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Incorporated incomplete answer into the question
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shane.orourke
shane.orourke
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EDIT: I posted the following observations on 16 April 2012 as an answer. I think I should really have made them a footnote to the question rather than an answer.

For what it's worth, I suspect the answer is no. This is essentially because admissibility is preserved by conjugation by upper triangular matrices, and non-identity diagonal matrices are not admissible. So suppose that $\iota$ is an embedding of the required sort. Then $\iota$ maps the subgroup $\mathrm{UT}(n,\mathbb{R})$ into $\mathrm{UT}(m,\mathbb{R})$, since these are the respective derived subgroups of the domain and codomain. Let $D^*$ denote the subgroup of $T^*(n,\mathbb{R})$ consisting of diagonal matrices. Since $\langle A,\mathrm{UT}(n,\mathbb{R})\rangle$ is soluble but not nilpotent for non-scalar $A\in D^*$, and $\iota$ is injective, we must have $\iota(A)\notin\mathrm{UT}(m,\mathbb{R})$ for non-trivial $A\in D^*$.

Now, on the assumption that there exists $M\in T^*(m,\mathbb{R})$ such that $\bar{A}=M^{-1}\iota(A)M$ is diagonal for all diagonal $A$ then on the one hand $\bar{A}$ is admissible since $\iota(A)$ is, by assumption, while on the other hand non-identity diagonal matrices are not admissible.

So does the $M$ in this argument necessarily exist? If the image of each diagonal matrix under $\iota$ is diagonalisable, then since the diagonal matrices commute, they are indeed simultaneously diagonalisable via a similarity matrix $M\in\mathrm{GL}(m,\mathbb{R})$. But I don't know whether the image of a diagonal matrix under $\iota$ is necessarily diagonalisable, nor do I see why $M$ can necessarily be chosen to be upper triangular...

EDIT: I posted the following observations on 16 April 2012 as an answer. I think I should really have made them a footnote to the question rather than an answer.

For what it's worth, I suspect the answer is no. This is essentially because admissibility is preserved by conjugation by upper triangular matrices, and non-identity diagonal matrices are not admissible. So suppose that $\iota$ is an embedding of the required sort. Then $\iota$ maps the subgroup $\mathrm{UT}(n,\mathbb{R})$ into $\mathrm{UT}(m,\mathbb{R})$, since these are the respective derived subgroups of the domain and codomain. Let $D^*$ denote the subgroup of $T^*(n,\mathbb{R})$ consisting of diagonal matrices. Since $\langle A,\mathrm{UT}(n,\mathbb{R})\rangle$ is soluble but not nilpotent for non-scalar $A\in D^*$, and $\iota$ is injective, we must have $\iota(A)\notin\mathrm{UT}(m,\mathbb{R})$ for non-trivial $A\in D^*$.

Now, on the assumption that there exists $M\in T^*(m,\mathbb{R})$ such that $\bar{A}=M^{-1}\iota(A)M$ is diagonal for all diagonal $A$ then on the one hand $\bar{A}$ is admissible since $\iota(A)$ is, by assumption, while on the other hand non-identity diagonal matrices are not admissible.

So does the $M$ in this argument necessarily exist? If the image of each diagonal matrix under $\iota$ is diagonalisable, then since the diagonal matrices commute, they are indeed simultaneously diagonalisable via a similarity matrix $M\in\mathrm{GL}(m,\mathbb{R})$. But I don't know whether the image of a diagonal matrix under $\iota$ is necessarily diagonalisable, nor do I see why $M$ can necessarily be chosen to be upper triangular...

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shane.orourke
shane.orourke
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Free affine actions of Borel subgroups

Call an upper triangular $m\times m$ matrix $A$ admissible if the lowest non-zero entry of $A-I$ lies in the last column, and is strictly lower than any other non-zero entry of $A-I$. I'll also regard $A=I$ itself as admissible. Let ${T^{*}}=T^{*}(n,\mathbb{R})$ denote the group of upper triangular matrices with real entries and positive entries on the main diagonal, and with the last diagonal entry equal to 1. This group has a natural affine action on $\mathbb{R}^{n-1}$ where the last column (apart from the bottom entry) of an element of $T^*$ acts by translation after the top left-hand $(n-1)\times (n-1)$ part has acted linearly.

(If $T^*$ seems artificial or awkward to work with, note that it embeds in the group $T(n+1,\mathbb{R})$ of upper triangular $(n+1)\times(n+1)$ matrices with positive diagonal entries (that is, the identity component of the Borel subgroup of $\mathrm{GL}(n+1,\mathbb{R})$), and that $T(n+1,\mathbb{R})$ embeds in $T^*=T^*(n+2,\mathbb{R})$, so $T^*$ may be replaced by $T$ in much of what follows.)

John Milnor in a well-known paper of 1977 proves that a soluble, connected and simply connected Lie group $G$ admits a free affine action on $\mathbb{R}^n$ for some $n$. By Lie's theorem, such a $G$ can be embedded in some $T^*$ as above, and the given action of $G$ can be obtained by restricting the natural action of $T^*$. What I'd like to know is whether we can arrange for elements of the embedded image to be admissible.

Does every $T^*(n,\mathbb{R})$ admit an (abstract group theoretic) embedding $\iota$ in $T^*(m,\mathbb{R})$ for some $m$ such that each $\iota(A)$ is admissible?

Equivalently, does the identity component of the Borel subgroup of each $\mathrm{GL}(n,\mathbb{R})$ admit an embedding $\iota$ in $T^*(m,\mathbb{R})$ for some $m$ such that each $\iota(A)$ is admissible?

I'm interested in these questions since groups that admit such embeddings admit free affine actions on $\Lambda$-trees. (See http://de.arxiv.org/PS_cache/arxiv/pdf/1112/1112.4832v2.pdf.) I should point out that I asked a ''nilpotent'' version of this question (involving $\mathrm{UT}(n,\mathbb{R})$ instead of $T^*(n,\mathbb{R})$) on the group pub forum some time ago, and received a very helpful reply from Karel Dekimpe: using his reply I can show that the answer is yes in this case. This approach appears to make essential use the theory of nilpotent Lie algebras, hence my question here.