Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For $A \in SU(2,C)$, it is clear that $A$ is completly determined by its first row (well any row or column, but let's say first column). In the general $SU(n,C)$-case this is no longer true. In fact, it seems that for every complex $n$-vector of unit norm, there exists a family of matrices for which the vector is the first column. By working with messy simultaneous equations I seem to have shown that every two elements of this family differ by a multiple of $SU(n-1)$ embedded in the bottom right hand corner with 1 in the first entry of the first column and zeros everywhere else. I suspect that there is a bijective correspondence between elements of this family and $SU(n-1)$ but can't prove it.

Can anyone give a simple slick reworking of all of this?

share|improve this question
    
I suspect this is known and in books if it holds up - for a start, what do you mean by "differ"? Also, the statement in your third sentence is almost immediate from general geometric principles. –  Yemon Choi Feb 4 '10 at 10:34
    
By differ I mean one is equal to the other multiplied by some element of $SU(n-1)$ (multiplication in the sense explained above). –  Mihail Matrix Feb 4 '10 at 10:39
    
... if it is known in books I'd be delighted with a reference. –  Mihail Matrix Feb 4 '10 at 10:40
    
Right, I misunderstood what "this family" was. See Thorny's answer below for the relevant details (I've definitely seen something like this for SO(n,R) in books on Lie groups, for instance). –  Yemon Choi Feb 4 '10 at 10:42
add comment

2 Answers 2

up vote 3 down vote accepted

Use the fact that matrices act on vectors. $SU(n)$ acts transitively on the space of unit-length vectors; the stabilizer of a point is $SU(n-1)$ by Thorny's argument. For example, for the vector $(1,0,...,0)$ the stabilizer is the subgroup $\left(\begin{matrix}1&0\\\\ 0&A\end{matrix}\right)\approx SU(n-1)$. Now by the orbit-stabilizer theorem, the space of unit-length vectors is identified with $SU(n)/SU(n-1)$. Fixing one vector $(1,0,...,0)$ fixes this identification, and then each other vector corresponds to a coset $gSU(n-1)$ which is the family you describe.

This isn't really using much about matrices or geometry; I referred to this as the "orbit-stabilizer theorem" above, but it is really just the basic structural feature of group actions. It's certainly something you can understand by yourself; if it's not immediately obvious, you can think about some simpler examples. In the group of permutations $S_n$, consider the family of permutations that map $1\mapsto 3$ -- how do elements of this family differ from each other? You could also try extending your argument to understand the first column of matrices in $GL(n,\mathbb{R})$; you will of course find a similar answer, but the details are interestingly different. Another fun example is to consider linear functions from $\mathbb{R}\to\mathbb{R}$, and look at the family of linear functions taking $2\mapsto 7$.

share|improve this answer
    
Great, just what I was looking for. Thanks. –  Mihail Matrix Feb 4 '10 at 19:16
add comment

Think of $SU(n)$ as the space of orthonormed bases in $C^n$ with the additional property of the determinant being equal to $1$ out of all the unit length complex numbers. Then fixing one of the vectors restricts the others to the orthogonal hyperplane and the additional requirement remains intact for some parametrization of said hyperplane. In the coordinate system defined by such a base, all the others with the same first vector differ only by a matrix of the form $1 \oplus A$ with some $A \in SU(n-1)$.

It does sound too much like a homework question to me though.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.