The 3-sphere S³ can be seen as R³ plus a point at infinity. But here London Tsai shows the "Hopf fibration": S³ as a bundle of circles over the 2-sphere. Each point in S³ lies on one circle. The set of all these circles forms a 2-sphere.

S³ is an S¹ bundle over S².

(1/n)

But the 3-sphere S³ is also a group! It's called SU(2): the group of 2×2 unitary matrices with determinant 1.

So we can see the group SU(2) as an S¹ bundle over S². But in fact we can build *many* groups from spheres!

(2/n)

Let's try SU(3). This acts on the unit sphere in C³. C³ is 6-dimensional as a real space, so this sphere has dimension one less: it's S⁵. Take your favorite point in here; each element of SU(3) maps it to some other point. So SU(3) is a bundle over S⁵.

(3/n)

*Many* elements of SU(3) map your favorite point in S⁵ to the same other point. What are they like? They form a copy of SU(2), the subgroup of SU(3) that leaves some unit vector in C³ fixed.

So SU(3) is an SU(2) bundle over S⁵.

(4/n)

SU(3) is an SU(2) bundle over S⁵. But SU(2) is itself a sphere, S³. So:

SU(3) is an S³ bundle over S⁵.

In other words, you can slice SU(3) into a bunch of 3-spheres, one for each point on the 5-sphere. Kinda like a higher-dimensional version of this picture.

(5/n)

How about SU(4), the 4×4 unitary matrices with determinant 1? We can copy everything: this group acts on C⁴ so it acts on the unit sphere S⁷. The elements mapping your favorite point to some other form a copy of SU(3).

So, SU(4) is an SU(3) bundle over S⁷.

(6/n)

We've seen:

SU(4) is an SU(3) bundle over S⁷.
SU(3) is an S³ bundle over S⁵.

So SU(4) is a S³ bundle over an S⁵ bundle over S⁷!

Maybe you see the pattern now. We can build SU(n) groups as "iterated sphere bundles".

(7/n)

For example, SU(5) is an S³ bundle over an S⁵ bundle over an S⁷ bundle over S⁹.

As a check, you can compute the dimension of SU(5) somehow and show that yes, indeed

dim(SU(5)) = 3 + 5 + 7 + 9

(8/n)

Even better, the group U(5) of *all* unitary 5×5 matrices is an S¹ bundle over an S³ bundle over an S⁵ bundle over an S⁷ bundle over S⁹. The S¹ here comes from the choice of determinant.

So:

dim(U(5)) = 1 + 3 + 5 + 7 + 9 = 5²

and this pattern works in general.

(9/n)

It's easy to see that the sum of the first n odd numbers is n². But we've found a subtler incarnation of the same fact! We've built U(n) out of the first n odd-dimensional spheres, as an iterated bundle.

What do you get for O(n)?

(10/n, n = 10)