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The de-Rham complex in one dimension describes phenomena that can be described in terms of ordinary differential equations. The de-Rham complex in three dimensions can be used to describe classical results in vector analysis.

Whereas the cochain morphism from the de Rham complex to complexes build out of scalar and vector fields in dimension $1$ and $3$ is very canonical, I don't know how to "interpret" the de-Rham complex in two dimensions.

For a demonstration, let the domain be $\mathbb R^n$ and let $\mathcal S$ be the set of smooth scalar fields, $\mathcal V$ the set of smooth vector fields.

The (smooth) de Rham complex in two dimensions reads

$\lambda^0 \overset{d_0}{\longrightarrow} \lambda^1 \overset{d_1}{\longrightarrow} \lambda^2$

as usual. We want to translate this into terms of classical vector analysis. The spaces $\lambda^0$ and $\lambda^2$ are canonically isomorphic to scalar fields. How do we translate $\lambda^1$ into vector fields?

If we translate it into vector fields by the harmonic isomorphism, and set the second differential from $\mathcal V$ to $\mathcal S$ to be the divergence $\operatorname{div} = \partial_x + \partial_y$, then we must set the first isomorphism to $(\partial_y,-\partial_x)$ or $(-\partial_y,\partial_x)$.

If instead we set the first differential to be the gradient $\operatorname{grad} = (\partial_x,\partial_y)$ from $\mathcal V$ to $\mathcal S$, then the second differential can either be $-\partial_y+\partial_x$ or $\partial_y-\partial_x$.

so we have complexes

$\mathcal S \overset{(-\partial_y,\partial_x)}{\longrightarrow} \mathcal V \overset{\partial_x + \partial_y}{\longrightarrow} \mathcal S$

and

$\mathcal S \overset{(\partial_x,\partial_y)}{\longrightarrow} \mathcal V \overset{-\partial_y+\partial_x}{\longrightarrow} \mathcal S$

In either case, the concatenation $d_1 \circ d_0$ reads

$\operatorname{div} \begin{bmatrix} 0 & 1 \cr -1 & 0 \end{bmatrix} \operatorname{grad}$.

Therefore I ask for help to understand the following issues

  1. Which choice of chain morphism from the de Rham complex into a vector analytic setting is the "right" one? Note that if you dualize one of these vector-analytic complexes, the codifferentials turn out to be the morphisms of the other complex, with signs and arrow directions switched.
  2. The matrix $\begin{bmatrix} 0 & 1 \cr -1 & 0 \end{bmatrix}$ is a prototypical example of a sympletic matrix. Does there exist an relation in terms of sympletic geometry?

Thank you very much.

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I'm not sure I understand the question. I guess, you can view $$d(fdx+gdy) = (\partial g/\partial x-\partial f/\partial y)dx\wedge dy$$ as the $z$-component of the curl. –  Donu Arapura Jul 12 '11 at 16:53
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After reading your question a bit more carefully, I realize you are unsure of the signs. To make identifications, you need to choose a metric on $\mathbb{R}^2$ and an orientation. Pick the Euclidean metric oriented so that $dx\wedge dy$ is the volume form. Then this will fix the choice of isomorphisms between $1$-forms and vector fields, and $2$-forms and scalars. –  Donu Arapura Jul 12 '11 at 17:08

3 Answers 3

up vote 3 down vote accepted

After writing this, I noticed that Donu Arapura made essentially the same point in a comment. But perhaps my added detail will be of use.

There are two missing ingredients in what you wrote.

First, when you identify 1-forms with vector fields, you have implicitly chosen a bundle isomorphism $T\mathbb{R}^n \cong T^*\mathbb{R}^n$. Such a choice is the same thing as a choice of a Riemannian metric.

Second, when you identify n-forms on $\mathbb{R}^n$ with functions you are choosing an isomorphism between the exterior bundle $\bigwedge^n(\mathbb{R}^n)$ and a line bundle over $\mathbb{R}^n$. This is the same thing as a choice of orientation.

It is generally fine to leave the choice of standard Euclidean metric and standard orientation on $\mathbb{R}^n$ as implicit, but in this case these two choices are enough to determine a specific way to identify the rest of the De Rham complex. There are undoubtedly other ways to do it (particularly in low dimensions), but the way I will explain generalizes to any dimension (indeed, any oriented Riemannian manifold), it gives the right answers on $\mathbb{R}^3$, and it is compatible with Stokes' theorem.

Here's how it works. The Riemannian metric and the orientation determine a volume form $vol_n$ on $\mathbb{R}^n$; if $x_1, \ldots, x_n$ is a global oriented orthonormal coordinate system then we can take $vol_n = dx_1 \wedge \ldots \wedge dx_n$. There is an isomorphism $\star: \bigwedge^k(\mathbb{R}^n) \to \bigwedge^{n-k}(\mathbb{R}^n)$ (called the Hodge star operator) which is determined by the equation $\alpha \wedge \star(\alpha) = vol_n$. We will use this isomorphism to reduce high degree forms to low degree forms.

On $\mathbb{R}^2$ it's all very simple. Using the standard (oriented) coordinates $(x,y)$, the volume form is $dx \wedge dy$. The Euclidean metric identifies a vector field $(f, g)$ with the 1-form $f dx + g dy$. We have $d(f dx + g dy) = (f_x dx + f_y dy) \wedge dx + (g_x dx + g_y dy) \wedge dy = (g_x - f_y) dx \wedge dy$. The Hodge star operator identifies this 2-form with the 0-form $g_x - f_y$. Note that this is indeed the correct expression from the point of view of Stokes' theorem on the plane (AKA Green's theorem).

ADDED One more comment. To give a geometric interpretation of your other choice of map from vector fields to functions, note that a choice of Riemannian metric on $\mathbb{R}^n$ yields a metric on the exterior algebra bundle of $\mathbb{R}^n$ and hence an inner product on the space of forms. If $V$ is a vector field and $\alpha$ is the corresponding 1-form, then the divergence of $V$ is nothing more than the function $d^\ast \alpha$ where $d^\ast$ is the adjoint of $d$ relative to this inner product. Up to a sign which I do not remember this adjoint is given by $d^\ast = \star d \star$, and this explains why the divergence on $\mathbb{R}^3$ can be regarded either as $d$ acting on 2-forms or as $d^\ast$ acting on 1-forms.

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I can't think of a good answer to your second question. Given that the cotangent bundle of a manifold carries a canonical symplectic structure, it would not surprise me if there is something useful to be said. –  Paul Siegel Jul 13 '11 at 13:50

In $\mathbb R^3$ if you interpret $1$-forms as vector fields in a certain obvious way then the de Rham operator on $0$-forms becomes grad. If you now interpret $2$-forms as vector fields in a certain way then the de Rham operator on $1$-forms becomes curl and the de Rham operator on $2$-forms becomes div.

Generally, in $\mathbb R^n$, if you interpret $1$-forms as vector fields in a certain obvious way then the de Rham operator on $0$-forms becomes grad and if you interpret $(n-1)$-forms as vector fields in a certain obvious way then the de Rham operator on $(n-1)$-forms becomes div.

In particular in $\mathbb R^2$ if you interpret $1$-forms as vector fields in one way then the de Rham operator on $0$-forms becomes grad, while if you interpret $1$-forms as vector fields in a certain other way then the de Rham operator on $1$-forms becomes div. But these two ways of making a $1$-form in the plane correspond to a vector field differ by a $90$ degree rotation.

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Just a comment on Siegel comment

Well, considering $dx\wedge dy$ as a symplectic, rather than a volume form, one has the classical musical isomorphism $\sharp:\Omega^p\mathbb M\to {\cal X}^pM$ from $p$-forms to $p$-multivector fields, which intertwines (in fact it is an isomorphism of complexes) de Rham cohomology on forms with Poisson cohomology on multivector fields.

Here one needs to fix some signs, since one can consder the Poisson cohomology operator either as $[\Pi,-]$ or $[-,\Pi]$, where $\Pi$ is the Poisson bivector. In $\mathbb R^2$ everything gets very easy and in the usual basis the $\sharp$ isomorphism is exactly given by the elementary symplectic matrix: $dx\mapsto -\partial_y$, $dy\mapsto \partial_x$. In this case the Poisson cohomology complex is the first one listed in the question.

If one then combines this with Poincarè duality on one of the two sides then it is possible to get the second complex. This is nothing different from the two previous comments but allows for generalization to a 2-dim symplectic manifold rather than to a Riemannian. Im my opinion this is an explanation of the appearance of the symplectic matrix in the concatenation (which may be seen to be zero as a consequence of the Jacobi identity for the standard Poisson bracket, in the "Poisson cohomology" interpretation).

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