How to prove that SU(3) is compact

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How to prove that SU(3) is compact

Postby amit28it » Sat Nov 12, 2011 8:56 am UTC

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

Postby jestingrabbit » Sat Nov 12, 2011 9:05 am UTC

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

Postby Talith » Sat Nov 12, 2011 1:33 pm UTC

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

Postby amit28it » Mon Nov 14, 2011 5:18 am UTC

Thanks for giving me the answer of my 1st question but please help me in clearing my doubt in 2nd question ???????????? :roll: :?: :?: :?: :?: :?: :(
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Re: How to prove that SU(3) is compact

Postby mfb » Mon Nov 14, 2011 1:47 pm UTC

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

Postby Talith » Mon Nov 14, 2011 2:10 pm UTC

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

Postby amit28it » Tue Nov 15, 2011 6:34 am UTC

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

Postby Talith » Tue Nov 15, 2011 11:56 am UTC

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

Postby amit28it » Wed Nov 16, 2011 10:19 am UTC

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

Postby Cleverbeans » Wed Nov 16, 2011 5:03 pm UTC

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)?
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