How to prove $GL^+ (2, \mathbb{R})$ is connected

Question

I am looking to prove that the set of $2\times 2$ matrices with positive determinant is connected. I understand that the set of invertible $2\times 2$ matrices is however disconnected since the determinant can not equal zero.

I know That my aim is to prove that $GL^+ (2,\mathbb{R})$ cannot be written as a disjoint Union of two open and non-empty subsets. How do I go about this?

Thanks

Answer 1

In case you want an explicit path :

let $M \in GL^+(2,\mathbb{R}), M = \pmatrix{ a & b\\ c & d}$

My goal is to prove that any matrix of $GL^+(2,\mathbb{R})$ can be connected to $I$, i.e. that $\forall M \in GL^+(2,\mathbb{R}), \exists f(t) \in C^0, \forall t \in [0;1] f(t)\in GL^+(2,\mathbb{R}), f(0) = M, f(1)=I$

We know that $\det(M)=ad-cb>0$

So $ad > 0$ or $cb < 0$

We now have two case to consider :

1) if $ad > 0$

let $M' = \pmatrix{ a & 0\\ 0 & d}$ and $f(t)=tM+(1-t)M'$

$M' \in GL^+(2,\mathbb{R})$ and $f(t) \in GL^+(2,\mathbb{R}) \forall t \in [0;1]$

1.1) If $a>0$ then $d>0$ and $M'$ is easily connected to $I$ through $g(t)=tM'+(1-t)I$

1.2) If $a<0$ then $d<0$ then let $\epsilon >0$ and $M_\epsilon=\pmatrix{ a & \epsilon\\ -\epsilon & d}$ which is connected to $M'$, $M_\epsilon$ can be connected to $I$ using the instruction in the section 2

2) If $cb <0$

let $M'=\pmatrix{ 0 & b\\ c & 0}\in GL^+(2,\mathbb{R})$ which is connected to $M$ through $f(t) = tM+(1-t)M'$

then consider $M''=\pmatrix{ 1 & b\\ c & 1}\in GL^+(2,\mathbb{R})$ which is connected to $M'$ through $g(t) = tM'+(1-t)M''$

In the end $M''$ is connected to $I$ through $h(t)=tI+(1-t)M''$

Conclusion

If $A$ is connected to $B$ and $B$ is connected to $C$ then $A$ is connected to $C$. Which means that any matrix can be connected to $I$. All in all, $GL^+(2,\mathbb{R})$ is connected

Answer 2

If $det(M)= det(N)$, then $M= SN$ for some $S\in SL(2,\mathbb{R})$.

In particular, given $M\in GL^+(2,\mathbb{R})$ with determinant $D$, $M = S \pmatrix{ 1 & 0\\ 0 & D}$ for some $S\in SL(2,\mathbb{R})$. As $D>0$, there is a path $\gamma$ in $\mathbb{R}_+^*$ connecting $D$ and $1$. Hence $t\mapsto S \pmatrix{ 1 & 0\\ 0 & \gamma(t)}$ is a path in $GL^+(2,\mathbb{R})$ connecting $M$ and $S$.

Now $SL(2,\mathbb{R})$ is spanned by transvection matrices $I_2 + \alpha E_{i,j}$ for $i\neq j$. Let's denote with $T_{i,j}(\alpha)$ such a matrix. Then clearly, $t\mapsto T_{i,j}(t\alpha)$ is a path in $SL(2,\mathbb{R})$ connecting $I_2$ and $T_{i,j}(\alpha)$. So any product of those matrices is connected to $I_2$ in $SL(2,\mathbb{R})$.

In particular, any matrix in $SL(2,\mathbb{R})$ is connected in $SL(2,\mathbb{R})(\subset GL^+(2,\mathbb{R}))$ to $I_2$. It's the case for $S$. Therefore $M$ is also connected to $I_2$. Therefore $GL^+(2,\mathbb{R})$ is path-connected, hence connected.

Of course this generalizes to $GL^+(n,\mathbb{R})$

Answer 3

The Gram-Schmidt process allows us to make a deformation retract from $GL^+(2,\mathbb{R})$ to $SO(2)$. Since $SO(2) \simeq S^1$, it follows that $SO(2)$ is path-connected and thus $GL^+(2,\mathbb{R})$ must also be.

This argument can be generalized to show that $GL^+(n,\mathbb{R})$ is connected, using the fact that $SO(n)$ is connected (which can be inferred, for instance, from the fact that the spheres are connected, but this is a separate argument).