\ll\lambda$ for all positive $\lambda\in\Lambda_0$. Can $\zeta$ be extended to a homomorphism $\Lambda_{\Omega}\to\Lambda_{\Omega}$ that also satisfies $\zeta(\lambda)\ll\lambda$ for all positive $\lambda\in\Lambda_{\Omega}$?

I don't see why such an extension should exist in such generality.

For a heuristic explanation, let's consider a related question. Let $\Omega$ be a finitely generated (abelian) group generated by a set $S$. Let $\Lambda^\prime_\Omega$ be the ordered preserving group consisting of those order preserving group homomorphisms $(\lambda_\omega)_{\omega\in\Omega}$ with well ordered support (this is a subgroup of $\Lambda_\Omega$). I claim that a homomorphism $\zeta:\Lambda_S\to\Lambda_\Omega$ such that $\zeta(\lambda)\ll\lambda$ for positive $\lambda$, and such that if $\lambda\ll\lambda^\prime$, then $\zeta(\lambda)\ll\zeta(\lambda^\prime)$, cannot necessarily be extended to a homomorphism $\Lambda_\Omega\to\Lambda_\Omega$ that satisfies $\zeta(\lambda)\ll\lambda$ for those $\lambda$. Note that $\Lambda_S$ is a subgroup of $\Lambda_\Omega^\prime$, which in turn is a subgroup of $\Lambda_\Omega$. I'm therefore unable to see a reason for why this should be true for a general subgroup of a Hahn group.

To see this, choose any element $(\lambda_\omega)_{\omega\in\Omega}\in\Lambda^\prime_\Omega$. Since $S$ generates $\Omega$, we can write $\omega = \sum_i n_is_i$ for some finite $\{s_i\}\in S$, and since every $\lambda$ is a group homomorphism, we find that $(\lambda_\omega)_{\omega\in\Omega} = \sum_in_i(\lambda_{s_i})_{s_i\in S}$ where $(\lambda_{\omega_0})_{\omega_0\in\Omega}+(\lambda_{\omega_1})_{\omega_1\in\Omega} = (\lambda_{\omega_0+\omega_1})_{\omega_0,\omega_1\in\Omega}$ i.e. the expected definition. Now, let $\overline{T}_\lambda$ consist of those $s_i$ for which $\lambda_i<0$ (note the strict inequality), and let $\overline{(\lambda_{s_i})_{s_i\in\overline{T}_\lambda}}$ consist of $\lambda$ restricted to $\overline{T}_\lambda$ and then extended to $S$ by defining it to be zero everywhere else. Then $(\lambda_\omega)_{\omega\in\Omega} - \sum_in_i\overline{(\lambda_{s_i})_{s_i\in\overline{T}_\lambda}}$ is positive. Because $\zeta$ is a group homomorphism, we find that $\zeta(\lambda_\omega)_{\omega\in\Omega} + \sum_in_i\zeta(-\overline{(\lambda_{s_i})_{s_i\in\overline{T}_\lambda}}) \ll (\lambda_\omega)_{\omega\in\Omega} + \sum_in_i(-\overline{(\lambda_{s_i})_{s_i\in\overline{T}_\lambda}})$. If the first nonzero element of $(\lambda_\omega)_{\omega\in\Omega}$ is positive, so that $(\lambda_\omega)_{\omega\in\Omega}$ is positive, then $\sum_in_i(-\overline{(\lambda_{s_i})_{s_i\in\overline{T}_\lambda}})\ll (\lambda_\omega)_{\omega\in\Omega}$. This means that $\sum_in_i\zeta(-\overline{(\lambda_{s_i})_{s_i\in\overline{T}_\lambda}})\ll \zeta(\lambda_\omega)_{\omega\in\Omega}$, so it is not necessarily true, even in this ideal scenario, that $\zeta(\lambda_\omega)_{\omega\in\Omega}\ll(\lambda_\omega)_{\omega\in\Omega}$.

• Many thanks for your answer. There are some things I don't follow in it though, possibly because I'm missing something obvious. For one thing, 'we find that $(\lambda_\omega)_{\omega\in\Omega} = \sum_in_i(\lambda_{s_i})_{s_i\in S}$' -- should the left-hand side here be just $\lambda_\omega$, that is, the image of $\omega$ under the homomorphism $\lambda\in\Lambda'_{\Omega}$? More fundamentally, you claim that a homomorphism ( + extra properties) with domain $\Lambda_S$ can't necessarily be extended to a homomorphism with domain $\Lambda_{\Omega}$... Aug 15, 2016 at 22:18
• ...what's wrong with just putting the homomorphism equal to zero on elements supported in $\Omega\backslash S$? The only structure on $\Omega$ that is required for my purposes is the ordered one, not any additive structure that it may also be endowed with. Aug 15, 2016 at 22:19
• @shane.orourke Since $\omega = \sum n_i s_i$ and $\lambda$ is a homomorphism, for every $\omega\in\Omega$, $\lambda_\omega = n_i\lambda_{s_i}$, so that $\lambda$ itself is specified by $(\lambda_\omega)_{\omega\in\Omega} = \sum_i n_i(\lambda_{s_i})_{s_i\in S}$. As for the second question (which I have hopefully understood correctly), if $\zeta$ could be extended by just letting it be zero on $\Lambda_\Omega$ except on $\Lambda_S$, then $\zeta(\lambda)$ isn't even positive for positive $\lambda$.
– user97187
Aug 15, 2016 at 23:36
• But $\zeta(\lambda)$ isn't required to be positive for positive $\lambda$; if it were, then $\zeta$ would be an order-preserving homomorphism. Aug 15, 2016 at 23:52
• @shane.orourke In the question, you defined what $\lambda\ll\lambda^\prime$ means for positive $\lambda,\lambda^\prime$. Of course, this can be defined for nonnegative $\lambda,\lambda^\prime$, but I thought the symbol $\ll$ was only defined for positive maps $\Omega\to\mathbf{R}$. (Also, if $\zeta(\lambda)$ is positive for positive $\lambda$, why should it be order-preserving?)
– user97187
Aug 16, 2016 at 0:21