# Open condition for groups

1. Nov 26, 2013

### gentsagree

What does it mean for a condition to be "open"? E.g. it is said that det(A)≠0 is an open condition for a matrix group.

Furthermore, this implies that GL(n) has the same dimensions as the group of all nxn matrices as, and I quote, "the subgroup of matrices with det(A)=0 is a subset of measure zero".

1. Could you explain why this is true?
2. Could you shed light on what kind of measure we are talking about here (Lebesgue, Dirac, or else)?

Thank you.

2. Nov 27, 2013

### Office_Shredder

Staff Emeritus
mxn matrices are trivially isomorphic to [itex] \mathbb{R}^{m+n}[/tex] and have the Lebesgue measure induced from this. det(A) is a polynomial, so all they are using is that the set defined by p(x) = 0 for some polynomial p is a measure zero subset of Euclidean space.

3. Nov 27, 2013

### gentsagree

I don't really understand the explanation that Wikipedia gives of a Lebesgue measure-zero set. I understand intuitively that subsets consisting of single points are of measure zero, but the proof evades me. =(

any intuitive explanations, or a short proof perhaps?

4. Nov 27, 2013

### jbunniii

Intuitively, a set has measure zero if it can be covered by a collection of open "intervals" whose total "size" is arbitrarily small. In $n$ dimensions, "interval" means $n$-dimensional box, and "size" means the $n$-dimensional volume of that box, i.e. the product of its $n$ side lengths. We specify open boxes simply to avoid degenerate boxes where one or more sides have length zero.

If we have a set $A$ consisting of one point $x$, then surely it can be covered by a single open box centered at $x$, with all of its sides having length $\epsilon$. Thus the box has volume $\epsilon^n$. We can do this for arbitrarily small $\epsilon$, so $A$ has measure zero.

The same argument works if $A$ consists of a finite number of points, and it can be modified to work if $A$ is countably infinite.

Of course, there are uncountably many matrices with determinant zero, so a different argument is needed, but the intuitive idea is still the same.

5. Nov 27, 2013

### gentsagree

Thank you, very clear. However, do we need to establish a comparison between the subset of matrices with determinant zero and that (complementary) of matrices with determinant not equal zero, in order to say that the first has, comparatively, measure-zero? Or is the measure absolutely true?

6. Nov 27, 2013

### jbunniii

I'm not sure that comparing the set with its complement will be fruitful. The complement will have infinite measure, but that alone doesn't imply anything. It's easy to find examples of sets $A$ where $A^c$ has infinite measure, but $A$ can have zero, positive, or infinite measure.

Probably the way to proceed is what Office_Shredder suggested. For $n \times n$ matrices, the determinant is a polynomial function in $n^2$ variables. If you can prove that in general, the zero set of a polynomial function has measure zero, then you're done. This is clearly true if $n=1$, because then there are only finitely many zeros. You could proceed by induction from here, but you'll probably need some theorems from measure/integration theory to get very far.

7. Nov 27, 2013

### R136a1

It also follows from Sard's theorem: http://en.wikipedia.org/wiki/Sard's_theorem But the proof of Sard's theorem is a bit involved. However, the proof of this specific result requires just a special case of Sard's theorem, which is not as difficult as the full theorem. Jbunniii's approach is probably easiest though.