Abelianization of Lie groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:37:47Z http://mathoverflow.net/feeds/question/2677 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2677/abelianization-of-lie-groups Abelianization of Lie groups Theo Johnson-Freyd 2009-10-26T20:50:59Z 2009-11-13T03:08:53Z <p>If <em>G</em> is a group, its <strong>abelianization</strong> is the abelian group <em>A</em> and the map <em>G</em> &rarr; <em>A</em> such that any map <em>G</em> &rarr; <em>B</em> with <em>B</em> abelian factors through <em>A</em>. Abelianization is a functor, and in general a very lossy operation. The map <em>G</em> &rarr; <em>A</em> is always a surjection/quotient, because we can construct <em>A</em> by dividing <em>G</em> by the minimal normal subgroup that contains all conjugations <em>ghg<sup>-1</sup>h<sup>-1</sup></em> for <em>g,h</em>&isin;<em>G</em>.</p> <p>If <em>V</em> is a finite-dimensional (super)vector space over a field <em>K</em>, then the abelianization of GL(<em>V</em>) is isomorphic to the multiplicative group <em>K</em><sup>*</sup> of non-zero numbers in <em>K</em>. Indeed, the determinant exhibits the desired isomorphism.</p> <p>Here are two questions I'm curious about:</p> <ol> <li>What can be said about the abelianizations of other (finite-dimensional) Lie groups?</li> <li>If <em>V</em> is an infinite-dimensional vector space, what can be said about the abelianization of GL(<em>V</em>)? Most infinite-dimensional vector spaces have some analytic structure, e.g. topological vector spaces, and so it's reasonable to ask that the operators in GL(<em>V</em>) should preserve that structure; you are welcome to take your favorite type of infinite-dimensional vector space and your favorite type of GL(<em>V</em>), if you want.</li> </ol> http://mathoverflow.net/questions/2677/abelianization-of-lie-groups/2679#2679 Answer by Eric Wofsey for Abelianization of Lie groups Eric Wofsey 2009-10-26T21:10:59Z 2009-10-26T21:10:59Z <p>I don't have anything to say about specific examples, but here are some general remarks. A way to construct the abelianization of any compact group is to consider its image under the product of all its 1-dimensional unitary representations. This is because a compact abelian group is characterized by its set of characters by Pontrjagin duality. More generally, you can construct the double Pontrjagin dual of a locally compact group to get its locally compact abelianization as a subgroup of a space of maps to U(1) with the compact-open topology.</p> http://mathoverflow.net/questions/2677/abelianization-of-lie-groups/5308#5308 Answer by Jason DeVito for Abelianization of Lie groups Jason DeVito 2009-11-13T03:08:53Z 2009-11-13T03:08:53Z <p>(In some sense, this is just a restatement of what Eric said above....)</p> <p>For compact groups, quite a lot can be said. Every compact group H' has a finite cover H which is Lie group isomorphic to $T^{k} \times G$, where $G$ is compact and simply connected.</p> <p>Then, one can easily show that [H,H] = {$e$}$\times G$ and hence that the abelianization of $H$ (which, as in the finite case is H/[H,H]) is $T^{k}$.</p> <p>The same holds true of the original group $H'$: the abelianization is $H'/[H',H']$ and is isomorphic to $T^{k}$.</p>