0
$\begingroup$

Here, $\chi$ is the character of the sub-representation, i.e., Given

$\rho : G \to GL(V)$ is a representation, then the function $\chi_{\rho}: G \to \mathbb{C}: \chi_{\rho}(g) \to Tr(\rho_g)$.

I do not fully understand the notation here, as it was just sort of thrown at me without any explanation. Above is what I want to prove, that is:

$\chi_{V_1 \otimes V_2} (g) = \chi_{V_1} (g) \cdot \chi_{V_2} (g)$.

I know that since the trace map is invariant for similar matrices, the choice off basis does not affect the answer. If anyone would be so kind as to point me to how to prove this, it would be nice. I'm not asking for the full proof, but at least a starting point.


My thinking is that we need to choose some sort of matrix representation for $\rho$ and look at what the tensor product does to two randomly chosen matrix representations $V_1$ and $V_2$.

Thanks!

$\endgroup$
  • $\begingroup$ Read about Kronecker Products of Matrices $\endgroup$ – Geoff Robinson Apr 13 '14 at 17:52
  • 2
    $\begingroup$ Really want you want to show is that $Tr(A\otimes B) = Tr(A)Tr(B)$ as this does all of the work for you. $\endgroup$ – Cameron Williams Apr 13 '14 at 17:52
2
$\begingroup$

One thing I have noticed is that representation theorists love to be loose with notation and definitions (e.g. calling a $G$-invariant subspace a subrepresentation and the restriction of a representation to a $G$-invariant subspace a subrepresentation). It bothers me a lot. This is such an instance, where they have replaced $\rho_1\otimes\rho_2$ with $V_1\otimes V_2$. This does not match with the notational definition you've been given even though a representation theorist would likely argue that it's clear what is meant. I'm of the opinion that notation should be unambiguous and completely consistent throughout.

Suppose you have two representations $\rho_1:G\rightarrow GL(V_1)$ and $\rho_2:G\rightarrow GL(V_2)$. Better notation is perhaps $\chi_{\rho_1\otimes\rho_2}$ in place of $\chi_{V_1\otimes V_2}$. Piggybacking off of my comment, what this then says is that you want to show that the following is true:

$$\chi_{\rho_1\otimes\rho_2}(g) = \chi_{\rho_1}(g)\chi_{\rho_2}(g).$$

Or equivalently:

$$\operatorname{Tr}((\rho_1\otimes\rho_2)(g)) = \operatorname{Tr}(\rho_1(g))\operatorname{Tr}(\rho_2(g)).$$

$\endgroup$
  • $\begingroup$ This makes is much more clear. Thank you. I completely agree that people that use representation theory tend to not care about the notation, and that makes it extremely difficult to follow. $\endgroup$ – Calculus08 Apr 13 '14 at 18:09
  • 1
    $\begingroup$ Yeah I had plenty of issues with this. It's really troubling because it's a really nice branch of mathematics but there are some major miscues here and there (from my perspective).. $\endgroup$ – Cameron Williams Apr 13 '14 at 18:10
  • $\begingroup$ In my experience the notation is only confusing in the beginning. Anyway, the claimed fact is not actually a fact of representation theory but one of linear algebra; ${\rm tr}(A\otimes B)={\rm tr}(A){\rm tr}(B)$ holds for any matices. This can be done by determining an eigenbasis for $A\otimes B$ via eigenbases of $A$ and $B$ (even if our scalars are not algebraically closed, we can extend them to accomplish this task). $\endgroup$ – anon Apr 13 '14 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.