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Let $M$ be a riemannian manifold with metric $\langle \cdot, \cdot \rangle$ and riemannian connection $\nabla$. For a fixed vector field $V \in \mathfrak{X}(M)$, define the tensor $\beta_V = \beta : \mathfrak{X}(M) \times \mathfrak{X}(M) \to C^{\infty}(M)$ by

$$ \beta(X, Y) = \langle \nabla_X V, Y \rangle. $$

It is well known that $V$ is a Killing vector field if and only if $\beta_V$ is skew-symmetric. I was wondering: what if $\beta$ is symmetric instead? What constraints must $V$ obey for $\beta_V$ to be symmetric? Is there a name for these vector fields?

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    $\begingroup$ Hint: think about the one-form dual to $V$. $\endgroup$ – Anthony Carapetis Jul 29 '16 at 2:37
  • $\begingroup$ @AnthonyCarapetis If $\theta_V$ is this one-form, I found that $\beta_V$ is symmetric iff $\theta_V$ is closed :) $\endgroup$ – Eduardo Longa Jul 29 '16 at 2:58
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Thanks to the comment of Anthony Carapetis, I am posting what I found.

Let $\theta_V : \mathfrak{X}(M) \to C^{\infty}(M)$ be the 1-form dual to $V$, defined by $$ \theta_V(X) = \langle V, X \rangle. $$ The exterior derivative is given by \begin{align} d \theta_V(X, Y) &= X \theta_V(Y) - Y \theta_V(X) - \theta_V([X, Y]) \\ &= \langle \nabla_X V, Y \rangle + \langle V, \nabla_X Y \rangle - \langle \nabla_Y V, X \rangle - \langle V, \nabla_Y X \rangle - \langle V, [X, Y] \rangle \\ &= \beta_V(X, Y) - \beta_V(Y, X) + \langle V, \nabla_X Y - \nabla_Y X - [X,Y] \rangle \\ &= \beta_V(X, Y) - \beta_V(Y, X). \end{align}

Thus, $\beta_V$ is symmetric if and only if $d \theta_V = 0$, i.e., $\theta_V$ is closed.

Notice that if $V$ is a gradient vector field, with $V = \operatorname{grad} f$, then $\theta_V = df$ is closed. Conversely, if $\theta_V$ is closed and $H^1(M) = 0$, then $\theta_V = df$ for some function $f$, which shows that $V = \operatorname{grad} f$.

Hence,

$$\{ V \in \mathfrak{X}(M) : V = df, \, f \in C^{\infty}(M) \} \subseteq \{V \in \mathfrak{X}(M) : \beta_V \text{ is symmetric} \} $$

and equality occurs if, for example, $H^1(M) = 0$. I don't know if there is more we can say if $H^1(M) \neq 0$. Anyways, let me know if you agree with the above.

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