A nice proof that the irreducible characters form a basis for class functions


Warning: This is a very shoddily organized post and can be vastly improved. The method of proof is still nice however.

Let G be a finite group of size g and k an algebraically closed field such that g \neq 0 in it. Let V_1,\dots, V_n be the irreducible representations of G over k and \chi_1,\dots,\chi_n the corresponding characters.

A class function f: G \to k is a set function such that f is constant on conjugacy classes. That is, for all g,h \in G, f(h^{-1}gh) = f(g). It is easy to see that any character is a class function by cyclicity of the trace function.

Let us denote by H the vector space of class functions and W the vector space of functions spanned by the \chi_i within H. It is natural to wonder about the size of W relative to H. In fact, the two vector spaces are equal!

To prove this, let us consider the algebra k[G]. By our hypothesis and the usual argument of Maschke’s theorem, this is a semisimple ring. That is, k[G] is semisimple as a G representation. Let r_G be the character attached to this so called regular representation. A simple computation shows us that r_1 = g and r_G = 0 otherwise.

Moreover, for any irreducible representation V_k, we see from last time that:

Hom_G(V_k,k[G]) = \frac{1}{g}\sum_g r_G(g^{-1})\chi_k(g) = \chi_k(1) = \dim V_k.

Therefore, by semisimplicity

k[G] \cong \oplus_k V_k^{\dim V_k}

as representations. Moreover, the orthogonality relations from last time also show that the \chi_k are linearly independent. That is, \dim W = \# irreducible characters.

Taking the endomorphism ring on both sides (as modules over k[G]), we obtain:

k[G] \cong M_{\dim V_k}(k)

as algebras. Considering dimensions, we have incidentally shown that g = \sum_k \dim V_k^2. Moreover, let us consider the central elements in k[G]. They are easily seen to exactly be the elements of the form \sum_g f(g)g for $f:G \to k$ a class function.

On the other hand, the central elements in M_n(k) are always just k. Therefore, the dimension of the center of k[G] (= vector space of class functions) is equal to the number of irreducible representations. Therefore, \dim W = \dim H and we are done.

The above map can be thought of as the representation of a class function in the basis defined by the functions c_k\overline{\chi_k} for some appropriate constants c_k. This is because the element:

\lambda_i = \sum_{g}\overline{\chi_i}g \in k[G]

maps to a a diagonal element in M_{\dim V_j}(k) that can be calculated by taking the trace. That is to say, the image of \lambda_i in $latex M_{\dim V_j}(k) is given by:

\frac{1}{\dim V_k}\sum_{g}\overline{\chi_i(g)}\chi_j(g) = \frac{g}{\dim V_k}\delta_{ij}

by the orthogonality relations. We see that c_k = \frac{\dim V_k}{g}.




Schur’s Lemma and the Schur Orthogonality Relations.

Both Schur’s Lemma and the Schur Orthogonality relations are part of the basic foundation of representation theory. However, the connection between them is not always emphasized and the Orthogonality relations are proven more computationally.

The standard proofs of the relations never made sense to me, however there is very direct way to derive them from Schur’s Lemma (which makes perfect sense to me!) and simple facts about projections on vector spaces. More importantly, it gives a categorical interpretation of the inner product. I think this approach should be emphasized way more than it currently is and I hope this post will go a tiny way towards fixing that.


Continue reading

Congruent Numbers and Elliptic Curves

A congruent number n is a positive integer that is the area of a right triangle with three rational number sides. In equations, we are required to find rational positive numbers a,b,c such that:

\displaystyle a^2+b^2 = c^2    and    \displaystyle n = \frac12 ab.                       (1)

The story of congruent numbers is a very old one, beginning with Diophantus. The Arabs and Fibonacci knew of the problem in the following form:

Find three rational numbers whose squares form an arithmetic progression with common difference k.

This is equivalent to finding integers X,Y,Z,T with T\neq 0 such that Y^2 - X^2 = Z^2 - Y^2 = k which reduces to finding  a right triangle with rational sides

\displaystyle \frac{Z+X}{T}, \frac{Z-X}{T}, \frac{2Y}{T}

with area k. This is the congruent number problem for k. The Arabs knew several examples of congruent numbers and Fermat stated that no square is a congruent numbers. Since we can scale triangles to assume that n is square free, this is equivalent to saying that 1 is not a congruent number.

As with many other problems in number theory, the proof of this statement had to wait four centuries for Fermat. The problem led Fermat to discover his method of infinite descent.

In more recent times, the problem has been fruitfully translated into one about Elliptic Curves. We perform a rational transformation of the defining equations (1) for a congruent number in the following way. Set x = n(a+c)/b and y = 2n^2(a+c)/b. A calculation shows that:

\displaystyle y^2 = x^3 - n^2x.                                          (2)

and y \neq 0. If y = 0, then a=-c and b = 0 but then n = \frac12 ab = 0. Conversely, given x,y satisfying (2), we find a = (x^2-y^2)/y, b = 2nx/y and c = x^2+y^2/n and one can check that these numbers satisfy (1).

The projective closure of (2) defines an elliptic curve that we will call E_n. We are interested in finding rational points on it that do not satisfy y=0. I will prove that n is a congruent number precisely when E_n has positive rank.

The proof is an interesting use of Dirichlet’s Theorem on Arithmetic Progressions and some neat ideas about Elliptic Curves and their reductions modulo primes. I will essentially assume the material in Silverman’s first book and the aforementioned Dirichlet’s Theorem.


Continue reading

Descent on Vector Spaces and Cohomology

It is quite often of interest to study the properties of some variety {X/\mathbb Q}. However, it is generally much easier to study varieties over algebraically closed fields and so we need some way of translating a property of {X_{\overline{\mathbb Q}}/\overline{\mathbb Q}} to {X/\mathbb Q} . This idea is known as descent and in this post, I would like to say a little bit about the simplest example of descent – over vector spaces.

Let {L/K} be an extension of fields and {V} a vector space over {K} . Consider {W = V\otimes_K L } . The Galois Group {G = \mathop{Gal}(L/K)} acts on {W} through the second factor (not by linear actions but by semi-linear ones – see below). One can consider the {K} -vector space {W^G} . This is the vector space fixed by {G} .

Theorem 1 (Descent of Vector Spaces) The natural map {W^G\otimes_K L \rightarrow W} is an isomorphism. In particular, if {W} is finite dimensional, then {\dim_K W^G = \dim_L W} .

It is not hard to prove this theorem directly (Theorem 2.14 in these notes of K. Conrad) but I would like to relate it to another theorem. This is also well known and is a generalization of the famous Hilbert’s Theorem 90. Let {{GL}_n(L)} be the group of invertible {n} -dimensional matrices over {L} and consider the cohomology {H^1(G,{GL}_n(L))} . This is not a group unless {n=1} since {{GL}_n(L)} is non-commutative in general. However, it is a pointed set and we have the following theorem:

Theorem 2 (Hilbert’s 90) \displaystyle H^1(G,{GL}_n(L)) = \{0\}.

Hilbert stated the above theorem (in a disguised form) for {n=1} and {L/K} a finite cyclic extension. Noether generalized the theorem to arbitrary extensions. I do not know who is responsible for the generalization to general linear groups but I saw this theorem first in Serre’s “Galois Cohomology”.

In this post, I will show that the above theorems are equivalent in the following sense:


Continue reading