Timeline for Intuition behind the definition of quantum groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2015 at 17:00 | comment | added | Theo Johnson-Freyd | ... the "quantum plane" above I imagine as having spectrum $\mathbb R$ (or maybe $\mathbb C$). Then again, coordinates satisfying $XY = qYX$ have funny behavior along the axes... I guess the main difference is that $xp - px = i\hbar$ is natural if you have some translation-invariance, and for QM on a line we do want $x \mapsto x+x_0$ to be a symmetry. Whereas $XY = qYX$ is natural if you want an action by scaling --- if you're trying to do "quantum linear algebra". | |
| Jun 30, 2015 at 16:57 | comment | added | Theo Johnson-Freyd | @AndreKornell Good question. The word "plane" is, I hope, self-explanatory --- this is some kind of 2-dimensional linear space. Probably this use of "quantum" is nothing better than an abbreviation for "noncommutative". In the quantum mechanics of a particle on the line, the phase space is a plane with coordinates $x,p$, but those satisfy $xp - px = i\hbar$. Setting $q = e^{i\hbar}$, $X = e^x$ and $Y = e^Y$ does give coordinates satisfying $XY = qYX$ (or perhaps I'm off by a sign), but those should have spectrum $\mathbb R_+$ (or maybe $\mathbb C^\times$), whereas the coordinates on ... | |
| Jun 30, 2015 at 16:30 | comment | added | Andre Kornell | Theo, how does one motivate the term "quantum plane" here? | |
| May 7, 2015 at 3:37 | comment | added | Theo Johnson-Freyd | In the case $x_ix_j = qx_jx_i$, you discover that quantum matrices have coordinates $a_{i,j}$ such that each $2\times 2$ submatrix satisfies the equations $(\star,\star\star)$ that I wrote. The determinant is something like $\sum_{\sigma \in S_n} (-q)^{\ell(\sigma)}a_{i,\sigma(i)}$. | |
| May 7, 2015 at 3:34 | comment | added | Theo Johnson-Freyd | Of course, actually for any matrix $(q_{ij})$ which is "log antisymmetric" in the sense that $q_{ji} = q_{ij}^{-1}$, you can write down a quantum $n$-space with $x_ix_j = q_{ij}x_jx_i$, and I think you can still get a monoid of quantum matrices. I don't remember if there's a central "quantum determinant". | |
| May 7, 2015 at 3:31 | comment | added | Theo Johnson-Freyd | @NoahSnyder You have more more expertise than I have --- certainly my impression was that SL(n) also arises in this way. I learned the story from Matt Tucker-Simmons during Kolya's 2009 quantum groups class. As far as I know, the story is due to Manin --- I don't know where to read about it. What I think you can do is to define quantum $n$-space $\mathbb A^n_q$ as having coordinates $x_1,\dots,x_n$ with $x_ix_j = qx_jx_i$ if $i<j$. Then quantum $n\times n$-matrices, quantum SL(n), and quantum GL(n) arise as various (semi)groups of transformations of these. | |
| May 6, 2015 at 18:05 | comment | added | Noah Snyder | I knew this story for SL(2) from Kassel's book, but I had been under the impression that there wasn't such a nice story for higher SL(n). But you seem to be saying there is such a story. Where can I read about it? | |
| Apr 28, 2015 at 4:08 | history | answered | Theo Johnson-Freyd | CC BY-SA 3.0 |