## How to prove that SU(3) is compact

**Moderators:** gmalivuk, Moderators General, Prelates

### How to prove that SU(3) is compact

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???

- jestingrabbit
- Factoids are just Datas that haven't grown up yet
**Posts:**5959**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

### Re: How to prove that SU(3) is compact

A subset of R^n or C^n is compact if its closed and bounded. So, if you have a Cauchy sequence of matrices in SU(3), is their limit in SU(3)? Is there an upper bound on the entries in an element of SU(3)? If the answer to both of those is yes, then you have a compact group.

To answer your second question, no idea.

To answer your second question, no idea.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

- Talith
- Proved the Goldbach Conjecture
**Posts:**848**Joined:**Sat Nov 29, 2008 1:28 am UTC**Location:**Manchester - UK

### Re: How to prove that SU(3) is compact

The continuous image of a compact subset of a topological space is itself compact. Try and find some compact set S in R^n for some n, and some map f:R^n -> M(3,C) where C is the complex numbers. If you restrict this map to S, the map should have SU(3) as its image. From this, it follows that SU(3) is a compact subset of M(3,C).

Hint, it might be easier to think of R^n as being C^m for some m, seeing as the variables in the elements of SU(3) are complex, you can then just associate your map f(x) as f(g(x)) where g is your standard homeomorphism from R^2m to C^m.

Hint, it might be easier to think of R^n as being C^m for some m, seeing as the variables in the elements of SU(3) are complex, you can then just associate your map f(x) as f(g(x)) where g is your standard homeomorphism from R^2m to C^m.

### Re: How to prove that SU(3) is compact

Thanks for giving me the answer of my 1st question but please help me in clearing my doubt in 2nd question ????????????

### Re: How to prove that SU(3) is compact

My first guess: You cannot do regular quantum field theory if the symmetry group is not compact.

If it is not bounded, you will run into some strange infinities, and if it is not closed, you cannot use it as a symmetry group.

As I said, just a guess and probably wrong in some way.

If it is not bounded, you will run into some strange infinities, and if it is not closed, you cannot use it as a symmetry group.

As I said, just a guess and probably wrong in some way.

- Talith
- Proved the Goldbach Conjecture
**Posts:**848**Joined:**Sat Nov 29, 2008 1:28 am UTC**Location:**Manchester - UK

### Re: How to prove that SU(3) is compact

I suppose it's convenient that compact manifolds can be covered in a finite number of coordinate charts so they have a finite atlas. SU(3) is a compact manifold as a consequence of the construction I hinted to in my last post.

### Re: How to prove that SU(3) is compact

Can you tell me how SU(3) is a compact manifold in quark field as you have told plzzzz explain it well .

- Talith
- Proved the Goldbach Conjecture
**Posts:**848**Joined:**Sat Nov 29, 2008 1:28 am UTC**Location:**Manchester - UK

### Re: How to prove that SU(3) is compact

We're not here to do your homework for you. You should have more than enough information in the above thread, on wikipedia and in your course texts.

### Re: How to prove that SU(3) is compact

Sorry , but i was not clear so i had asked for it .Now 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?

- Cleverbeans
**Posts:**1378**Joined:**Wed Mar 26, 2008 1:16 pm UTC

### Re: How to prove that SU(3) is compact

amit28it wrote:Sorry , but i was not clear so i had asked for it .Now 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?

If you're using the sum of the moduli as your norm you can conclude it's bounded after your calculation. However these matrices are not in R^9, perhaps you meant living in C^9 (or R^18)?

"Labor is prior to, and independent of, capital. Capital is only the fruit of labor, and could never have existed if labor had not first existed. Labor is the superior of capital, and deserves much the higher consideration." - Abraham Lincoln

### Who is online

Users browsing this forum: measure and 7 guests