Geometric Representation Theory (Lecture 7)
Posted by John Baez
Today in the Geometric Representation Theory seminar, Jim continues to explain Hecke operators.
In his last talk, he proved a wonderful but actually very easy theorem. Whenever a group acts on a set $X$, we get a representation of that group on the vector space $\mathbb{C}^X$ in an obvious way. This is called a permutation representation. And the theorem says: if $G$ is a finite group acting on finite sets $X$ and $Y$, we get a basis of intertwining operators
$\mathbb{C}^X \to \mathbb{C}^Y$
from atomic invariant relations between $X$ and $Y$: that is, relations that are invariant under the action of our group, and can’t be chopped into a logical ‘or’ of smaller invariant relations.
The intertwining operators we get this way are called ‘Hecke operators’. They’re very easy to get ahold of, because we see atomic invariant relations all the time in geometry: a typical example is ‘the point $x \in X$ lies on the line $y \in Y$’.
So: Hecke operators link the world of geometry to the world of group representations.
This time, in lecture 7, Jim explains two nice applications of Hecke operators. In the second one, he starts using Hecke operators to grind down the permutation representations of the group $n!$ into irreducible pieces, one for each $n$box Young diagram. This process involves thinking of the representations of $n!$ as forming a ‘2Hilbert space’ — a categorified Hilbert space. In fact, the category of representations of any compact topological group is a 2Hilbert space. In this talk, Jim keeps the concept of 2Hilbert space quite loose and informal. Since I’ve written a paper on this subject, I’ll include a link to that if you want a precise definition.

Lecture 7 (Oct. 18) 
James Dolan on applications of Hecke operators. Theorem: if
a finite group $G$ acts in a doubly transitive way on a finite
set $X$, then the resulting permutation representation of $G$ on $\mathbb{C}^X$ is the direct sum of two irreducible representations, one being the trivial representation. Proof: every permutation representation contains the trivial representation, and
there are only two Hecke operators from $\mathbb{C}^X$ to itself.
Lemma: if $G$ is a finite group, $Rep(G)$ is a 2Hilbert space with
the irreducible representations of $G$ as an orthonormal basis.
(This is a combination of Schur’s Lemma and Maschke’s Theorem.)
Another application: using GramSchmidt orthonormalization to take the permutation representations of $G = n!$ coming from $n$box Young diagrams and turn them into an “orthonormal basis” of $Rep(G)$: that is, a complete collection of irreducible representations. Beginning of an explicit calculation for $n = 4$.

Streaming
video in QuickTime format; the URL is
http://mainstream.ucr.edu/baez_10_18_stream.mov  Downloadable video
 Lecture notes by Alex Hoffnung

Streaming
video in QuickTime format; the URL is
 Supplementary reading: John Baez, Higherdimensional algebra II: 2Hilbert spaces.
Re: Geometric Representation Theory (Lecture 7)
A comment about doubly transitive $G$sets $X$.
$X \times X$ has two orbits, the diagonal and the undiagonal, as pointed out in the lecture, and these give a basis for $Hom_{G}(\mathbb{C}[X],\mathbb{C}[X])$. But I think the diagonal gives the identity operator rather than the projection onto the trivial representation, as claimed. I’ll explain, and please correct me if I’m wrong.
If $X=\{x_1, ..., x_n\}$, then a basis for $Hom(\mathbb{C}[X],\mathbb{C}[X])$ is given by things of the form $x^{i} \otimes x_{j}$, where $x^{i}(x_{j})=\delta_ij$. The diagonal orbit then gives the operator $1=\sum_{i} x^{i}\otimes x_{i}$, right?
On the other hand, the undiagonal gives the operator $\sum_{i \neq j} x^{i} \otimes x_{j}$.
The sum of these two is the projection onto the trivial, I think.