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Let the [[finite group]] in question be G.
Let the [[finite group]] in question be G. All the linear reps in this article are finite dimensional.


[[Schur's lemma]]:
[[Schur's lemma]]:
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Invariant here means, if you think about it, the linear transformation commutes with &rho;(g) for all g in G. Suppose we have an invariant linear transformation A which is NOT a multiple of the identity. Let &lambda; be one of its eigenvalues. Then, A-&lambda;<b>1</b> is also invariant. Look at its kernel. It can't be zero dimensional since its determinant is zero, but it can't be the entire vector space either because A is NOT a multiple of the identity. The kernel is obviously a proper subrep. Hence, the rep is reducible.
Invariant here means, if you think about it, the linear transformation commutes with &rho;(g) for all g in G. Suppose we have an invariant linear transformation A which is NOT a multiple of the identity. Let &lambda; be one of its eigenvalues. Then, A-&lambda;<b>1</b> is also invariant. Look at its kernel. It can't be zero dimensional since its determinant is zero, but it can't be the entire vector space either because A is NOT a multiple of the identity. The kernel is obviously a proper subrep. Hence, the rep is reducible.

If we have a subrep of a subrep, then the quotient space is another rep.


[[Group algebra]]:
[[Group algebra]]:

Revision as of 02:46, 3 August 2004

Let the finite group in question be G. All the linear reps in this article are finite dimensional.

Schur's lemma:

If ρ is a rep of G, then all linear transformations mapping the rep to itself which are invariant under the action of G are multiples of the identity if and only if the rep is irreducible.

Invariant here means, if you think about it, the linear transformation commutes with ρ(g) for all g in G. Suppose we have an invariant linear transformation A which is NOT a multiple of the identity. Let λ be one of its eigenvalues. Then, A-λ1 is also invariant. Look at its kernel. It can't be zero dimensional since its determinant is zero, but it can't be the entire vector space either because A is NOT a multiple of the identity. The kernel is obviously a proper subrep. Hence, the rep is reducible.

If we have a subrep of a subrep, then the quotient space is another rep.

Group algebra: Let C[G], the group algebra of G be the associative algebra generated the elements of g subject to g1g2 (algebra product)=g1g2 (group product). It is a |G| dimensional algebra.

C[G] is an algebra representation of G. (See Peter-Weyl for topological groups).

C[G] as a vector space is a rep of G×G. More specifically, if g1 and g2 are elements of G and h is an element of C[G] corresponding to the element h of G,

(g1,g2)[h]=g1h g-12.

In particular, C[G] (as a linear rep) decomposes into the direct sum of n copies of each irrep from each equivalence class where n is the dimension of the irrep.

If ρ is an n dimensional irrep of G, then the subspace of C[G] consisting of the elements for all y in the irrep and x in the dual irrep is a n×n dimensional irrep of C[G] as a rep of G×G, which of course would be n direct sum copies of equivalent n dimensional irreps of G.

There is a mapping from G to the complex numbers for each rep called the character given by the trace of the linear transformation upon the rep generated by the element of G in question. All elements of G belonging to the same conjugacy class have the same character.

Tr[ρ(ghg-1)]=Tr[ρ(g)ρ(h)ρ(g)-1]=Tr[ρ(h)] (cyclic property of the trace)

The number of conjugacy classes of G is the same as the number of inequivalent irreps of G.

The characters are orthogonal to each other.