Question in the title, ie why is Tr(T_{a_1}T_{a_2}...T_{a_n}) independent of which representation we choose, where the Ts are a matrix representation of the group generators.
For example, if you pick a positive integer m, there is a representation where every group element has trace m: this representation sends every group element to the identity matrix acting on an m-dimensional space.
And given any matrix representation \rho of a group, one can construct a new representation \rho' by
<br />
\rho'(g) = \left[ \begin{matrix}{\rho(g) & 0 \\ 0 & \rho(g)} \end{matrix} \right]<br />
and under this representation, Tr\, \rho'(g) = 2 Tr \, \rho(g).
You're making some extra, relevant assumptions -- what are they?
#3
Bobhawke
142
0
Ah I see. But couldn't we also have
<br />
\rhosingle-quote(g) = \left[ \begin{matrix}{\rho(g) & 0 \\ 0 & 0} \end{matrix} \right]<br />
as a higher dimensional rep with the same trace?
I think what I meant is that given a 2 different matrix reps of the same dimension, the casimirs are the same. But they can change when you change the dimension of the rep, thus they can be used to label reps. Thanks
Different representations with the same dimension can still have that trace be different. For example, consider
\rho'(g) = \rho(g)^2
and
\rho'(g) = (\det g)^{-1}\rho(g)
Anyways, I went to look up Casimir on wikipedia; if that's what you're talking about, then I think you are misunderstanding things. The Casimir invariant is an element of the universal enveloping algebra of the Lie algebra, and its representation under \rho is a matrix.
Wikipedia states that by Schur's lemma, for any irreducible representation, \rho(\Omega) is proportional to the identity matrix. That constant of proportionality can be computed with a formula involving traces; maybe that's what you're thinking of?
#5
Bobhawke
142
0
Yeah I think I am confusing things a little.
Casimirs can be constructed from the generators by:
But is is their eigenvalues that label the reps. So I think what I really meant to ask is why can the eigenvalues of the casimirs can be used to label reps of different dimension.
##\textbf{Exercise 10}:##
I came across the following solution online:
Questions:
1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##"
2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho.
In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states
"Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels).
Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product
$$
\langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$
where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...