Symmetry condition of $\nabla X$

Question

I'm currently calculating a bit in local coordinates and tensor calculus notation, which I'm new to.

I am given the vector field $X$ and the associated one form $\phi$ on a riemannian manifold with Levi-Civita connection. I want to show that $\nabla X$ is a symmetric if and only if $d\phi=0$.

My idea was to write the statements in local coordinates and simply transform the equations until i have a sequence of equivalece transformations. I got $$ (\nabla_iX)^j=\partial_i X^j +\Gamma^j_{ik}X^k$$ and from $d\phi=0$ i can extract the following by using that the connection is metric. \begin{align} \partial_i X_l=\partial_jX_i&=\partial_j(g_{ik}X^k)\\ &=(\partial_j g_{ik})X^k+g_{ik}\partial_j X^k \\ &=(\Gamma ^l_{ji}g_{lk}+\Gamma^l_{jk}g_{il})X^k+g_{jk}\partial_j X^k \end{align} I tried to play around with these term, but it seems like I'm missing something crucial.

Answer 1

Using the derivative of the metric only complicates the matters. All we need is its symmetry which leads to the symmetry of the Christoffel symbols in their two lower indices.

\begin{align} d\phi&=\partial_iX_j\,dx^i\wedge dx^j=\Big(\partial_iX_j-\partial_jX_i\Big)\,dx^i\otimes dx^j\,. \end{align} This vanishes identically if and only if all terms $$ \partial_iX_j-\partial_jX_i $$ vanish. These terms are equal to $\nabla_iX_j-\nabla_jX_i$ because $$ \nabla_iX_j=\partial_iX_j-\Gamma^k_{ij}X_k $$ and $\Gamma^k_{ij}$ is symmetric in $i,j$ (Levi-Civita connection).

In words: $d\phi$ vanishes identically if and only if $\partial_iX_j$ is symmetric or equivalently $\nabla_iX_j$ is symmetric.

Answer 2

@KurtG.'s answer perfectly answers your concern. I would like to give a coordinate-free approach to the question.

$\DeclareMathOperator{\d}{d} \newcommand{\dx}{\d \!}$ If $\phi$ is a $1$-form, its exterior derivative $\dx\phi$ satisfies $$ \dx\phi(Y,Z) = Y\phi(Z) - Z\phi(Y) -\phi([Y,Z]), $$ where $Y$ and $Z$ are vector fields. If $\phi = g(X,\cdot)$ for a vector field $X$, recalling that the Levi-Civita connection is torsion-free, one has \begin{align} \dx\phi(Y,Z) &= Yg(X,Z) - Zg(X,Y) -g(X,[Y,Z]) \\ &= g(\nabla_YX,Z) + g(X,\nabla_YZ) -g(\nabla_ZX,Y) - g(X,\nabla_ZY) -g(X,[Y,Z]) \\ &= g(\nabla_YX,Z) - g(\nabla_ZX,Y) +g(X,\nabla_YZ-\nabla_ZY-[Y,Z]) \\ &= g(\nabla_YX,Z) - g(\nabla_ZX,Y). \end{align} Hence, \begin{align} \dx\phi= 0 &\iff \forall Y,Z, \quad g(\nabla_YX,Z) = g(\nabla_ZX,Y) \\ &\iff \nabla X \text{ is symmetric.} \end{align}

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I would like to add a short digression. Like any other endomorphism, $\nabla X$ uniquely decomposes as $\nabla X = \nabla^+ X + \nabla^- X$, with $\nabla^+ X$ symmetric and $\nabla^- X$ skew-symmetric. For vector fields $Y$ and $Z$, they read \begin{align} 2g(\nabla^+_YX,Z) &= g(\nabla_YX,Z) + g(Y,\nabla_ZX), \\ 2g(\nabla^-_YX,Z) &= g(\nabla_YX,Z) - g(Y,\nabla_ZX). \end{align} Playing around with the properties of the Levi-Civita connection as above, one easily verifies that \begin{align} 2g(\nabla^+_YX,Z) &= (\mathcal{L}_Xg)(Y,Z), \\ 2g(\nabla^-_YX,Z) &= \dx\phi^X(Y,Z), \end{align} where $\mathcal{L}_X$ is the Lie derivative in the direction of $X$ and $\phi^X$ is the dual form of $X$. In particular, one has $$ 2g(\nabla_{\cdot} X,\cdot) = \mathcal{L}_Xg + \dx\phi^X, $$ which turns out to be... exactly Koszul's formula. I always found this formula very obscure until I realized that.