Manifold of SU(3) group

Question

I understand that matrix elements of SU(2) can be arranged on the 3-sphere, thanks to the relations among the matrix coefficients. If I repeat the same reasoning for SU(3) which kind of manifold do I obtain? I tried to do myself but got lost. Excuse me for the sloppiness but I am a physicist not expert in group theory or differential geometry.

Answer 1

This is a partial answer rather than a complete one; it's long, but I hope it's of some use:

The smaller "classical groups" like $SO(2)$ turn out to be easy to understand ($SO(2)$ is diffeomorphic to a circle, for instance); the larger ones get more complicated. $SO(3)$, for instance, is the same as $RP^3$, which is the quotient of a 3-dimensional sphere by the relation where $x \sim -x$ for every point $x \in S^3$. That's harder to understand/visualize than was $SO(2)$. Eventually many of us think of those classical groups as "simple" objects in and of themselves, so that finding some manifold "looks like" $SO(3) \times U(4)$ might be regarded as "having gotten a grip on what it's like."

But there are some basic ideas that are helpful, as @Kevin suggests in his answer. One is the idea of a bundle: a way in which some space "locally looks like a product". The classic example here is the Mobius band: it "looks like" $S^1 \times I$, locally, but not globally. When you have a Lie group like $SU(n)$, it turns out that if there's a local bundle structure near the identity element, then there's a bundle structure overall.

The way we like to describe bundle structures is to have a map from our space to a "base space"; in the case of the Mobius band, the base space is the core circle, and the map is "project inward towards the core circle." If you imagine building the mobius strip from a long strip of paper,

  |---------------v--------------------|
  |               |                    |
  |....................................|
  |               |                    |
  |---------------^--------------------|

the base space is the dotted centerline (which becomes a circle when we glue the ends), and the "projection" is vertical projection, as I've indicated for one vertical bit. The preimage of a point in the circle ends up being the vertical line segment passing through that point; we call this the "fiber". Thus the mobius band is ALMOST the cartesian product of the base and the fiber...but not quite.

Looking at $SU(3)$, the map $p$ that takes a matrix to its first column is a nice projection. (You actually have to prove that this map is "nice" in a technical sense; I'm just asserting it here). The image of this is ... well, the first column of a matrix in $SU(3)$ is a complex 3-vector of length $1$, so the image $p(SU(3))$ consists of all unit vectors in complex 3-space, or (easier for me to see), real 6-space. Hence the image is 5-sphere. So we have $p : SU(3) \to S^5$.

The preimage of one particular point in $S^5$ is generally tough to see, but if we take a particular point, $(1,0,0,0,0,0) \in \Bbb R^6$, i.e. $c = (1 + 0i, 0 + 0i , 0 + 0i) \in \Bbb C^3$, then the preimage consists of all matrices in $SU(3)$ whose first column is $c$. Such a matrix looks like $$\pmatrix{ 1 & 0 & 0 \\ 0 & z & -\bar{w}\\ 0 & w & \bar{z} \\ } $$ it turns out, and by ignoring the first row and column, we can get a diffeomorphism from this set of matrices to $SU(2) = S^3$.

In short, $SU(3)$ "looks like" $S^5 \times SU(2)$, locally. The only question that remains is "does that 'looks like' extend to global product or not?" (For the Mobius band, we see that the local product structure suggests something like $S^1 \times I$, but in fact there's a "twist" that makes the Mobius band different from a cylinder.

So is $SU(3) = S^5 \times SU(2)$, or is it something different? Well, it turns out that if you have a bundle over $S^5$, it's always a product when you look just at, say, the northern hemisphere (points of $S^5$ with $x_1 \ge 0$), and similarly for the southern hemisphere.

(The same's true for $S^1$: each "half" of the mobius band looks like the cartesian product of half-circle with the interval $[-1, 1].$)

The subtlety comes when you glue together these two halves.

(In the case of the mobius band, at one of the two gluing points, we match $[-1, 1]$ with $[-1, 1]$ by the map $x \mapsto x$; for the other, use use $x \mapsto -x$; if we'd used the same map at both points (whether it was $x \mapsto x$ or $x \mapsto -x$), we'd have gotten a cylinder.)

The gluing takes place along the common boundary of the two hemispheres, i.e., for $SU(3)$, we glue along the equatorial $S^4$. So at each point of $S^4$, we have to say how to match the $SU(2)$ for the upper hemisphere to the $SU(2)$ for the lower hemisphere; the "gluing" at a point $p \in S^4$ can be done with a map of the form $M \mapsto A(p)M$, where $M \in SU(2)$ is any matrix in the fiber, and $A(p) \in SU(2)$ is a matrix that varies with $p$.

(In the Mobius band example, we "glue" along $S^0$, which consists of two points. The "A" matrix at one point is $[1]$; at the other point, it's $[-1]$.)

In fact, everything about the topology of $SU(2)$ is summarized in that function $A$ that says, for each point on the equator, how to glue the "upper hemisphere" fiber to the corresponding "lower hemisphere" fiber. It's called the "clutching function."

Let's look at $A$. It's a map from $S^4$ to $SU(2) = S^3$. In the language of homotopy theory, it's an element of $\pi_4(S^3)$. (And in fact, the topological structure of the bundle depends only on the homotopy class of $A$, so thinking of it this way is appropriate.)

Fortunately for us, someone has already computed the group $\pi_4(S^3)$; it turns out to be $\Bbb Z / \Bbb 2Z$, i.e., there's an identity element and one other element.

If our clutching map $A$ is the identity element, then $SU(3)$ really is the product $S^5 \times SU(2) = S^5 \times S^3$; if the clutching map is the nontrivial element, then $SU(3)$ is ... well, an $S^3$ bundle over $S^5$.

My recollection --- and in this I could easily be wrong --- is that the actual result is the latter, that $SU(3)$ is a twisted product of $SU(2)$ and $S^5$, but it's been too long since I had these facts at my fingertips.

I hope that the analogy with the Mobius band has been of some use to you, but the short answer is "this stuff is complicated, which is why we have a whole field called 'topology' ". :)

Answer 2

Try to think about the map from SU($n$) to $C^n$, mapping each matrix to the first column.

What then, is the image and what is the preimage of each point in the image?

Hint: confirm that it suffices to figure out the pre-image of $(1,0,...,0)$ and all other preimages are essentially the same.