### Rotations in the 7th Dimension

#### Posted by John Baez

Squark emailed me an interesting question about spin groups, which I take the liberty of reproducing here:

Dear John,

I would appreciate to hear your thoughts on the following matter.

In dimensions $\le 6$ we have the exceptional isomorphisms. In complex form:

$Spin(2) \cong GL(1)$ $Spin(3) \cong SL(2)$ $Spin(4) \cong SL(2) \times SL(2)$ $Spin(5) \cong Sp(2)$ $Spin(6) \cong SL(4)$

In dimension 8 the triality isomorphisms play the same role, in a way:

$Spin(8) \cong Spin(8)$

What about dimension 7? Does anything of the sort happen there?

Best regards,

Squark

I don’t known an exceptional isomorphism involving $Spin(7)$. But, I know a way to build a ‘squashed 7-sphere’ using $Spin(7)$. So, I’ll talk about that.

In fact there are 4 different ways to build the 7-sphere as a homogeneous space, which give it 4 different Riemannian metrics. We can describe these all using spin groups. Unlike Squark, I’ll always be using the *compact real* forms of the spin groups.

The boring way to build a 7-sphere is this:

$S^7 \cong Spin(8)/Spin(7)$

This gives the usual round 7-sphere, where we think of $Spin(8)$ as acting on unit vectors in $\mathbb{R}^8$, so that $Spin(7)$ is the subgroup fixing a unit vector. Of course it’s overkill to use spin groups here; we could have just said $SO(8)/SO(7) \cong S^7$. But that would spoil the beauty of the pattern to come.

The next way to build a 7-sphere is this:

$S^7 \cong Spin(7)/G_2$

This gives a squashed 7-sphere, where we think of $Spin(7)$ as acting on unit spinors so that $G_2$ is the subgroup fixing a unit spinor. Remember: spinors in 7 dimensions form an 8-dimensional real vector space! The cool thing is that $Spin(7)$ acts transitively on the unit sphere in this 8d vector space.

Then there’s this:

$S^7 \cong Spin(6)/SU(3)$

This gives an more squashed 7-sphere, where we think of $Spin(6) \cong SU(4)$ as acting on unit vectors in $\mathbb{C}^4$. It’s pretty easy to see that $SU(4)$ acts transitively on the unit vectors in $\mathbb{C}^4$, and that $SU(3)$ is the subgroup fixing a unit vector. This gives us the isomorphism above.

Finally, there’s this:

$S^7 \cong Spin(5)/SU(2)$

This gives an even more squashed 7-sphere, where we think of $Spin(5) \cong Sp(2)$ as acting on unit vectors in $\mathbb{H}^2$, with $SU(2) \cong Sp(1)$ is the subgroup fixing a unit vector.

See? The reals, complexes, quaternions and octonions all show up in the construction of these 4 squashed 7-spheres. I explained more about what’s secretly going on here in week195. It boils down to this: in 8 dimensions we have left- and right-handed spinors, both forming the space $\mathbb{R}^8$. In 7 dimensions we have $\mathbb{R}^8$ as our space of spinors, in 6 dimensions we have $\mathbb{C}^4$, and in 5 dimensions we have $\mathbb{H}^2$. We keep getting the same 8-dimensional real vector space equipped with more and more extra structure. So, the unit vectors keep forming a 7-sphere, but with less and less symmetry — more and more squashed.

And here’s something about $Spin(6) \cong SL(4,\mathbb{C})$ which I wrote a long time ago, on a faraway planet. It’s phrased in terms of real forms, and in terms of Lie algebras…

I remember why

$su(4) \cong so(6)$

and

$su(2,2) \cong so(2,4)$

so I might as well say why before I forget.

The basic idea is to start with $\mathbb{C}^4$ and give it the inner product

$\langle (x_1,...,x_4),(y_1,...,y_4)\rangle = \overline{x}_1 y_1 + ... + \overline{x}_4 y_4$

where the overline means complex conjugation. The group of transformations preserving this pairing is $SU(4)$, whose Lie algebra is $su(4)$. Using our familiar tricks we get a Hodge star operator on the (complex-valued) 2-forms on $\mathbb{C}^4$. I’ll call this operator $\star$. This Hodge star operator is conjugate-linear and satisfies $\star^2 = 1$.

Now, whenever we have a conjugate-linear operator $\star$ that satisfies $\star^2 = 1$, we can use it to split this vector space into a “real part” and an “imaginary part”. That is, we can write any vector as

$v = Re(v) + i Im(v)$

where

$Re(v) = (v + \star v)/2$ $Im(v) = (v - \star v)/2i$

and all sorts of familiar properties hold.

So far so good. Now, to get our isomorphism

$su(4) = so(6)$

here is what we do. The 2-forms on $\mathbb{C}^4$ have an inner product arising naturally from the inner product on $\mathbb{C}^4$. Because it’s defined

naturally, without any arbitrary choices, this inner product is preserved by the action of $\mathrm{U}(4)$ on the 2-forms. The subgroup of $\mathrm{U}(4)$ that also preserves a volume form is $SU(4)$. Since the Hodge star operator is defined using only the inner product and a volume form, it commutes with the action of $SU(4)$ on 2-forms. This means that $SU(4)$ acts on any “real” 2-form to give another “real” one.The space of “real” 2-forms is a 6-dimensional

realvector space equipped with arealinner product coming from the inner product on 2-forms. The above paragraph thus gives us a homomorphism$SU(4) \to SO(6)$

One can check that this is 2-1. So, we get a 1-1 homomorphism of Lie algebras

$su(4) \to so(6)$

Since both these spaces are 15-dimensional, this is an isomorphism. Since both $SU(4)$ and $Spin(6)$ are simply-connected, we thus get an isomorphism:

$SU(4) \cong Spin(6)$

The same sort of trick gives us isomorphisms

$su(2,2) \cong so(4,2)$

and

$SU(2,2) \cong Spin(4,2)$

except that here we should start with a complex inner product of signature (2,2) on $\mathbb{C}^4$:

$\langle (x_1,...,x_4),(y_1,...,y_4)\rangle = \overline{x}_1 y_1 + \overline{x}_2 y_2 - \overline{x}_3 y_3 - \overline{x}_4 y_4$

which gives a sesquilinear form of signature $(2,4)$ on the 2-forms.

This is important because $so(4,2)$ is the Lie algebra of the group of conformal transformations of 4-dimensional Minkowski space.

## Re: Rotations in the 7th Dimension

The first answer that comes to mind is No there is not an exceptional isomorphism. The best I can do is the homomorphism $G_2$ to $Spin(7)$.