# The last accidental spin groups

For dimensions $n\le 6$ there are accidental isomorphisms of spin groups with other Lie groups: $\DeclareMathOperator{Spin}{\mathrm{Spin}}$

$$\begin{array}{|l|l|} \hline \Spin(1) & \mathrm{O}(1) \\ \hline \Spin(2) & \mathrm{SO}(2) \\ \hline \Spin(3) & \mathrm{Sp}(1) \\ \hline \Spin(4) & \mathrm{Sp}(1)\times\mathrm{Sp}(1) \\ \hline \Spin(5) & \mathrm{Sp}(2) \\ \hline \Spin(6) & \mathrm{SU}(4) \\ \hline \end{array}$$

The definition of $\Spin(n)$ is the double cover of $\mathrm{SO}(n)$. There is a trivial double cover $\mathrm{SO}(n)\times C_2$ but that is disconnected. However for $n=1$ this trivial cover is the spin group - do we make this choice because we want to have a $\Spin(1)$ in the family and there is no other option?

My symplectic groups $\mathrm{Sp}(n)$ are invertible $n\times n$ quaternionic matrices acting from the left by matrix multiplication on column vectors in $\mathbb{H}^n$. If we view $\mathbb{H}^n$ as a right vector space over $\mathbb{H}$, this action is in fact linear. The standard sesquilinear form on $\mathbb{H}^n$ is given by

$$\langle\mathbf{u},\mathbf{v}\rangle=\overline{u_1}v_1+\cdots+\overline{u_n}v_n.$$

(We are using the physics convention of conjugation on the left vector.) Note $\overline{u}$ denotes quaternionic conjugation. Then $\mathrm{Sp}(n)$ is precisely the invertible quaternionic matrices that preserve this sesquilinear form. In particular, $\mathrm{Sp}(1)$ is unit quaternions under multiplication.

To make an isomorphism $\Spin(n)\cong G$, one exhibits a double covering $G\to\mathrm{SO}(n)$. I know:

1. The map $\mathrm{O}(1)\to\mathrm{SO}(1)$ is just the trivial map $C_2\to1$.
2. The map $\mathrm{SO}(2)\to\mathrm{SO}(2)$ is just the squaring map $x\mapsto x^2$.
3. Conjugating a pure imaginary quaternion in $\mathbb{R}\mathbf{i}\oplus\mathbb{R}\mathbf{j}\oplus\mathbb{R}\mathbf{k}\subset \mathbb{H}$ by a unit quaternion $\cos(\theta)+\sin(\theta)\mathbf{u}$ rotates it around the directed axis $\mathbb{R}\mathbf{u}$ by $2\theta$ according to the right-hand rule. This creates the desired map $\mathrm{Sp}(1)\to\mathrm{SO}(3)$.
4. Multiplying a quaternion on the left and right by unit quaternions achieves a rotation of four-dimensional space, hence a map $\mathrm{Sp}(1)\times\mathrm{Sp}(1)\to\mathrm{SO}(4)$.

(Of course, one must do some work to prove that #3 and #4 are double coverings.)

However, I don't know the maps $\mathrm{Sp}(2)\to\mathrm{SO}(5)$ or $\mathrm{SU}(4)\to\mathrm{SO}(6)$, and some cursory googling didn't give me an answer. (My informal class is reading through Stillwell's Naive Lie Theory which doesn't cover spin groups, although it is where I learned the accidental isomorphisms for $n=3,4$.) Can anybody explain to me what they are, or at least tell me?

For $n=5$, here are my thoughts. We know $\mathrm{Sp}(2)$ acts on $\mathbb{H}^2$, so we want to find a special copy of $\mathbb{R}^5$ or $\mathbb{S}^5$ in $\mathbb{H}^2$ that can be defined with the inner product for $\mathrm{Sp}(2)$ to act on. My first idea was

$$A=\{(\mathbf{u},\mathbf{v})\in\mathbb{H}^2~\mathrm{s.t.}~\|\mathbf{u}\|=\|\mathbf{u}\|=1~\mathrm{and}~\mathbf{u}\perp\mathbf{v}\}$$

The orthonormality condition can be phrased in terms of the sesquilinear form, so $\mathrm{Sp}(2)$ does act on our subset $A$. However, the first coordinate projection yields a fibration $A\to\mathbb{S}^3$ with fibers $\simeq\mathbb{S}^2$ (I think), which means we have a fiber bundle $\mathbb{S}^2\to A\to\mathbb{S}^3$, but $\mathbb{S}^5$ is not in any of the four sphere-sphere-sphere fiber bundles that Adam's theorem says exist, so $A\not\simeq\mathbb{S}^5$ and this doesn't work.

My second idea is taking $\mathbb{S}^7\subset\mathbb{H}^2$ (defined by $\|\mathbf{u}\|^2+\|\mathbf{v}\|^2=1$), having $\mathrm{U}(1)\times \mathrm{U}(1)$ act from the right (interpret $\mathbb{C}\subset\mathbb{H}$ as a subset), and then having $\mathrm{Sp}(2)$ act from the left on the quotient space I abbreviate $B=\mathbb{S}^7/(\mathbb{S}^1)^2$. This space also has the correct dimension $5$, but I don't know if it's $\mathbb{S}^5$.

(Another possibility is I am not on a fruitful path to discovering the map $\mathrm{Sp}(2)\to\mathrm{SO}(5)$ at all.)

Finally, Wikipedia (on the spin group article) says

There are certain vestiges of these isomorphisms left over for n = 7, 8.

I don't see any vestiges of the $n\le 6$ maps for $n=7,8$ reading through the linked triality article, although honestly I don't understand its preamble. Is it possible to explain how triality should be interpreted as "vestiges" here without knowing about Lie algebras and Dynkin diagrams?

Added: I had some further thoughts. My third idea for a $5$-dimensional subspace of $\mathbb{H}^2$ was the real subspace of ordered pairs $(\mathbf{u},\mathbf{v})\in\mathbb{H}^2$ for which $\langle\mathbf{u},\mathbf{v}\rangle\in\mathbb{R}$. This is a copy of $\mathbb{R}^5$. $\def\acts{\curvearrowright}$

Also, I realized a couple issues in my first two ideas. If I am to look for a sphere in $\mathbb{H}^2$, I should be looking for $\mathbb{S}^4$ not $\mathbb{S}^5$! (Derp.) Another potential issue is that $\mathrm{Sp}(2)$ acts faithfully on $\mathbb{H}^2$, but in order for $\mathrm{Sp}(2)\to\mathrm{SO}(5)$ to be $2$-to-$1$ we need $-I_2$ to act as the identity.

A fourth idea I should have had earlier is to consider projective spaces. If we consider the standard representation $\mathrm{SO}(2)\acts\mathbb{R}^2$ we get an induced action $\mathrm{SO}(2)\acts\mathbb{P}^1(\mathbb{R})\cong\mathbb{S}^1\subset\mathbb{R}^2$ which yields the nontrivial double covering $\mathrm{SO}(2)\to\mathrm{SO}(2)$. We can do the same thing over $\mathbb{C}$: $\mathrm{SU}(2)\acts\mathbb{C}^2$ induces $\mathrm{SU}(2)\acts\mathbb{P}^1(\mathbb{C})\cong\mathbb{S}^2\subset\mathbb{R}^3$ which might extend to $\mathrm{SU}(2)\to\mathrm{SO}(3)$. I haven't checked if this matches my understanding of $\mathrm{Sp}(1)\to\mathrm{SO}(3)$.

(Note $\mathbb{H}$ as a right vector space over $\mathbb{C}$ allows us to interpret left-multiplication-by-quaterions as linear maps, which induces a ring embedding $\mathbb{H}\to M_2(\mathbb{C})$ which restricts to $\mathrm{Sp}(1)\xrightarrow{\sim}\mathrm{SU}(2)$.)

Presumably, we can have $\mathrm{Sp}(2)\acts \mathbb{H}^2$ induce an action $\mathrm{Sp}(2)\acts\mathbb{P}^1(\mathbb{H})\cong\mathbb{S}^4\subset\mathbb{R}^5$ which might extend to a $\mathrm{Sp}(2)\to\mathrm{SO}(5)$ that we desire. Finally, I have heard from a couple sources that the "Klein quadric" explains $\mathrm{Spin}(6)$, I will research this rumor and explore the projective space ideas.

• The "choice" of $\text{Spin}(1)$ is actually not really a choice - there is a (convenient in spin geometry) definition of $\text{Spin}(n)$ as a certain subgroup of the invertible elements of the Clifford algebra $\text{Cliff}(n)$. This definition is valid for all $n$ and gives $\text{Spin}(1) \cong \Bbb Z/2$ as desired.
– user98602
Oct 10, 2015 at 1:16
• @MikeMiller Thanks, that clears $\Spin(1)$ up. Oct 10, 2015 at 1:20
• Side note: the only $S^2$ bundle over $S^3$, up to fiber-preserving diffeomorphism, is $S^2 \times S^3$.
– user98602
Oct 10, 2015 at 1:35
• Nov 17, 2016 at 21:56
• maybe you knew this math.stackexchange.com/q/3607058/141334 ? Apr 3, 2020 at 0:53

The double cover $SU(4) \to SO(6)$ is obtained as follows. By definition, $SU(4)$ has a distinguished $4$-dimensional unitary representation $V$. Its exterior square $\Lambda^2(V)$ is a complex $6$-dimensional representation, and $SU(4)$ respects the wedge product pairing

$$\Lambda^2(V) \otimes \Lambda^2(V) \to \Lambda^4(V) \cong \mathbb{C}$$

where the last isomorphism is an isomorphism of $SU(4)$ representations. This pairing is nondegenerate and symmetric, and hence we get a map $SU(4) \to O_6(\mathbb{C})$. It remains to explain why $SU(4)$ actually acts by real matrices with respect to an appropriate basis. At this point I don't know the full details, but the point is that there is a second pairing

$$\Lambda^2(V) \otimes \Lambda^2(V) \to \mathbb{C}$$

induced this time by the inner product on $V$. We should get a real structure on $\Lambda^2(V)$ by comparing the two pairings carefully.

The double cover $Sp(2) \to SO(5)$ is similarly obtained using the exterior square. One way to define $Sp(2)$ is that it is the intersection of $U(4)$ and $Sp(4, \mathbb{C})$ inside $GL_4(\mathbb{C})$; that is, it's the group of automorphisms of a $4$-dimensional complex vector space $V$ equipped with both a complex inner product and a complex symplectic form. The symplectic form gives a nonzero $Sp(2)$-invariant map

$$\Lambda^2(V) \to 1$$

and the kernel of this map is a complex $5$-dimensional representation of $Sp(2)$. Again the action of $Sp(2)$ preserves a bilinear form and an inner product on this representation, and so as above we get a real $5$-dimensional representation with an invariant inner product.

• Do you know somewhere this is written down?
– user98602
Oct 10, 2015 at 1:29
• @Mike: nope. There's a discussion of the exceptional isomorphisms in Lawson-Michelson (Theorem 8.1) but the proof there is totally different and uses some facts about Clifford algebras. Oct 10, 2015 at 1:34
• How do we get a ($SU(4)$-invariant) real structure on $\Lambda^2\Bbb C^4$ from the two pairings? Jul 27, 2016 at 3:16
• @arctic: I haven't written out the details so I could be mistaken, but I had in mind the following: in an appropiate basis, one of the pairings should look like $v \cdot w$ where $v, w \in \mathbb{C}^n$ and by $\cdot$ I mean the complex dot product (which is not an inner product), and the other one should look like $\overline{v} \cdot w$. They differ by complex conjugation, and that difference lets you see the standard real structure of $\mathbb{C}^n$. Jul 27, 2016 at 7:53
• The hodge star on $\Lambda^2\Bbb R^4$ has $\pm1$ eigenspaces $\Lambda^2_L,\Lambda^2_R$ of left and right isoclinic rotations (using the identification $\Lambda^2V\cong{\frak so}(V)$). This can be extended to an antilinear operator on $\Lambda^2\Bbb C^4$; then ${\rm SL}_4(\Bbb C)$ respects the pairing $(\Lambda^2\Bbb C^4)\times(\Lambda^2\Bbb C^4)\to\Bbb C$ and ${\rm U}(4)$ commutes with the hodge star operator, so ${\rm SU}(4)$ acts on the new $\pm1$ eigenspaces, which are $\Lambda_L^2+i\Lambda_R^2$ and $\Lambda_R^2+i\Lambda_L^2$, each having signature $(6,0)$ with respect to the pairing. Jan 2, 2017 at 2:49