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Is it possible for a vector field on a smooth manifold $M$ to be a gradient with respect to a Riemannian metric $g$, but not a gradient with respect to a different Riemannian metric $h$?

For completeness: the gradient of a smooth function $f:M\to \mathbb{R}$ with respect to the metric $g$ is the unique smooth vector field $\text{grad}_g\, f$ on $M$ such that for all smooth vector fields $Y$ on $M$, $$g(\text{grad}_g\, f, Y) = df(Y).$$

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Consider the vector field $$X=(y-10x)\partial_x-x\partial_y$$ it is not a gradient vector field with respect to the standard Riemannian metric of $\mathbb{R}^2$ but it is a gradient vector field with respect to the Riemannian metric $$g=5dx\otimes dx-dx\otimes dy -dy\otimes dx +5dy\otimes dy$$

For this metric we have $X=grad_g f$ where $f(x,y)=-1/2(49x^2-10xy+y^2)$ The motivation for consideration of a big number as $10$ in $X$ is the following:

Note: If we choose the vector field $$X=(y-\epsilon x)\partial_x-x\partial_y$$ where $|\epsilon|<2$ then there is no an analytic Riemannian metric $g$ with $X=\text{grad}_g$. Because the orbits tend to the singularity at the origin, spirally. This means that the singularity is a focus singularity. On the other hand the Riemannian version of the Thom gradient conjecture says that if an orbit of an analytic gradient vector field tends to an isolated singularity then it must approach to the singularity in a specific direction. Of course a focus singularity violate this condition.

Shahshahani Gradient: As an interesting example, please search "Shahshahani Gradient". This gradient corresponds to a Riemannian metric on the phase space of certain vector field which is not a gradient system with respect to the standard metric but is a gradient vector field with respect to the Shahshahani Riemannian metric.

See here for some explanation on Shahshahani metric.

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    $\begingroup$ Thanks for the answer and for the reference to the Shahshahani metric/gradient. I hadn't heard of that before. $\endgroup$ – Matthew Kvalheim Jun 3 '18 at 7:28
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    $\begingroup$ @MatthewKvalheim You are well come! $\endgroup$ – Ali Taghavi Jun 3 '18 at 7:44
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First note that a vector field $v$ can be written as a gradient of a function wrt. to a metric $g$ iff the corresponding one-form $\flat_g v$ is exact, i.e. $\flat_g v=df$ for some smooth function $f$, where $\flat_g$ is the isomorphism $TM\to T^{\ast}M$ induced by the metric $g$. If $M$ is contractible, any differential form $\omega$ is exact iff it is closed, i.e. $d\omega=0$.

Thus, take any vector field $v$ on, say, $\mathbb{R}^2$ such that $d(\flat_g v)=0$. It is easy to perturb $g$ into another metric $h$ such as to achieve $d(\flat_h v)\neq 0$.

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  • $\begingroup$ That's a nice explanation! It seems obvious now when you put it that way. Thank you. I wish I could accept more than one answer. $\endgroup$ – Matthew Kvalheim Jun 3 '18 at 8:14

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