Timeline for Generalization of Schur's lemma (Update)
Current License: CC BY-SA 2.5
15 events
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| Aug 21, 2010 at 0:30 | history | edited | Yemon Choi | CC BY-SA 2.5 |
tweaked the LaTeX formatting to try and imrove readability
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| Aug 12, 2010 at 9:22 | answer | added | Paul Zinn-Justin | timeline score: 2 | |
| Aug 5, 2010 at 21:17 | comment | added | Peter Shor | I guess it depends on how much you need to solve the problem, and whether $U(2)$ will help you, or whether you need $U(N)$. For $U(2)$, I suspect that if you worked out a bunch of small examples, and looked at them, you could figure out the pattern. And $U(2)$ has enough nice structure that it should be possible to prove your answer. In general, however, it's a lot hairier doing things explicitly in $U(N)$. | |
| Aug 5, 2010 at 20:20 | comment | added | Alireza | Thanks for the comments. Yes, It looks much more complicated than I first thought. If it is hopeless for you mathematicians, maybe I should give up. :( | |
| Aug 5, 2010 at 4:04 | comment | added | Peter Shor | This is clearly basis-dependent (I used the $z$-momentum basis, which is the standard basis for $SU(2)\,$), and it isn't ending up looking very nice. I suppose it might be nicer in a different basis. I suspect you could work out everything for the group $U(2)$, but it won't be easy. And $U(N)$ for larger $N$ are going to be a lot worse. | |
| Aug 5, 2010 at 3:57 | comment | added | Peter Shor | I've looked at the previous example, and unless I did something wrong (quite possible) it looks to me like the terms add up (and cancel) in rather unexpected ways. I'll group them according to $(l_1,l_2)$. For $l_1=2$, $l_2=2$, there are six non-zero terms, each contributing an $I_2=1/12$. The same holds for $(l_1,l_2) \in \{(2,3),(3,2),(3,3)\}$ If you take $l_1=1$, $l_2=1$, you get one term contributing $1/4$, and the same holds for $(l_1,l_2) \in \{(1,1),(1,4),(4,1),(4,4)\}$. For the remaining pairs $(l_1, l_2)$, you get various terms of $\pm \sqrt{3}/12$, which all end up canceling. | |
| Aug 2, 2010 at 20:31 | comment | added | Peter Shor | If you want a simpler example to think about, you could take ${\bf r}=2$, ${\bf r}' = 2$ and ${\bf r}'' = 4$. Then the tensor products decompose as $R_2 \otimes R_2=R_1\oplus R_3$ and $R_2\otimes R_4=R_3\oplus R_5$. | |
| Aug 2, 2010 at 20:29 | comment | added | Peter Shor | I don't think your conjecture can be right. Look at $SU(2)$. Suppose ${\bf r}=3$, ${\bf r}'=5$ and ${\bf r''=7}$. Then the tensor products decompose into irreducible representations as: $R_3\otimes R_5=R_3\oplus R_5 \oplus R_7$ and $R_3\otimes R_7=R_5\oplus R_7\oplus R_9$, where $R_n$ is the irreducible representation of dimension $n$. When you sum over your indices $i_{1,2},\, j_{1,2},\, k_{1,2},\, l_{1,2}$, you should get $12=5+7$, not $0$, since these overlap in $R_5$ and $R_7$. How this sum of $12$ is distributed over $i_{1,2},\,j_{1,2},\,k_{1,2},\,l_{1,2}$ is a mystery to me. | |
| Aug 1, 2010 at 20:01 | history | edited | Alireza | CC BY-SA 2.5 |
added 1318 characters in body; edited title
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| Aug 1, 2010 at 18:24 | comment | added | Alireza | Yes. In my problem, I have 3 distinct representations. There is no typo error. However, my conjecture for the answer is that $I_2=0$ if $\mathbf{r}^{\prime} \neq \mathbf{r}^{\prime \prime}$. | |
| Aug 1, 2010 at 15:46 | comment | added | Peter Shor | Do you really want two ${\bf r}$'s, one ${\bf r'}$ and one ${\bf r''}$ in your expression? Or is there a typo? | |
| Jul 31, 2010 at 21:01 | comment | added | Alireza | Thanks for the hint. p.s. The mathematicians I have talked to so far call I_1 the Schur's lemma. | |
| Jul 30, 2010 at 23:32 | comment | added | José Figueroa-O'Farrill | I don't know the answer to your question, but this seems to me like the sort of calculation that lattice gauge theorists might know how to do. | |
| Jul 30, 2010 at 23:30 | comment | added | José Figueroa-O'Farrill | Why do call this a generalisation of the Schur's Lemma? It seems to me like a generalisation of the Peter-Weyl theorem instead. | |
| Jul 30, 2010 at 23:21 | history | asked | Alireza | CC BY-SA 2.5 |