Harmonic Expansion - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:51:26Z http://mathoverflow.net/feeds/question/16553 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16553/harmonic-expansion Harmonic Expansion Anirbit 2010-02-26T20:11:02Z 2010-02-28T11:25:52Z <p>This is an expansion which frequently occurs in the papers of Camporesi and Higuchi but I couldn't find a derivation of it in either their review papers or in standard books on homogenous vector bundles.</p> <p>Consider the lie group $G$ and its subgroup $H$ and a homogeneous vector bundle be defined on $G/H$ corresponding to a representation $\tau$ of $H$ and $\sigma$ be a section of the principle bundle formed by $G$ over $G/H$. Let $\psi$ be a section of this homogeneous vector bundle and $\psi ^a$ be the component of the section along some basis labeled by $a$. So $a$ runs through all natural numbers till the dimension of $\tau$. This basis is defined at every fiber of the bundle by pushing forward by $\sigma$ a chosen basis in the fiber at identity of $G/H$.</p> <p>Then $\psi ^a$ has the following expansion known as the "Harmonic Expansion",</p> <p><code>$\psi ^{a}(x) = \sum _{\lambda} \sum _I \sum _{\xi} U^{\lambda}((\sigma (x))^{-1})^{a\xi}_{I} \psi ^{I}_{\lambda \xi}$</code></p> <p>where $\lambda$ runs over representatives from all equivalence classes of representations of $G$ which when restricted to the subgroup $H$ of $G$ contain $\tau$ in them. $I$ runs over the dimension of $\lambda$ and $\xi$ runs over natural numbers till the multiplicity of $\tau$ in the above restriction. $U^{\lambda}(g)^{a\xi}_{I}$ is the matrix entry of the $\lambda$ representation of $g$ between the basis field $I$ (of the vector space on which $\lambda$ is) and $a\xi$ (basis vector labeled by $a$ in the copy $\xi$ of the representation $\tau$ )</p> <p>I would like to know the derivation of the above expansion, especially how apparently the Frobenius Reciprocity theorem plays role in it?</p> <p>And in this expansion which is the "harmonic"?</p> <p>By "harmonic" we should have a quantity here which is an eigen-function of some laplacian. Which is it here and of which laplacian?</p> <p><em>Like to put in one more aspect here,</em></p> <p>In a homogeneous vector bundle on $G/H$, $G$ naturally has a representation on the space of sections of the homogeneous bundle. It seems that sections in the irreducible representations of this are also called "Harmonics".</p> <p>What is the rationale for this terminology? </p> http://mathoverflow.net/questions/16553/harmonic-expansion/16560#16560 Answer by Marty for Harmonic Expansion Marty 2010-02-26T22:38:10Z 2010-02-27T15:13:02Z <p>I'd look in books on representation theory rather than books on homogeneous vector bundles for this sort of thing. Perhaps George Mackey's famous book "Unitary group representations in physics, probability, and number theory" would be of interest, and cover such facts.</p> <p>To answer your question, it's probably better to consider the finite group case, before the more general Lie group case. So let $H$ be a subgroup of a finite group $G$. Let $(\tau, W)$ be a finite-dimensional irreducible representation of $H$ (it's helpful to give the vector space its own name). Let $\psi$ be a section of the homogeneous bundle over $G/H$ associated to $\tau$. In this finite context, this simply means that $\psi: G \rightarrow W$ is a function which satisfies the identity: $$\psi(gh) = [ \tau(h^{-1})] (\psi(g)).$$</p> <p>In other words, $\psi$ is simply a vector in the induced representation $Ind_H^G W$. One may decompose this induced representation into irreducible representations of $G$: $$Ind_H^G W = \bigoplus_{\lambda \in \hat G} V_\lambda^{m_\lambda},$$ where $(\lambda, V_\lambda)$ runs over the irreducible representations of $G$, and $$m_\lambda = dim(Hom_G(V_\lambda, Ind_H^G W)) = dim(Hom_H(V_\lambda, W))$$ is the multiplicity of $V_\lambda$ in the induced representation, or equivalently (by Frobenius reciprocity) $m_\lambda$ is the multiplicity of $W$ in the restriction of $V_\lambda$.</p> <p>Given such a decomposition, we may decompose the vector $\psi \in Ind_H^G W$ as a tuple (or finite sequence of tuples): $$\psi = (v_\lambda^{(1)}, \ldots, v_\lambda^{(m_\lambda)})_{\lambda \in \hat G}.$$</p> <p>This is the same, in spirit, to the formula you mentioned. The main difference is that I am averse to choosing bases (unlike many physicists). In fact, I think my dislike of choosing bases (or slight laziness) prevents me from deriving your formula on the nose. Sorry!</p> <p>But to finish my answer, I should say that the entire argument above can be carried out with few changes, when working with a compact Lie group $G$ and closed subgroup $H$, and unitary representations throughout. (This situation, I would bet, is covered by Mackey). A classic example would be when $G = SO(n)$ and $H = SO(n-1)$, so that $G/H$ is homeomorphic to the $(n-1)$-dimensional sphere $S^{n-1}$.</p> <p>There, when one works with finite-dimensional unitary representations, the Lie group action yields a natural action of the Lie algebra, and hence the universal enveloping algebra. In other words, the representation spaces also have actions of differential operators arising from the groups.</p> <p>In particular, if one takes an irreducible subrepresentation of $Ind_H^G W$ as before, one gets a space of functions from $G$ to $W$, which is an irreducible representation of $G$ by translation. It follows that the center of the universal enveloping algebra of $G$ acts via scalars on this space of functions. Hence, if $G$ is a simple compact connected Lie group, the Casimir operator (generalization of the Laplacian) acts via a scalar on this space of functions -- i.e., the functions satisfy a nice second order differential equation. This is the source of "spherical harmonics", for example, and the reason for the word harmonic in this context.</p>