# Gradient in local coordinates on a manifold with Riemannian metric

Let $M$ be a smooth manifold with a Riemannian metric g : $TM\otimes TM$ -> R

If f is a smooth function from M to R, the gradient of f with respect to g is the vector field $\nabla f$ defined by $df$=$g(\nabla f, *)$

(1) In local coordinates {$x^i$}, compute $\nabla f$ in terms of local coordinates.

(2) Now consider $p \in M$. Show that if $V \in T_p M$ satisfies $df_p(V)>0$, then there exists a Riemannian metric $g$ on $M$ with $\nabla f(p)$=$V$

I'm having trouble with how to represent $g(\nabla f, *)$ in terms of local coordinates.

Any help would be appreciated.

Thanks.

Point (1) : Since $g$ is non degenerate, it induces two musical isomorphisms :

• $\flat : TM \to T^*M ; \quad \flat(X):=g(X,\cdot)$
• $\sharp : T^*M\to TM ; \quad \sharp := \flat^{-1}$

In local coordinates $\{x^i\}$ we have $g = g_{ij}\mathrm d x^i \otimes \mathrm d x^j$ and the musicalities are given by :

• $\flat(\partial_i) = g_{ij}\mathrm d x^j$
• $\sharp(\mathrm d x^i) = g^{ij}\partial_j$

where $[g^{ij}] = [g_{ij}]^{-1}$. The gradient $\nabla f$ is then : $$\nabla f = \sharp(\mathrm d f) = \sharp(\partial_i f \mathrm d x^i) = (\partial_i f) g^{ij}\partial_j$$ So here you have the gradient of $f$ in local coordinates : $\nabla f = (\partial_i f) g^{ij}\partial_j$.

Point (2) : This problem is an example of a more generic problem : given $\alpha\in T_p^*M$ and $v\in T_p M$ such that $\alpha(v)>0$, show there exists $g$ on $M$ such that $\sharp(\alpha) = v$. Since a metric defined at a point can be extended to a global metric, all we need to do is show that there exists a scalar product $g_p : T_pM\otimes T_pM\to \mathbb R$ such that $g_p(v,\cdot) = \alpha$. Now, $\alpha(v)>0$ implies $v\ne 0$ and $\alpha \ne 0$. So we can choose coordinates $\{x^i\}$ around $p$ such that $v = \partial_1$. Now, in those coordinates, we are looking for a symmetric positive definite matrix $[g_{ij}]$ such that $$g_{1j} = \alpha_j$$ Now, $\alpha(v)>0$ implies $\alpha_1>0$. So we are looking for a symmetric positive matrix $[g_{ij}]$ such that $g_{11}>0$ and such $g_{1j} = \alpha_j$. Such a matrix exists.