Prove that SU(n) is closed and bounded

[RickilusSmith]

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}.

To prove the closed portion, I tried mapping a sequence in SU(n) to a sequence in R^{{2n}^2}. 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 R^{{2n}^2}.

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!

 

[arkajad]

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?

 

[RickilusSmith]

Oh, so since SU(n) consists of n \times n matrices with complex entries, I first mapped each with points in C^{n^2}.

And consequently, each of the points in C^{n^2} can be mapped to another point in R^{{2n}^2} since each entry of SU(n) can be written as a + bi. So we have coordinates of (a_{1,1}, b_{1,1}, ..., a_{i,j} , b_{i,j}, ...) and 2n of such so... each point is mapped in R^{{2n}^2}

 

[arkajad]

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.