$GL^+(2, \mathbb R)$ is homeomorphic to $S^1$ × $\mathbb R^3$

Question

Can anyone give me an idea of how to prove that $GL^+(2)$ is homeomorphic to $S^1$ × $\Bbb R^3$ , where $$GL^+(2) = \left\{A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\bigg| \det A > 0\right\}.$$

Any help will be appreciated, thanks a lot!

Answer 1

$GL^+(2)\cong \Bbb R^+ \times SL(2)$ via the isomorphism $A\longmapsto (\det A, \frac{1}{\sqrt{\det A}}A)$.

Since $SL(2)\cong \Bbb R^2\times S^1$ you have

$$GL(2)\cong \Bbb R^+\times \Bbb R^2 \times S^1 \cong \Bbb R^3 \times S^1$$

where $\cong$ means homeomorphic.

EDIT: Why $SL(2)\cong \Bbb R^2\times S^1$? This is a non trivial fact known as Iwasawa decomposition; which state that:

Let $K= \left\{\begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\ \end{pmatrix}\right\}$; $\quad A=\left\{\begin{pmatrix} r & 0\\ 0 & \frac{1}{r}\ \end{pmatrix}: r>0\right\}$; $\quad N=\left\{ \begin{pmatrix} 1 & x\\ 0 & 1\ \end{pmatrix}\right\}$.

Then $SL(2)=KAN$: every matrix $U\in SL(2)$ has a unique decomposition as $U=kan$, where $k\in K$, $a\in A$ $n\in N$.

It's easy to see that $K,A,N$ are one dimensional subspace of $M_2(\Bbb R)$, and that $K\cong S^1$, $A\cong \Bbb R^+\cong \Bbb R$ and $N\cong \Bbb R$. Thus $SL(2)$ is a $3$-dimensional subspace of $M_2(\Bbb R)$ homeomorphic to $\Bbb R^2\times S^1$.

Answer 2

It is a consequence of Gram-Schmidt. The topological space $\mathrm{GL}^+(2)$ can be seen as the set of linearly independent $v_1,v_2\in\mathbb R^2$ such that $\det(v_1\ v_2)>0$. Let $\mathrm{Orth}_2$ be the set of orthogonal pairs $(v_1,v_2)$ with $\det(v_1\ v_2)>0$.

Gram-Schmidt gives the homeomorphism: $$\mathrm{GL}^+(2)\xrightarrow\sim\mathrm{Orth}_2\times\mathbb R:(v_1,v_2)\mapsto ((v_1,v_2-\mathrm{proj}_{v_1}v_2),\langle v_1,v_2\rangle).$$ Now, there is also a homeomorphism $$\mathrm{SO}(2)\times\mathbb R^2\xrightarrow{\sim}\mathrm{Orth}_2:((v_1,v_2),(x,y))\mapsto (e^xv_1,e^yv_2),$$ where $\mathrm{SO}(2)$ is the special orthogonal group (i.e., the set of orthonormal pairs $(v_1,v_2)$ such that $\det(v_1\ v_2)=1$). It is well-known that $\mathrm{SO}(2)=S^1$, so that $\mathrm{GL}^+(2)\simeq S^1\times\mathbb R^3$.

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The advantage of this approach is that it generalizes readily to any other dimension: $$\mathrm{GL}^+(n)\simeq\mathrm{Orth}_n\times\mathbb R^{n(n-1)/2}\simeq \mathrm{SO}(n)\times\mathbb R^{n(n+1)/2},$$ so for instance, $\mathrm{GL}^+(3)\simeq\mathbb{RP}^3\times\mathbb R^6$ and $\mathrm{GL}^+(4)\simeq(S^3\times S^3)/\{\pm1\}\times\mathbb R^{10}$.