$SO(3)$ group is defined as linear transformations that preserve euclidean norm. Now if I identify $\phi$ (angle in $xy$ plane whose axis of rotation is $z$-axis) $0$ to $2\pi\beta$ where $\beta$ can vary from $0$ to $1$. What happens to our spherical symmetry? Will $SO(3)$ group still preserve the euclidean norm?
According to me as soon as I picked the $xy$ plane ($z$-axis of rotation) the spherical symmetry is broken symmetry due to both polar angle as well as azimuthal angle is lost. Is my reasoning right?
Edit: My terminology might be non-standard so what I meant by angle defect(or deficit) is $2\pi(1-\beta)$
In my case, it will look like
with the open section stuck together. Also since all the line element for above case is $$ds^2=d\theta ^2+\beta^2\sin^2(\theta)d\phi^2$$. It's killing vector, which forms the generator of $SO(3)$ will be same as that of $S^2$ with $\phi$ replaced by $\beta\phi$ i.e.
$$K_1=\cos(\beta\phi)\frac{\partial}{\partial \theta}-\cot\theta\sin(\beta \phi)\frac{\partial}{\partial(\beta\phi)}$$ $$K_2=-\sin(\beta\phi)\frac{\partial}{\partial(\theta)}-\cot(\theta)\cos(\beta\phi)\frac{\partial}{\partial(\beta\phi)}$$ $$K_3=\frac{1}{\beta}\frac{\partial}{\partial \phi}$$
And oddly enough they do satisy their usual commutation relation $$[K_i,K_j]=\epsilon_{ijk}K_k$$ So there seems to be no issue with spherical symmetry since commutation relation is smoking gun to fire away any issue of coordinate representation. But clearly looking at the would-be distorted image of $\beta$ sphere is a clear no to spherical symmetry? What is going on? Is our intuition of spherical symmetry failing in the above case?