<s>[INST] Rotation matrix with deficit angle I need to find the rotation matrix for a space with a deficit angle. The question is as pictured

The following is my answer to the question:
If $\theta$ could vary between $0$ and $2 \pi$,
            $$ 
    R(\theta) =
    \begin{pmatrix}
     \cos(\theta) && \sin(\theta) \\
        -\sin(\theta) && \cos(\theta)
    \end{pmatrix}
   $$
            In this space, instead of rotating $2 \pi$ to get to the same point, we rotate $2 \pi - \phi$. So a rotation of $2 \pi$ (full circle) in this funny space is equivalent to a rotation of $2 \pi - \phi$ in ordinary space. So a rotation of $\theta$ in the ordinary space is equivalent to a rotation of $\frac{\theta}{1 - \frac{\phi}{2 \pi}}$ in the funny space. Thus, with the new metric, we let $ \theta \rightarrow \frac{\theta}{1-\frac{\phi}{2 \pi}}$ and we have
            $$
    R(\theta) =
    \begin{pmatrix}
     \cos\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big) && \sin\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big) \\
     -\sin\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big) && \cos\Big(\frac{\theta}{1-\frac{\phi}{2 \pi}}\Big)
    \end{pmatrix}
   $$
            $$
    \therefore
    R(0) = 
    \begin{pmatrix}
    1 && 0 \\
    0 && 1
    \end{pmatrix}
   $$
            and 
            $$
    R(2 \pi - \phi ) = 
    \begin{pmatrix}
    1 && 0 \\
    0 && 1
    \end{pmatrix}
   $$
            This satisfies the requirement that $R(0) = R(2 \pi - \phi) = I_{2} $.
            Is this the correct rotation matrix and are my steps logical? Thank you.
 [/INST] I think that your metric is not correct. why ?
your new polar coordinates are:
$x=r\cos \left( {\frac {2\pi \,\theta}{2\,\pi -\phi}} \right) $
$y=r\sin \left( {\frac {2 \pi \,\theta}{2\,\pi -\phi}} \right) $
The Jacobi matrix is:
$J=\left[ \begin {array}{cc} \cos \left( 2\,{\frac {\pi \,\theta}{2\,
\pi -\phi}} \right) &-2\,r\sin \left( 2\,{\frac {\pi \,\theta}{2\,\pi 
-\phi}} \right) \pi  \left( 2\,\pi -\phi \right) ^{-1}
\\ \sin \left( 2\,{\frac {\pi \,\theta}{2\,\pi -\phi
}} \right) &2\,r\cos \left( 2\,{\frac {\pi \,\theta}{2\,\pi -\phi}}
 \right) \pi  \left( 2\,\pi -\phi \right) ^{-1}\end {array} \right] 
$
and the metric :
$g=J^T\,J=\left[ \begin {array}{cc} 1&0\\ 0&\,{\frac {4{\pi }^{2}}{ \left( 2\,\pi -\phi \right) ^{2}}r^2}\end {array} \right] 
$
If you know the  equations for $x$ and $y$ you can calculate the transformation matrix $R$ with this equation:
$J=R\,H$ , with the matrix $H_{i,i}=\sqrt{g_{i,i}}\,,H_{i,j}=0$
$H= \left[ \begin {array}{cc} 1&0\\ 0&2\,{\frac {\pi \,
r}{2\,\pi -\phi}}\end {array} \right] 
$
$R=J\,H^{-1}$ 
$R=\left[ \begin {array}{cc} \cos \left( 2\,{\frac {\pi \,\theta}{2\,
\pi -\phi}} \right) &-\sin \left( 2\,{\frac {\pi \,\theta}{2\,\pi -
\phi}} \right) \\ \sin \left( 2\,{\frac {\pi \,
\theta}{2\,\pi -\phi}} \right) &\cos \left( 2\,{\frac {\pi \,\theta}{2
\,\pi -\phi}} \right) \end {array} \right] 
$
This is your transformation matrix.
Remark: I use symbolic Program MAPLE to do the calculation
</s>