Classification of (compact) Lie groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:42:59Z http://mathoverflow.net/feeds/question/6079 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6079/classification-of-compact-lie-groups Classification of (compact) Lie groups Gian Maria Dall'Ara 2009-11-19T09:16:12Z 2010-05-19T22:16:59Z <p>I would like to study/understand the (complete) classification of compact lie groups. I know there are a lot of books on this subject, but I'd like to hear what's the best route I can follow (in your opinion, obviously), since there are a lot of different ideas involved. </p> http://mathoverflow.net/questions/6079/classification-of-compact-lie-groups/6095#6095 Answer by Jason DeVito for Classification of (compact) Lie groups Jason DeVito 2009-11-19T13:46:53Z 2010-05-19T20:26:21Z <p>First, here's a rough outline of how the classification works:</p> <ol> <li><p>Prove that if G and H are simply connected and have the same Lie algebra, then G and H are isomorphic as Lie groups.</p></li> <li><p>Prove that if G is any Lie group, its universal cover $\tilde{G}$ inherits a natural Lie group structure for which G = $\tilde{G}/Z$ where $Z\subseteq Z(\tilde{G})$.</p></li> </ol> <p>This reduces classification to a) understanding the Lie algebras and b) understanding the centers of simply connected Lie groups.</p> <p><code>3.</code> Classify (simple) Lie algebras. This is done via root diagrams (Dynkin diagrams).</p> <p><code>4.</code> For each simply connected compact Lie group, compute its center.</p> <p>For references, I'd check out Fulton and Harris' book "Representation Theory". I'm not sure if it actually does 4., but that's a fairly easy exercise afterwards (except for perhaps the exceptional groups).</p> http://mathoverflow.net/questions/6079/classification-of-compact-lie-groups/6129#6129 Answer by Charles Rezk for Classification of (compact) Lie groups Charles Rezk 2009-11-19T16:50:42Z 2009-11-19T16:50:42Z <p>There's a concept called a <a href="http://en.wikipedia.org/wiki/Root_datum" rel="nofollow">root datum</a>, which consists of an integer lattice and its dual, together with distinguished subsets (roots and coroots) satisfying some simple properties.</p> <p>The theorem (I think!) is: isomorphism classes of compact connected Lie groups are in one-to-one correspondence with isomorphism classes of root data.</p> <p>In particular, you can read off all sorts of information about your Lie group from it's root datum: Weyl group, Dynkin diagram, center, fundamental group, etc.</p> <p>Unfortunately, root data only ever seem to be discussed in the context of algebraic groups; I've never seen an elementary treatment which talks about root data for compact Lie groups without reference to the algebraic group case.</p> http://mathoverflow.net/questions/6079/classification-of-compact-lie-groups/6151#6151 Answer by Theo Johnson-Freyd for Classification of (compact) Lie groups Theo Johnson-Freyd 2009-11-19T19:04:06Z 2009-11-19T19:04:06Z <p>I don't know if it's appropriate to link to self-advertise here. So at the risk of a minor faux pas, my edited notes from the Lie Groups class taught by Prof. Mark Haiman are available <a href="http://math.berkeley.edu/~theojf/LieGroupsBook.pdf" rel="nofollow">here</a>. The bulk of the notes is the classification of complex semisimple Lie groups. For compact ones, follow the same argument, but add one fact: a simple group over R is compact iff the Killing form is negative definite.</p> <p>In case one should not post one's own notes, <a href="http://math.berkeley.edu/~anton/written/LieGroups/LieGroups.pdf" rel="nofollow">here are some by Anton from the previous year</a>. These include a bit more on real forms, and a bit less on the non-semisimple groups.</p> http://mathoverflow.net/questions/6079/classification-of-compact-lie-groups/16269#16269 Answer by BCnrd for Classification of (compact) Lie groups BCnrd 2010-02-24T15:31:07Z 2010-02-24T15:31:07Z <p>Compact Lie groups may not be connected, and the question did not assume connectedness whereas all of the other answers did. If $G$ is a linear algebraic group over $\mathbf{R}$ then $G(\mathbf{R})$ has finite component group which may be non-trivial even when $G$ is connected in the sense of the Zariski topology, such as $G = {\rm{GL}} _n$; see end of Borel's book on linear algebraic groups for a proof. And orthogonal groups for non-degenerate real quadratic forms of various signatures are very useful, but (in their incarnation as linear algebraic groups over $\mathbf{R}$) are even disconnected for the Zariski topology. So there is good reason to be interested in real Lie groups with nontrivial but finite component group (e.g., finite groups!) And in Hochschild's book "Structure of Lie Groups" he proves that the theory of maximal compact subgroups "works" in the case of finite component groups too: they're all conjugate and meet every component. </p> <p>This is all setup for what I really wanted to say, which is that there is an awesome result, largely due to Chevalley, which beautifully "explains" the essentially algebraic nature of the category of compact Lie groups (without connectedness hypotheses). This concerns the functor $G \rightsquigarrow G(\mathbf{R})$ from linear algebraic groups over $\mathbf{R}$ to real Lie groups with finite component group. To appreciate the result, we need some preliminary observations, as follows. If $G(\mathbf{R})$ is to be compact then of course $G$ cannot contain as an $\mathbf{R}$-subgroup either $\mathbf{G}_a$ or $\mathbf{G}_m$. In particular, $G$ has to be reductive (maybe disconnected) since $\mathbf{R}$ has characteristic 0. [Side remark: for reductive groups over any field $k$ whatsoever, if there is a $\mathbf{G}_a$ as a $k$-subgroup then there <em>must</em> be a $\mathbf{G}_m$ as a $k$-subgroup; i.e., it is $k$-isotropic. See Corollary C.2.9 in "Pseudo-reductive groups" for a generalization.] It is a general fact (over any local field $k$ at all, whether archimedean or not) that a reductive $k$-group has compact group of $k$-points if and only if it contains no $\mathbf{G}_m$ as a $k$-subgroup (i.e., is $k$-anisotropic). Also, if $G$ is disconnected then it can happen that some connected components of $G$ have no $\mathbf{R}$-point (e.g., kernel of ${\rm{det}}^3$ on ${\rm{GL}} _n$ for any $n$), but the union of those components which do contain an $\mathbf{R}$-point is an open $\mathbf{R}$-subgroup which is "all" one could hope to ever recover from $G(\mathbf{R})$. So we may as well focus attention on those $G$ for which each connected component of $G$ does have an $\mathbf{R}$-point.</p> <p>OK, so now we can state Chevalley's result. The preceding remarks show that formation of real points (as a Lie group) is a functor from the category of $\mathbf{R}$-anisotropic reductive $\mathbf{R}$-groups to the category of compact real Lie groups. (Beware it is not obvious if $G(\mathbf{R})$ is also connected for such a $G$ that is connected.) Chevalley proved that this is an <em>equivalence</em> of categories. More specifically, given a compact real Lie group $K$, he showed how to use the representation theory of $K$ to functorially construct a linear algebraic $\mathbf{R}$-group $K^{\rm{alg}}$ whose Lie group of $\mathbf{R}$-points is naturally isomorphic to $K$ (so $K^{\rm{alg}}$ must be reductive and $\mathbf{R}$-anisotropic, for reasons noted above). The construction shows also that every connected component of $K^{\rm{alg}}$ has an $\mathbf{R}$-point, and clearly $K^{\rm{alg}}$ is connected if $K$ is connected. That much is proved in the book by Brocker and tom Dieck. Using some input from the algebraic side (especially the fact that a semisimple Lie subalgebra of the Lie algebra of a linear algebraic group over a field $k$ of characteristic 0 <em>is</em> the Lie algebra of a unique connected closed $k$-subgroup which moreover is semisimple, so over $\mathbf{R}$ these "exponentiate" to <em>closed</em> $\mathbf{R}$-subgroups), one can show that it really inverts the functor of $\mathbf{R}$-points on the full subcategory of $\mathbf{R}$-anisotropic reductive $\mathbf{R}$-groups whose connected components all have $\mathbf{R}$-points. </p> <p>To summarize, incorporating topological aspects: </p> <p>Theorem (Chevalley) The category of compact Lie groups is equivalent to the category of $\mathbf{R}$-anisotropic reductive $\mathbf{R}$-groups whose connected components have $\mathbf{R}$-points, and if $G$ is such an $\mathbf{R}$-group then $G^0(\mathbf{R}) = G(\mathbf{R})^0$. The $\mathbf{R}$-group $G$ is semisimple if and only if $G(\mathbf{R})$ has finite center, and in such cases $G^0$ is simply connected in the sense of algebraic groups if and only if $G(\mathbf{R})^0$ is simply connected in the sense of topology. </p> <p>Remark: The anisotropicity hypothesis is crucial. For example, if $n > 1$ is odd then ${\rm{SL}} _n \rightarrow {\rm{PGL}} _n$ is a degree-$n$ isogeny of $\mathbf{R}$-groups (so not an isomorphism) which induces an isomorphism on $\mathbf{R}$-points.</p> <p>For any $G$ as in the Theorem, one can show that $G(\mathbf{C})$ contains $G(\mathbf{R})$ as a maximal compact subgroup, and that this is a "complexification" of $G(\mathbf{R})$ in the sense of being initial among complex Lie groups equipped with a homomorphism from $G(\mathbf{R})$. Using this and passing to the connected case, one shows that every $\mathbf{R}$-split connected reductive $\mathbf{R}$-group admits a <em>unique</em> "$\mathbf{R}$-anisotropic form" (usually called "compact form"), and this correspondence is also <em>functorial</em> if we keep track of a choice of suitable maximal torus. But the algebraic theory provides an equivalence between the category of pairs $(G,T)$ consistng of split connected reductive groups $G$ equipped with a choice of split maximal torus $T$ over any field, with isogenies as morphisms, and the category of root data with (a suitable notion of) "isogenies" as morphisms. This recovers exactly the classification of compact connected Lie groups (equipped with a maximal torus) in terms of root data as was mentioned in the other answers.</p> <p>Of course, this is a much longer route to the punchline, and I am not recommending it as a good way to learn the classification of compact Lie groups in terms of root data (though it would not be circular to do so). But there is something remarkable about the direct link between compact Lie groups and algebraic groups (allowing disconnectedness as well, and no specified maximal torus), not defined by going through the crutch of root data and Lie algebras. Historically the case of compact groups was a very important guide for Borel and Tits and others when developing the structure theory for connected reductive groups, and the above result "explains" a posteriori why this case was such an excellent guide to the general case. </p> http://mathoverflow.net/questions/6079/classification-of-compact-lie-groups/16325#16325 Answer by Jim Humphreys for Classification of (compact) Lie groups Jim Humphreys 2010-02-24T23:38:00Z 2010-02-24T23:38:00Z <p>The answers given are useful. Naturally some care always has to be taken with the connectedness question, since finite groups might be regarded as compact Lie groups or as algebraic groups. Then a full classification becomes unreasonable. On the other hand, some compact groups or algebraic groups occur most naturally as disconnected groups with an interesting component group.</p> http://mathoverflow.net/questions/6079/classification-of-compact-lie-groups/25240#25240 Answer by Jeffrey Adams for Classification of (compact) Lie groups Jeffrey Adams 2010-05-19T15:47:27Z 2010-05-19T19:25:14Z <p>There is a clear, self-contained classification of compact, connected Lie groups in "Lie Groups: An Approach through Invariants and Representations" by Claudio Procesi. See Chapter 10, Section 7.2, Theorem 4, page 380. It says such a group is of the form $K_1\times\dots\times K_n\times T/Z$ where each $K_i$ is connected, compact, and simply connected, $T=(S^1)^m$ is a compact torus, and $Z$ is a finite subgroup of the center satisfying $Z\cap T=1$. Furthermore two such groups $K_1\times\dots\times T/Z$ and $K'_1\times\dots\times T'/Z'$ are isomorphic if and only if there is an isomorphism $K_1\times\dots\times T\rightarrow K'_1\times\dots\times T'$ taking $Z$ to $Z'$. Note that the numerator is specified by a list of types $A_n,B_n,\dots, E_8$ and the integer $m$. The result is obtained from a bijection between compact connected and reductive algebraic groups, and is equivalent to the classification by root data.</p> <p>For example there are precisely $3$ compact groups of rank 2 and semisimple rank 1: </p> <p>$SU(2)\times S^1$ ($Z=1$)</p> <p>$SO(3)\times S^1$ ($Z=\langle -I,1\rangle$)</p> <p>$U(2)=SU(2)\times S^1/\langle (-I,-1)\rangle$.</p> <p>For another example the following two groups are not isomorphic, where $\zeta=e^{2\pi i/5}$:</p> <p>$SU(5)\times S^1/\langle(\zeta I, \zeta)\rangle \simeq U(5)$</p> <p>$SU(5)\times S^1/\langle(\zeta I,\zeta^2)\rangle$</p>