Why are characters so well-behaved? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:39:13Z http://mathoverflow.net/feeds/question/2795 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved Why are characters so well-behaved? Saul Glasman 2009-10-27T09:26:25Z 2011-06-29T15:48:30Z <p>Last year I attended a first course in the representation theory of finite groups, where everything was over C. I was struck, and somewhat puzzled, by the inexplicable perfection of characters as a tool for studying representations of a group; they classify modules up to isomorphism, the characters of irreducible modules form an orthonormal basis for the space of class functions, and other facts of that sort.</p> <p>To me, characters seem an arbitrary object of study (why take the trace and not any other coefficient of the characteristic polynomial?), and I've never seen any intuition given for their development other than "we try this, and it works great". The proof of the orthogonality relations is elementary but, to me, casts no light; I do get the nice behaviour of characters with respect to direct sums and tensor products, and I understand its desirability, but that's clearly not all that's going on here. So, my question: is there a high-level reason why group characters are as magic as they are, or is it all just coincidence? Or am I simply unable to see how well-motivated the proof of orthogonality is?</p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2808#2808 Answer by Noah Snyder for Why are characters so well-behaved? Noah Snyder 2009-10-27T13:28:04Z 2009-10-27T13:28:04Z <p>Orthogonality makes sense without character theory. There's an inner product on the space of representations given by dim Hom(V, W). By Schur's lemma the irreps are an orthonormal basis. This is "character orthogonality" but without the characters.</p> <p>How to recover the usual version from this conceptual version? Notice dim Hom(V,W) = dim Hom(V \otimes W*, 1) where 1 is the trivial representation. So in order to make the theory more concrete you want to know how to pick off the trivial part of a representation. This is just given by the image of the projection 1/#G \sum_g g. The dimension of a space is the same as the trace of the projection onto that space, so dim Hom(V \otimes W*, 1) = tr(1/#G \sum_g \rho_{V \otimes W*}(g)) = 1/#G \sum_g \chi_V(g) \chi_W(g^-1) using the properties of trace under tensor product and duals.</p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2815#2815 Answer by D. Savitt for Why are characters so well-behaved? D. Savitt 2009-10-27T14:00:50Z 2009-10-27T14:00:50Z <p>As to "why take the trace and not any other coefficient of the characteristic polynomial", note that for completely elementary reasons the trace <b>of the whole representation</b> still knows the characteristic polynomial of each individual element: for instance the second-from-top coefficient of the characteristic polynomial of rho(g) is 1/2(tr(rho(g))^2 - tr(rho(g^2))). Writing down the formula for subsequent coefficients is an exercise with symmetric functions. On the other hand, the higher coefficients of the characteristic polynomial do lose information -- e.g. non-isomorphic representations rather often have the same determinant. </p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2817#2817 Answer by Ben Webster for Why are characters so well-behaved? Ben Webster 2009-10-27T14:10:42Z 2009-10-27T18:25:20Z <p>Maybe it's silly to add this as a separate answer, since Noah basically said what I would have, but he didn't emphasize the one-sentence version of the answer: <em>"the dimension of a subspace is the trace of the projection to that subspace."</em> </p> <p>That and linearity are what's special about trace. </p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2819#2819 Answer by Qiaochu Yuan for Why are characters so well-behaved? Qiaochu Yuan 2009-10-27T14:28:02Z 2009-10-27T14:28:02Z <p>For what it's worth, I had exactly the same question as you and worked out the proof that Noah sketches in some detail <a href="http://qchu.wordpress.com/2009/08/30/the-orthogonality-relations-for-representations-of-finite-groups/" rel="nofollow">here</a>, although I don't use the word "projection" explicitly.</p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2820#2820 Answer by David Speyer for Why are characters so well-behaved? David Speyer 2009-10-27T14:52:21Z 2009-10-27T14:52:21Z <p>Here is another answer: the representation ring of G and the center of k[G] are both commutative algebras. When your ground field is C, these are both isomorphic to each other, and are both isomorphic to C^{r}. The rows and columns of the character table give this isomorphism explicitly. </p> <p>When doing representation theory in other contexts, most statements about characters turn into either statements about the representation ring or statements about the center of k[G]. But the lack of such an explicit presentation of the ring means that you have to work harder.</p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2821#2821 Answer by Dave Penneys for Why are characters so well-behaved? Dave Penneys 2009-10-27T14:57:22Z 2009-10-27T14:57:22Z <p>A bit unrelated, but nonetheless interesting:</p> <p>A finite von Neumann algebra is a von Neumann algebra together with a trace, i.e., a linear functional which is</p> <ul> <li>positive ($tr(a^*a)\geq 0$ for all $a$)</li> <li>faithful ($tr(a^*a)=0$ implies $a=0$)</li> <li>normalized ($tr(1)=1$)</li> <li>tracial ($tr(ab)=tr(ba)$)</li> </ul> <p>This trace does almost everything for a finite von Neumann algebra. It gives us the standard representation of $M$ on $L^2(M)$, the predual $L^1(M)$, and the noncommutative $L^p$-spaces. In a finite factor (trivial center), it gives us a total ordering on projections (for a $II_1$-factor, it gives a "continuous" version of what Ben is saying). If we have an inclusion of finite von Neumann algebras $N\subset M$, we get a canonical conditional expectation $E\colon M\to N$, which is one of the basic tools. If they are subfactors, if we get a trace on $M_1$, we can iterate the basic construction (nicely that is). The list goes on...</p> <p>So what does this all have to do with groups? Well von Neumann algebras are exactly the commutants of (unitary) group representations. Moreover, from a unitary group representation, we can form its left regular von Neumann algebra $L(G)$, which is the easiest way to construct an example of a $II_1$-factor (the group must have all infinite conjugacy classes other than that of the identity). Moreover, the study of $II_1$-subfactors generalizes the study of representation theory, including induction-restriction, etc. (see <a href="http://mathoverflow.net/questions/544/why-are-subfactors-interesting" rel="nofollow">Kate's question</a>).</p> <p>Basically, every time I see an algebra, I wonder if it has a trace. So from my point of view, or with this hindsight I should say, it's no surprise to me that characters do everything!</p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2825#2825 Answer by Gian Maria Dall'Ara for Why are characters so well-behaved? Gian Maria Dall'Ara 2009-10-27T15:17:10Z 2009-10-27T15:17:10Z <p>More a question than an answer, but anyway: It's plain that if one wants to study representations up to equivalence, one would like to attach to every matrix an invariant by conjugation. One could then look for polynomial functions on End(V) or GL(V) (V a finite dimensional vector space), which display such an invariance. We are really looking for the fixed orbits of the linear action of GL(V) on P(End(V)) (which incidentally, is itself an infinite dimensional representation of an algebraic group). In fact this is a subalgebra of P(End(V)) and maybe one can determine a set of generators. David Savitt observation could lead to the question: does tr(A^n) suffice to generate our algebra?</p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2827#2827 Answer by Terry Tao for Why are characters so well-behaved? Terry Tao 2009-10-27T15:20:39Z 2009-10-27T18:51:12Z <p>The trace is about the strongest general way we have to linearly project a non-abelian situation (matrices) to an abelian situation (scalars): tr(AB)=tr(BA). By using the trace, the representation theory of non-abelian groups begins to resemble the representation theory of abelian groups, i.e. Fourier analysis. (Note though that the correspondence becomes less tight when considering triple products: tr(ABC) != tr(CBA). For related (though not identical) reasons, the theory of tensor products of representations is far richer in the nonabelian world (Littlewood-Richardson coefficients, etc.) than it is in the abelian world (convolution), and characters aren't always the best way to proceed here.)</p> <p>This of course raises the question of why Fourier analysis is so miraculous, but I tend to take that as axiomatic. :-) </p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/2865#2865 Answer by Ilya Nikokoshev for Why are characters so well-behaved? Ilya Nikokoshev 2009-10-27T18:42:19Z 2009-10-27T18:42:19Z <p>There are lots of good points in other answers, so I want to to add one specific thing about why representations are uniquely defined by their characters.</p> <p>Irreducible representations are uniquely determined by <strong>highest weight</strong>, therefore all representations can be described by functions <code>highest weight -&gt; ZZ</code>. Now going from highest weight to the character and back is an invertible transformation on the space of such functions (moreover, the matrix of this transformation is uppertriangular).</p> <p>This construction can be encountered in several generalizations, e.g. perverse sheaves in <strong>geometric representation theory</strong>, so it appears to be a fundamental representation-theoretical fact.</p> http://mathoverflow.net/questions/2795/why-are-characters-so-well-behaved/69124#69124 Answer by Phil Isett for Why are characters so well-behaved? Phil Isett 2011-06-29T15:43:15Z 2011-06-29T15:48:30Z <p>To add to D. Savitt's comment: Given $g \in G$ ($G$ a finite group), the element $\rho(g) \in GL(V)$ is diagonalizable over the complex numbers since its minimal polynomial is separable. More specifically, it is a divisor of $x^n - 1$ for some $n$ -- once you have the minimal polynomial, you can compute the projection operators to the various eigenspaces as polynomials in $\rho(g)$, but generally this will require complex coefficients and working in a field which is not algebraically complete will only allow you to project to the kernel of irreducible factors.</p> <p>Now, a diagonal matrix is determined up to conjugacy by its eigenvalues. And as was pointed out $\mbox{tr } \rho(g^k)$ gives the sum of $k$th powers of the eigenvalues. From these numbers, one can actually recover the set of eigenvalues of the matrix. You can prove this with symmetric functions, but you can also view the spectrum as a measure, and view $\mbox{tr } \rho(g^k)$ as the k'th (complex) moment / Fourier coefficient of the measure. These numbers, together with their complex conjugates give all the Fourier coefficients. They are periodic, so you do not need all of them (you could apply Stone Weierstrass, but that would be very wasteful). We already know the finitely many (say, $n$) candidate eigenvalues from the fact that they satisfy $z^n - 1 = 0$, so we can easily design a polynomial which vanishes on all but one of the eigenvalues and hence determine the multiplicity of each eigenvalue. Namely: $f(z) = \prod_{i \neq j} (z - \lambda_i)$. Then $\mbox{tr } f( \rho(g) )$ (the trace of the projection operator up to a constant) tells you the multiplicity of $\lambda_j$. (Is this exactly the proof through symmetric functions? If not, what is the relationship?)</p> <p>Inspecting the above argument, it does not seem like the complex numbers play too large a role. But even for a more general unitary representation, the eigenvalues live on a circle and we can view them as a measure $\mu$. We can still obtain $\int z^k d\mu$ for all integers $k$. By Stone-Weierstrass or Fourier inversion, we get the whole spectrum in this way.</p> <p>This is the extent to which I've thought about the above things so I'd be really happy to know if anyone can say more about it or give more points of view. For example, what very critical type of information do you miss out on by only considering one cyclic subgroup at a time? What about other unitary representations, possibly infinite groups or possibly infinite dimensional? (Should I start a separate thread?)</p>