## September 26, 2007

### 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 real vector space equipped with a real inner 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.

Posted at September 26, 2007 9:42 PM UTC

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### 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 $Spin(7)$.

Posted by: Bruce Westbury on September 26, 2007 11:08 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

So the homomorphism G2 → SO(7) lifts to Spin(7)?

Consider a 7-dimensional Minkowski spacetime M. Denote X the set of null lines passing though a fixed point p (the image of the light cone in the 6-dimensional projective space). Equip G2 with real structure so that we have a real homomorphism

G2 → Spin(6, 1)

Does G2 act transitively on X?

Do the same trick with a spacetime of dimension/signature 5+2. What is the answer now?

Posted by: Squark on September 27, 2007 4:42 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

There is a problem with the question. There are two real forms of $G_2$ compact and split. The split form is the automorphism group of the split octonions. The inner product has type 4+4. Taking the imaginary octonions gives a map from $G_2$ to $Spin(3,4)$; the one case you don’t mention.

Posted by: Bruce Westbury on September 27, 2007 10:31 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

So the homomorphism $G_2 \to SO(7)$ lifts to $Spin(7)$?

The answer is yes, but the nice way to see it doesn’t even mention $SO(7)$.

The idea is to use triality. $\Spin(8)$ has three 8-dimensional irreps: the left-handed spinor rep $S_+$, the right-handed spinor rep $S_-$, and the vector rep $V$. There’s a map

$m: V \otimes S_- \to S_+$

coming from the action of the Clifford algebra on spinors. This map is secretly octonion multiplication… but we only see that if we choose unit vectors in $S_+, S_-$ and $V$ to be the identity element of the octonions!

More precisely: first pick a unit vector $v \in V$. The subgroup of $Spin(8)$ that fixes $v$ is clearly $Spin(7)$. Then pick a unit spinor $s \in S_+$. The subgroup of $Spin(8)$ that fixes $v$ and $s$ is, by definition, called $G_2$. Having chosen $v \in V$ and $s \in S_+$, we get isomorphisms

$V \cong S_+ \cong S_-$

where we use

$m(v \otimes -): S_+ \to S_-$

as our isomorphism between $S_+$ and $S_-$, and use

$m(- \otimes s) : V \to S_-$

as our isomorphisms between $V$ and $S_-$. One needs to check that these maps are indeed 1-1. Moreover, $m(v \otimes s)$ is a unit vector in $S_-$.

So, $V$, $S_-$ and $S_+$ are all isomorphic as representations of $G_2$, and we can call them all the octonions. Then $m$ becomes multiplication of octonions, and the chosen unit vector becomes the identity. By the way we’ve played this game, the automorphism group of the octonions is $G_2$.

As a spinoff — pardon the pun — we get inclusions

$G_2 \subseteq Spin(7) \subseteq Spin(8)$

In the current version of my original post (I keep making it bigger), you’ll see how these inclusions give us the variously squashed 7-spheres:

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

(which is unsquashed),

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

(which is a bit squashed), and

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

(which is more squashed).

Posted by: John Baez on September 28, 2007 4:22 AM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

So the homomorphism G2 → SO(7) lifts to Spin(7)?

Though John’s answer is more illuminating, I’ll just mention the simple fact that G→ H automatically lifts to G~ → H~, where G~ is a lazy way of denoting the universal cover of a connected group. Here G2 is already simply connected, so one doesn’t notice the G~ side of things.

Posted by: Allen Knutson on September 28, 2007 2:12 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

With regard to G2 etc, this paper recently came up in physics discussions.

Posted by: Kea on September 26, 2007 11:23 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

I don’t see the relevance of this reference. If you are interested in the spin representation of $Spin(7)$ then have a look at this paper.

I am sensitive about your reference because there is a priority issue. Could you please also refer to my paper on the Sextonions published in the Journal of the L.M.S?

Posted by: Bruce Westbury on September 27, 2007 7:22 AM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

I would say that thinking Spin(7) should be anything like Spin(8) or Spin(6) is a bit of a red herring. They’re not considered the same series of groups for a reason. If we had discovered the classical groups by messing around with Dynkin diagrams, rather than tying them to a particular representation in our mind, it never would have occurred to us to associate the odd and even spin groups.

Posted by: Ben Webster on September 27, 2007 1:41 AM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Then the real question is to make a pattern

Spin(3) = SL(2)
Spin(5) = Sp(2)
Spin(7) = ?

?

The two places to look for this pattern are clear: what are Spin(1) and Spin(9)?

Posted by: Theo on September 27, 2007 7:46 AM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Ben wrote:

If we had discovered the classical groups by messing around with Dynkin diagrams, rather than tying them to a particular representation in our mind, it never would have occurred to us to associate the odd and even spin groups.

There’s some truth to that. But, just to be contrary, let me point out that we can always ‘fold’ a Dynkin diagram using any symmetry of that diagram, and get a new Dynkin diagram. The Dynkin diagram for $Spin(2n)$ always has a 2-fold symmetry:

              o
/
o---o---o---o
\
o


If we fold it with respect to this symmetry we get the diagram for $Spin(2n-1)$:

o---o---o---o=>=o


The two dots on the ‘fishtail’ end of the diagram for $Spin(2n)$ get folded together into a single dot. This is related to how lefy-handed and right-handed spinor representations of $Spin(2n)$ both restrict to the spinor representation of $Spin(2n-1)$.

Posted by: John Baez on September 28, 2007 1:55 AM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Of course the 7th dimension is unusual for spin. John Baez has published in depth about this.

To quote from wikipedia for a moment:

“A cross product for 7-dimensional vectors can be obtained in the same way by using the octonions instead of the quaternions. See seven dimensional cross product for the main article. The nonexistence of cross products of two vectors in other dimensions is related to the result that the only normed division algebras are the ones with dimension 1, 2, 4, and 8.”

“In general dimension, there is no direct analogue of the binary cross product. There is however the wedge product, which has similar properties, except that the wedge product of two vectors is now a 2-vector instead of an ordinary vector. The cross product can be interpreted as the wedge product in three dimensions after using Hodge duality to identify 2-vectors with vectors.”

To put it another way, as Ram Kakarala writes:

There is a cross product of two vectors, satisfying all the usual
assumptions, only in dimensions 3 and 7. In dimension 7 one can
define for e1, e2, …, e7

e1 x e2 = e4, e2 x e4 = e1, e4 x e1 = e2
e2 x e3 = e5, e3 x e5 = e2, e5 x e2 = e3
.
.
e7 x e1 = e3, e1 x e3 = e7, e3 x e7 = e1

and ei x ej = -ej x ei. The above multiplication table can be condensed
into the form

e_i x e_{i+1} = e_{i+3}

where the indices i,i+1,i+3 are permuted cyclically among themselves
and computed modulo 7.

This cross product of two vectors in R^7 satisfies the usual rules

(a x b).a = 0, (a x b).b = 0 orthogonality
(a x b)^2 + (a.b)^2 = a^2 b^2 Pythagoras/Lagrange

where the second rule can also be written as |a x b| = |a| |b| sin(a,b).
A cross product of two vectors satisfying the above rules (orthogonality
and Pythagoras/Lagrange) is unique to dimensions 3 and 7; it does not
exist in any other dimensions. It can be also defined by quaternion
or octonion products as a x b = _1 where _1 means taking the
1-vector part (or pure=imaginary part) of the product ab of two vectors
a and b in R^n, n=3 or n=7, with R+R^n being either the quaternions H
or the octonions O.

Contrary to the usual 3-dimensional cross product this 7-dimensional
cross product does not satisfy the Jacobi identity

(a x b) x c + (b x c) x a + (c x a) x b = 0

(but satisfies the so called Malcev identity, a generalization of Jacobi).
The 3-dimensional cross product is invariant under all rotations of SO(3),
while the 7-dimensional cross product is not invariant under all of SO(7),
but only under the exceptional Lie group G_2, a subgroup of SO(7).
In R^3 the direction of a x b is unique, up to orientation having two
possibilities, but in R^7 the direction of a x b depends on vectors
defining the cross product; namely (expressed with the contraction “_|”):

a x b = (a ^ b) _| e123 in R^3 when e123 = e1^e2^e3 but

a x b = (a ^ b) _| (e124+e235+e346+e457+e561+e672+e713) in R^7.

Also in R^7 there are other planes than the linear span of a and b
giving the same direction as a x b.

The image set of the simple bivectors a ^ b, where a,b in R^7, is a
manifold of dimension 2.7-3=11 > 7 in the linear space of bivectors
(of dimension 7(7-1)/2=21) while the image set of a x b is just R^7.
So the “identification”

a x b = (a ^ b) _| (e124+e235+e346+e457+e561+e672+e713)

is not a 1-1 correspondence (just a method of associating a vector to a
simple bivector).

Now, who volunteers to rephrase this in n-category ways?

Posted by: Jonathan Vos Post on September 27, 2007 8:42 AM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

The current September (!!) 2008 Notices 2008 has a very good article on G2 which references John’s Bull AMS article: “so-called (I wish editors would catch that misleading word) “vector cross-products”), as is explained with great care in the article by J. Baez.”

Btw, it was Yuri Rainich’s discussion of cross-products that initiated my interest in associativity and ‘the rest is history’.

Posted by: jim stasheff on August 16, 2008 1:20 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

You say that cross products satisfying the orthogonality and Pythagorean relations exist only in dimensions 3 and 7, but one exists also in dimension 1; it’s just identically 0.

Posted by: L Spice on May 1, 2019 9:08 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Squashed 7-spheres? What’s that?

Posted by: Squark on September 27, 2007 4:30 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Hi Squark, you can see what squashed 7-spheres are in pages 263-274 of the book “Analysis, Manifolds and Physics - part II” by Y. Choquet Bruhat and C. DeWitt-Morette

If you google the phrase “isomorphism spin(7)” then you will see papers which explain what spin(7) is isomorphic to. If you don’t have access to these papers then I could look them up for you if they are available via the Harvard or MIT libraries.

Posted by: Charlie Stromeyer Jr on September 27, 2007 5:35 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Squark wrote:

Squashed 7-spheres? What’s that?

By now I’ve expanded this blog entry to include an explanation of the squashed 7-spheres. Briefly, there are 4 different ways to see the 7-sphere as a homogeneous space $G/H$ admitting a $G$-invariant metric. The obvious way is the usual ‘round’ 7-sphere, where the symmetry group $G$ is as big as possible. The other 3 ways are less symmetrical, hence the name ‘squashed’.

This stuff is a beautiful spinoff of relations between the even Clifford algebras in 5, 6, 7, and 8 dimensions, all of which have 8-dimensional representations.

It also touches on many of the Lie group coincidences you mentioned!

Posted by: John Baez on September 28, 2007 6:00 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Out of curiosity, are all of these 7-spheres diffeomorphic? Seven is, after all, the first dimension in which you can find exotic spheres, so it would be pretty neat if we could describe some of them just by standard Lie theory.

Posted by: David Speyer on September 27, 2007 11:34 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

David wrote:

Out of curiosity, are all of these 7-spheres diffeomorphic?

Yes: they’re just the unit sphere in an 8-dimensional vector space — but we think of this space as $\mathbb{R}^8, \mathbb{C}^4, \mathbb{H}^2$ and $\mathbb{O}$, enabling us to put different metrics on the unit sphere — each with a transitive group of isometries, but with different-sized isometry groups.

Of course this is about rotations in the 8th dimension, not the 7th — despite the title of the blog entry!

Posted by: John Baez on September 28, 2007 6:02 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

“In fact there are 4 different ways to build the 7-sphere as a homogeneous space, which give it 4 different Riemannian metrics…”

So the squashed spheres are unit spinors in dimensions 5, 6, 7 (the usual sphere being spinors in dimension 8). What are the Riemannian metrics on them? It appears to me that since the spinors in all of those dimensions, regarded as a real vector space, carry an inner product,
the usual Riemannian metric on S^7 is Spin(n) invariant for n = 5, 6, 7, 8. So I guess there should be different Riemannian metrics, invariant _only_ under Spin(n), each for its own n?

Posted by: Squark on September 29, 2007 8:08 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

Squark writes:

So the squashed spheres are unit spinors in dimensions 5, 6, 7 (the usual sphere being spinors in dimension 8).

Right! The spin group only acts transitively on real unit spinors up to a certain dimension, which I can never remember… not much more than 8.

What are the Riemannian metrics on them?

Umm… err… this is the part that always gets me confused.

It appears to me that since the spinors in all of those dimensions, regarded as a real vector space, carry an inner product, the usual Riemannian metric on $S^7$ is $Spin(n)$ invariant for $n = 5, 6, 7, 8$.

I think you’re right.

So I guess there should be different Riemannian metrics, invariant only under $Spin(n)$, each for its own $n$?

That’s what they say. I guess I’ve always hoped that given a compact Lie group $G$ and a compact Lie subgroup $H$, there’s a canonical way to put a $G$-invariant Riemannian metric on $G/H$, built from the unique left- and right-invariant metric on $G$.

Then, we might get four different metrics on the 7-sphere from thinking of it as a quotient space like this in four different ways. But, we’d have to check that these metrics really are different, with none of them ‘more symmetrical than necessary’. Apparently it’s true.

As Charlie said, the second volume of Choquet–Bruhat is a good source of info on this subject. Unfortunately I’m at home and my copy is at work.

Posted by: John Baez on September 29, 2007 10:04 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

G-invariant Riemannian metrics on G/H is the same as H-invariant real inner products on g/h. Shouldn’t be too hard to classify for the squashed spheres, at least in cases other than Spin(7)/G_2. Maybe I’ll do it when I’m not feeling lazy. In any case, the invariant metrics form a convex space which contains a point corresponding to the usual S^7 metric.

Posted by: Squark on October 1, 2007 10:53 PM | Permalink | Reply to this

### Re: Rotations in the 7th Dimension

arXiv:0709.4594 [ps, pdf, other]
Title: Flows of Spin(7)-structures
Authors: Spiro Karigiannis
Comments: 12 pages. Based on a talk given at the 10th International Conference on Differential Geometry and its Applications, in honour of the 300th anniversary of the birth of Leonhard Euler, Czech Republic
Subjects: Differential Geometry (math.DG)

We consider flows of Spin(7)-structures. We use local coordinates to describe the torsion tensor of a Spin(7)-structure and derive the evolution equations for a general flow of a Spin(7)-structure on an 8-manifold M. Specifically, we compute the evolution of the metric and the torsion tensor. We also give an explicit description of the decomposition of the space of forms on a manifold with Spin(7)-structure, and derive an analogue of the second Bianchi identity in Spin(7)-geometry. This identity yields an explicit formula for the Ricci tensor and part of the Riemann curvature tensor in terms of the torsion.

Posted by: Jonathan Vos Post on October 1, 2007 5:46 PM | Permalink | Reply to this

### 8-dimensional crystallography; Re: Rotations in the 7th Dimension

As an open question in the 8-dimensional spaces, has anyone since 2001 calculated the number of abstract 8-dimensional crystallographic point groups?

For n-dimensions, the sequence in Slaone is A006226.

n a(n)
1 1
2 2
3 9
4 18
5 118
6 239
7 1594

W. Plesken and T. Schulz, Counting crystallographic groups in low dimensions, Experimental Mathematics, 9 (No. 3, 2000), 407-411.

Posted by: Jonathan Vos Post on October 5, 2007 6:40 PM | Permalink | Reply to this

### Spin(7) holonomy metrics; Re: Rotations in the 7th Dimension

I don’t understand this, but it seems on-topic:

Title:
A Kahler-Einstein inspired anzatz for Spin(7) holonomy metrics and its solution
Authors:
Santillan, O. P.
Publication:
eprint arXiv:hep-th/0609088
Publication Date:
09/2006
Origin:
ARXIV
Keywords:
High Energy Physics - Theory, Mathematics - Differential Geometry
Comment:
Some formulas corrected

We construct propose an anzatz for Spin(7) metrics as an R-bundle over closed G2 structures. These G2 structures are R3 bundles over 4-dimensional compact quaternion Kahler spaces. The inspiration for the anzatz metric comes from the Bryant-Salamon construction of G2 holonomy metrics and from the fact that the twistor space of any compact quaternion Kahler space is Kahler-Einstein. The reduction of the holonomy to a subgroup of Spin(7) gives non linear system relating three unknown functions of one variable. We obtain a particular solution and we find that the resulting metric is a Calabi-Yau cone over an Einstein-Sassaki manifold which means that the holonomy is reduced to SU(4). Another coordinate change show us that our metrics are hyperkahler cones known as Swann bundles, thus the holonomy is reduced to Sp(2) and the cone is tri-Sassakian. We revert our argument and state that the Swann bundle define a closed G2 structure by reduction along an isometry. We calculate the torsion classes for such structure explicitly.

Posted by: Jonathan Vos Post on October 10, 2007 2:42 AM | Permalink | Reply to this

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