I am thinking about the 2 dimensional vector calculus operation "grad perp" $\nabla^\perp := (-\partial_y , \partial_x)$.

According to these notes, this can be defined using the Hodge star operator and exterior derivative $\nabla^\perp := \sharp \star \boldsymbol{\mathrm{d}}f$.

This is fine, but it is not perfectly satisfying to me, since $\nabla^\perp$ is essentially the Hamiltonian vector field, and $\sharp, \star$ are dependent on a Riemannian metric. I want to do something with symplectic structures only, and then show that there are "nice cases" where things collapse down to what would one expect should one chose to work with a Kahler manifold, where both options are present.

What I had in mind was to consider the symplectic manifold $(\mathbb{R}^2, \omega := \boldsymbol{\mathrm{d}} x \wedge \boldsymbol{\mathrm{d}}y)$. Since $\omega$ is a nondegenerate billinear form, I ought to be able to define sharps and Hodges w.r.t $\omega$. Even better, $\mathbb{R}^2$ is a Kahler manifold, the compatability of $\omega$ with $g = \boldsymbol{\mathrm{d}}x \otimes \boldsymbol{\mathrm{d}}x + \boldsymbol{\mathrm{d}}y \otimes \boldsymbol{\mathrm{d}}y$ using the almost complex structure defined by $\perp $ and I ought to convert the "symplectic gradient = Hamiltonian vector field" into the "metric gradient" by translating the formulas using $\sharp_\omega, \star_\omega$ into those defined by $\sharp_g, \star_g$ using some identities found here.

I thought I had this cracked until I tested my example:

$\boldsymbol{\mathrm{d}}f = \frac{\partial f}{\partial x}\boldsymbol{\mathrm{d}}x + \frac{\partial f}{\partial y}\boldsymbol{\mathrm{d}}y, \quad \star_{\boldsymbol{\mathrm{d}}x \wedge \boldsymbol{\mathrm{d}}y} \boldsymbol{\mathrm{d}}f = \frac{\partial f}{\partial x}\boldsymbol{\mathrm{d}}y - \frac{\partial f}{\partial y}\boldsymbol{\mathrm{d}}x, \quad \sharp_{\boldsymbol{\mathrm{d}}x \wedge \boldsymbol{\mathrm{d}}y} \star_{\boldsymbol{\mathrm{d}}x \wedge \boldsymbol{\mathrm{d}}y} \boldsymbol{\mathrm{d}}f = \frac{\partial f}{\partial x} \frac{\partial }{\partial x} + \frac{\partial f}{\partial y} \frac{\partial }{\partial y} = \nabla f \neq \nabla^\perp f$

Why is this happening? and what's the correct way to make $\nabla^\perp$ using $\omega$ only. I am certain this is all well known somewhere but most resources I've found on Kahler manifolds are not very interested in computing stuff. If anyone can provide references as well as resolving my confusion that would be much appreciated.

  • $\begingroup$ Just dualize $df$ using $\omega$. $\endgroup$ Commented Dec 22, 2023 at 16:53
  • $\begingroup$ @MoisheKohan I agree this makes sense for the symplectic gradient, but then if call $\nabla^\perp := \sharp_\omega \boldsymbol{\mathrm{d}}f$ and use $\sharp_g = \perp \sharp_\omega \perp$ (I believe you apply the almost complex structure both sides) I cannot get this to agree with the "metric" definition of $\nabla^\perp$ since if you calculate this out in coordinates the two perps kill eachother to a minus sign. Wouldn't this mean $\nabla^\perp = \nabla$, unless my "sharp converting identity" is wrong. $\endgroup$ Commented Dec 22, 2023 at 17:11
  • $\begingroup$ Perhaps I am wrong. Is it true that $M$ is Kahler, and $J$ is the almost complex structure, that $\sharp_\omega = J \circ \sharp_g$? This would immediately give me what I want. $\endgroup$ Commented Dec 22, 2023 at 17:15
  • $\begingroup$ Ok, I believe this to be true: $\sharp_\omega = J \circ \sharp_g \implies \flat_g \circ -J$, applying this to a vector gives $g(-J X, \cdot) = g(X, J \cdot) = \omega(X, \cdot) = \flat_\omega (X)$, inverting gives what is required. $\endgroup$ Commented Dec 22, 2023 at 17:31

2 Answers 2


The correct definition as described by MoisheKohan is to use $\sharp_\omega$ only and no Hodge star (from either $g, \omega$). This will give the skew gradient in coordinates, and in the case that a compatible metric structure exists, one may turn this into "perp of the gradient" by the identity $\sharp_\omega = J \circ \sharp_g$.

I believe the presence of Hodge star in the original notes is a quirk of $J$ coinciding with $-\star_g$ on 1-forms in $\mathbb{R}^2$. Since the author is only interested in doing calculus with a metric, it is easier to avoid mentioning $J$ and to use this coincidence to allow working only with one structure.


The Hamiltonian equations of motion are $\omega(v,\cdot)=dH$ as 1 forms where $v$ is the time evolution vector field with Hamiltonian function $H$. Using the symplectic form $dxdy$, $v$ is related to $H$ exactly as you want (I may be messing up signs). It is an implicit equation, but it is solvable because $\omega$ is assumed to be nondegenerate, so when it is written as a matrix, it has an inverse at every point.


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