The group

$$ Sp(2)\cdot Sp(1) := \big( Sp(2) \times Sp(1)\big)/\big\{(1,1), (-1,1)\big\} $$

is canonically a subgroup of $Spin(8)$ in three different ways, these representing three distinct conjugacy classes of subgroups, as indicated here:

quaternionic subgroups of Spin8

Under the action of the triality group

$$ Out(Spin(8)) \;\simeq\; S_3 $$

these three conjugacy classes are permuted into each other.

The relevant definitions and references for these statements are collected on the nLab here.

This means that there are automorphisms making the outer two circles in the following diagram commute with respect to the given subgroup inclusions:

triality on quaternionic subgroups of Spin8

My question is about the inner circle:

Each of the above quaternionic subgroups canonically contains the further subgroup

$$ Sp(1) \cdot Sp(1) \cdot Sp(1) \;:=\; \big( Sp(1) \times Sp(1) \times Sp(1) \big)/ \big\{ (1,1,1), (-1,-1,-1) \big\} $$

Moreover, this subgroup has an evident automorphism action of the symmetric group $S_3$ on it, given by permuting its dot-factors.

This suggests that

  1. the middle automorphism circle above also commutes with the inner one,

  2. the relevant automorphisms of the inner circle are these dot-factor permutations,

  3. and maybe that $Sp(1)\cdot Sp(1) \cdot Sp(1)$ is something like the homotopy-fixed locus of triality.

I can see that a) and b) are true at least for the left sector of the above diagram, and it seems it must be at least close to being true also in the other two sectors, but I haven't tried yet to make that a full proof.

Is this known?

  • $\begingroup$ Isn't this $Sp(1)^3$ just the one spanned by the three outer nodes on the Dynkin diagram? That $Sp(1)^3$ is acted on nontrivially by triality. The action is by outer automorphisms, so remains highly nontrivial even if you look at the classifying space $BSp(1)^3$. Or do you have a different $Sp(1)^3$ in mind? $\endgroup$ Apr 12, 2019 at 15:32
  • 1
    $\begingroup$ The non-homotopy fixed points of triality (in its usual manifestation) is the exceptional group $G_2$. This is not contained in $Sp(1)^3$, so $Sp(1)^3$ definitely is not the anything fixed points of triality. It probably doesn't matter, but $G_2$ does contain an $Sp(1)^2$ centralizing an element of order $2$. I write "in its usual manifestation" because implicitly I splitting of the map $Aut(Spin(8)) \to Out(Spin(8)) = Aut(Spin(8))/Inn(Spin(8))$. IIRC, there are inequivalent conjugacy classes of such a splitting, but there is a standard splitting coming from the Chevalley presentation. $\endgroup$ Apr 12, 2019 at 15:37
  • $\begingroup$ Incidentally, in the group theory papers I have read, the central product $(G \times H) / Z$, where $Z = Z(G) = Z(H)$, is usually written $G \circ H$. The group theorists seem to usually use $G \cdot H$ for a (usually non-split) extension with normal subgroup $G$ and quotient $H$. $\endgroup$ Apr 12, 2019 at 15:38
  • $\begingroup$ @TheoJohnson-Freyd sure, the three Sp(1) "dot-factors" want to be thought of as associated with the three outer nodes of the Dynkin diagram, but does that alone serve to settle the question? Probably, if one chases through details carefully. Hopefully somebody has written that out? Or can write it out on the spot, if it's really easy? $\endgroup$ Apr 12, 2019 at 16:18
  • 1
    $\begingroup$ I didn't know those papers. The open circle is used for the central product in the literature I've seen on finite groups, eg the Atlas. $\endgroup$ Apr 12, 2019 at 19:16

1 Answer 1


This isn't quite an answer to your question so much as an elaboration of my comments.

Each Dynkin diagram, or equivalently Cartan matrix $a_{ij}$, determines a presentation of the corresponding complex Lie algebra $\mathfrak{g}$ (which is much the same data as the compact Lie group). I have the name ``Chevalley'' associated to this presentation, but perhaps Cartan and Serre are also important. In any case, the presentation is the following. For each node $i$ of the Dynkin diagram, you introduce three generators $h_i,x^+_i,x^-_i$. You declare that the $h_i$s commute among themselves, that $[h_i, x^{\pm}_j] = \pm a_{ij}x^\pm_j$, and that $[x_i^+,x_j^-] = \delta_{ij}h_i$, so that for fixed $i$, $\{h_i,x_i^+,x_i^-\}$ forms an $\mathfrak{sl}(2)$-triple. So far this presents an infinite-dimensional Lie algebra with a unique finite-dimensional quotient. To cut down to the quotient, you need also to impose the Serre relations which say that, for $\epsilon \in \{+,-\}$, $(\operatorname{ad} x^\epsilon_i)^{1-a_{ij}} x^\epsilon_j = 0$, where of course $(\operatorname{ad} x)(y) = [x,y]$. This presentation of $\mathfrak{g}$ is detailed in various textbooks, for instance Section 5.6 of my lecture notes.

This presentation has various advantages and disadvantages. It makes manifest the sub Lie groups corresponding to sub diagrams, and obscures most other subgroups. In particular, it makes manifest an action on $\mathfrak{g}$ by the diagram automorphisms — it is a standard, although nontrivial, result that the group of diagram automorphisms is isomorphic via this action to the outer automorphism group $\operatorname{Out}(\mathfrak{g}) = \operatorname{Aut}(\mathfrak{g})/\operatorname{Inn}(\mathfrak{g})$ of $\mathfrak{g}$ — and obscures other possible splittings of the map $\operatorname{Aut}(\mathfrak{g}) \to \operatorname{Out}(\mathfrak{g})$.

In particular, taking the $D_4$ diagram and its subdiagram of the three outer nodes, we find a manifest triality-fixed (in the setwise sense) subgroup $\mathrm{Sp}(1)^3 \subset \mathrm{Spin}(8)$. Well, at the Lie algebra level it is $\mathfrak{sl}(2)^3 = \mathfrak{sp}(1)^3 \subset \mathfrak{spin}(8) = \mathfrak{so}(8)$; at the Lie group level one should take some central quotient. Clearly triality acts to permute the three $\mathrm{Sp}(1)$s, and so its fixed locus is a single copy of $\mathrm{Sp}(1)$. Note that the $\mathfrak{sl}(2)$-triple corresponding to the central node in the $D_4$ diagram is also fixed (pointwise!) by triality. Together, this fixed $\mathfrak{sl}(2)$ and the diagonal $\mathfrak{sl}(2) \subset \mathfrak{sl}(2)^3$ generate the full fixed subgroup $G_2$ — indeed, they correspond to the two nodes in the $G_2$ dynkin diagram. (The "diagonal" $\mathfrak{sl}(2)\subset \mathfrak{sl}(2)^3$ is "three times longer" than any coordinate, explaining why $G_2$ has roots of different lengths.)

Actually, there is a copy of $\mathfrak{sl}(2) \subset \mathfrak{g}$ associated to each root, not just the simple ones. (This is a slightly subtle point. Given a root $\alpha$, you can take $x^\pm$ to be generators of the $\pm\alpha$ root spaces, and $h = [x^+,x^-]$. But the normalization of $x^\pm$ is ambiguous, so the root $\alpha$ does not uniquely determine a homomorphism $\mathfrak{sl}(2) \to \mathfrak{g}$, but rather determines a subalgebra of $\mathfrak{g}$ abstractly-isomorphic to $\mathfrak{sl}(2)$. Actually, using the theory of vertex algebras, you can almost specify the normalization of $x^\pm$, but there is a fundamental sign ambiguity. This sign ambiguity is reflected by the fact that the the Weyl group is not typically a subgroup of the Lie group, but that there is a typically-nonsplit extension of shape $2^{\text{rank}}.\text{Weyl}$, called the "Tits lift of the Weyl group", which is a subgroup of the Lie group.)

In particular, you can take the three outer nodes of the $D_4$ Dynkin diagram together with the highest root, and these will together produce a subgroup $\mathrm{Sp}(1)^4 \subset \mathrm{Spin}(8)$ (I continue to ignore central quotients), which turns out to be maximal and to be a centralizer in $\mathrm{Spin}(8)$.

You actually know this $\mathrm{Sp}(1)^4$. Identify $\mathrm{Spin}(4) = \mathrm{Sp}(1)^2$; but of course $\mathrm{Spin}(4)^2 \subset \mathrm{Spin}(8)$, up to central factors. This lets you see exactly how the vector and spin representations of $\mathrm{Spin}(8)$ decompose. Indeed, the vector $V^8$ decomposes over $\mathrm{Spin}(4)^2$ as $V^4 \otimes 1 \oplus 1 \otimes V^4$, where $1$ denotes the trivial module, and the isomorphism $\mathrm{Spin}(4) = \mathrm{Sp}(1)^2$ identifies $V^4$ with the tensor product of the two 2-dimensional modules, so all together $$ V^8|_{\mathrm{Sp}(1)^4} = 2 \otimes 2 \otimes 1 \otimes 1 + 1 \otimes 1 \otimes 2 \otimes 2.$$ On the other hand, the full spin module for $\mathrm{Spin}(m+n)$ decomposes over $\mathrm{Spin}(m)\times \mathrm{Spin}(n)$ as the product of the two full spin modules. In even dimensions, the full spin module breaks as a sum of half-spin modules. The result is that the half-spin module $S^8_+$ decomposes over $\mathrm{Spin}(4)^2$ as $S^4_+ \otimes S^4_+ \oplus S^4_- \otimes S^4_-$, where $S^4_\pm$ are the two 2-dimensional half-spin modules for $\mathrm{Spin}(4)$, whereas $S^8_-$ decomposes as $S^4_+ \otimes S^4_- \oplus S^4_+ \otimes S^4_-$. Under the identification $\mathrm{Spin}(4) = \mathrm{Sp}(1)^2$, the half-spin modules are identified with $2\otimes 1$ and $1\otimes 2$, respectively, and so $$ S^8_+ |_{\mathrm{Sp}(1)^4} = 2 \otimes 1 \otimes 2 \otimes 1 + 1 \otimes 2 \otimes 1 \otimes 2,$$ $$ S^8_- |_{\mathrm{Sp}(1)^4} = 2 \otimes 1 \otimes 1 \otimes 2 + 1 \otimes 2 \otimes 2 \otimes 1.$$ Triality permutes the modules $V^8 \mapsto S^8_+ \mapsto S^8_-$, and so permutes the three ways of pairing off these four $\mathrm{Sp}(1)$s.

The four $\mathrm{Sp}(1)$s in $\mathrm{Sp}(1)^4 \subset \mathrm{Spin}(8)$ all look the same from this perspective. What we did when we fixed the presentation (aka Dynkin diagram) was to choose one of them to correspond to the highest root, and the other three to be the outer nodes of the dynkin diagram. The specific triality automorphism depended on this choice. (Different choices are related by inner automorphisms. In this case, if you chose a different $\mathrm{Sp}(1)$ from $\mathrm{Sp}(1)^4$ as your highest root, your choice would be related to mine by an element of the Weyl group. This is because implicitly we chose the Cartan of $\mathrm{Spin}(8)$ to be the Cartan of $\mathrm{Sp}(1)^4$.) I mention this to emphasize the subtleties when talking about "fixed points of triality".

Maybe I'll end with one more comment, which is that I mentioned $G_2$ contains an $\mathrm{Sp}(1)^2$ (up to central blah blah). We can recognize is at spanned by the (fixed) "highest" $\mathrm{Sp}(1)$, which might as well be the first of the $\mathrm{Sp}(1)^4$, and a diagonal copy inside the other three. Then over this, how do the above 8-dimensional modules break? The answer is clear: $2\otimes 1\otimes 1$ restricts over the diagonal to $2$, whereas $1 \otimes 2 \otimes 2$ restricts as $3+1$. So we get $$ 8 = 2 \otimes 2 + 1 \otimes 3 + 1\otimes 1.$$ The first two of these are together the 7-dimensional representation of $G_2$, and the third is the trivial rep.

  • 1
    $\begingroup$ 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. $\endgroup$ Apr 13, 2019 at 7:07
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 14, 2019 at 3:57
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 14, 2019 at 4:00
  • 1
    $\begingroup$ 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 $L$. $\endgroup$ Apr 14, 2019 at 4:02
  • 1
    $\begingroup$ 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 $[x^+,x^-] = h$. $\endgroup$ Apr 14, 2019 at 22:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.