Recall
Definition 34.1. The character of a representation is the function , .
is a class function on , i.e. only depends on the conjugacy class of .
Example 34.2.
- hence .
- For a permutation representation associated to acting on , .
Orthogonality of characters
Recall:
If is a representation of , the invariant part is .
Proposition 34.3. is a projection onto : and .
So:
If are representations of , then , so
but if and are irreducible, then by Schur’s lemma,
Definition 34.4. Define a Hermitian inner product on the space of class functions by
For characters of representations, by the above,
Theorem 34.5. The characters of irreducible representations of are orthonormal for :
This implies the characters of irreducible representations are linearly independent class functions!
Corollary 34.6. The number of irreducible representations of is at most the number of conjugacy classes of . (We’ll see later that they are in fact equal.)
Corollary 34.7. Every representation of is completely determined by its character: denoting the irreducible representations by , any representation , where
Corollary 34.8. For any representation , , and is irreducible iff .
This is useful because, given a representation , it gives information about the number of irreducible summands making up . E.g.:
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is |
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irreducible |
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a direct sum of different irreducibles |
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a direct sum of different irreducibles |
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either a sum of different irreducibles, or twice the same one |
The regular representation
We now apply this to the regular representation: = the vector space with basis , on which acts by permuting the basis vectors by left multiplication: .
Now let be the irreducible representations of , and write . What are the ?
Since acts by permutation matrices, ; but unless there are no fixed points, so
So
Hence each appears times in the regular representation .
And now the third corollary above (the one saying ) implies
where by direct calculation: .
Corollary 34.9. The irreducible representations of satisfy .
At this point we actually have a lot of information about the irreducible representations of and their characters.
The character table of
Example 34.10 (). The conjugacy classes are: , of size ; transpositions, size ; -cycles (); -cycles (); pairs of transpositions ().
We know irreducible representations: = trivial, = alternating, = standard. Character table:
- Row : acts by , so .
- Row : .
- Row : to find this one, = the permutation representation , and , so .
Quick check: these are indeed orthonormal!
However: , so there are other irreducible representations! In fact:
- The first corollary above (the bound by the number of conjugacy classes) says we’re missing at most two ( irreducible representations conjugacy classes );
- since we’re missing , which is not a square, we’re missing exactly two, of dimensions and (so that ).
How do we build the missing entries? Start by looking at tensor products of known representations.
For a start, the tensor product of an irreducible representation with a -dimensional representation is still irreducible ( a -dimensional representation has the “same” invariant subspaces), so we can look at (twist the standard representation by ), which has
This is indeed irreducible () and different from !
We have one last -dimensional irreducible representation to find!
Since is also a -dimensional irreducible representation, necessarily . This implies , i.e. on the odd conjugacy classes ( and ).
The orthogonality relations allow us to find the rest of without having constructed it!
Note that means the eigenvalues are and ! (They are roots of unity summing to .)
This gives a big clue about : the normal subgroup is in the kernel of , i.e. factors through the quotient . (Recall: acts on the set of splittings of into pairs; there are of those.)
Under this quotient, transpositions and -cycles transpositions, -cycles -cycles, and the character becomes
This is the standard representation of , “pulled back” to by !
The other option to construct is to look at : . We have
so contains (of dimension ), and this leaves us one copy of the missing irreducible . So
(and we can find by subtracting the others from ).
The character table of
Example 34.11 (, the alternating subgroup of ). This has conjugacy classes:
(The -cycles are one conjugacy class in but split in ; see Lecture 23.)
- We can start by restricting to the irreducible representations of . Some become isomorphic (e.g. the alternating representation has elements of acting by , so it becomes trivial); others might become reducible. This is feasible but tricky (largely ’s fault).
- Or we can go at it directly! We know there are at most irreducible representations, with , including the trivial representation of dimension , so the only option is .
The three -dimensional representations correspond to , which contains the trivial homomorphism (the trivial representation) and two other elements… Observe that is a normal subgroup, with , so this gives the answer:
Concretely, let ; then the rank representations are below (the act by ), and the last one, by orthogonality, is :
This is the restriction to of the standard representation of !
(Note: .)
Notes not reached in this lecture, covered at the start of {{LINK:L35}}
Characters form a basis of the class functions
Next, we want to prove: the characters of irreducible representations are actually an orthonormal basis (for ) of the space of class functions .
The proof uses a more general averaging/projection formula. Last time we saw that
is the projection onto the invariant subspace ( the trivial summands in ).
Proposition 34.12. Given any class function and any representation of , let
Then is -linear (equivariant).
Theorem 34.13. The characters of the irreducible representations of form an orthonormal basis (for ) of the space of class functions , and irreducible representations conjugacy classes.
Proof. To show the characters of the irreducible representations span all class functions, it suffices to show:
Given any class function and an irreducible representation , consider as above. Then by Schur’s lemma, , where , . So
So if for all irreducible , then for all ; hence, by considering direct sums, for all representations of , in particular for the regular representation of (the permutation representation for left multiplication on ).
So, for the regular representation (with the identity element),
Since the are linearly independent, this implies for all , i.e. . ∎
Along the way, we found: for irreducible, look at for . Then , where
Proposition 34.14. If is any representation of and is its decomposition into irreducibles, then
is the projection onto the summand (i.e. the identity on that summand, on the others).
(The case of the trivial representation is our previous projection formula for .)