- #1
AxiomOfChoice
- 531
- 1
(At least, I think it's simple.)
Disclaimer: I'm approaching this subject from the vantage point of a chemist, so be careful with how much lingo/jargon/rigor you lay on me
The claim is that if you have two representations of a group, \Gamma_1 and \Gamma_2, with bases \{ f_i \} and \{ g_j \}, then the set of products \{f_i g_j\} is a basis for the direct product representation. Fine. But in the process of showing this, the following assertion is made: If \hat R is an element of the group, then
<br /> \hat R(f_i g_j) = \hat R(f_i) \hat R(g_j).<br />
...huh? How does that work? How do I know that if I apply a group operation to a product of basis elements, then it's just the product of the operation applied to the individual basis elements? I missed something there, because (a priori) that move is just as suspect as saying that x(yz) = (xy)(xz) = x^2yz for x,y,z\in \mathbb R.
Disclaimer: I'm approaching this subject from the vantage point of a chemist, so be careful with how much lingo/jargon/rigor you lay on me
The claim is that if you have two representations of a group, \Gamma_1 and \Gamma_2, with bases \{ f_i \} and \{ g_j \}, then the set of products \{f_i g_j\} is a basis for the direct product representation. Fine. But in the process of showing this, the following assertion is made: If \hat R is an element of the group, then
<br /> \hat R(f_i g_j) = \hat R(f_i) \hat R(g_j).<br />
...huh? How does that work? How do I know that if I apply a group operation to a product of basis elements, then it's just the product of the operation applied to the individual basis elements? I missed something there, because (a priori) that move is just as suspect as saying that x(yz) = (xy)(xz) = x^2yz for x,y,z\in \mathbb R.