# Subgroups of $\text{Spin}(7)$.

I am interested in subgroups of $$\text{Spin}(7)$$ and identifying certain elementary properties that they have. Specifically relating to commutativity. Let me apologise at this point for my ignorance and sloppy presentation below. I know little about this subject, which has recently come up in my work.

I write down the extended Dynkin diagram $$\widetilde B_3$$ corresponding to the Lie algebra of $$\text{Spin}(7)$$

with appended long root $$\alpha_0$$, and short root $$\alpha_3$$.

Removing the short node $$\alpha_3$$ I get $$A_3$$ coresponding to the $$SU(4)\cong \text{Spin}(6)$$ subgroup. Cutting out $$\alpha_1$$ leaves something with a fold symmetry that gives the $$G_2\leq \text{Spin}(7)$$ subgroup inclusion.

On the other hand, if we take out $$\alpha_2$$ then we are left with three $$A_1$$'s. It's pretty clear that the pair given by $$\alpha_0$$, $$\alpha_1$$ are the two commuting $$\text{Spin}(3)$$-subgroups that together constitute the canonical $$\text{Spin}(4)\leq \text{Spin}(7)$$. The final $$A_1$$ corresponds to the short root $$\alpha_3$$, and is what is confusing me. My intuition is telling me that it is a third copy of $$\text{Spin}(3)$$, covering the $$SO(3)\leq SO(7)$$ which is complementary to the $$SO(4)$$ subgroup.

As such this third $$A_1$$ should be another $$\text{Spin}(3)$$ which commutes with the $$\text{Spin}(4)$$ identified previously. However, my understanding runs out here, and the presence of the half-length root indicates to me that what I have actually identified is in fact an $$(S^3\times S^3\times S^3)/\mathbb{Z}_2$$ subgroup containing a $$\text{Spin}(4)$$ subgroup which does not commute with the other factor.

Have I identified a commuting $$\text{Spin}(3)$$ and $$\text{Spin}(4)$$ subgroups, or is it rather the second option $$(S^3\times S^3\times S^3)/\mathbb{Z}_2$$?

Moreover, the diagram indicates to me the presence of three distinguished homotopy classes of maps $$S^3\rightarrow \text{Spin}(7)$$. The first two come from $$\text{Spin}(4)$$, with $$\alpha_0$$ corresponding to the canonical generator $$i_0:S^3\cong \text{Spin}(3)\hookrightarrow \text{Spin}(7)$$, and $$\alpha_1$$ its negative, $$i_1=-i_0$$. Since $$\frac{|\alpha_0|^2}{|\alpha_3|^2}=2$$, the corresponding homotopy class of $$\alpha_3$$ should be $$i_3=2\cdot i_0$$, which would follow from elementary properties of the Dynkin index. I'm not entirely sure this is what I want to see.

Am I correct about the identifications here?

• When you remove $\alpha_3$, don't you rather get $A_3$ ($=D_3$) [not $\tilde A_2$], which corresponds to a $SU(4)$ [not $SU(3)$] $\simeq Spin(6)$ subgroup? Jun 4 at 15:59
• @TorstenSchoeneberg yes, I imaging that was a typo I left for myself. Jun 4 at 18:42

I don't know if you still need an answer to this question, and I don't know enough about roots to answer the question using that approach, but I can confirm a positive answer to your questions. That is, there is (up to conjugacy) a unique subgroup of $$Spin(7)$$ which is locally isomorphic to $$SU(2)^3$$. This subgroup is isomorphic to $$(SU(2))^3/\langle -(I,I,I)\rangle$$. Second, the inclusions of the $$SU(2)$$ factors into $$Spin(7)$$ have Dynkin index $$2$$, $$1$$, and $$1$$.

I'll sketch proofs of all this below.

First, let's work in $$SO(7)$$. A finite-to-one homomorphism $$f:SU(2)^3\rightarrow SO(7)$$ is nothing but a real $$7$$-dim representation of $$SU(2)^3$$, so let's try to understand those.

We will need the following facts:

1. For each integer $$n \geq 1$$, there is a unique $$n$$-dimensional complex irreducible representation of $$SU(2)$$. It has real type if $$n$$ is odd, and symplectic type if $$n$$ is even.

2. Irreducible representations of a product of Lie groups are given by tensor products of irreducible representations of the factors. The tensor product of two real representations or two symplectic representations is real.

3. A representation $$V$$ is real iff it can be decomposed into irreps $$V = \oplus V_i \oplus (W_j \oplus \overline{W}_j)$$ where all $$V_i$$ are real and $$\overline{W}_j$$ indicates the complex conjugate of $$W_j$$.

Proposition: There is a unique real $$7$$-dimensional representation of $$SU(2)^3$$ with the property that the restriction to each $$SU(2)$$ factor is non-trivial.

Proof: I'll use notation $$m\otimes n\otimes p$$ to refer to the tensor product of the $$m$$, $$n$$, and $$p$$-dimensional irreps of $$SU(2)$$.

We first list all the irreps of $$SU(2)^3$$ which have dimension at most $$7$$.

They are, up to permutations, $$1\otimes 1\otimes p$$ for $$p\leq 7$$, $$1\otimes 2\otimes 2$$, $$1\otimes 2\otimes 3$$.

Existence: the representation $$(3\otimes 1\otimes 1)\oplus (1\otimes 2\otimes 2)$$ is real, and non-trivial on all three $$SU(2)$$ factors.

Uniqueness: We show no other combination works. If our $$7$$-dim representation has $$1\otimes 2\otimes 3$$ as a subrepresentation, then the complmentary subrepresentation has dimension $$1$$, so is trivial. Thus, such a representation is trivial on some $$SU(2)$$ factor, ruling out this case.

Likewise, if our $$7$$-dim representation contains $$1\otimes 1\otimes p$$ for $$p\geq 4$$, then the complementary subrepresentation has dimension $$7-p < 4$$, so can't be injective on the other two $$SU(2)$$ factors.

If our $$7$$-dim representation doesn't contain $$1\otimes 2\otimes 2$$, and is non-trivial on all $$SU(2)$$ factors, then, up to permutation, it must be $$(2\otimes 1\otimes 1)\oplus(1\otimes 2\otimes 1)\oplus (1\otimes 1\otimes 2)\oplus (1\otimes 1 \otimes 1)$$ or $$(3\otimes 1\otimes 1)\oplus(1\otimes 2\otimes 1)\oplus (1\otimes 1\otimes 2)$$. In both these cases, the representation is complex, not real.

So, our $$7$$-dim representation is either $$(1\otimes 2\otimes 2)\oplus (2\otimes 1\otimes 1)\oplus (1\otimes 1\otimes 1)$$, or it is $$(1\otimes 2\otimes 2)\oplus (3\otimes 1\otimes 1)$$. But the first of these is not real, and the second is the one I gave in the "existence" portion of the proof. $$\square$$

A result of Malcev (and I can try to get the precise reference if you need it) indicates equivalent representations have conjugate images in $$SO(7)$$. (The actual result is that this is true for all classical groups except $$SO(even)$$).

So, up to conjugacy, there is a unique homomorphism $$f:SU(2)^3\rightarrow SO(7)$$ with finite kernel. This is the one with image given by the block embedding of $$SO(3)\times SO(4)$$. (I'm sure you were already aware that there were commuting $$SO(3)$$ and $$SO(4)$$ in $$SO(7)$$. The point of all this is that we've established that any subgroup of $$SO(7)$$ which is locally $$SU(2)^3$$ is this one, up to conjugacy. Thus, any subgroup of $$Spin(7)$$ which is locally $$SU(2)^3$$ must be a lift of this one.)

Now we'll try to lift all this to $$Spin(7)$$. Of course, the map $$f$$ lifts since $$SU(2)^3$$ is simply connected, but what is the kernel of the lift $$\tilde{f}$$? Obviously, $$\ker \tilde{f}\subseteq \ker f = \langle (-I,I,I), (I,-I,-I)\rangle$$, but which subgroup is it?

I claim that it's $$\langle -(I,I,I)\rangle$$, as you guessed. To see this, simply observe that the maps $$SO(3),SO(4)\rightarrow SO(7)$$ can't lift because they're non-trivial on $$\pi_1$$. Thus, the $$\ker \tilde{f}$$ is either trivial or it's $$\langle -(I,I,I)\rangle$$, but it can't be trivial because $$\pi:Spin(7)\rightarrow SO(7)$$ has a kernel of order $$2$$, while $$f$$ has a kernel of order $$4$$.

What about the corresponding elements of $$\pi_3(SO(7))$$ (which is, of course, isomorphic to $$\pi_3(Spin(7))$$?

Well, for the map $$SU(2)\rightarrow SO(3)\times \{I\}\subseteq SO(3)\times SO(4)\subseteq SO(7)$$, the inclusion $$SO(3)\subseteq SO(7)$$ has Dynkin index $$2$$.

To see this, first observe that $$SO(5)/SO(3)$$ is the unit tangent bundle of $$S^4$$, $$T^1S^4$$, so $$H^4(SO(5)/SO(3))\cong \mathbb{Z}/2\mathbb{Z}$$. On the other hand, since $$Spin(5) = Sp(2)$$, we can rewrite this as $$SO(5)/SO(3) = Sp(2)/\Delta Sp(1)$$. Since the symplectic groups are $$2$$-connected, we see that $$\pi_3(Sp(2))/\pi_3(Sp(1))\cong \pi_3(T^1 S^4)\cong H_3(T^1 S^4)\cong H^4(T^1 S^4)$$, so the Dynkin index of $$SO(3)$$ in $$SO(5)$$ is $$2$$. On the other hand, the Dynkin index of $$SO(5)$$ in $$SO(7)$$ is $$1$$ since $$SO(7)/SO(5) = Spin(7)/Spin(5) = T^1 S^6$$ which is $$4$$-connected.

Lastly, the Dynkin index of either $$SU(2)\subseteq SO(4)$$ in $$SO(7)$$ is $$1$$. To see, note that using the previous paragraph, it is sufficient to show that the Dynkin index of either $$SU(2)$$ in $$SO(5)$$ is $$1$$. Regardless of which of the two $$SU(2)$$ we pick, $$SO(5)/SU(2)$$ is double covered by $$Sp(2)/SU(2) = S^7$$, which is $$6$$-connected. So the Dynkin index of $$SU(2)$$ in $$SO(5)$$ is $$1$$.

• I don't know if I still needed an answer to this question, but I was delighted to see yours. There was originally a bounty attached to the question, and I'll try to sort something out for you tomorrow when I have more time to go through everything you wrote. Jul 20 at 22:45
• Don't worry about a bounty. I don't use the reputation privileges I have anyway. Consider this a thanks for all the homotopy theory I've learned from your wonderful answers! Jul 20 at 23:02