How to prove that SU(3) is compact

[zahero_2007]

How to prove that SU(3) is compact?I have no idea how to do this . And What is the significance of The compactness of SU(3) on the quark model?

 

[Jamma]

Can't you just show that its closed and bounded in R^9?

 

[zahero_2007]

Can't you just show that its closed and bounded in R^9?

Unfortunately no , My knowledge in Topology is rudimentary .

 

[Tinyboss]

Jamma has exactly the right plan. Here are hints.

Bounded: What do you know about the magnitudes of entries in a unitary matrix?

Closed: Determinant is a polynomial function whose variables are matrix entries.

 

[zahero_2007]

Jamma has exactly the right plan. Here are hints.

Bounded: What do you know about the magnitudes of entries in a unitary matrix?

Closed: Determinant is a polynomial function whose variables are matrix entries.

Thanks jamma , Regarding the first question I have Computed the sum of the moduli of the elements of the unitary matrix I found it to be 3 . So I guess if the SU(3) matrices are living in R^9 they occupy the Surface of an 8-sphere whose radius is 3. So I conclude that it is bounded in R^9 . Is this right?

 

[lavinia]

How to prove that SU(3) is compact?I have no idea how to do this . And What is the significance of The compactness of SU(3) on the quark model?

I think this works.

The unitary group U(3) acts transitively on the 5 sphere and the stabilizer of a point is isomorphic to U(2) which itself is homeomorphic to S^{3}

Thus U(3)/U(2) is homeomorphic to S^{5} and U(2) is compact.

So U(3) is compact. But SU(3) is a closed subset of U(3)

 

[dextercioby]

I may be wrong, but isn't the condition for determinant =1 enough to prove that the group's manifold is homemorphic to a sphere in R^(2n+1), thus compact ? And this should be valid for al su(n), regardless of n ?

 

[Jamma]

Unfortunately no , My knowledge in Topology is rudimentary .

Heh, well if it is too rudimentary, then you shouldn't be asking yourself to show SU(3) is compact! Start of with the basics, I say.

Anyway, I think the following is the easiest way to see things:

The unitary group is a sort of complex analogue of the orthogonal group- it's a sort of rigid rotation possibly with a flip of the vector space C^n.

You can also view it as a change of orthonormal basis- the columns will form an orthonormal basis. So it's not too hard to see this thing is compact, consider the map from the matrices into C^n which just maps onto the i'th column. It's image is the sphere of radius 1 because the columns have norm one (the sphere is closed since it is the preimage of the value 1 of the map C^n -> R taking the norm). Since these sets are preimages of closed sets, they are closed.

Then the preimages of these things are closed, and their intersection is a finite intersection of closed sets, and hence closed and is, by construction, all the unitary matrices. They are clearly also bounded in C^n since the columns all have norm one and hence no entry can have absolute value larger than 1.

Since they are closed and bounded in C^n, that means they are a compact subset of it (see Heine-Borel theorem).

However, you asked not about the unitary matrices, but the special unitary matrices. These are just those unitary matrices with norm one. To see these are also compact, we simply note that they are the preimage of the value one of the continuous map taking the unitary matrices into the complex numbers by taking the determinant (which is continuous being a polynomial in the entries of the matrix). Being the preimage of the closed set {1}, they are a closed set of the compact set of unitary matrices, and hence are also compact.

This is the gritty, intuitive way to do it, whereas Lavinia's method is probably a bit more slick.

 

[Jamma]

I may be wrong, but isn't the condition for determinant =1 enough to prove that the group's manifold is homemorphic to a sphere in R^(2n+1), thus compact ? And this should be valid for al su(n), regardless of n ?

Unfortunately not, consider just the 2 by 2 matrices with determinant one. Consider just the diagonals matrices, in fact. Then putting value x in the top left and 1/x in the bottom right, you have a matrix of determinant one, but clearly this will generate a non-compact set- consider the sequence of increasing x.

I think the group you are referring to is the special linear group, and it is non-compact (for dimensions above 1). Being unitary is the important part (in the analogous way that the set of orthogonal matrices is quite clearly compact, and restricting to det 1 just gives us a closed subset of this compact set). The special linear group will just be "volume preserving" matrices, so you can see why this isn't compact- you can "stretch" arbitrarily much as long as you "squeeze" in the other direction. Orthogonality/being unitary disallows this sort of "squashing".

 

[dextercioby]

Ok, I understand. Indeed the det condition alone leads to SL(n,C) which is non-compact.

 

[lavinia]

Thanks jamma , Regarding the first question I have Computed the sum of the moduli of the elements of the unitary matrix I found it to be 3 . So I guess if the SU(3) matrices are living in R^9 they occupy the Surface of an 8-sphere whose radius is 3. So I conclude that it is bounded in R^9 . Is this right?

SU(3) is not the 8 sphere but it is an 8 dimensional manifold. I think it is an S3 bundle over S5.

 

[Jamma]

Yes, I believe that is correct (although I'd need to check- so there is an obvious action of SU(3) on the 5 sphere embedded in C^3, right? And the stabilizer of a point will locally look like a way of rotating the 5 dimensional space, so will correspond to SU(2) which is a 3 sphere?) Could you explain to me more simply why this is true?

It certainly isn't the 8 sphere though- that's not even paralizable (non of the even dimensional spheres are). In fact, you can't put a group structure on any of the spheres except for dimension 0,1,3 and 7.

 

[lavinia]

Yes, I believe that is correct (although I'd need to check- so there is an obvious action of SU(3) on the 5 sphere embedded in C^3, right? And the stabilizer of a point will locally look like a way of rotating the 5 dimensional space, so will correspond to SU(2) which is a 3 sphere?) Could you explain to me more simply why this is true?

It certainly isn't the 8 sphere though- that's not even paralizable (non of the even dimensional spheres are). In fact, you can't put a group structure on any of the spheres except for dimension 0,1,3 and 7.

SU(3) preserves the Euclidean length in R^{6} so it maps the 5 sphere into itself. To stabilize the north pole,(1,0,0), the first column of the matrix must be (1,0,0) but since its inverse is its conjugate transpose its first row must also be (1,0,0). This shows that the stabilizer is isomorphic to SU(2).

The equator of this 5 sphere is a 4 sphere and the stabilizer of the north pole maps this equator into itself. In fact, it preserves a 3 sphere embedded in it. This 3 sphere is the intersection of the 5 sphere with the the 2 dimensional complex plane that is orthogonal to the polar axis. In terms of matrices, this plane is just the plane preserved by the lower 2x2 block of a stabilizer matrix.

Translate this plane from the equator to the north pole and take its intersection with the tangent 4 sphere at the pole. This intersection is a tangent 3 sphere. So the 5 sphere has a naturally determined tangent 3 sphere at any point and the union of all of these 3 spheres forms the total space of a 3 sphere bundle over the 5 sphere.

Each element of SU(3) is a differentiable map of the 5 sphere into itself and the derivatives of these maps determine an action of SU(3) on this tangent 3 sphere bundle. This derivative action is not only transitive but has no fixed points since any tangent 3 sphere that is mapped into itself is rotated by a quaternionic multiplication. Thus it defines a homeomorphism between SU(3) and this tangent 3 sphere bundle.

The homotopy exact sequence of a fibration shows that this bundle has third homotopy group isomorphic to Z. Thus it can not be the 8 sphere whose third homotopy group is zero. It seems that this type argument restricts which spheres can be the total space of a sphere bundle over another sphere.

BTW: This same type argument can be used in other ways. For instance the action of SO(3) on the tangent circle bundle of the 2 sphere is transitive and without fixed points which shows that the tangent circle bundle of the 2 sphere is homeomorphic to real projective 3 space. This gives another proof that the 2 sphere has no vector fields without zeros since the second real cohomology of real projective 3 space is zero.

 

[Jamma]

Nice description, thanks!

 

[Sina]

I will give you an easy exposition.

Seeing SO(3) in M(3) (matrices of 3 dimension) which is isomorphic to R⁹ then you can show its compactness by boundedness and closedness. Or else you have to show that it is totally bounded and complete which is not so fun.

You first construct a continuous map that takes all of SO(3) to the let's say element (I,1)
and such that the inverse image of (I,1) under this map is again SO(3). Here I is 3x3 identity matrix and 1 is 1 :p And this resides in the space M(n)xR and is a single point there. Thus it is closed so, its inverse image SO(3) is closed.

This map should suffice I think: T(A) = [A^TA,det(A)]. Both determinant and composition is continous. Inverse image of (I,1) is SO(3)

Boundedness easily comes from the fact that it is orthonormal. Given any orthonormal matrix, all of its rows constitute an orthonormal set of vectors. Thus if the norm of the matrix is the usual norm (that is taking square of all entries summing and taking square root again) then by the fact that its rows are orthonormal, all SO(3) matrices can be put inside a sphere of appropriate size (I think its 3 but just to make sure take it 9 heh :p) in R⁹.

 

[Jamma]

What do you mean by "takes all of SO(3) to the let's say element (I,1)"? What is (I,1) an element of, and why would mapping all of SO(3) to a single point give us any information on it (you could map any topological space into any other by mapping it to a point)?

Also, he asked about SU(3), not SO(3). SU(3) is the sort of complex analogue of SO(3)- it is the group of all linear maps from C^3 to C^3 which preserves the canonical inner product on C^3 and have determinant one (just as SO(3) is the group of linear maps from R^3 to R^3 which preserves the canonical inner product on R^3 which have determinant one). Please correct me if I'm wrong.

 

[Sina]

Oops sorry I have seen it as SO(3).

If I find a continuous map that maps SU(3) to a single point such that it is also the inverse image of that point than it is closed because that point is closed.

The proof is a verbatim repeat of what I said above for the SU(3) case.

About your comment, it doesn't help arbitrarly mapping SU(3) to a single point because if I choose it as my ambient topological space then it is both closed and open :) I see it is a subset of M(C,3) and I find a continuous map on whole of that space, that maps the SU(3) to a single point such that SU(3) is also the inverse image of that point. Then it is closed.

 

[Sina]

By the way the technique above can also be used to show that SU(3) is a closed embedded submanifold in M(C,n) since it is a level set by the map I have constructed.

 

[Jamma]

Let me pick up on a mistake that I made earlier:

I said that "...their intersection is a finite intersection of closed sets, and hence closed and is, by construction, all the unitary matrices."

This isn't true- they are closed but they will include things which aren't unitary matrices. However, taking the inverse image of the things with determinant modulus one will be (I've included too many things in here, there will be matrices with repeated columns which we don't want) and together with the above proves that this thing is closed. I'm sure there's a much nicer and more obvious way, yet still simple and intuitive (which is what I was intending rather than being elegant).

 

[Jamma]

Oops sorry I have seen it as SO(3).

If I find a continuous map that maps SU(3) to a single point such that it is also the inverse image of that point than it is closed because that point is closed.

The proof is a verbatim repeat of what I said above for the SU(3) case.

I see. What you mean is that you mapped all of the 3x3 matrices into this other space, and that the group SU(3) is precisely the set which is the inverse image of this point?

Can you explain though still? What is it you are mapping the matrices to and what is this map, precisely?

 

[Sina]

Sorry I am writing to hastily. The map, I have given just after that exposition :) It is

M(A) = (A^TA, det(A)) from M(C,n) to (M(C,n),R). This is continuous on M(C,n) and SU(N) is just the inverse image of the singlet (I,1). Hence closed.

And to the previous message I have added this incase if you haven't seen

"About your comment, it doesn't help arbitrarly mapping SU(3) to a single point because if I choose it as my ambient topological space then it is both closed and open :) I see it is a subset of M(C,3) and I find a continuous map on whole of that space, that maps the SU(3) to a single point such that SU(3) is also the inverse image of that point. Then it is closed."

 

[Jamma]

Yes, I didn't realize that you were mapping all the matrices, it read like you were just mapping SU(3).

I understand your proof then, it's quite neat actually! You can do the same for just U(n), you'd need to take the inverse image of the circle of modulus one (or, actually, the whole of (I,C) because it can be shown that the determinant of a unitary matrix is one). And, as you said, showing that the set is bounded is simple.

I don't quite follow how you conclude that SU(3) is the level set of a function though- don't you need a map into just the real numbers for a level set- at the moment you have a map into M(C,n)xC.

 

[Jamma]

Actually, I think I've just read that it can be a map R^n into R^k for any k, not just one (I don't know why I expected anything else).

For level sets to be manifolds, you need the map to be smooth, but also for the differential to be well behaved i.e. of maximal rank so that the map is a submersion at each x or that each x is a regular point.

Otherwise, a result of Whitney says that for any closed set X of R^n, there exists a smooth map R^n -> R for which the inverse image of 0 is the set X.

 

[Sina]

Level sets can be defined as the inverse image of any fixed point in any space not just reals. If it bothers you that it is a matrix you can map (M(n,C),R) to (Rⁿ²,R) if you want :)

But you are right about the fact that map should have additional properties. It is enough to be the level set of a submersion. And there seems to be a problem with that requirement because the image space has larger dimension :) So we have to modify the map. Instead now I define the map

F(A) = A^TA

You can show that I is a regular value of this map and that it is surjective. Then inverse image of I is U(n) which is a submanifold. Now WITHIN U(n) you defined the map T(A) = det(A) and SU(N) is the level set of 1 which makes it a submanifold with in U(N).

 

[Jamma]

Level sets can be defined as the inverse image of any fixed point in any space not just reals.

Yes, but to apply theorems about the things being matrices we really need maps R^n into R^k (and we'll need k < n for any hope of applying said theorems). This is so we have a notion of smoothness and regular points (actually, we can generalise a bit and use smooth manifolds instead of just vector spaces).

It's important to note that the product of a matrix with its complex transpose is a real matrix. I'm sure it's not too hard to show that all points are then regular, and indeed this will prove that U(n) is a manifold (and, as you said, you can then easily show that SU(n) is a closed submanifold of that).

 

[Sina]

I think for any map M: A->B where A and B are some smooth manifolds, it is enough for U<M to be a manifold that it is the inverse image of some fixed point p in B and that M is a surjection (or has rank equal to dimension of B). That is because you can compose M with the coordinate charts of B which are full rank :)