MathOverflow is a question and answer site for professional mathematicians. 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

Hi I just started working on degeneration and contractions, I would like to know: why no Lie algebra degenerate to a rigid algebra?(rigid algebra:an algebra whose orbit is zariski open) Why the closure of a rigid Lie algebra forms the irreducible component of variety of Lie algebras? Thank you

share|cite|improve this question
The first question follow immediately from the definitions involved. The second is not quite correctly stated (you mean the orbit of a rigid Lie algebra) and is pretty similarly easy. You should probably read the FAQ and/or ask on obe of the sites the FAQ suggests, – Mariano Suárez-Alvarez Dec 7 '12 at 3:01
up vote 1 down vote accepted

For the second question suppose that $L$ is a geometrically rigid Lie algebra, i.e., the orbit $O(L)$ is open. The algebra $L$, and hence $O(L)$ is contained in some irreducible component $\mathcal{C}$. Since a nonemty open subset in an irreducible space is dense, the closure of $O(L)$ equals $\mathcal{C}$. The first question follows similarly.

It might also be helpful for understanding to use a more intuitive definition of rigidity. Call $L$ formally rigid, if $L$ does not admit any non-trivial formal deformation. Over fields of characteristic zero, geometrical rigidity is equivalent to formal rigidity (otherwise this is not true). Assume that we have two Lie algebras $L_0$ and $L$ of the same dimension over a field of characteristic zero. Suppose that $L_0$ degenerates properly to $L$, i.e., $L$ lies in the boundary of the orbit $O(L_0)$. Then this degeneration directly gives a non-trivial deformation of $L$, so $L$ cannot be formally rigid, and hence not geometrically rigid.

share|cite|improve this answer
Thank you for your reply – user29742 Mar 23 '13 at 9:42

Your Answer


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.