Difference between revisions of "Size-degree-weighted characters are algebraic integers"

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(Proof)
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==Statement==
 
==Statement==
  
Suppose <math>k</math> is an algebraically closed field of characteristic zero, and <math>G</math> is a [[finite group]]. Let <math>\rho</math> be an irreducible linear representation of <math>G</math> over <math>k</math>, and <math>\chi</math> be the character corresponding to <math>\rho</math>. Let <math>c</math> be a conjugacy class in <math>G</math> and <math>g \in c</math> be an element. Then:
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Suppose <math>k</math> is an algebraically closed field of characteristic zero, and <math>G</math> is a [[finite group]]. Let <math>\rho</math> be an [[fact about::irreducible linear representation]] of <math>G</math> over <math>k</math>, and <math>\chi</math> be the character corresponding to <math>\rho</math>. Let <math>c</math> be a conjugacy class in <math>G</math> and <math>g \in c</math> be an element. Then:
  
 
<math>\frac{|c|\chi(g)}{\chi(1)}</math>
 
<math>\frac{|c|\chi(g)}{\chi(1)}</math>
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==Related facts==
 
==Related facts==
  
* [[Characters are cyclotomic integers]]: This statement holds in much greater generality. In particular, it holds over any field.
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* [[Characters are cyclotomic integers]]: This statement holds in much greater generality. In particular, it holds over any field and it does not require the linear representation to be irreducible.
 
* [[Characters are algebraic integers]]
 
* [[Characters are algebraic integers]]
  
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Let <math>G</math> be the cyclic group of order three and <math>\R</math> be the field. <math>G</math> has an irreducible two-dimensional linear representation over <math>\R</math> given by rotation by multiples of <math>2\pi/3</math>. For a non-identity element <math>g</math> of <math>G</math>, <math>\chi(g) = -1</math> for the corresponding character, while <math>\chi(1) = 2</math>. Thus, the expression works out to <math>-1/2</math>, which is not an algebraic integer.
 
Let <math>G</math> be the cyclic group of order three and <math>\R</math> be the field. <math>G</math> has an irreducible two-dimensional linear representation over <math>\R</math> given by rotation by multiples of <math>2\pi/3</math>. For a non-identity element <math>g</math> of <math>G</math>, <math>\chi(g) = -1</math> for the corresponding character, while <math>\chi(1) = 2</math>. Thus, the expression works out to <math>-1/2</math>, which is not an algebraic integer.
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===Breakdown for a representation that is not irreducible===
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The same example as the above (the one for breakdown over a field that is not algebraically closed) works. Specifically, the irreducible representation over <math>\R</math> can be viewed as a reducible representation over <math>\mathbb{C}</math>.
  
 
==Proof==
 
==Proof==

Revision as of 22:33, 22 February 2011

Statement

Suppose k is an algebraically closed field of characteristic zero, and G is a finite group. Let \rho be an Irreducible linear representation (?) of G over k, and \chi be the character corresponding to \rho. Let c be a conjugacy class in G and g \in c be an element. Then:

\frac{|c|\chi(g)}{\chi(1)}

is an algebraic integer.

Related facts

Applications and further results

Breakdown for a field that is not algebraically closed

Further information: cyclic group:Z3

Let G be the cyclic group of order three and \R be the field. G has an irreducible two-dimensional linear representation over \R given by rotation by multiples of 2\pi/3. For a non-identity element g of G, \chi(g) = -1 for the corresponding character, while \chi(1) = 2. Thus, the expression works out to -1/2, which is not an algebraic integer.

Breakdown for a representation that is not irreducible

The same example as the above (the one for breakdown over a field that is not algebraically closed) works. Specifically, the irreducible representation over \R can be viewed as a reducible representation over \mathbb{C}.

Proof

The proof is based on the idea of the convolution algebra on conjugacy classes.

Description of the convolution algebra on conjugacy classes

Let C(G,\mathbb{Z}) be a \mathbb{Z}-subalgebra of the group ring \mathbb{Z}(G) defined as follows: as a group, it is the free Abelian group on all indicator class functions for conjugacy classes. In other words, for each conjugacy class, we have a free generator that corresponds to the sum of elements of that conjugacy class.

The structure constant for multiplication of elements of C(G,\mathbb{Z}) is defined as follows: given conjugacy classes c_1, c_2, c_3, the coefficient of the c_3-indicator function in the product of the c_1-indicator function and the c_2-indicator function is the number of ways of writing g_1g_2 = a where g_i \in c_i, and a a fixed element of c_3.

Note that all the structure constants are integers.

A homomorphism from this convolution algebra to the matrix ring

The representation \rho gives rise to a homomorphism from C(G,\mathbb{Z}) to the matrix ring M_n(k). The indicator function for a conjugacy class c goes to the matrix given by:

\sum_{g \in c} \rho(g).

This sum commutes with \rho(h) for all h, and thus, by Schur's lemma, the sum is a scalar matrix. The trace of the sum is |c|\chi(g), so the sum must be a scalar matrix with scalar entry:

\frac{|c|\chi(g)}{\chi(1)}.

Thus, the set of scalar matrices with entries described as above additively generate a group that is a ring under multiplication. The structure constants for this ring are the same as the structure constants for the convolution algebra. A result from algebraic number theory now tells us that this forces the entire ring to be a ring of algebraic integers, and in particular, the generating elements are algebraic integers.