So what is rep theory? It’s taking a group \(G\) and then associating linear maps (also known as matrices if you are willing to fix a basis) to the group elements in a “structure-preserving” way. More specifically, it is a homomorphism \[\phi: G \to GL(V)\] This means \(\phi(gh) \cdot v = \phi(g)\phi(h) \cdot v.\)
IN this blog post I hope to introduce you to some of the important theorems. But really if you want to learn rep theory you gotta work some examples yourself.
Example. find all the representations of \(S_n\) up to isomorphism.
- For \(n=3\).
- For all \(n\)
ok so first, some definitions:
Definition. \(G\)-invariant subspace: subspace fixed by action of \(G.\) i.e. \(W\subset V\) such that \(\phi(g) W = W\) for all \(g\in G\).
begin defin irreducible rep: no proper \(G\)-invariant subspace end defin
Remark. In this blog post all groups are assumed to be finite, and all vector spaces are over the complex numbers. Some stuff breaks if we don’t have these assumptions.
shur
Lemma. Let \(\phi: G\to GL(V), \phi': G\to GL(V')\) be irreducible \(G\)-representations. Let \(T:V\to V'\) be a map that commutes with the reps, i.e. \(T\circ \phi = \phi'\circ T.\) Then \(T\) is a scalar.
Proof. \(\ker T\) is \(G\)-invariant subspace, but not proper subspace (none exist due to the irreducibility of the representations). so the map is the zero-map or an isomorphism.
Take an eigenvector \(\lambda\) of \(T\) : this must exist due to us working over the complex numbers. Then we see that \(T=\lambda I\).
maschke
Theorem. And representation can be decomposed as a direct sum of irreps
Proof. Define a \(G\)-invariant hermetian form on the space. Then, if \(W\) is \(G\) invariant, \(W^\perp\) is also \(G\)-invariant. Thus, the rep decomposes into blocks as \(W\oplus W^\perp\)
main
character: trace character inner-product: hermetian thing
Theorem. - num irreps (up to iso) is num conj classes - irreps are orthonormal - \(\sum d_i^2 = |G|, d_i | |G|\) where \(d_{i}\) is the dimension of the \(i\)-th irrep.
Proof. class functions: constant on conj classes. characters are basis for class functions.
bunch of complicated stuff
regular rep
some random facts that are kind of cool
can we read the normal subgroups of a group off of its character table? good question! one idea is, well normal subgroups are unions of conjugacy classes, so take a bunch of unions of conjugacy classes. Lagrange’s theorem them (subgroups have order dividing order of the group). Then you would really have to check if it’s closed as far as I can tell.
ok but here’s a better way.
Theorem. Normal subgroups appear as kernels of irreps
EDIT: this is actually false Some normal subgroups will appear as kernels of irreps, but in general there may be many more normal subgroups than can possibly occur as kernels of irreps.
Lemma. For character \(\chi_\rho\) of a dimension \(d\) irrep \(\rho\), \(\chi(g)=\chi(e)=d \iff \rho_g = Id.\)
Proof. Say that we have a representation \(g\mapsto R_g\) where \(g^k=e\) and thus of course \(R_g^k = I\). Clearly any eigenvector \(v\) for \(R_g\) satisfies \(R_g v = \lambda \implies R_g^k v = \lambda^k v = v \implies \lambda^k=1.\) so \(\lambda\) is a \(k\)-th root of unity. The character of a \(d\)-dimensional irrep must be the sum of \(d\) such eigenvalues.
In order for the sum to be \(d\) it must be the sum of all \(1\)’s. Only \(Id\) has all eigenvalues being \(1\).
Proof. Let \(H\) act on \(\mathbb{C}\) by the trivial action (doing nothing!). Now, consider the induced representation \(\rho\) on \(G\) from \(H\) (look down to see what this means.)
the following stuff is kind of false; I have yet to completely understand why We claim that \(\ker \rho = H\). We can see this by Mackey’s formula, and the fact that \(H\) is normal so conjugation fixes \(H\). Oh and we are also applying the lemma here.
Ok, then by Maschke’s theorem, we can write \(\rho\) as a direct sum of irreps. But to be in the kernel of a direct sum you must be in the kernel of each (non-zero) summand.
Hence \(H\) is in the kernel of some non-zero number of the irreps.
frobenius
induction: build a rep from a rep of a subgroup. To do this you basically make a bunch of copies of the rep on the subgroup and move around between them. restriction: build a rep of a subgroup from a rep on the group. you just take the restriction
\[Hom(\texttt{Res}_H^G V, W)^H \cong Hom(V, \texttt{Ind}_H^G W)^G\]
And the formula also holds for characters.
Also it is probably equivalent in some sense to Mackey’s formula \[\chi_{ind}(g) = \frac{1}{|H|}\sum_{x\in G, xgx^{-1}\in H} \chi(xgx^{-1}).\]
how many MIT students does it take to prove Frobenius Reciprocity? apparently more than 6.
specht modules
ok we can construct the irreps of \(S_n\) by defining - young diagram: partition of n arranged into a left justified square diagram - tableux: put numbers in young diagram - tabloid: we don’t care about numbers order within rows - polytabloid: linear combo of tabloids
and then apparently polytabloids with the obvious group action of \(S_n\) corresponds to irreps of \(S_n.\) crazy!