Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is rather "simple" to ask : what is the dimension of the quotient variety $GL_3/U$, where $U$ is the (closed) group of upper triangular unipotent matrices (= upper triangular matrices with 1's on the diagonal).

Before working on that quotient variety, I've worked on $SL_2/U$ and I could determine the dimension thanks to the isomorphism $SL_2/U \simeq \mathbb{A}^2 \setminus \{(0,0)\}$ (I did use the fact that an open set of an affine variety has the same dimension, is that true?).

Thank you for your help !

share|cite|improve this question
The big Bruhat cell $\overline{P} \times U$ is an open subvariety of $G$ (determined by the non-vanishing of the top left $1 \times 1$ and $2 \times 2$ minors). Since $\overline{P}$ is obviously 6-dimensional, aren't we done? – David Loeffler Jan 12 '12 at 14:59
Your statement about non-empty open subsets of algebraic varieties is correct. – Hagen Knaf Jan 12 '12 at 15:34
Thank you for the confirmation Hagen. Regarding what you said David, could you explain it a little bit more ? I'm not familiar with the term "big cell" for Bruhat cells, nor am I with the notation $\bar{P}$. – ng_th Jan 12 '12 at 16:27
Here $\overline{P}$ is just the natural "complement" of $U$, the subgroup of lower-triangular matrices. This statement goes by the name of "LU decomposition" in undergraduate linear algebra textbooks. – David Loeffler Jan 12 '12 at 16:36

Your Answer


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

Browse other questions tagged or ask your own question.