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Nicola Ciccoli
Nicola Ciccoli
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Let me consider the finite-dimensional case (as YCOR was commenting it is basically impossible to say something in the infinite-dimensional one unless you clarify in a more detailed manner your definitions).

From a geometrical point of view a ``rigid Lie algebra'' is an open subset in the variety of all Lie algebras such that all Lie algebras in this set are isomorphic one to the other. A non rigid Lie algebra which can be deformed into a rigid one is simply a Lie algebra which sits in the closure of this open set. Can two different open sets share the same closure.? Sure. So it is possible that the same Lie algebra has different deformations giving rise to non isomorphic rigid Lie algebras.

The fact that the second cohomology of the Lie algebra with values in itself gives the tangent space to the variety of Lie algebras at a given point, quotiented by the tangent space of trivial deformations tells you that $\dim H^2(\mathfrak g,\mathfrak g)$ measures ``exactly'' how many nonisomorphic deformations a fixed Lie algebra may (infinitesimally) have: more than one of this can be rigid.

I wrote ``exactly'' in quotation marks because understanding which infinitesimal deformations are integrable, whether you look for rigid Lie deformations or infinitesimally rigid ones etc... is the kind of questions that make a big difference here and that are much context dependent.

Let me consider the finite-dimensional case (as YCOR was commenting it is basically impossible to say something in the infinite-dimensional one unless you clarify in a more detailed manner your definitions).

From a geometrical point of view a ``rigid Lie algebra'' is an open subset in the variety of all Lie algebras such that all Lie algebras in this set are isomorphic one to the other. A non rigid Lie algebra which can be deformed into a rigid one is simply a Lie algebra which sits in the closure of this open set. Can two different open sets share the same closure. Sure. So it is possible that the same Lie algebra has different deformations giving rise to non isomorphic rigid Lie algebras.

The fact that the second cohomology of the Lie algebra with values in itself gives the tangent space to the variety of Lie algebras at a given point, quotiented by the tangent space of trivial deformations tells you that $\dim H^2(\mathfrak g,\mathfrak g)$ measures ``exactly'' how many nonisomorphic deformations a fixed Lie algebra may (infinitesimally) have: more than one of this can be rigid.

I wrote ``exactly'' in quotation marks because understanding which infinitesimal deformations are integrable, whether you look for rigid Lie deformations or infinitesimally rigid ones etc... is the kind of questions that make a big difference here and that are much context dependent.

Let me consider the finite-dimensional case (as YCOR was commenting it is basically impossible to say something in the infinite-dimensional one unless you clarify in a more detailed manner your definitions).

From a geometrical point of view a ``rigid Lie algebra'' is an open subset in the variety of all Lie algebras such that all Lie algebras in this set are isomorphic one to the other. A non rigid Lie algebra which can be deformed into a rigid one is simply a Lie algebra which sits in the closure of this open set. Can two different open sets share the same closure? Sure. So it is possible that the same Lie algebra has different deformations giving rise to non isomorphic rigid Lie algebras.

The fact that the second cohomology of the Lie algebra with values in itself gives the tangent space to the variety of Lie algebras at a given point, quotiented by the tangent space of trivial deformations tells you that $\dim H^2(\mathfrak g,\mathfrak g)$ measures ``exactly'' how many nonisomorphic deformations a fixed Lie algebra may (infinitesimally) have: more than one of this can be rigid.

I wrote ``exactly'' in quotation marks because understanding which infinitesimal deformations are integrable, whether you look for rigid Lie deformations or infinitesimally rigid ones etc... is the kind of questions that make a big difference here and that are much context dependent.

Source Link
Nicola Ciccoli
Nicola Ciccoli
  • 3.4k
  • 19
  • 24

Let me consider the finite-dimensional case (as YCOR was commenting it is basically impossible to say something in the infinite-dimensional one unless you clarify in a more detailed manner your definitions).

From a geometrical point of view a ``rigid Lie algebra'' is an open subset in the variety of all Lie algebras such that all Lie algebras in this set are isomorphic one to the other. A non rigid Lie algebra which can be deformed into a rigid one is simply a Lie algebra which sits in the closure of this open set. Can two different open sets share the same closure. Sure. So it is possible that the same Lie algebra has different deformations giving rise to non isomorphic rigid Lie algebras.

The fact that the second cohomology of the Lie algebra with values in itself gives the tangent space to the variety of Lie algebras at a given point, quotiented by the tangent space of trivial deformations tells you that $\dim H^2(\mathfrak g,\mathfrak g)$ measures ``exactly'' how many nonisomorphic deformations a fixed Lie algebra may (infinitesimally) have: more than one of this can be rigid.

I wrote ``exactly'' in quotation marks because understanding which infinitesimal deformations are integrable, whether you look for rigid Lie deformations or infinitesimally rigid ones etc... is the kind of questions that make a big difference here and that are much context dependent.