Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

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...

share|improve this question
add comment

1 Answer

Yes.

Start by embedding $\mathrm{UT}(n,\mathbb{R})$ in $\mathrm{UT}(k+1,\mathbb{R})$ via $u\mapsto\left(\begin{array}{c|c}\varphi_0(u) & b(u)\\ \hline 0 & 1\end{array}\right)$ so that each element of the image is admissible (this is possible using the last paragraph before the edit, and I will use properties of the particular embedding I have in mind shortly: for starters, $k=n(n-1)/2$).

Now $T^*(n,\mathbb{R})$ is isomorphic to the semidirect product $D^*(n,\mathbb{R})\ltimes\mathrm{UT}(n,\mathbb{R})$ where $D^*$ denotes diagonal matrices with positive diagonal entries. For diagonal $d$, map $\bar{\varphi}:d\mapsto\left(\begin{array}{l|l|l}\bar{d} & 0 & 0\\ \hline 0 & I_n& \log(d)\\ \hline 0 & 0 & 1\end{array}\right)$ where $\log(d)$ denotes the column vector consisting of $\log(d_i)$ and $d_i$ ranges through the respective diagonal entries of $d$, and $\bar{d}$ is a suitably chosen diagonal $k\times k$ matrix: if $\varphi$ arises from a suitable affine structure on $\mathrm{UT}(n,\mathbb{R})$ (that is, if the latter arises from the natural left symmetric structure on the upper zero triangular $n\times n$ matrices) we can take the diagonal entries of $\bar{d}$ to have the form $d_i/d_j$ ($1\leq i<j\leq n$).

Meanwhile, the map $\bar{\varphi}$ restricted to $\mathrm{UT}(n,\mathbb{R})$ simply maps $\bar{\varphi}:u\mapsto\left(\begin{array}{c|c|c}\varphi_0(u) & 0 & b(u)\\ \hline 0 & I_n& 0 \\ \hline 0 & 0 & 1\end{array}\right)$.

This gives rise to an embedding $\iota=\bar{\varphi}$ of $T^*(n,\mathbb{R})$ in $T^*(k+n+1,\mathbb{R})$: the main challenge is to show that $\bar{\varphi}(dud^{-1})=\bar{\varphi}(d)\bar{\varphi}(u)\bar{\varphi}(d^{-1})$. And while I won't present this here, suffice it to say that it can be done, at least with the $\varphi$ and $\bar{d}$ I have in mind. It's not hard to show that the image $\bar{\varphi}(T^*(n,\mathbb{R}))$ is admissible.

See, for example, the paper of Dekimpe and Malfait `Affine structures on a class of virtually nilpotent groups', Topology Appl., 73 no. 2 (1996) 97-119 for more details on affine structures and how they relate to left symmetric structures.

Note that the image of $\bar{\varphi}(d)$ is not diagonalisable (for non-identity diagonal $d$): this sidesteps the misgivings expressed in the edit above.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.