# why no Lie algebra degenerate to a rigid algebra? Why the closure of a rigid algebra forms the irreducible component of variety of Lie algebras?

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

-
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

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.