Consider a smooth 2d-manifold $M$ and let $g$ be a smooth $(0,2)$-tensor satisfying $rk(g)\geq1$ everywhere. Obviously if $rk(g)=2$ at a point $p\in M$ then $g$ is locally diagonalisable (i.e. there exists a local coordinate system in which $g$ is diagonal). The same conclusion holds if $rk(g)=1$ in a neighbouhood of a point. But:


**Question**: If $rk(g)=1$ at a point $p\in M$ but $rk(g)=2$ elsewhere, is it possible to find a system a local coordinate system such that $g$ is diagonal, i.e.
$$
g=\begin{pmatrix}\lambda\_1&0\\0&\lambda\_2\end{pmatrix}
$$
with $\lambda\_1(p)=0$ and $\lambda\_2(p)\neq0$.



Local diagonalisation of a degenerated 2d metric tensor
The answer is 'yes'. Here is how one can see this: Suppose that $g$ is a $(0,2)$ form on a neighborhood of the origin in the $xy$-plane such that the rank of $g$ is $1$ at the origin and $2$ everywhere else. Let $h = \mathrm{d}x^2 + \mathrm{d}y^2$ and let
$$
g = E(x,y)\,\mathrm{d}x^2 + 2 F(x,y)\,\mathrm{d}x\,\mathrm{d}y + G(x,y)\,\mathrm{d}y^2.
$$
The symmetric matrix
$$
S(x,y) = \begin{pmatrix} E(x,y) & F(x,y)\\ F(x,y) & G(x,y)\end{pmatrix}
$$
then has the property that it has two distinct eigenvalues at the origin $(x,y) = 0$ and hence has two distinct eigenvalues on an open disk $U$ containing the origin. Consequently, the eigenvalues are smooth functions of $(x,y)$ on $U$, as are their corresponding eigenvectors, which can be thought of as orthogonal unit vector fields with respect to $h$. Thus, we can write
$$
g = \mu\_1(x,y)\,\alpha^2 + \mu\_2(x,y)\,\beta^2
$$
where the functions $\mu\_i$ on $U$ are the eigenvalues of $S$, with the property that $\mu\_1(0,0) = 0$, but $\mu\_2(0,0)\not=0$ and $\alpha$ and $\beta$ are nonvanishing $1$-forms on $U$ such that $\alpha^2 + \beta^2 = h$.


Now, by shrinking $U$ if necessary, we can find nonvanishing functions $u$ and $v$ on $U$ such that $\alpha = p\,\mathrm{d}u$ and $\beta = q\,\mathrm{d}v$. (This is the classical fact that every nonvanishing $1$-form on the plane is locally the multiple of an exact form, i.e., the existence of 'integrating factors'.) Then we have
$$
g = \lambda\_1\,\mathrm{d}u^2 + \lambda\_2\,\mathrm{d}v^2,
$$
where $\lambda\_1 = p^2\mu\_1$ and $\lambda\_2 = q^2\mu\_2$, as desired.


