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Thursday, January 26, 2012

Suspicious Characters, Part 2

It's trace operator time!  For those who don't know already, the trace of a matrix is the sum of the entries on the main diagonal.

Why is it importart?  We won't be proving this fact today, but the trace of a matrix \rho_g is also the sum of its eigenvalues!

Eigenvalues?  Wündebar!  Time to make de magics.

The trace operator also plays well with the composition of spaces!  If I have a linear transformation \rho_g of a vector space V, and another linear transformation \sigma_h of a vector space W, the corresponding linear transformation \rho_g \oplus \sigma_h of the space V \oplus W has this property:
Tr(\rho_g \oplus \sigma_h) = Tr(\rho_g) + Tr(\sigma_h)
Which is nice.  So if we had some sort of decomposition of a space V into some sort of irreducible components W_1, W_2,... W_m and a family of linear transformations \rho_g^{W_1}, \rho_g^{W_2},... \rho_g^{W_m} on those spaces...

V=\displaystyle \bigoplus_{i=1}^m W_i  \, \, \Rightarrow \, \, Tr\left( \bigoplus_{i=1}^m \rho_g^{W_i}\right) = \sum_{i=1}^m Tr(\rho_g^{W_i})

Neat!  So if we have a representation \rho: G \rightarrow V, a decomposition of V into G-invariant subspaces also induces a sort of decomposition on the values of Tr(\rho_g).  But to study Tr(\rho(g)) as a function on G, we're going to face some clunky notation.  So let's give this function its own name, \chi_\rho .

This map we've created, \chi_\rho : G \rightarrow \mathbb{C} such that \chi_\rho(g) = Tr(\rho(g)) \, \, \forall \, g \in G, is called the character of \rho.

Hence why this stuff is called "Character Theory."  And now we can make this come together around some very neat properties of \chi_\rho when \rho is an irreducible representation!  Remember that proposition we proved at the end of the last installment?  We're going to use that result.

Proposition:
If \rho: G \rightarrow V is an irreducible representation, and we write each \rho_g as  matrix in the basis \{ \vec{e_1}, \vec{e_2}, ... \vec{e_n}\} of V, with g_{ij} to denote the entry in the ith row and jth column,
\frac{1}{|G|}\displaystyle \sum_{g\in G} g_{ij} = \left\{\displaystyle \begin{array}{lc} \delta_{ij} & \mbox{ if } \rho \mbox{ is the trivial representation} \\ 0 & \mbox{ otherwise} \end{array} \right.   
Where \delta_{ij} is the Kroenecker delta.
Proof:
If \rho is the trivial representation (i.e. \rho_g is the identity matrix \, \forall \, g \in G) then the above sum is just the entry in the ith row and jth column of the identity matrix. 
If \rho is not trivial, define u_k: V \rightarrow \mathbb{C} as \hspace{2mm} u_i \left( \displaystyle \sum_{j=1}^n c_j \vec{e_j} \right) = c_i \hspace{2mm} and rewrite the sum,
\frac{1}{|G|} \displaystyle \sum_{g\in G}g_{ij}= \frac{1}{|G|} \displaystyle \sum_{g\in G} u_i \left( \rho_g (\vec{e_j}) \right)
But by the proposition at the end of the last blogpost, the expression on the right can only be nonzero if \rho is trivial.  Thus it is zero in these nontrivial cases.
\Box
Now let's use this result with the trace operator:
\frac{1}{|G|}\displaystyle\sum_{g\in G}Tr(\rho_g) = \frac{1}{|G|}\displaystyle\sum_{i=1}^n\left(\sum_{g\in G} g_{ii}\right)
When \rho is irreducible and nontrivial, this sum we made is zero.  When \rho is irreducible and trivial, this sum becomes 1.  And... remember that additive bit about the trace operator earlier?  When \rho is not irreducible, this sum is the 'multiplicity' of trivial subrepresentations in the decomposition of V and \rho.
If only there were similarly powerful techniques
for keeping track of dressmaking supplies...

Just you wait, we'll get some orthogonality relations out of this...

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