4
$\begingroup$

EDIT: I rephrased the claim for clarity.

Let $k$ be a field (that we may assume to be algebraically closed, but I don't think it is necessary). Let $n\geq 1$ and $T$ denote the subgroup of $GL_n$ consisting of invertible upper triangular matrices.

Let $1\leq r \leq n$ and consider a sequence $a_1,\ldots,a_r$ of positive numbers such that $a_1+\ldots+a_r=n$. Consider the subgroup $P_{(a_1,\ldots,a_r)}$ of $GL_n$ consisting exactly of all matrices $M$ of the form $$M = \begin{bmatrix} M_1 & * & \dots & * \\ 0 & M_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & * \\ 0 & \dots & 0 & M_r \end{bmatrix}$$

with $M_i\in GL_{a_i}(k)$ for all $i$, and the $*$'s being any elements of $k$ (or rather, any matrices with coefficients in $k$ and of appropriate dimensions). In other words, $P_{(a_1,\ldots,a_r)}$ consists of all invertible upper-triangular by blocks matrices with diagonal blocks being squares of dimensions $a_1,\ldots,a_r$. The claim I am considering is the following:

Any subgroup $P$ of $GL_n$ containing $T$ must have the form $P=P_{(a_1,\ldots,a_r)}$ for some $r$ and $a_1,\ldots,a_r$.

I suspect that this result may be true, however I can't find a way to prove it. In particular, given a group $P$ containing $T$, I have trouble seeing how I could characterize $r$ and the (ordered !) sequence $a_1,\ldots,a_r$ solely in terms of $P$.

The motivation behind this lies in the theory of algebraic groups. We know that $T$ is a connected closed solvable subgroup of $GL_n$. With the above result, I could deduce that $T$ is maximal with respect to such properties, because all subgroups described above with $r<n$ are unsolvable.

$\endgroup$
9
  • $\begingroup$ Can you explain why you think it's true? $\endgroup$ – yona tan Dec 25 '19 at 7:06
  • $\begingroup$ @yonatan Well, that is for a very pragmatic reason actually. This result seems to be claimed as true in the following paper www-math.mit.edu/~dav/glnpar.pdf . It is stated around the end of page 2, start of page 3, in the case of $k=\mathbb C$. At the moment I have no other incentive for it to be true though. $\endgroup$ – Suzet Dec 25 '19 at 7:11
  • $\begingroup$ $T$ itself is a subgroup containing $T$, so your claim can't be true. $\endgroup$ – Bungo Dec 25 '19 at 8:03
  • 1
    $\begingroup$ Ah, I see, that sounds more plausible. (Not sure if it's true, will think about it..) $\endgroup$ – Bungo Dec 25 '19 at 8:11
  • 2
    $\begingroup$ See Theorem 16.5 in the following paper Parabolic Subgroups $\endgroup$ – Nourddine Snanou Feb 9 '20 at 23:32
1
$\begingroup$

What you are heading for is classification of parabolic subgroups which can be done (and your statement is true) but requires deep theory of reductive algebraic groups (see, e.g. Borel's Linear algebraic groups or Springer's Linear algebraic groups or Milne's Algebraic groups or Algebraic groups and arithmetic groups).

However, the result that the group $B$ of upper triangular matrices is the Borel subgroup of $GL_n$ (that is, a maximal Zariski closed connected solvable subgroup) can be proven in a much easier way (and indeed it one of the very first steps towards classifying parabolic subgroups). The key fact to prove here is the Lie–Kolchin theorem which says that every connected solvable subgroup $H$ of $GL_n$ is upper-triagularisable: there is a basis such that all matrices from $H$ have upper-triangular shape w.r.t this basis. The proof originates in the work of Sophus Lie (and indeed, he proved it for Lie algebras), and was later extended to algebraic groups by Ellis Kolchin. A nice exposition can be found here (Theorem 5.2.5).

$\endgroup$
1
  • $\begingroup$ Thank you a lot for this detailled answer. It's a great motivation for me to pursue my reading of Springer's book, as there is now a concrete problem I would like to solve. By the way, proving that $T$ is a Borel subgroup of $GL_n$ actually is given as an exercise in this book before Lie-Kolchin theorem is stated. That is why I thought there may be another down-to-earth way to prove it, but I was unsuccessful. Anyways, thanks a lot for the clarification! It helped a lot =) $\endgroup$ – Suzet Dec 25 '19 at 23:20
1
$\begingroup$

I'd like to show an incomplete elementary approach. It's enough to find $a_1$ such that $M_1\in GL_{a_1}(k)$ and all entries below $M_1$ are zero. We choose $a_1$ such that for alle $M = (m_{ij})\in P$ we have $m_{i1}=0 \forall i>a_1$ and $m_{a_1 1}\neq 0$ for some $M\in P$. Because of $T\subset P$ we can freely add rows/columns from left to right and bottom to top.

With this we can freely choose values for $m_{11},\ldots,m_{a_1}1$ with $m_{a_1 1}\neq 0$. This can be used to show that all entries below $M_1$ are zero. (Assume an entry isn't and the multiply the matrix with well chosen $m_{11},\ldots,m_{a_1 1}$ to get a result where an $m_{k1}\neq 0,k> a_1$.)

That's the essential step to proof $P\subset P_{(a_1,\ldots,a_r)}$ and $P_{(a_1,\ldots,a_r)}$ is minimal with respect to inclusion.

For $P_{(a_1,\ldots,a_r)}\subset P$ I have only an incomplete idea. Basically with adding rows/columns we can show that $M_1$ can be chosen as a permutation matrix $S$ which maps $1$ to $n$. If we could show that the group generated by $S$ and $T_{a_1}$ is alread $GL_{a_1}(k)$ it would proof the other inclusion. That's currently missing.

$\endgroup$

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.