Timeline for Is Sp(1).Sp(1).Sp(1) the homotopy-fixed locus of Triality?
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18 events
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| Apr 15, 2019 at 20:43 | comment | added | მამუკა ჯიბლაძე | Beautiful, thanks! | |
| Apr 14, 2019 at 22:37 | comment | added | Theo Johnson-Freyd | Ah, but now the point is that $z$, which is essentially canonical, is pretty much the same as a reduction from a $U(1)$ ambiguity in the choice of $x^\pm$ to a sign ambiguity. See, you can normalize $h$ pretty easily, and then you can normalize $x^+$ up to sign by requiring that $x^- = (x^+)^z$ and that $[x^+,x^-] = h$. | |
| Apr 14, 2019 at 22:34 | comment | added | Theo Johnson-Freyd | Ok, so $z$ has order 2. The next fact is that all lifts of $z$ are conjugate, and this is easy: $tz = t^{1/2} t^{1/2} z = t^{1/2} z t^{-1/2}$ for any choice of square root $t^{1/2}$ of $t \in T$. The choice is not canonical of course (there are $2^{rank}$ choices), but it exists. So up to inner autoromosphisms, $z$ is canonical. | |
| Apr 14, 2019 at 22:33 | comment | added | Theo Johnson-Freyd | Which I guess is just the fact that we are in the adjoint form, because at least if you believe that all maximal tori are conjugate, then $z^2$ will necessarily be in the centre of $G$. Note that the assumption of being adjoint form is required: in $SU(2)$, the central $-1 \in Weyl$ lifts to an element of order 4. | |
| Apr 14, 2019 at 22:31 | comment | added | Theo Johnson-Freyd | The first nontrivial statement is that $z$ has order $2$. I don't know a clean way to prove this --- it follows as a special case of a result of Lepowsky's about VOAs, and I'm sure it was known earlier than that to Tits and other. Oh, I should mention that all lifts of $z$ have the same order. Indeed, a different lift is $tz$, but $(tz)^2 = t t^z z^2$, where $t^z$ means the conjugation of $t$ by $z$, which is just the action of $-1$ on $t$, so that $t^z = t^{-1}$, and so $(tz)^2 = z^2$. So all you need to do is convince yourself that there's no special nontrivial element "$z^2"$ of $T$. | |
| Apr 14, 2019 at 22:27 | comment | added | Theo Johnson-Freyd | @მამუკაჯიბლაძე There's a simpler story worth mentioning. Let's work with $G = Aut(\mathfrak{g})$, which is the adjoint form extended (semidirectrly) by the outer automorphism group. Take for granted that the normalizer of the maximal torus $T \subset G$ is an extension of shape $T.O(L)$, where $O(L)$ is the group of automorphisms of the root lattice (the lattice dual to $T$). The thing to observe is that $O(L)$ contains a canonical central element, which I will call $-1$, because that's how it acts on $L$. Choose any lift $z \in T.O(L)$ of this element. | |
| Apr 14, 2019 at 15:01 | comment | added | მამუკა ჯიბლაძე | Ok thanks a lot for explanations. I will also read that book with great interest, although vertex algebras seem to be very carefully hidden there... | |
| Apr 14, 2019 at 4:03 | comment | added | Theo Johnson-Freyd | The sign ambiguity gets "resolved" when you choose the double cover of $L$, I mean when you choose a cocycle representing the cohomology class. | |
| Apr 14, 2019 at 4:02 | comment | added | Theo Johnson-Freyd | Then a version of Noether's theorem implies that the Lie algebra of derivations of that VOA is equal, as a vector space, to $L \otimes \mathbb{C} \oplus$ a vector space with a basis consisting of the length-2 vectors in $L$. | |
| Apr 14, 2019 at 4:00 | comment | added | Theo Johnson-Freyd | But to give a sense of the answer here, given a positive-definite even lattice $L$, the mod-2 reduction of the lattice pairing represents a class in $H^2(L, \mathbb{Z}/2)$, and so a double cover of $L$; given a choice of such double cover (which is determined up to isomorphism, but not up to unique isomorphism), you can functorially construct a vertex operator algebra. | |
| Apr 14, 2019 at 3:57 | comment | added | Theo Johnson-Freyd | @მამუკაჯიბლაძე Since I've already rambled too long, I'm tempted to just point you to section 8.2 of my book. (That chapter is based on lectures of R. Borcherds. Chapter 5 that I mentioned earlier is based on lectures of M. Haiman.) The canonical reference for the full details is the book by Frenkel, Lepowsky, and Meurman. | |
| Apr 14, 2019 at 3:53 | comment | added | Theo Johnson-Freyd | What you can see from this is that the $Spin(3) \cong Sp(1)$ that you are asking about is none of the $Sp(1)$s in the end of my story. Rather, it is the diagonal copy inside the first two. | |
| Apr 14, 2019 at 3:53 | comment | added | Theo Johnson-Freyd | @UrsSchreiber Come to think of it, I'm now a bit concerned. The vector rep $V^8$ of $Spin(8)$ breaks over $Spin(3) \times Spin(5)$ as $8 = 3 \otimes 1 + 1 \otimes 5$. That 5 breaks over $Sp(1) \times Sp(1)$ as $2\otimes 2 + 1 \otimes 1$, so all together we get $3 \otimes 1 \otimes 1 + 1 \otimes 2 \otimes 2 + 1 \otimes 1 \otimes 1$. | |
| Apr 14, 2019 at 3:42 | comment | added | Theo Johnson-Freyd | @UrsSchreiber I admit I wasn't quite sure what your question was. In any case, I'm sure I didn't answer it --- I wanted to elaborate on my comments a bit, and the "answer" spot was the best place I could find to do so. | |
| Apr 13, 2019 at 11:02 | comment | added | Urs Schreiber | I agree that Dynkinology could well provide the answer that I am after: One would need to consider the trisectorial diagram that I drew above and replace the Lie subgroups with their corresponding Lie algebra generators. It would be sufficient then to see that the diagram commutes on the level of sets of generators. For the outer circle this is trivial, for the inner circle this is what you are describing here. To answer the question from this approach, one will need to identify the generators for the groups in the middle circle and check that the inclusions and automorphisms commute. | |
| Apr 13, 2019 at 7:07 | comment | added | Urs Schreiber | Thanks, but do you see how this relates to the question I asked? The question was if it is known that this Sp(1).Sp(1).Sp(1) subgroup in Spin(8) together with the triality action on it by permutation of dot factors - which you recall can be associated with the shape of the Dynkin diagram - is compatible with its canonical inclusions into a) Sp(1).Sp(2), b) Sp(2).Sp(1), c) Spin(5).Spin(3) and their permutation among each other by triality. I sure expect it is, but it would be nice to have a definite argument or reference. | |
| Apr 12, 2019 at 21:33 | comment | added | მამუკა ჯიბლაძე | I realize that your answer is already long, but could you at least briefly elaborate on how the theory of vertex algebras helps in normalizing $x^\pm$? | |
| Apr 12, 2019 at 21:21 | history | answered | Theo Johnson-Freyd | CC BY-SA 4.0 |