# homeomorphism of a subset of $GL_3(\mathbb{R})$ with $GL_2(\mathbb{R})$ and connectedness

Suppose I denote $G_3$ be the set of all $3\times 3$ matrices with positive determinant, and consider the map $\pi:G_3\rightarrow \mathbb{R}^3\setminus\{0\}$ define by $\pi(g)=ge_1$ where $e_1=(1,0,0)$ then i want to know about the set $\{g:ge_1=e_1\}$, is it connected? is this set homeomorphic to $G_2\times\mathbb{R}^2$? well, is the map surjective submersion?

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Do you mean homeomorphic in the sense of albebra or in the sense of topology? If the latter, what metric do you want to use on $G_2$? –  Karolis Juodelė Jul 19 '12 at 11:05
in the sense of topology as this is a subgroup of a topological group $GL_2(\mathbb{R})$ any suitable metric let say. –  La Belle Noiseuse Jul 19 '12 at 11:06
@DavideGiraudo: Sure about convexity? Let $G = \{g \in G_3 \mid \pi(g) = e_1\}$ We have $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \in G$, $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \in G$, but $\frac 12(A + B) \not\in G_3$ (and hence $\not\in G$). –  martini Jul 19 '12 at 11:19
@martini: I should have seen that it's the $g$ in $G_3$, not in the set of matrices. –  Davide Giraudo Jul 19 '12 at 11:20

Your set is homeomorphic to $G_2\times \mathbb{R}^2$ (and hence connected). Notice that $ge_1 = e_1$ emplies that the first column of $g$ is $e_1$, but the other entries are free, modulo the fact that the bottom right $2\times 2$ matrix must have positive determinant.
Then a map from your set to $G_2\times \mathbb{R}^2$ is given by mapping the bottom $2\times 2$ matrix to $G_2$ and the top middle and top right entries to $\mathbb{R}^2$.
In the "usual" topology on $G_3$ (obtained by viewing it as an open subset of $\mathbb{R}^9$), this map is easily seen to be a homeomorphism.
Further, the map is surjective. To see this, pick any $v\in \mathbb{R}^3-\{0\}$ and extend it to a positively oriented basis of $\mathbb{R}^3$. Then there is a unique linear transformation mapping map $e_1$ to $v$, and $e_i$ to the other basis vectors you chose. Then this transformation is in $G_3$ and sends $e_1$ to $v$.