Is GL2(R) an Open Subspace, Compact, or Connected?

[mathshelp]

I've come across this question during revision and don't really know what you would say? Any help?

Regard a 2 x 2 matrix A as a topological space by considering 2x2 matrices as vectors (a,b,c,d) as a member of R⁴. Let GL₂(R) c R⁴ be the subset of the 2x2 matrices A which are invertible, i.e. such that ad does not equal bc.

Consider the following, giving reasons:

(i) Is GL₂(R) c R⁴ an open subspace?

(ii) Is GL₂(R) compact?

(iii) Is GL₂(R) connected?

 

[rasmhop]

\det : GL_2(\mathbb{R}) \to \mathbb{R} is continuous. If you don't know this fact you can see it by writing it in terms of continuous functions:
\det(M) = \pi_4(M) \times \pi_1(M) - \pi_2(M) \times \pi_3(M)
where:
\pi_i\left(\begin{array}{cc} x_1 & x_2 \\ x_3 & x_4 \end{array}\right) = x_i
which are projections and therefore continuous.

i) GL_2(\mathbb{R}) is connected if every invertible matrix is an interior point, i.e. every matrix M has a neighborhood contained in GL_2(\mathbb{R}).

For some invertible matrix M consider the open interval I = (0,2\det(M)), then \det^{-1}(I) is open, contains M and is a subset of GL_2(\mathbb{R}). Remember that a subspace is open if and only if every set is an interior point.

EDIT: Actually it's much easier to note that (-\infty,0) \cup (0,\infty) = \mathbb{R}\setminus\{0\} is open, so,
\det^{-1}\left(\mathbb{R}\setminus\{0\}\right) = GL_2(\mathbb{R})
is open.

ii) Can an open subset be compact?

iii) If it's connected then \det(GL_2(\mathbb{R})) = \mathbb{R} \setminus \{0\} is connected since continuous functions take connected spaces to connected spaces.

 

[wofsy]

the determinant answer is right and works for nxn matrices not only 2x2.

the determinant is a polynomial in the entries of the matrix and so is continuous in any dimension.

moreover, a polynomial is differentiable. What are the critical values of the determinant?
Answering this question will give you a picture of the general linear group and some of its subgroups as geometric objects.

If the general linear group were compact then the range of values of the determinant would be bounded - which it is not.

On the other hand the inverse image of a single number - e.g. all matrices of determinant equal to 1 - is compact. Why is that?