Prove that SU(n) is closed and bounded

In summary, to prove that SU(n) is closed and bounded, we first map it to a subset of R^{2n}^2 using the complex entries of each n \times n matrix. This can be done by mapping each complex entry to a point in C^{n^2} and then mapping those points to R^{{2n}^2}. The equations for SU(n) are the same as those for U(n), with the additional restriction that the determinant of the matrix must be equal to 1.
  • #1
RickilusSmith
3
0

Homework Statement



Prove that SU(n) is closed and bounded

Homework Equations


The Attempt at a Solution



So in order to prove this, I first mapped SU(n) to be a subset of [tex]R^{{2n}^2}[/tex].

To prove the closed portion, I tried mapping a sequence in SU(n) to a sequence in [tex]R^{{2n}^2}[/tex]. However, I have trouble showing that the limit of that sequence in SU(n) is still within SU(n).

For the bounded portion, I got to the point in needing to find a radius, r, such that SU(n) is a subset of that ball of radius around the origin in [tex]R^{{2n}^2}[/tex].

However, its at these points that I'm having trouble for both problems in finding the intuition to solve them

Thanks in advance for the help!
 
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  • #2
RickilusSmith said:

Homework Statement



Prove that SU(n) is closed and bounded

Homework Equations





The Attempt at a Solution



So in order to prove this, I first mapped SU(n) to be a subset of R^{2n}^2.

And how did you do it? Can you describe this subset explicitly?
 
  • #3
Oh, so since SU(n) consists of [tex] n \times n [/tex] matrices with complex entries, I first mapped each with points in [tex]C^{n^2}[/tex].

And consequently, each of the points in [tex]C^{n^2}[/tex] can be mapped to another point in [tex]R^{{2n}^2}[/tex] since each entry of SU(n) can be written as [tex]a + bi[/tex]. So we have coordinates of [tex](a_{1,1}, b_{1,1}, ..., a_{i,j} , b_{i,j}, ...)[/tex] and 2n of such so... each point is mapped in [tex]R^{{2n}^2}[/tex]
 
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  • #4
And what are the equations that describe SU(n) as a set of matrices. What are the equations for U(n) and what is the equation that adds the "S" in front.
 

Related to Prove that SU(n) is closed and bounded

1. What does it mean for a group to be closed and bounded?

Being closed and bounded means that the group contains all of its limit points and that the group elements are limited in magnitude.

2. How do you prove that SU(n) is closed and bounded?

To prove that SU(n) is closed and bounded, we need to show that for any sequence of elements in the group, the limit point also belongs to the group and that the elements are limited in magnitude.

3. Why is it important for SU(n) to be closed and bounded?

It is important for SU(n) to be closed and bounded because it ensures that the group operations are well-defined and that the group is a compact topological space.

4. Can you provide an example of a group that is not closed and bounded?

Yes, the group of real numbers under addition is not closed and bounded. The limit point of the sequence of natural numbers, which are all elements of the group, is not a real number and the elements of the group are not limited in magnitude.

5. How does the proof of SU(n) being closed and bounded relate to topology?

The proof of SU(n) being closed and bounded relies on the concept of sequential compactness, which is a topological property. This property states that any sequence in a compact topological space has a convergent subsequence whose limit point is also in the space.

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