Re: dimension of an irreducible representation
- From: José Carlos Santos <jcsantos@xxxxxxxx>
- Date: Mon, 23 Apr 2007 17:33:40 +0100
On 23-04-2007 13:26, oferlock@xxxxxxxxx wrote:
for a finite group why does the dimension of the representation divide
the order of the group? Any hints welcome.
What do you mean by "the representation"? What happens is that for a
finite group, the dimension of every _irreducible_ representation
divides the order of the group. This is a theorem due to Burnside.
Here's a sketch of the proof. Suppose that your group G acts on a
finite-dimensional complex vector space V and that this action is
irreducible. Let C be a conjugacy class in G. Then the map m_C from V
into itself defined by m_C(v) = sum_{g in C} g.v is such that, for each
g in G and each _v_ in V, g.m_C(v) = m_C(g.v). So, since the action is
irreducible, Shur's lemma tells us that m_C is the product by a scalar
s_C. Therefore,
trace(m_C) = s_C.dim V.
But if X_V(C) is the character of the action of G on V computed at one
(and therefore any) element _g_ of C, then you also have:
trace(m_C) = #C.X_V(C).
So, this proves that
s_C.dim V = #C.X_V(g).
On the other hand,
|G| = sum_{h in G}|X_V(h)|^2
= sum_C #C.|X_V(C)|^2
= sum_C #C.X_V(C).conj(X_V(C))
= sum_C s_C.dim(V).conj(X_V(C)).
So, |G|/dim V = sum_C s_C.conj(X_V(C)). But |G|/dim V is a rational
number and it can be proved that sum_C s_C.conj(X_V(C)) is an algebraic
integer. So, |G|/dim V is an integer.
A generalization of this theorem is: the dimension of an irreducible
representation of V divides the index of the center of G.
Best regards,
Jose Carlos Santos
.
- References:
- dimension of an irreducible representation
- From: oferlock
- dimension of an irreducible representation
- Prev by Date: Re: Fermat's Last theorem short proof
- Next by Date: Re: Fermat's Last theorem short proof
- Previous by thread: Re: dimension of an irreducible representation
- Next by thread: Formula to combine physical measurements to one?
- Index(es):
Relevant Pages
|
