9
$\begingroup$

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.

$\endgroup$
6
  • 3
    $\begingroup$ 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. $\endgroup$
    – user98602
    Oct 10, 2015 at 1:16
  • $\begingroup$ @MikeMiller Thanks, that clears $\Spin(1)$ up. $\endgroup$
    – whacka
    Oct 10, 2015 at 1:20
  • $\begingroup$ Side note: the only $S^2$ bundle over $S^3$, up to fiber-preserving diffeomorphism, is $S^2 \times S^3$. $\endgroup$
    – user98602
    Oct 10, 2015 at 1:35
  • $\begingroup$ Related: math.stackexchange.com/q/193546/11127 $\endgroup$
    – Qmechanic
    Nov 17, 2016 at 21:56
  • $\begingroup$ maybe you knew this math.stackexchange.com/q/3607058/141334 ? $\endgroup$ Apr 3, 2020 at 0:53

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ Do you know somewhere this is written down? $\endgroup$
    – user98602
    Oct 10, 2015 at 1:29
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 10, 2015 at 1:34
  • $\begingroup$ How do we get a ($SU(4)$-invariant) real structure on $\Lambda^2\Bbb C^4$ from the two pairings? $\endgroup$ Jul 27, 2016 at 3:16
  • $\begingroup$ @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 $\mathbb{C}^n$. $\endgroup$ Jul 27, 2016 at 7:53
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 2, 2017 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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