is given by if the natural diagram commutes:
such a map is called called as an intertwining map or an equivariant map.
V --f--→ V
W --f'-→ W
§ Schur's lemma
The only equivariant maps between irreducible representations is either the
zero map or a scalar multiple of the identity map. This is stronger than
saying that the equivariant map is a diagonal matrix; scalar multiple of
identity implies that all dimensions are scaled uniformly.
The main idea of the proof is to show that the kernel and image of the
intertwining map is an irreducible subspace of retrospectively. Since
the maps are irreducible, we must have the the intertwining is either the zero
map, or a map into the full group. This forces the map to be zero or a scalar
multiple of the identity.
One way to look at this is that for irreps and
, the dimension of is either 0 or 1 (scalings of identity).
§ Schur orthogonality relations
We consider representations "one matrix index" at a time, and show that
the matrix entries of irreducible representations is going to be orthogonal
The proof is to consider representations
, and an intertwining map .
How do we involve all of at once? Recall that since is an intertwining, we must have:
Now, since is invertible (it must be since it's a member of ), I can rewrite
the above as:
This needs that is an intertwining map. Can we generalize this
to any linear map? Suppose that is a linear map, not
necessary intertwining. Let's induce an intertwining map from :
We average the intertwining condition of to produce an appropriate .
Is this an intertwining? Yes, because when we compute ,
the averaging trick winds up shifting the index, exactly as it does for the inner product:
Thus, for every linear map , if the representation is not
isomorphic to the representation , then , or:
The above equality holds for all indexes and for all choices
of (since can be any linear map). In particular,
we can choose for arbitrary .
This gives us the equation:
This tells that we can choose any index and index and these will be orthogonal,
when viewed as vectors "along" the set of matrices.
If the representation is a one-dimensional representation/character, then we have no freedom
in indexing, and the above becomes:
Thus, different characters are all orthogonal.
§ Inner product of class functions
we impose an inner product relation on the space of class functions (complex valued functions
constant on conjugacy classes) , given by
where is the complex conjugate.
Using the Schur orthogonality relations, we immediately deduce that the inner product
of two irreducible characters can be viewed as the schur orthogonality applied to their
(only) matrix entry at location (1, 1). Thus, irreducible characters will be orthogonal,
and equal characters will have inner product 1.
§ Regular representation
The "Cayley-style" representation one would naturally dream up. For a group ,
build a vector space whose basis is given by elements of . Have
act on by seding to . Ie, act with as a permutation on .
This gives us a "large" representation. For example, the permutation group of
letters will have a regular representation of basis vectors.
This representation contains every irrep. The idea is to show that the dot
product of the trace of the regular representation with every other irrep is
nonzero. Furthermore, since the regular representation has finite dimension,
this tells us that there are only finitely many irreps: the irreps correspond
to subrepresentations, and a finite representation only has finitely many
subrepresentations. This makes the idea of classifying irreps a reasonable task.
§ Character of the regular representation
Theorem: The character of the regular representation is given by ,
- The matrix for the identity element is the identity matrix, and the sizeof the matrix is the size of the vector space, which is sincethere's a basis vector for each element of . Thus, .
- For any other element , the regular representation will be a permutation matrixwith no fixed points. Thus, the diagonal of the matrix is all zeros, and hence .
§ Regular representation contains all other irreps
The inner product of the character of the regular representation with any other
irrep is going to be:
Thus, the regular rep contains the other irreps, since the character of the regular rep has non-zero
inner product with irrep, and irrep characters are all orthogonal.
§ Abelian groups are controlled by characters
Since abelian groups map to automorphism that all commute with each other, we can
simultaneously diagonalize these matrices. Thus, we only need to consider
the data along each diagonal, which is independent. This reduces the representation
to a direct sum of scalars / 1D representations / characters.