16

This question, or rather any answer that it might receive, would probably belong to the realm of Awfully sophisticated proof for simple facts. Still, I claim that I have quite serious motivation for it.

First, the question itself. Consider the space of half-spinors in the 10-dimensional (complex) vector space $V=\mathbb C^{10}$. One of its descriptions is this. Take some non-degenerate quadratic form $q$ on $V$; let $W\subset V$ be an isotropic $5$-dimensional subspace. It generates a subalgebra in the Clifford algebra $\operatorname{Cl}_q(V)$ isomorphic to the exterior algebra $\bigwedge^*(W)$. This, if I am not mistaken, is called the algebra of pure spinors, and splits into the direct sum of two $16$-dimensional subspaces $S_\pm$ called spaces of half-spinors, namely, the even degree part $\bigwedge^{2*}(W)$ and the odd degree part $\bigwedge^{2*+1}(W)$.

The question is whether there exists a $\textsf{natural}$ isomorphism $\mathbb C^4\otimes\mathbb C^4\to S_+$ (or $\to S_-$). By "natural" you may understand what you want, but I mean something invariant, for example, not requiring to choose bases or, say, invariant under the actions of as large a group as possible.

A lightweight version: find a $\textsf{natural}$ bijection between $\{1,2,3,4\}\times\{1,2,3,4\}$ and subsets of $\{0,1,2,3,4\}$ with even number of elements.

Now, that awfully sophisticated motivation. One of the four (up to conjugation) automorphisms of order four of the simple Lie algebra $\mathfrak e_8$ is such that with respect to the induced $\mathbb Z/4\mathbb Z$-grading $\mathfrak g^1\oplus\mathfrak g^i\oplus\mathfrak g^{-1}\oplus\mathfrak g^{-i}$ of $\mathfrak e_8$, the algebra $\mathfrak g^1$ of fixed points of this automorphism is of type $\mathfrak{so}_{10}\oplus\mathfrak{sl}_4$ and its action on $\mathfrak g^i$ (which is $64$-dimensional) is via the tensor product $\mathbf{16}\otimes\mathbf 4$ of the half-spinor representation of $\mathfrak{so}_{10}$ and the standard representation of $\mathfrak{sl}_4$. What I really want to understand is this: when one classifies orbits of the action of the corresponding group on $\mathfrak g^i$, what does one really classify? If there would be some "natural" isomorphism $\mathbf 4\otimes\mathbf 4\to\mathbf{16}$, then each element of $\mathbf{16}\otimes\mathbf 4$ would give rise to some binary operation on $\mathbf 4$, and one could start studying properties of those operations which arise in this way. But maybe there is something even more interesting and less straightforward, I don't know.

მამუკა ჯიბლაძე
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    I confidently posted two wrong candidates for the lightweight isomorphism. In lieu of a third wrong candidate, I wonder if the lightweight fact can be viewed as a manifestation of some accidental isomorphism involving $\mathbb F_5$ and $\mathbb F_4$? Or perhaps of the fact that two different presentations yield the same group of order 16? – LSpice Dec 12 '20 at 17:56
  • Instead of formulating a paradox, you may try to make sense. – Wlod AA Dec 12 '20 at 18:45
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    @LSpice Really intriguing suggestion. I wonder whether $SL_2(4)\cong SL_2(5)$ might get involved... – მამუკა ჯიბლაძე Dec 12 '20 at 18:46
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    That's exactly what first came to mind, although I can't see how to exploit it; the orders seem wrong to give interesting information about 16-element sets. (Dare I imagine some kind of metaplectic group getting involved?) – LSpice Dec 12 '20 at 18:48
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    @WlodAA, do you see a paradox in the question? It seems more like an $\mathbb F_1$-style desideratum, where one sees some structure that demands explanation and seeks the explanation …. In any case, though I don't know an answer, it certainly seems to make sense to me. – LSpice Dec 12 '20 at 18:57
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    This sounds like something that could be found in the book Mathematics Made Difficult by Carl Eric Linderholm (1937−2002; 1963 Ph.D. under Halmos) [Wolfe Publishing (1971) and World Publishing Company (1972), 207 pages; MR 58 #26623; Zbl 217.00102]. For an indication of the book's spirit, see here. However, as far as I can tell, the identity $4^2 = 2^4$ is not parodied in Linderholm's book. – Dave L Renfro Dec 12 '20 at 19:00
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    @DaveLRenfro Yes, I started by admitting that it sounds like that – მამუკა ჯიბლაძე Dec 12 '20 at 19:19
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    One more thought: a Sylow 2-subgroup of $\mathrm S_6$ has order 16. Could it be made to act simply transitively in some natural way on the 2-tuples in $\overline4$ and, possibly by virtue of some of the exotic embeddings of $\mathrm S_5$, on the even-order subsets of $\overline5$? – LSpice Dec 12 '20 at 19:44
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    risking to point out the obvious: a natural bijection for the lightweight version cannot exist in the sense that $\mathfrak S_4$ acts diagonally on the pairs and naturally on the subsets (fixing $0$), because the characters are different. – Martin Rubey Dec 13 '20 at 20:32
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    @MartinRubey In case the non-lightweight version has any satisfactory solution, this might be related to the fact that $\mathfrak S_5$ has a $4$-dimensional irrep which is not permutational... – მამუკა ჯიბლაძე Dec 13 '20 at 21:29
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    Some thoughts on the lightweight version of your question. We clearly need some structure on the sets, but how much do we need? E.g. we can do it given a distinguished partition of the 5-element set into a 1-element set Z and two 2-element sets A and B: Even-sized subsets of a 5-element set are the same as partitions of the 5-element set into two (since exactly one of each partition will be even-sized); such a partition corresponds to a choice of A1 <= A and B1 <= B such that Z u A1 u A2 is one of the sets in the partition. Thus even-sized subsets correspond to a choice of subsets of A and B. – Alec Edgington Dec 17 '20 at 21:10
  • @AlecEdgington Interesting! This comes quite close, except that on the $4\times4$ side arbitrary tuples (including the diagonal) are needed... – მამუკა ჯიბლაძე Dec 18 '20 at 10:14
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    The closest I can find to this is: there is a pseudo-Levi subalgebra of $\mathfrak{so} _{10}$ isomorphic to $\mathfrak{so}_6\oplus\mathfrak{so}_4\cong\mathfrak{sl}_4\oplus\mathfrak{sl}_2\oplus\mathfrak{sl}_2$, and the spin representation decomposes as $V_4\otimes V_2^{(1)}\oplus V^*_4\otimes V_2^{(2)}$ where $V_4$, resp. $V_2^{(1)}$, $V_2^{(2)}$ is the natural module for $\mathfrak{sl} _4$, resp. the first, second copy of $\mathfrak{sl}_2$. – Paul Levy Oct 24 '21 at 12:44

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