"Lie algebra" for a general group ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:11:30Z http://mathoverflow.net/feeds/question/94860 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94860/lie-algebra-for-a-general-group "Lie algebra" for a general group ? Dushya 2012-04-22T17:39:13Z 2012-06-22T00:58:22Z <p>Is there analog of Lie algebra for the case of topological groups which are not necessarily differentiable manifolds, and in particular for finite groups? here by "analog" i mean that it should have similar kind of relations to given group as of a finite dimensional Lie algebra with its corresponding Lie group. for example there should be some analog of "Exponential map"; Every linear representation of group should also be a linear representation of its algebra, and this should be in some sense compatible with the exponential map between the two; etc</p> <p>thanks,</p> http://mathoverflow.net/questions/94860/lie-algebra-for-a-general-group/94869#94869 Answer by Igor Rivin for "Lie algebra" for a general group ? Igor Rivin 2012-04-22T19:05:05Z 2012-04-22T19:05:05Z <p>This is studied by Helge Glockner in <a href="http://dl.dropbox.com/u/5188175/Glockner.pdf" rel="nofollow">this nice paper</a> (which has good references, too).</p> http://mathoverflow.net/questions/94860/lie-algebra-for-a-general-group/94870#94870 Answer by Marc Palm for "Lie algebra" for a general group ? Marc Palm 2012-04-22T19:20:15Z 2012-04-28T19:05:39Z <p>Every locally compact group $G$ contains an open, closed, almost connected subgroup. This is a variant of van Dantzig's theorem. Pull back an open, compact subgroup in the totally disconnected group $G/G_0$ along the surjection $G \twoheadrightarrow G/G_0$, where $G_0$ is the connected component of the identity.</p> <p>An almost connected, locally compact group $G'$ admits a net of normal compact subgroups $\cap N = \{ 1 \}$ and such that $G'/N$ is isomorphic to a Lie group, i.e. $G'$ is a projective limit of Lie groups:</p> <p>$$G' = \lim\limits_{\leftarrow} G'/N.$$</p> <p>This is the solution to Hilbert's 5th problem.</p> <p>Now, one is able to define everything via inductive and projective limits.</p> <p>Edit: As far as I understand, these are the systems of proejctive Lie groups also considered in the refernce of Igor Rivin. Terrence Tao has written something about this on his blog: <a href="http://terrytao.wordpress.com/2011/10/08/254a-notes-5-the-structure-of-locally-compact-groups-and-hilberts-fifth-problem/" rel="nofollow">http://terrytao.wordpress.com/2011/10/08/254a-notes-5-the-structure-of-locally-compact-groups-and-hilberts-fifth-problem/</a> One personal note: You do not necessarily need to understand the proofs, to apply the theorems.</p> http://mathoverflow.net/questions/94860/lie-algebra-for-a-general-group/94915#94915 Answer by mt for "Lie algebra" for a general group ? mt 2012-04-23T08:08:09Z 2012-04-23T20:39:27Z <p>For finite $p$-groups one has the Lazard correspondance. This gives an equivalence of categories between $p$-groups of class less than $p$ ("Lazard groups") and Lie rings of class less than $p$ whose additive groups are $p$-groups. Given such a $p$-group $G$, its Lazard corespondent is a Lie ring defined on the same underlying set as $G$, with addition and bracket product defined in terms of the group operations as follows:</p> <p>$$x +_L y = x y [x,y]^{-1/2} [y,[x,y]]^{7/12}[x,[y,x]]^{5/12} [y,[y,[y,x]]]^{5/8} [y,[x,[y,x]]]^{1/2}[x,[x,[y,x]]] ^{3/8}\ldots$$</p> <p>$$[x,y]_L = [x,y][y,[x,y]]^{-1/2}[x,[y,x]]^{1/2}[y,[y,[y,x]]]^{-1/3}[y,[x,[y,x]]]^{-1/4}[x,[x,[y,x]]]^{-1/3}\ldots$$</p> <p>(these are related to the Baker-Campbell-Hausdorff formula). Example: for $G$ the extra-special $p$-group of order $p^3$ and exponent $p>2$, the corresponding Lie ring is the 3-dimensional Heisenberg Lie algebra over $\mathbb{F}_p$. See for example Khukhro's book on $p$-automorphisms of $p$-groups or Lazard's Sur les groups nilpotents et les anneaux de Lie.</p> <p>There is a correspondence of representations here, but only in small dimension. Any characteristic $p$ rep can be thought of as a homomorphism into the group of upper unitriangular $n\times n$ matrices. For $n\leq p$ this has class $\leq p-1$, and its associated Lie ring is isomorphic, via the log map, to the Lie algebra of strictly upper triangular matrices. So functoriality gives a correspondence of representations of dimension $\leq p$.</p> <p>It is worth pointing out that in characteristic p, it is not usually possible to find a Lie algebra with the same representation theory as a given group. This is because all Lie algebras have their cohomology finitely generated over the subring generated by degree two elements,while few groups have this property. However there are miraculous special cases such as the dihedral group of order eight whose modular group algebra is actually isomorphic to a universal enveloping algebra.</p> http://mathoverflow.net/questions/94860/lie-algebra-for-a-general-group/95086#95086 Answer by Ralph for "Lie algebra" for a general group ? Ralph 2012-04-24T22:39:20Z 2012-04-24T22:39:20Z <p>If $G$ is a group and $G = \gamma_{0}^p(G) \supseteq ... \supseteq \gamma_{k}^p(G) \supseteq \gamma_{k+1}^p(G)\supseteq ...$ its p-lower central series then we have the associated Lie algebra over $k := \mathbb{F}_p$:</p> <p>$$\operatorname{gr}^p(G) = \bigoplus_{k\ge 0}\gamma_{k}^p(G) / \gamma_{k+1}^p(G)$$ those Lie bracket is given by the commutator map: $[\bar{x},\bar{y}] := \overline{[x,y]}$. </p> <p>We can associate $G$ with another Lie algebra: Let $I$ be the augmentation ideal of $kG$. We have the graded ring $\operatorname{gr}(kG):= \bigoplus_{k\ge 0} I^k/I^{k+1}$. Then the map $$\operatorname{gr}^p(G) \to \operatorname{gr}(kG),\; \bar{x} \mapsto \overline{x-1}$$ is a homomorphism of Lie $k$-algebras. </p> <p>Now by a celebrated theorem of Quillen the induced map $$U(\operatorname{gr}^p(G) \otimes_{\mathbb{Z}} k) \to \operatorname{gr}(kG)$$ is an isomorphism of $k$-algebras $(U$ denotes the universal envelope$)$. </p> <p>As an application, one obtains $H^\ast(G;k) \cong H^\ast(\operatorname{gr}^p(G);k)$ if $G$ is a $p$-group. Hence one can compute group cohomology by Lie algebra cohomology in this case. </p> http://mathoverflow.net/questions/94860/lie-algebra-for-a-general-group/95133#95133 Answer by Adrien for "Lie algebra" for a general group ? Adrien 2012-04-25T08:52:17Z 2012-06-22T00:58:22Z <p>If $G$ is a finitely generated group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.</p> <p>Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r>0$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.</p> <p>Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.</p> <p>Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.</p>