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I am studying deformation (as it is introduced in https://arxiv.org/pdf/math/0611793.pdf or http://web.cs.elte.hu/~fialowsk/pubs-af/condefnew2.pdf) and rigidity of some infinite dimensional Lie algebras which are defined on field with characteristic zero which its commutation relations are:

$$[J_m,J_n]=(m-n)J_{m+n},$$

$$[J_m,P_n]=(m-n)P_{m+n},$$

$$[P_m,P_n]=0.$$

In this connection, I infinitesimally deform the last commutator(the ideal of algebra) by adding some terms in its RHS and reach to an infinite dimensional rigid Lie algebra.The first commutator is a subalgebra which is known as Witt algebra and will remain rigid by deformation procedure. But I am not sure about the second commutator which can be deformed or not. This problem has connection with this question: Is the rigid Lie algebra I mentioned above "unique" and independent of how I deform the initial Lie algebra? In fact, if I started with other commutation relations would I reach to another rigid Lie algebra? Is there any theorem about "uniqueness" of rigid Lie algebra which is derived in deformation procedure?

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  • $\begingroup$ Could you be a little more specific about the setting, so as to clarify the meaning of "deformation"? Lie algebras over which field? deformation in a topological sense? (I imagined so but you also tagged "algebraic-geometry", so...) and topological deformation is sensitive to choices of topology in infinite dimension. $\endgroup$ – YCor Mar 12 '18 at 7:41
  • $\begingroup$ As I understand deformation of Lie algebra is related to its second adjoint cohomology, so if this cohomology is nontrivial Lie algebra would admit a deformation. Lie algebras I am working on are defined on fields with characteristic zero (for example real numbers). $\endgroup$ – Hamidreza Safari Mar 12 '18 at 7:53
  • $\begingroup$ I pointed the issue of infinite dimension. Intuition and analogies are one thing, but at some point you need to define objects (deformations, (co)homology...) and this will be sensitive to the setting. At this point all is very vague: I have no idea if you have in mind abstract Lie algebras over arbitrary fields, continuous brackets on Banach spaces, anything in relation with the tag "algebraic geometry"... $\endgroup$ – YCor Mar 12 '18 at 8:22
  • $\begingroup$ @YCor OP is a physicist, I believe. My guess is that he will be happy with any reasonable choices for topological issues. Analytic questions about Virasoro are myriad, exciting, important, and mostly ignored by physicist practitioners. $\endgroup$ – Theo Johnson-Freyd Mar 13 '18 at 4:56
  • $\begingroup$ Specifically, my guess is that for the purpose of the question, the Witt algebra could be any semisimple Lie algebra $\mathfrak{g}$, and OP's algebra is $\mathfrak{g} \ltimes \mathfrak{g}_{adj}$, where by $\mathfrak{g}_{adj}$ I mean the adjoint $\mathfrak{g}$-module (thought of as an abelian Lie algebra). $\endgroup$ – Theo Johnson-Freyd Mar 13 '18 at 4:58
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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.

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