Prove that the cone given by $σ(u,v) = (\cos(u)v, \sin(u)v, v)$ with $0 < u < 2\pi$ and $0 < v < \infty$ is isometric to (part of) the plane.
Ok so far I have the first fundamental form $\sigma$ which is $ds^2$ $= v^2 du^2 + 2dv^2$
However, to my understanding, the first fundamental form of a plane is $ds^2$ $= du^2 + dv^2$.
The two are not equivalent, so how is $\sigma$ isometric to the plane? I thought the first fundamental forms must be equal to have an isometry?
They are indeed equivalent, via a transformation of coordinates. You have to find an appropriate smooth change of coordinates. You can do it in two steps. Observe that the first form for the cone is $$ ds^2=v^2du^2+2dv^2 $$ If you take $\hat u=\sqrt{2}u$ and $\hat v=v/\sqrt{2}$ this becomes $$ ds^2=\hat v^2d\hat u^2+d\hat v^2 $$ which is precisely the first fundamental form of the plane in polar coordinates, that is if you now make the transformation $x=\hat v\cos \hat u$ and $y=\hat v\sin \hat u$, in the coordinates $x,y$ the form is $$ ds^2=dx^2+dy^2. $$ which is the Euclidean structure of the plane. By the nature of the transformation the origin of the plane and say the polar axis are excluded so the transformation is a diffeomorphism. We all know that a cone can be made out of a circular sector without deformation.