# Why do sl(2) and so(3) correspond to different points on the Vogel plane?

Vogel assigns to every simple metric Lie algebra (and more generally to every simple metric Lie algebra object in a symmetric monoidal category) a point in the orbifold $\mathbb{P}^2/S_3$ (where $S_3$ acts by permuting the 3 projective coordinates) based on the value of the Casimir on the various summands of the symmetric square of the adjoint representation. These three numbers are only defined up to permutation (as there's no natural way to specify which rep is which) and rescaling (because the Casimir itself is only well-defined up to rescaling).

Under this assignment we have $\mathfrak{sl}_2$ and $\mathfrak{so}_3$ going to different points. How is this possible? My best guess is that $\mathfrak{sl}_2$ and $\mathfrak{so}_3$ are different as metric Lie algebras, but that also seems weird.

In the conventions of this paper $\mathfrak{sl}_2$ corresponds to the point $(-1:1:1)$ while $\mathfrak{so}_3$ corresponds to the point $(-1:2:-1)$ and these are different points in $\mathbb{P}^2/S_3$. You can also easily check that the points are different in other convetions.

This question was originally asked in comments by Scott Carnahan, but I wanted to move it up to the main page.

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Could it have something to do with the fact that over R, they are different? –  Ben Webster May 4 '11 at 19:33
But they are different as metric Lie algebras: the inner product is definite in $\mathfrak{so}_3$ and lorentzian in $\mathfrak{sl}_2$, assuming I'm understanding the question. –  José Figueroa-O'Farrill May 4 '11 at 19:36
Perhaps I'm using the word "metric" wrong here. Vogel says "pseudo-quadratic" rather than "metric." The point is that you have (in addition to the bracket and the crossing) an adjoint pair of maps of representations $\mathfrak{g} \otimes \mathfrak{g} \rightarrow \mathbf{1}$ and $\mathbf{1} \rightarrow \mathfrak{g} \otimes \mathfrak{g}$. But at any rate, its a bilinear form, not a Hermitian one. –  Noah Snyder May 4 '11 at 19:57
Even more surprising: $\mathfrak{sl}_3$ and $\mathfrak{so}_8$ appear twice (they also appear in the exceptionnal series in math.tamu.edu/~jml/LMunivpub.pdf). –  DamienC May 4 '11 at 20:20
@DamienC: Those two "different" points for $\mathfrak{sl}_3$ are actually the same in the quotient $\mathbb{P}^2/S_3$. So that's a different issue. –  Noah Snyder May 5 '11 at 0:04

I take it you mean $\mathfrak{sl}_2$ is on the line $\mathfrak{sl}_n$ and $\mathfrak{so}_3$ is on the line $\mathfrak{so}_n$. There is no contradiction because there is a whole line for this metric Lie algebra. The symmetric square of the adjoint decomposes as the trivial representation (which is accounted for by the Killing form) and one other irreducible (of dimension 5). This means you only have one of the three coordinates. Usually the coordinates are the values of the quadratic Casimir on the three non-trivial factors of the symmetric square of the adjoint representation. However looking at the adjoint representation you also know the sum of the three coordinates. This determines a line in the Vogel plane.
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? –  Noah Snyder May 4 '11 at 23:46
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. –  Bruce Westbury May 20 '11 at 20:39