Lie group examples - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:36:13Z http://mathoverflow.net/feeds/question/61315 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61315/lie-group-examples Lie group examples Richard Borcherds 2011-04-11T18:15:12Z 2012-05-20T10:24:13Z <p>I'm looking for interesting applications of Lie groups for an introductory Lie groups graduate course. In particular I'd like to hear of non-standard examples that at first sight do not seem to be related to Lie groups (so please don't suggest well-known things like Clifford algebras or triality that appear in standard Lie groups texts such as Fulton and Harris). Here are some examples of the sorts of things I'm looking for:</p> <p>*The cohomology of a compact Kaehler manifold is a representation of SL2, so the Hopf manifold cannot be Kaehler.</p> <p>*q-binomial coefficients are unimodal, as they are characters of representations of SL2</p> <p>*Hilbert's theorem on the finite generation of rings of invariants can be proved using invariant integration on compact Lie groups. </p> <p>*Holomorphic modular forms are really highest weight vectors of discrete series representations of certain Lie groups. </p> <p>*Most closed 3-manifolds are quotients of SL2(C) by discrete subgroups. </p> <p>*Bessel functions cannot be expressed using elementary functions and indefinite integration. (Differential Galois theory was one of Lie's original motivations, but seems to have been eliminated from texts on Lie theory.)</p> <p>*Classifying manifolds up to cobordism uses orthogonal groups. </p> http://mathoverflow.net/questions/61315/lie-group-examples/61319#61319 Answer by Qiaochu Yuan for Lie group examples Qiaochu Yuan 2011-04-11T18:38:54Z 2011-04-11T18:38:54Z <p>The sum identity</p> <p>$$\frac{1}{k} \sum_{i=0}^{k-1} \left( 2 \cos \frac{2 \pi i}{k} \right)^{2n} = {2n \choose n}$$</p> <p>(where $k > 2n$) can be explained combinatorially as follows: the adjacency matrix $A_k$ of the cycle graph of size $k$ has eigenvalues $2 \cos \frac{2 \pi i}{k}, 0 \le i \le k-1$, and the sum of the $2n^{th}$ powers of the eigenvalues of the adjacency matrix is the total number of closed walks of length $2n$ on the graph. For $k > 2n$ this is easily seen to be $k {2n \choose n}$ (where the coefficient of $k$ comes from the choice of starting vertex), and dividing by $k$ we get the identity above.</p> <p>Letting $k \to \infty$ the above becomes a Riemann sum and we obtain the integral identity</p> <p>$$\int_0^1 (2 \cos 2 \pi x)^{2n} \, dx = {2n \choose n}$$</p> <p>which is fairly straightforward to prove but not quite as straightforward to interpret directly, since it's not obvious that the argument above about adjacency matrices generalizes.</p> <p>This identity can be explained combinatorially using the representation theory of the circle group $\text{SO}(2)$. Associated to (say) any compact Lie group $G$ and representation $V$ of $G$ there is a graph $\Gamma_G(V)$, the <em>principal graph</em>, whose vertices are the irreducible representations of $G$, and where the number of edges from a representation $A$ to a representation $B$ is $\dim \text{Hom}_G(A \otimes V, B)$. </p> <p>The principal graph of the standard representation of $\text{SO}(2)$ is precisely the Cayley graph of $\mathbb{Z}$ with generators $\pm 1$, which one can think of as the "limit" of the cycle graphs above (the Cayley graphs of the finite, rather than infinite, cyclic groups) in some appropriate sense. It follows that the number of closed walks from the origin to itself on $\mathbb{Z}$ of length $2n$ is, on the one hand, clearly ${2n \choose n}$ and, on the other hand, is $\dim \text{Hom}_G(V^{\otimes n}, 1)$, or the dimension of the invariant subspace of $V^{\otimes n}$, and this quantity can be computed by character theory in a way that exactly generalizes the eigenvalue computation above.</p> <p>The principal graph of the standard representation of $\text{SU}(2)$ is similar, but is infinite in one direction rather than two. This gives a corresponding integral identity for the Catalan numbers, and in order to get the Riemann sum version of this integral identity one must pass to the representation theory of <a href="http://mathoverflow.net/questions/17197/how-does-this-relationship-between-the-catalan-numbers-and-su2-generalize" rel="nofollow">quantum groups at roots of unity</a>, as I learned in the linked MO question. </p> http://mathoverflow.net/questions/61315/lie-group-examples/61325#61325 Answer by Xiaolei Wu for Lie group examples Xiaolei Wu 2011-04-11T18:47:59Z 2011-04-11T18:47:59Z <p>Classifying of fiber bundles is dependent on its structure group, which is a Lie group, essentially it is how those fibers are pasted together. For any Lie Group $G$, there is a classifying space $BG$ associated to this group. And the theorem says, bundles over a space $X$ (good enough) with given fiber $F$ (with good enough action) is one to one correspondence with the homotopy classes from $X$ to $BG$. This also leads to the definition of Characteristic classes, which is in some sense, just the pull back of the generator of the cohomology Ring for $BG$.</p> http://mathoverflow.net/questions/61315/lie-group-examples/61328#61328 Answer by Faisal for Lie group examples Faisal 2011-04-11T19:20:04Z 2011-04-11T19:20:04Z <p>Here are three examples.</p> <ol> <li><p>A finite field extension of <code>$\mathbb R$</code> must be quadratic. For if <code>$\mathbb R^n$</code> carries the structure of a field, then its group of units <code>$\mathbb R^n - \{0\}$</code> is an abelian Lie group. But <code>$\mathbb R^n - \{0\}$</code> is simply connected if <code>$n&gt;2$</code>, which means that <code>$\exp$</code> gives us an isomorphism of groups <code>$\mathbb R^n - \{0\} \cong \text{Lie}(\mathbb R^n - \{0\}) = \mathbb R^n$</code>, which is absurd.</p></li> <li><p>The complex structure on the complex grassmannian <code>$Gr(d,\mathbb C^n)$</code> is <em>locally rigid</em>. Roughly what this means is that if you have a smoothly varying family $X_t$ of complex manifolds, where the index $t$ takes values in an open connected subset of <code>$\mathbb{C}^N$</code> that contains $0$, and if <code>$X_0 = Gr(d,\mathbb C^n)$</code>, then one can find a neighborhood $U$ of $0$ such that $X_t$ is isomorphic to $X_0$ as a complex manifold for all $t \in U$. A theorem of Frölicher and Nijenhuis states that the complex structure of a compact complex manifold $X$ is locally rigid if <code>$H^1(X,T_X)=0$</code>, where <code>$T_X$</code> is the holomorphic tangent bundle of $X$. Using representation theory, Bott showed that <code>$H^q(X,T_X)=0$</code> for all $q\geq1$ if <code>$X=G/P$</code> for $G$ a complex semisimple Lie group and $P$ a parabolic subgroup, i.e., if $X$ is a "generalized flag variety." This establishes the local rigidity of the complex structure of generalized flag varieties, such as <code>$Gr(d,\mathbb C^n)$</code>.</p></li> <li><p>It's easy to believe that the combinatorics of integer partitions and Young diagrams is related to the representation theory of the symmetric group <code>$S_n$</code> (over <code>$\mathbb C$</code>, say). Schur--Weyl duality relates the latter to the representation theory of <code>$GL_m(\mathbb C)$</code>. This in turn can be related to the geometry of the flag varieties of <code>$GL_m(\mathbb C)$</code>. With these observations one can, for example, relate the combinatorics of Young diagrams to the multiplication in the cohomology ring of the grassmannian, which of course carries some kind of geometric information. This is quite remarkable, in my opinion.</p></li> </ol> http://mathoverflow.net/questions/61315/lie-group-examples/61542#61542 Answer by Allen Knutson for Lie group examples Allen Knutson 2011-04-13T12:57:39Z 2011-04-13T12:57:39Z <p>Let $C$ be a finite-dimensional field extension of the real numbers, so $(C \setminus 0)/{\mathbb R}_+$ </p> <p>is a compact abelian Lie group, and a sphere of dimension $\dim_{\mathbb R} C - 1$. If $C \neq {\mathbb R}$ so this group is connected, then it's a $K(\pi,1)$ (proved using the exponential map, which is a surjective group homomorphism). In particular, it can't be a sphere of dimension $\geq 2$, so $\dim C = 2$.</p> <p>(Anybody know who this proof is due to? I believe I heard Mazur or Gross, but I'm not sure.)</p> http://mathoverflow.net/questions/61315/lie-group-examples/62023#62023 Answer by Alexander Chervov for Lie group examples Alexander Chervov 2011-04-17T11:07:11Z 2011-04-17T11:07:11Z <p>Integrable system theory might be an example, however the examples I can propose requires some technicality level (which probably undesired)...</p> <p>The line of application is the following</p> <p>Questions (seemingly not related to Lie groups): Consider the differential operator - $H = \sum \partial_i^2 + \sum exp(x_i - x_{i+1} )$ </p> <p>Qeust 1: Can you find some differential operators which commute with H? Quest 2: Can you find eigenvectors for H ?</p> <p>(This is called Toda quantum system, Calogero and some other can be considered as well)</p> <p>Lie groups comes into the game like this:</p> <p>The main idea is that this differential operator comes from the Casimir of gl_n, so higher Casimirs (i.e. Z(U(gl_n)) will provide the commiting operators. What we need to do to obtain this operator from standard Casimir - is to make some reduction (integration) over some some subgroup. Eigenfucntions can be obtained by intgration of some characters of the representations, which comes from the fact that Casimirs acts by scalars on any irreps...</p> <p>The sl(2) example is simple technically and for me it was quite a beatiful, when it was explained to me... In sl(2) case we get Bessel functions as eigenfunctions, some properties like recurrent formulas for different "n" in Bessel can be derived from tensor decomposition of corresponding representations...</p> <p>If necessary I can provide refrences...</p> <p>--</p> <p>These are "quantum integrable" systems, similar can be done for classical ones - one can obtain solutions of diffurs.</p> http://mathoverflow.net/questions/61315/lie-group-examples/62034#62034 Answer by zroslav for Lie group examples zroslav 2011-04-17T13:38:18Z 2011-04-17T13:38:18Z <p>There is a jet group: the group of $\sum_{i>0}a_ix^i$ ($a_1\ne 0$) under the operation of composition. For example you can easily find every pair of f(x) and g(x) such that f(g(x))=g(f(x)) (for $a_1=1$ it is very easy)</p> http://mathoverflow.net/questions/61315/lie-group-examples/62076#62076 Answer by Daniel Moskovich for Lie group examples Daniel Moskovich 2011-04-18T02:40:07Z 2011-04-18T02:40:07Z <p>What could be a nicer example than <a href="http://en.wikipedia.org/wiki/Link_group#Milnor_invariants" rel="nofollow">Milnor invariants</a>? At first sight, these "higher linkage" invariants seem to have nothing to do with Lie groups at all; you can even define them using <a href="http://mathworld.wolfram.com/HabiroMove.html" rel="nofollow">Habiro moves</a> to make any connection to Lie algebras even more mysterious. Or, you can speak of the <a href="http://mathworld.wolfram.com/Baguenaudier.html" rel="nofollow">Baguenaudier puzzle</a>, whose solution involves Habiro moves in disguise, as <a href="http://arxiv.org/abs/math.GT/0007134" rel="nofollow">discussed by Przytycki and Sikora</a>.<br> And yet, the natural home for Milnor invariants is the group $D(H)$, which is the kernel of the left bracketting map $L(H)\otimes H \to L(H)$, where $L(H)$ denotes the free Lie algebra over $H$, the first homology of the link complement.<br> The group $D(H)$ also comes up in other contexts which at first sight don't seem to have anything to do with free Lie algebras. It's related to the rational homology of the outer automorphism groups of free groups, as first observed by Kontsevich. Morita, and Conant-Vogtmann, took this idea and ran with it. The group $D(H)$ was also used by Dennis Johnson to study the relative weight filtration of the mapping class group of a surface. </p>