10
$\begingroup$

Consider the group $T_p(n)$ of all non-singular upper triangular matrices with entries in $\mathbb{F}_p.$ Its commutator subgroup is $U_p(n)$ (all elements in $T_p(n)$ with $1$s on the main diagonal). The question is: is every element in $U_p$ a commutator of matrices in $T_p?$

$\endgroup$
10
  • 4
    $\begingroup$ Need $p>2$, to start. $\endgroup$ Apr 8 '16 at 13:03
  • 3
    $\begingroup$ I checked for $n=3$ with $p \le 13$, $n=4$ with $p \le 7$ and $n=5$, $p=3$, and the answer was yes, but that was using completely naive checks. $\endgroup$
    – Derek Holt
    Apr 8 '16 at 14:59
  • 4
    $\begingroup$ A theorem of Bier and Holubowski ("A note on commutators in the group of infinite triangular matrices over a ring", Linear and Multilinear Algebra 63 (2015) no. 11, 2301-2310) gives that the answer is "yes" for $n=2$ and $p$ odd; and for $n>2$, every element is a product of at most two commutators of matrices in $T_p(n)$. $\endgroup$ Apr 8 '16 at 17:16
  • 5
    $\begingroup$ This is easy if $p > n$. Choose $a$ to be a diagonal matrix whose diagonal entries are all distinct. Then, for any elementary matrix $I + e_{i,j}$, we have $[a, e_{i,j}] = I + \lambda_{i,j} e_{i,j}$ for some nonzero $\lambda_{i,j}$. Using this, we can obtain any desired element of $U_p$ as $[a, u]$, for some $u \in U_p$, by working upward from the main diagonal (or, in other words, by working down the central series of $U_p$). Later modifications to $u$ do not change the previously determined entries of $[a, u]$. $\endgroup$ Apr 8 '16 at 22:04
  • 2
    $\begingroup$ @DaveWitteMorris: This can also be seen as follows: The centralizer of such an $a$ in the matrix ring is all diagonal matrices and thus the centralizer in $U_p(n)$ is $1$, which implies $U_p = [a, U_p]$. Similarly, when $u \in U_p$ has minimal polynomial $(x-1)^n$, then its centralizer in the matrix ring consists of polynomials in $u$, and the centralizer of $u$ in $U_p$ has order $p^{n-1}$. Thus $|[u,U_p]| = |U_p|/ p^{n-1} = |U_p'|$, and thus $[u,U_p] = U_p'$. $\endgroup$ Apr 9 '16 at 20:21
8
$\begingroup$

Too long for a comment and only a partial answer for $n>2$. A note on commutators in the group of infinite triangular matrices over a ring, by Agnieszka Bier and Waldemar Holubowski, Linear and Multilinear Algebra 63 (2015) no. 11, 2301-2310; seems relevant. It includes results on finite matrices, despite the title.

Let $R$ be an associative ring with $1$, and let ${U}(R)$ be its group of invertible elements. $T(n,R)$ denotes the group of upper triangular $n\times n$ matrices; $UT(n,R)$ the set of upper unitriangular matrices ($1$s in the diagonal), $UT(n,m,R)$ the subgroup containing exactly all those matrices which have zero entries on the first $m$ super diagonals; and the basic commutators $c_k(x_1,\ldots,x_k)$ are defined inductively by $c_1(x_1)=x_1$, $c_{i+1}(x_1,\ldots,x_{i+1}) = [c_i(x_1,\ldots,x_i),x_{i+1}]$; let $\gamma_k(G)=G$ be the $k$th term of the lower central series of $G$. Finally, the Engel words $e_k(x,y)$ are given by $e_2(x,y)=[x,y]$, $e_{m+1}(x,y) = [e_m(x,y),y]$.

The authors prove:

Theorem 1.5 Let $R$ be an associative ring with $1$ such that $U(R)$ is commutative. Then:

  1. $\gamma_k(UT(n,R))=UT(n,k,R)$ and every element of $\gamma_k(UT(n,R))$ is a value of the basic commutator $c_k$.
  2. Every element of $\gamma_k(UT(n,R))$ is a value of the Engel word $e_k$.

Moreover, if $1$ is a sum of two invertible elements, then

  1. $[T(n,R),T(n,R)] = UT(n,R)$ whenever $n\geq 2$.
  2. Every element of $\gamma_2(T(2,R)) = UT(2,R)$ is a commutator.
  3. Every element of $\gamma_2(T(n,R)) = UT(n,R)$ with $n>2$ is a product of at most two commutators.

In particular, for $R=\mathbb{F}_p$ with $p>2$, then the answer is "yes" for $n=2$ by point 4; but for $n>2$ they only show every element is a product of at most two commutators.

Perhaps this was already known for the special case of matrices over commutative rings/fields?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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