Local diagonalisation of a degenerated 2d metric tensor 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 $\lambda_2(p)\neq0$.
 A: If you prefer vector fields instead of forms, here's an alternative way (but secretly the same as Robert Bryant's) of doing the construction, that works provided that $g$ has rank $\geq 1$ everywhere.
Given that $g$ has rank $\geq 1$, given any point $p$, there is a neighborhood $U$ and a vector field $v$ on $U$ such that $g(v)$ (thinking of $g$ as a mapping from vectors to covectors) is a non-vanishing one form. (Just choose a vector $v_p\in T_pM$ that does not sit in the kernel of $g$, then by continuity any smooth extension of $v_p$ to a vector field $v$ will have a small neighborhood such that $g(v) \neq 0$.)
Since $g(v)$ is non-vanishing, it has a one dimensional kernel, let $w$ be a non-vanishing vector field in this kernel (shrinking $U$ if necessary).
Now take the integral curves of $v$ and $w$. They are necessarily independent. Choose your "coordinates" $x$ and $y$ so that $x$ is constant on the integral curves of $v$ and $y$ is constant on the integral curves of $w$.
That $w$ is in the kernel of $g(v)$ ensures that $g$ is diagonalized w.r.t. $\{x,y\}$.
To ensure that $\{x,y\}$ are actually coordinates, choose them such that $w(x) \neq 0$ and $v(y)\neq 0$ (shrinking $U$ if necessary). This guarantees that $\partial_x$ is a non-zero multiple of $w$ and $\partial_y$ is a non-zero multiple of $v$, and so $g_{yy}\neq 0$, and $g_{xx}$ only vanishes where $g$ is rank 1.
A: 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.
