## Representation Theory: Basics and Heisenberg Representation (2) May 21, 2010

Posted by Sarah in Uncategorized.
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More notes from Shamgar Gurevich’s lectures bringing representation theory to the applied-math populace.

We start with an example. Let $G = S_4 = Aut\{a, b, c, d\}$. Denote the irreducible representations of G by $G^v$. There are 5 of these, one for each conjugacy class (in the case of the symmetric group, a conjugacy class is a partition.)

Constructions: $S_4$ goes into the complex numbers over the tetrahedron, which is $V_0 \oplus V_{const}$, where $V_0$ is the three-dimensional space of the front face of the tetrahedron. The representation is
$\rho_3: S_4 \to GL(V_0).$
Why is $\rho_3$ irreducible?

General tool: the intertwining number, $\langle \rho, \pi \rangle$.

Proposition: $\rho$ is irreducible iff $\langle \rho, \rho \rangle =1$.
Proof: $\rho = \oplus m_i \rho_i$.
$1 = = \sum m_i^2$
Since these are integers, the equation is true iff there is only one subrepresentation.

In our case,
$\rho = \rho_3 + \rho_1$
$\langle \rho, \rho \rangle = \langle \rho_3, \rho_3 \rangle + 2\langle \rho_3, \rho_1 \rangle +\langle \rho_1, \rho_1 \rangle$.
We have X, Y as G-sets: $latex (\pi_X, G, \mathbb{C}(X))$.

Proposition:
$\langle X, Y \rangle$ is the number of G orbits in $X \times Y$.
In our case, $\langle \rho, \rho \rangle= 2$ implies $\langle \rho_3, \rho_3 \rangle= 1$.

Proof of proposition:
$K: Hom(\mathbb{C}(X), \mathbb{C}(Y)) \to \mathbb{X \times Y}$
both groups are isomorphic to G. This implies
$Hom_G (\mathbb{C}(X), \mathbb{C}(Y)) \to \mathbb{C}(X \times Y)^G.$

Another 3-dimensional representation:
$\pi_3= sgn \otimes \rho_3: S_4 \to GL(V_0).$
Clearly this is also irreducible.
How do we know it’s not equivalent to $\rho_3$?

Well, take three function $f_1, f_2, f_3$ defined on the triangle (all entries have to add to 1.) $f_1$ is 0 on both vertices a, b, $f_2$ is 1 on both, and $f_3$ is 1 on a and -1 on b.
These form a basis. The trace of $\rho_3$ in this basis is 1 + 1 + -1 = 1, and the trace of $\pi_3$ in this basis is sign(a, b) = -1. So they are not identical.

(To be continued — I’m posting from Montreal, so I don’t have long blocks of computer time.)