How does the group algebra look as a Lie algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:30:04Z http://mathoverflow.net/feeds/question/60447 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60447/how-does-the-group-algebra-look-as-a-lie-algebra How does the group algebra look as a Lie algebra Klim Efremenko 2011-04-03T16:31:29Z 2011-05-07T17:04:55Z <p>It's probably a well known question, so it is just a reference question. Let $G$ be a finite group and let $C[G]$ be a group algebra. Then we can define a bracket on $C[G]$ by $[f,h]=f*h-h*f$. What does $C[G]$ look like as a Lie algebra? When is it solvable?</p> http://mathoverflow.net/questions/60447/how-does-the-group-algebra-look-as-a-lie-algebra/60450#60450 Answer by Johannes Ebert for How does the group algebra look as a Lie algebra Johannes Ebert 2011-04-03T16:53:40Z 2011-04-03T16:53:40Z <p>Assuming that your ground field $K$ has characteristic prime to the order of $G$. Then the group ring is a seminsimple algebra. Therefore, $C[G]=\bigoplus_{i=1}^{r} Mat_{n_i}(R_i)$ is a direct sum of matrix algebras, where $R_i$ is a finite-dimensional division ring over $K$. All this is very classical and nicely explained in Procesis book on Lie groups.</p> <p>Thus the Lie algebra is a sum of general linear Lie algebras over division rings. If $K$ is $\mathbb{C}$, then $R_i=\mathbb{C}$. So the Lie algebra is solvable iff all $n_i$ are $1$ (this happens iff $G$ is abelian).</p> <p>For $\mathbb{R}$, you also get the quaternions. For fields like $\mathbb{Q}$ or finite characteristic dividning the order of $G$, the story is going to be way more complicated and interesting.</p> http://mathoverflow.net/questions/60447/how-does-the-group-algebra-look-as-a-lie-algebra/60464#60464 Answer by Adrien for How does the group algebra look as a Lie algebra Adrien 2011-04-03T18:12:53Z 2011-04-03T18:12:53Z <p>Maybe you could look at the following paper:</p> <p>Ivan Marin, Group algebras of finite groups as Lie algebras</p> <p><a href="http://arxiv.org/abs/0809.0074" rel="nofollow">http://arxiv.org/abs/0809.0074</a></p> http://mathoverflow.net/questions/60447/how-does-the-group-algebra-look-as-a-lie-algebra/62874#62874 Answer by Salvatore Siciliano for How does the group algebra look as a Lie algebra Salvatore Siciliano 2011-04-25T00:02:46Z 2011-04-25T00:02:46Z <p>A full characterization of group algebras which are solvable or nilpotent as Lie algebras can be found in the paper [I.B.S. Passi - D. Passman - S.K. Sehgal: Lie solvable group rings, Can. J. Math. 25 (1973), 748-757].</p> http://mathoverflow.net/questions/60447/how-does-the-group-algebra-look-as-a-lie-algebra/63906#63906 Answer by Geoff Robinson for How does the group algebra look as a Lie algebra Geoff Robinson 2011-05-04T11:42:02Z 2011-05-04T12:59:13Z <p>There is an interesting proof by Brauer of the fact that when $G$ is a finite group and $F$ is an algebraically closed field of prime characteristic $p$, then the number of simple $FG$ modules is the number of conjugacy classes of elements of $G$ of order prime to $p.$ This proof has a definite Lie Algebras flavour to it (and can be found in Curtis and Reiner, Representation Theory of Finite Groups and Associative Algebras,Wiley,1962).</p> <p>In this proof, Brauer sets $A = FG$, and considers the $F$-subspace $K(A)$, which is the $F$-span of ${ab - ba: a,b \in A }.$ It is relatively easy to see that the codimension of $K(A)$ is the dimension of $Z(A)$, which is clearly the number of conjugacy classes of $G$. Then he defines $T(A)$ as the set of elements $x \in A : x^{p^{n}} \in K(A)$ for some $n$. It is not immediately obvious that $T(A)$ is a linear subspace, but it is. It is clear then that $T(A)$ contains the Jacobson radical $J(A)$, so it is then reasonably easy to calculate the codimension of $T(A)$ in two ways to get the result: one is by considering $T(A)/J(A)$ inside the semisimple algebra $A/J(A)$, and other is by thinking on terms of group theoretic information. A person who has developed this line of thinking further in recent years is Burkhard K\"ulshammer, who has found invariants by this sort of method which have recently been proved to be useful invariants under derived equivalence. </p> http://mathoverflow.net/questions/60447/how-does-the-group-algebra-look-as-a-lie-algebra/64190#64190 Answer by Pasha Zusmanovich for How does the group algebra look as a Lie algebra Pasha Zusmanovich 2011-05-07T13:17:32Z 2011-05-07T17:04:55Z <p>Concerning the situation in characteristic $p$: when $p$ divides the order of $G$, the case not covered by Maschke's theorem, the group algebra $KG$ is no longer semisimple. There is, however, a whole array of papers, started from J.D. Donald and F.J. Flanigan, A deformation theoretic version of Maschke's theorem for modular group algebras: the commutative case, J. Algebra 29 (1974), 98-102, DOI:10.1016/0021-8693(74)90114-8, aiming to prove a conjecture which can be considered as a modular analog of Maschke's theorem: the group algebra KG is <i>deformed</i> to a semisimple algebra. Most of these papers have a group-theoretic flavor, arguing in terms of blocks and other group-representation-theoretic data. Murray Gerstenhaber and Anthony Giaquinto have claimed in: Compatible deformations, Trends in the Representation Theory of Finite Dimensional Algebras (ed. E.L. Green and B. Huisgen-Zimmerman), Contemp. Math. 229 (1998), 159-168, that there is a counterexample to this conjecture: a 8-element quaternion group over a field of characteristic 2. This was believed to be true for a decade or so, after it has been proved wrong (N. Barnea and Y. Ginosar, A separable deformation of the quaternion group algebra, Proc. Amer. Math. Soc. 136 (2008), 2675-2681, DOI: 10.1090/S0002-9939-08-09480-X, arXiv:0704.1556). As far as I know, the Donald-Flanigan conjecture is still open.</p>