Let $T^\*$ denote upper triangular matrices (of the appropriate size) with positive diagonal entries and $\mathrm{UT}$ upper triangular matrices with all diagonal entries equal to 1.



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> Does every (abstract group) embedding $\varphi:\mathrm{UT}(n,\mathbb{R})\to\mathrm{UT}(m,\mathbb{R})$ extend to $\bar{\varphi}:T^\*(n,\mathbb{R})\to T^\*(m,\mathbb{R})$?
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(In the other direction, any $\bar{\varphi}:T^\*(n,\mathbb{R})\to T^\*(m,\mathbb{R})$ restricts to a homomorphism $\mathrm{UT}(n,\mathbb{R})\to\mathrm{UT}(m,\mathbb{R})$, since $\mathrm{UT}$ is the derived subgroup of $T^\*$.)


An affirmative answer to this question implies a relatively easy affirmative answer to [this question](https://mathoverflow.net/questions/93091/free-affine-actions-of-borel-subgroups). I've recently answered the latter independently, but wonder if there's a general result that could be used, and which I should know about. I asked the current question on Math.stackexchange but without success.


EDIT: Florian Eisele has shown that as stated above the answer to the original question is no. This makes me wonder if there's a reasonably natural reformulation for which the answer is yes. For the sake of asking a concrete question, let me hazard the following.



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> Does every embedding $\varphi:\mathrm{UT}(n,\mathbb{Q})\to\mathrm{UT}(m,\mathbb{Q})$ extend to $\bar{\varphi}:T^\*(n,\mathbb{Q})\to T^\*(m,\mathbb{Q})$?
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Does every embedding of one unipotent group (over R) in another extend to an embedding of the respective upper triangular matrix groups?
Here's a counterexample ($m=n=3$) which is a continuous homomorphism and actually probably also works for your second question with $\mathbf{Q}$.


In short: most automorphisms of $UT(3)$ do not extend to $T^\*(3)$.


Since I deal with continuous automorphisms, it boils down to a Lie algebra problem. An automorphism of $UT(3)$ induces an automorphism of its abelianized subgroup, giving a $2\times 2$ matrix and every invertible matrix occurs this way. But my claim is that if the automorphism extends to $T^\*(3)$ then (some power of) this matrix has to be diagonal.


Sketch of argument: I work upside down by considering an automorphism of $T^1(3)$, the group of determinant 1 matrices in $T^\*(3)$ (it's enough and more convenient to deal with it) and describe its restriction to $UT(3)$.


First, because $UT(3)$ is the derived subgroup, it is stable. Write $T^1(3)=D.UT(3)$ where $D$ is the group of diagonal matrices. Since $D$ is a Cartan subgroup in $T^1(3)$ (that is, its Lie algebra is nilpotent and self-normalized, see Bourbaki, Groups and Lie algebras) and is unique up to conjugacy, it is mapped to a conjugate subgroup. So after composing by a conjugation, we can suppose $D$ mapped into itself. Now any automorphism of $T^1(3)$ induces a finite order automorphism of $D$: indeed, any one-parameter group of automorphisms of $T^1(3)$ induces the identity on $D$, as we see using the fact that the derived subgroup of the semidirect product $\mathbf{R}\ltimes T^1(3)$ is nilpotent (hence contained in the nilpotent radical $UT(3)$ of $T^1(3)$). So after conjugation, some power of the automorphism acts on $D$ as the identity and therefore preserves the weight decomposition of the Lie algebra of $UT(3)$ induced by the action of $D$, and thus induces a diagonal automorphism of the derived subgroup of $UT(3)$. 


If you don't follow the argument you can also compute the group of continuous automorphisms of $T^\*(3)$ and check that no one maps a matrix $\pmatrix{1 & x & \* \cr 0 & 1 & y \cr 0 & 0 & 1\cr}$ to a matrix of the form $\pmatrix{1 & x+y & \* \cr 0 & 1 & y \cr 0 & 0 & 1\cr}$.


NB: it's unclear from your question if you mean "extends as an automorphism" or "extends as an endomorphism". However, it is not hard to show that an endomorphism of $T^1(3)$ that is injective on $UT(3)$ remains injective on $T^1(3)$, and it easily follows that an automorphism of $UT(3)$ extends as an endomorphism of $T^\*(3)$ iff it extends as an automorphism.


