# Tag Info

I'd say none. If you are talking about a Riemannian cylinder, its metric (say, induced from its embedding in $\Bbb{R}^{3}$) is $$g=dz^{2}+d\phi^{2}$$ If the two were conformally equivalent, a conformal transformation $(z,\phi)=(z(t,x),\phi(t,x))$ would pull back the metric as  g'=\bigg(\frac{\partial z}{\partial t}dt+\frac{\partial z}{\partial ...