# Why are Lie algebras “rigid” objects?

I read the following motivation for quantum groups on wikipedia:

The discovery of quantum groups was quite unexpected, since it was known for a long time that compact groups and semisimple Lie algebras are "rigid" objects, in other words, they cannot be "deformed".

Why is that? Could one point me towards the concerned theorems?

• Apparently someone already asked a similar question math.stackexchange.com/questions/876663/…, however it got an answer. – Anne O'Nyme Apr 4 '15 at 15:20
• I can't be sure exactly what the author meant, but given a semisimple Lie algebra with a root system, roots are only allowed to be at very particular angles to one another. See here. To me, this seems to be the "rigidity" mentioned in your quoted passage. – André 3000 Apr 4 '15 at 15:37

## 1 Answer

Here's an algebraic approach to deformations:

Definition. Let $({\mathfrak g},[-,-])$ be a finite-dimensional Lie algebra over a field ${\mathbb k}$. An infinitesimal deformation (of order $2$) of ${\mathfrak g}$ is a ${\mathbb k}[t]/(t^2)$-Lie algebra structure on ${\mathfrak g}\otimes_{\mathbb k} {\mathbb k}[t]/(t^2)$ restricting to ${\mathfrak g}$ upon annihilating $t$.

In other words, we seek for Lie brackets $\widetilde{[-,-]}$ on ${\mathfrak g}\otimes_{\mathbb k} {\mathbb k}[t]/(t^2)$ which have the form $$(\ddagger)\qquad\widetilde{[X,Y]_\psi}\ \ =\ \ [X,Y] + t\psi(X,Y)\quad\text{ for some }\quad \psi: {\mathfrak g}\otimes_{\mathbb k}{\mathfrak g}\to{\mathfrak g};$$ however, not any such $\psi$ will yield a Lie algebra structure, since the validity of antisymmetry and the Jacobi identity will put restrictions on it - see below.

Note that there is always the constant deformation given by the ${\mathbb k}[t]/(t^2)$-linear extension of $$\widetilde{[X,Y]}_\text{triv}\quad:=\quad [X,Y]\quad\text{ for }\quad X,Y\in{\mathfrak g}.$$

Definition. A morphism of infinitesimal deformations $$({\mathfrak g}\otimes_{\mathbb k}{\mathbb k}[t]/(t^2),\widetilde{[-,-]})\to ({\mathfrak g}\otimes_{\mathbb k}{\mathbb k}[t]/(t^2),\widehat{[-,-]})$$ is a homomorphism of ${\mathbb k}[t]/(t^2)$-Lie algebras restricting to the identity on ${\mathfrak g}$ upon annihilating $t$.

In particular, call an infinitesimal deformation trivial if it is isomorphic to the constant deformation.

Theorem: Any infinitesimal deformation of a semisimple Lie algebra is trivial.

This follows from Lie algebra cohomology once one has made the above notions more explicit.

1. A bilinear map $\psi: {\mathfrak g}\otimes_{\mathbb k}{\mathfrak g}\to{\mathfrak g}$ gives rise to a Lie algebra structure on ${\mathfrak g}\otimes_{\mathbb k}{\mathbb k}[t]/(t^2)$ via $(\ddagger)$ if and only if it is antisymmetric and a $2$-cocycle in the Cartan-Eilenberg complex computing the Lie algebra cohomology $\text{H}^{\ast}({\mathfrak g};({\mathfrak g},\text{ad}))$ of the adjoint representation.

2. An isomorphism of deformations between the constant deformation and $({\mathfrak g}\otimes_{\mathbb k}{\mathbb k}[t]/(t^2),\widetilde{[-,-]}_\psi)$ has the form $X\mapsto X + t\varphi(X)$ for some $\varphi: {\mathfrak g}\to{\mathfrak g}$, and this morphism being compatible with the bracket turns out to be equivalent to $\psi=\text{d}\varphi$ in the Chevalley-Eilenberg complex.

Hence, if $\text{H}^2({\mathfrak g};({\mathfrak g},\text{ad}))=0$, any infinitesimal deformation is trivial. For semisimple Lie algebras, this is the case by the following

Theorem (Whitehead). If ${\mathfrak g}$ is semisimple, then $\text{H}^2({\mathfrak g};V)=0$ for all f.d. representations $V$ of ${\mathfrak g}$.

• Great answer. I always wondered precisely how these deformations were defined. Is it obvious that the quantum enveloping algebras correspond to such an infinitesimal deformation? – Tobias Kildetoft Apr 4 '15 at 17:07
• @Hanno Very nice, though with my background the final argument related to Cartan Eilenberg complex and so are somewhat obscure... Btw how is the bracket $\widehat{[-,-]}$ defined? And referring to the remark above, I just though that one deformed the UEA instead of the algebra itself because that yielded non-trivial deformations? – Anne O'Nyme Apr 6 '15 at 8:24
• @AnneO'Nyme: Given some $\psi: {\mathfrak g}\otimes{\mathfrak g}\to{\mathfrak g}$, you can define a bracket from it via $(\ddagger)$, with its antisymmetry being equivalent to the antisymmetry of $\psi$ and with the Jacobi identity being equivalent to $\psi$ being a $2$-cocycle in the CE-complex. The brackets $\widehat{[-,-]}$ and $\widetilde{[-,-]}$ are just two so obtained brackets. – Hanno Apr 6 '15 at 8:26
• @AnneO'Nyme: Concerning last question, I agree: the quantum group is a deformation of the UEA as an algebra (not as a Hopf algebra), and in this context, nontrivial deformations do exist. – Hanno Apr 6 '15 at 8:26
• @TobiasKildetoft: Thank you! Yes I think so - the quantum group should correspond to some Hochschild cohomology class of the ordinary UEA, although I haven't checked that. – Hanno Apr 6 '15 at 8:27