Set $\mathcal{F}:=\{ A \in \text{SL}\_2(\mathbb{R}) \, | \, Ae\_1 \in \operatorname{span}(e\_1) \, \, \text{ and } \, \, A \, \text{ is not conformal} \,\}$, and


$\mathcal{NC}:=\{ A \in M\_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$.


By a non-conformal matrix, I mean a matrix whose singular values are distinct. (i.e. I allow non-zero singular matrices in $\mathcal{NC}$).



> 
> Is each connected component of $\mathcal{F}$ contractible in $\mathcal{NC}$?
> 
> 
> 


$\mathcal{F}$ has two connected components, both homeomorphic to an open half-plane with one point removed.


Indeed, $Ae\_1 \in \operatorname{span}(e\_1)$ and $A \in \text{SL}\_2(\mathbb{R})$ imply that


$$ A=\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix} \, \, \, \text{for some }\, \lambda \neq 0.$$ $A$ is conformal if and only if $\lambda=\pm 1$ and $y=0$. So, $A$ is **not** conformal if f $\lambda \neq 1,-1,0$ or $\lambda=\pm 1$ and $y \neq 0$. Thus, one connected component of $\mathcal{F}$ is homeomorphic to


$$\{ 0<\lambda \neq 1\} \times \mathbb{R} \cup \{1\} \times \mathbb{R}\setminus\{0\}.$$


(The second component corresponds to $\lambda <0$.)




---


*Here is what I know about the topology of $\mathcal{NC}$:*


Let $\mathcal D=\{ (\sigma\_1,\sigma\_2) \, | 0 \le \sigma\_1 < \sigma\_2\}$. Then the map


\begin{align\*}
\mu: SO\_2\times \mathcal D\times SO\_2\to \mathcal{NC}\\
(U,\Sigma,V)\mapsto U\Sigma V^T
\end{align\*}


is a $2$-fold smooth covering map\*. (i.e. $\mu(U,\Sigma,V)=\mu(-U,\Sigma,-V)$, and this is the only ambiguity in $U,V$ for a pre-image of a given point in $\mathcal{NC}$. 


Since $SO\_2 \cong \mathbb{S}^1$, and since [after identifying antipodal points](https://math.stackexchange.com/questions/3561814/what-is-the-surface-obtained-by-identifying-antipodal-points-of-mathbbs1-t/3563215#3563215) in $\mathbb{S}^1 \times \mathbb{S}^1$, we get the $2$-torus $\mathbb{T}^2$ again, it follows that $\mathcal{NC} \cong \mathbb{T}^2 \times D$.


\*I am not entirely sure regarding the behaviour at the boundary points where $\sigma\_1=0$, but I don't think this creates a serious problem.



Is this subset of matrices contractible inside the space of non-conformal matrices?
**Edited.** In the first version of the answer I was assuming that the space in which the contraction was taking place was not $\cal NC$ but the complement to non-conformal matrices in $SL(2,\mathbb R)$. I'll suggest a fix for this now.


Note, that we have a natural continuous map $u: {\cal NC}\to S^1=\mathbb RP^1$. Namely, to each matrix $A$ from ${\cal NC}$ we can associate the following one-dimensional subspace $u(A)\in \mathbb R^2$. Take the matrix $AA^{\*}$ and take the eigenspace corresponding to the maximal eigenvalue of $AA^\*$ (there will be two distinct eigenvalues since $A$ is not conformal). 


So, if we find a closed path $\gamma$ in $\cal F$, such that its image $u(\gamma)\subset S^1$ is not contractible, we are done. How to find such a path is explained in the previous answer to this question, which the path $\gamma(t)$ constructed in the previous answer below. ( I believe that what I suggest works for several reasons but I don't have time to work out all the details now. By the way, it is also funny that $\pi\_1(\cal NC)$ seem to be equal $\mathbb Z^2$, moreover it deformation retracts to $T^2$, I believe.)


**Previous answer.**


It is not contractible. Let us associate to each matrix $A\in SL\_2(\mathbb R)$ the following vector $v(A)$. Take an orthogonal matrix $O\in SO\_2(\mathbb R)$ such that $OA(e\_1)$ is proportional to $e\_1$ with a positive coefficient. Then set $v(A)=OA(e\_2)$. We get a map to the upper half plane:
$$V:SL(2,\mathbb R)\to \{y>0\}$$


Note that the image of confromal matrices is the point $(0,1)$, and the image of any component $\cal F$ is the complement to $(0,1)$. And so each component can be identified with this puncutred half-plane. Hence it is enough to construct a path in $\cal F$ whose image under $V$ is not contractible in $\{y>0\}\setminus \{(0,1)\}$. This is easy, just take a non-contractible path $\gamma(t)\subset \{y>0\}\setminus \{(0,1)\}$ (that has a non-zero winding number around $(0,1)$), and consider the unique path of matrices $A\_t\subset \cal F$ such that $A\_t(e\_2)=\gamma(t)$.


