Timeline for Why do sl(2) and so(3) correspond to different points on the Vogel plane?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 3 at 4:36 | comment | added | მამუკა ჯიბლაძე | (and both $e$, $f$ isotropic) | |
| Jan 26 at 5:15 | comment | added | მამუკა ჯიბლაძე | Sorry, I realize this question is eleven years old, but may I ask this: straightforward calculations show that the only symmetric bilinear forms $\left\langle-,-\right\rangle$ on $\mathfrak{sl}_2$ that satisfy $\left\langle[x,y],z\right\rangle=\left\langle x,[y,z]\right\rangle$ are scalar multiples of the one with $h$ orthogonal to both $e$ and $f$, and $\left\langle h,h\right\rangle=2\left\langle e,f\right\rangle$. Do you mean that all these scalar multiples are pairwise nonequivalent and produce that line? | |
| Jun 1, 2011 at 23:36 | comment | added | Noah Snyder | In general I'd expect the dimension formulas to be poorly behaved here. Think about what happens say for quantum su(2) at a root of unity. At first the dimension formulas make sense, then you get zero, and then after that there's no irreps coming from the generic q reps and various formulas break down. | |
| Jun 1, 2011 at 5:05 | comment | added | Bruce Westbury | Let's take the coordinate on this line to be $t$. Then there are a load of dimension formulae (not just in the tensor square) which give the dimension of a representation as a rational function of $t$. What I was wondering (and was too busy or lazy to work out myself) was if these rational functions are in fact constant. | |
| May 31, 2011 at 22:37 | comment | added | Noah Snyder | If you look at the dimensions in the symmetric square (taken from Vogel) then two of them are zero (and the third one gives the 5-dimensional rep). So these reps don't contribute to closed diagrams, but do contribute if you're looking at (1,1) tangles. | |
| May 20, 2011 at 20:40 | comment | added | Bruce Westbury | What happens to the dimension formulae on this line? | |
| May 20, 2011 at 20:39 | comment | added | Bruce Westbury | There is something I have never cleared up. What does it mean to "evaluate a diagram at a point in the Vogel plane". The way I understand it you evaluate a diagram at a point of $\mathrm{Spec}(\Lambda)$ where $\Lambda$ is Vogel's ring; this is not finitely generated and has zero divisors. The Vogel plane is $\mathrm{Spec}(R)$ for some other ring $R$. These are related but I think the connection is somewhat mysterious. | |
| May 20, 2011 at 20:11 | comment | added | Noah Snyder | I thought about this some more and talked to Dylan, and I still think that every point on this line will give the same value on closed diagrams, and you need to look at what they do to diagrams with boundaries before you see a difference. (This is because the reps which differ over different points on this line are all negligible.) | |
| May 5, 2011 at 14:46 | vote | accept | Noah Snyder | ||
| May 5, 2011 at 6:08 | comment | added | Bruce Westbury | I have not worked it through in detail but my understanding is that they are different. | |
| May 4, 2011 at 23:46 | comment | added | Noah Snyder | If I understand correctly the point is that if a certain projection is negligible (the projections corresponding to the two irreps that don't occur for $\mathfrak{sl}_2$) it doesn't necessarily follow that the Casimir acts by 0 on those projections. But there's one point I'm still confused about, do you actually get different finite type invariants (or, I think equivalently, different values on closed diagrams) for different points on the $\mathfrak{sl}_2$ line? Or is the point that the real configuration space is some sort of blow-down along this line? | |
| May 4, 2011 at 20:26 | history | answered | Bruce Westbury | CC BY-SA 3.0 |