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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.)

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