Decomposing symmetric tensor field into sum of metric and tensor product

Question

Consider a symmetric tensor field $K_{ij}$ on a 2-dimensional Riemannian manifold with metric $g_{ij}$. How does one show that there exists a (smooth) function $f$ and one-form $\phi$ such that $$K_{ij}=fg_{ij}+\phi_i\phi_j?$$ I tried raising the $j$ index (so the metric becomes the identity) and solving the equation directly, but couldn't get the algebra to work out, and it would almost certainly involve division somewhere so smoothness gets dicey.

Answer 1

I don't believe this is true in general.

If we, like you suggest, raise an index with $g$, then we are looking for a decomposition $$K^i{}_j = f \delta^i{}_j + \phi^i \phi_j .$$ Since the endomorphism $K^i{}_j - f \delta^i{}_j = \phi^i \phi_j$ has rank $\leq 1$, $f(p)$ is an eigenvalue of $K(p)^i{}_j$ at each point $x$ (since $K$ is symmetric, both of the eigenvalues of $K$ are real). Taking the trace gives $$K^i{}_i - 2 f = \phi^i \phi_i \geq 0 ,$$ and since $K^i{}_i$ is the sum of the two eigenvalues of $K^i{}_j$, nonnegativity pins down $f$: At each point $p$, $f(p)$ is the smallest of the eigenvalues.

On the other hand, the function that maps $p$ to the smallest eigenvalue of $K(p)^i{}_j$ need not be smooth. For example, if we take $g$ to be the flat metric $\bar g$ on $\Bbb R^2$ and $K_{ij}$ to be symmetric tensor $x \,dx^2 + y \,dy^2$, then $K^i{}_j$ is $x \,\partial_x \otimes dx + y \,\partial_y \otimes dy$. The eigenvalues of this endomorphism are $x, y$, but $\min\{x, y\}$ is not a smooth function on $\Bbb R^2$.

Probably this can be repaired if we eliminate the nonnegativity restriction, e.g., by replacing $\phi^i \phi_j$ with $h \phi^i \phi_j$ for some smooth function $h$. Then, finding a decomposition looks like it amounts to making a smooth choice $f(p)$ of eigenvalue of $K^i{}_j$.