Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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. –  Une Femme Douce Jul 19 '12 at 11:06
2  
@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

1 Answer 1

up vote 3 down vote accepted

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 $v$.

(I've run out of time, but I'll think and post on the submersion question later if no one else does)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.