5
$\begingroup$

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 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.

$\endgroup$
  • 2
    $\begingroup$ What is a conformal matrix? I thought fractional linear transformations are conformal but this is probably not what you mean. $\endgroup$ – Igor Belegradek Feb 26 at 17:09
  • 1
    $\begingroup$ A conformal matrix is a nonzero scalar multiple of an orthogonal matrix. $\endgroup$ – Moishe Kohan Feb 26 at 18:51
  • $\begingroup$ @MoisheKohan, this turns out not to be the OP's definition (see my answer below). $\endgroup$ – macbeth Mar 3 at 3:12
  • $\begingroup$ @macbeth: A matrix is a scalar multiple of an orthogonal matrix if and only if all its singular values are equal. $\endgroup$ – Moishe Kohan Mar 3 at 4:02
  • $\begingroup$ @macbeth Indeed, Moishe Kohan is right. (the two definitions coincide). In the case of $2 \times 2$ matrices, they are not conformal if and only the two singular values are different. $\endgroup$ – Asaf Shachar Mar 3 at 14:37
2
$\begingroup$

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 $A_t(e_2)=\gamma(t)$.

$\endgroup$
  • $\begingroup$ Hi, after some further thinking I am not so sure that this argument settles the question. The question was whether or not $F$ is contractible inside $\mathcal{NC}:=\{ A \in M_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$. But the point $(0,1)$ does lie in the image of your map $V:\mathcal{NC} \to M_2(\mathbb{R})$; take $Ae_1=2e_1, Ae_2=e_2$. (We allow matrices outside $\text{SL}_2(\mathbb{R})$ in $\mathcal{NC}$. I think that you have only proved that $\mathcal{F}$ is not contractible inside $\text{SL}_2(\mathbb{R}) \cap \mathcal{NC}$). $\endgroup$ – Asaf Shachar Mar 4 at 9:25
  • $\begingroup$ In other words, by restricting the domain of $V$ to $\text{SL}_2(\mathbb{R})$ your argument only applies to deformations of $\mathcal{F}$ in $\text{SL}_2(\mathbb{R}) \cap \mathcal{NC}$, while I am allowing arbitrary deformations in $\mathcal{NC}$. $\endgroup$ – Asaf Shachar Mar 4 at 9:32
  • $\begingroup$ I see Asaf. I read the question too quickly, I thought that you want to contract this in the complement to conformal matrices in $SL(2,\mathbb R)$. Let me think how to fix the argument. I believe this can be done. At last this can be done very easily if you replace $\cal NC$ by ${\cal NC}_+$ of matrices of positive determinant $\endgroup$ – Dmitri Panov Mar 4 at 10:19
  • $\begingroup$ Asaf, I added some details - not 100% detailed, but this should work I guess. Let me know if you are happy with this. $\endgroup$ – Dmitri Panov Mar 4 at 11:21
  • $\begingroup$ Thank you for this modification. I am not sure why the path you suggested in the first version should apply in this case as well. Can you say why do you think that this should be easier inside ${\cal NC}_+$ instead of ${\cal NC}$? $\endgroup$ – Asaf Shachar Mar 11 at 14:44
0
$\begingroup$

Yes, they are contractible in $\mathcal{NC}$, in fact even with $SL$ rather than $GL$. First simultaneously rotate $\mathcal{F}$, so the typical element is sent to $$ \begin{pmatrix} \cos a & -\sin a \\\ \sin a & \cos a \end{pmatrix}\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix}=\begin{pmatrix} \lambda \cos a & y\cos a-\lambda^{-1}\sin a \\\ \lambda \sin a & y\sin a+\lambda^{-1}\cos a \end{pmatrix} $$ for some small $a$. These are not conformal, since the $(2,1)$-entry is nonzero. Then simultaneously shrink all these matrices down to the $\lambda=\pm 1$, $y=0$ matrix (depending on the connected component), $$ \pm\begin{pmatrix} \cos a & -\sin a \\\ \sin a & \cos a \end{pmatrix}. $$

$\endgroup$
  • 2
    $\begingroup$ Thanks. However, by a non-conformal matrix, I refer to a matrix whose singular values are distinct. So $\lambda=\pm 1, y=0$ is exactly the kind of matrices I want to avoid. $\endgroup$ – Asaf Shachar Mar 2 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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