Decomposition of 1-forms in simply connected open subset of Euclidean space

Question

Let $U$ be a simply connected bounded open subset of $\mathbb{R}^n$.

Let $\Omega^k$ be the space of $k$-forms on $U$, and let $d_k: \Omega^k \to \Omega^{k+1}$ be the exterior derivative. Endow $U$ with a Riemannian metric and let $\delta_k: \Omega^k \to \Omega^{k-1}$ be the corresponding co-differential operator.

A 1-form $\omega \in \Omega^1$ is

• closed if $d_1{\omega} = 0 \in \Omega^2$, i.e. $\omega \in \ker{d_1}$
• exact if $\omega = d_0f$ for some $f \in \Omega^0$, i.e. $\omega \in \text{im}~{d_0}$
• co-closed if $\delta_1{\omega} = 0 \in \Omega^0$, i.e. $\omega \in \ker{\delta_1}$
• co-exact if $\omega = \delta_2{\beta}$ for some $\beta \in \Omega^2$, i.e. $\omega \in \text{im}~{\delta_2}$

Since $U$ is simply connected, by Poincaré lemma any 1-form is

• closed iff exact, and
• co-closed iff co-exact

Question: is it true that $\Omega^1$ admits the orthogonal direct sum decomposition $$\Omega^1 = \text{im}~{d_0} \oplus \text{im}~{\delta_2}$$

I think the answer is "no". If $\Omega^1$ was a finite dimensional vector space, then

$$ \begin{split} \Omega^{1} & = \ker{d_{1}} \oplus \left( \ker{d_{1}} \right)^{\perp} \\ & = \ker{d_{1}} \oplus \text{im}~{\delta_{2}} \\ & = \text{im}~{d_{0}} \oplus \text{im}~{\delta_{2}} \\ \end{split} $$ where

• the first equality is true by definition of direct sum and orthogonal complement, since $\ker{d_1}$ is a linear subspace;
• the second equality is true since $\delta_2$ is the adjoint of $d_1$;
• the last equality holds by Poincarè lemma.

But, $\Omega^1$ is a module over the algebra of smooth functions $C^{\infty}(U)$, so the above does not make sense - right?

Edit

Here is the proof that Hodge theorem holds true for a special Riemannian metric on the relative interior of the standard simplex.

$\newcommand{\rint}[1]{\mathring{#1}}$ $\newcommand{\X}{\mathcal{X}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\d}{d}$ $\newcommand{\pull}[1]{#1^{*}}$ $\newcommand{\topspace}{\R^{n+1}_{>0}}$

Prop. There exists a Riemannian metric such that Hodge theorem holds true for the space of 1-forms on the relative interior of the standard simplex that are well-behaved on the boundary.

Proof.

Consider the relative interior of the standard $n$-dimensional simplex seen as an immersed submanifold of the positive orthant $\topspace$:

$$\rint{\X} = \{x \in \topspace: \sum_{i = 0}^n x_i = 1\} \hookrightarrow \topspace$$

Consider the standard Euclidean coordinates and endow $\topspace$ with the following Riemannian metric (known in the literature as Shahshahani metric [1, 2]): $$g(x) = \sum_{i, j = 0}^n \frac{\delta_{ij}}{x_i} \, dx^i \otimes dx^j$$ The matrix representation of this metric is $g(x) = \text{diag}(x_0, \dots, x_n)$.

Considering the following diffeomorphism of $\topspace$ onto itself: $$F: \topspace \to \topspace$$ $$F(x) = 2\sqrt{x}$$ $$F^{-1}(y) = \frac{y^2}{4}$$ This is known as Akin transformation [3].

The image of $\rint{\X}$ under $F$ is the portion of the sphere of radius $2$ lying in the positive orthant: $$\rint{C} := F(\rint{\X}) = \{y \in \topspace: \sum_{i = 0}^n y_i^2 = 4 \}$$ Furthermore it's easy to check that the pull-back of the metric $g$ along $F^{-1}$ is the Euclidean metric on $\topspace$. Hence, $F$ restricted to $\rint{\X}$ is an isometry $F|_{\rint{\X}}: \rint{\X} \to \rint{C}$ between $\rint{\X}$ that inherits the Shahshahani metric from the ambient space, and $\rint{C}$ that inherits the Euclidean metric from the ambient space.

Let now $V$ be a 1-form defined on a tubolar neighborhood $U$ of $\rint{\X}$ (in particular, well-behaved on the boundary of $\rint{\X}$), and let $v$ be the restriction of $V$ to $\rint{X}$. In Euclidean coordinates $V$ can be expressed as $$V(x) = \sum_{i = 0}^n V_i(x) \, dx^i$$ where $V_i$ are smooth functions on $U$. Let $\omega = (F^{-1})^*{V}$ be the pull-back of $v$ on $F(U)$. By an explicit computation $$\omega(y) = \sum_{i = 0}^n y_i \, (V_i \circ F^{-1})(y) \, dy^i$$ Since both $F$ and the $V_i$-s are well-behaved on the boundary of $\rint{\X}$, the 1-form $\omega$ is well-behaved on the closure $C$ of $\rint{C}$.

Then, by the Extension Lemma for Smooth Functions [4, Lemma 2.26], there exists a 1-form $\tilde{\omega}$ defined on the whole sphere $S$ of radius $2$ that agrees with $\omega|_C$ on $C$ and that is zero outside an arbitrarily small neighborhood of $C$.

The sphere is a closed (compact, no boundary) oriented manifold that inherits the Euclidean metric from the ambient space; thus Hodge theorem applies to the $1$-form $\tilde{\omega}$:

$$\tilde{\omega} = d{f} + \delta{\beta} + \gamma$$ with $f$ smooth function on $S$, $\beta$ 2-form on $S$, and $\gamma$ harmonic 1-form on $S$. Notice that if $n \geq 2$ then $S$ is simply connected and the harmonic component vanishes identically.

The restriction to $\rint{C}$ gives

$$ \begin{split} \tilde{\omega}|_{\rint{C}} = \omega|_{\rint{C}} & = \left( d\psi + \delta\beta + \gamma \right)|_{\rint{C}} \\ & = d{\left( \psi|_{\rint{C}} \right)} + \delta{\left( \beta|_{\rint{C}} \right)} + \gamma|_{\rint{C}} \end{split} $$ Pulling $\omega|_{\rint{C}}$ back to $\rint{\X}$ gives

$$ v = \pull{F}{\omega|_{\rint{C}}} = \pull{F} \Big( \d{\left( \psi|_{\rint{C}} \right)} + \delta{\left( \beta|_{\rint{C}} \right)} + \gamma|_{\rint{C}} \Big) $$

The exterior derivative always commutes with the pull-back, and the codifferential commutes with the pull-back via isometries, so

$$ v = \d\Big( \pull{F}(\psi|_{\rint{C}}) \Big) + \delta\Big( \pull{F}(\beta|_{\rint{C}}) \Big) + \pull{F} \left( \gamma|_{\rint{C}} \right) $$ where

• $\pull{F}(\psi|_{\rint{C}})$ is a smooth function on $\rint{\X}$
• $\pull{F}(\beta|_{\rint{C}})$ is a 2-form on $\rint{\X}$
• $\pull{F} \left( \gamma|_{\rint{C}} \right)$ is a harmonic 1-form on $\rint{\X}$

q.e.d.

---

1. Shahshahani, S. A new mathematical framework for the study of linkage and selection. (American Mathematical Soc., 1979).
2. Mertikopoulos, P. & Sandholm, W. H. Riemannian game dynamics. Journal of Economic Theory 177, 315–364 (2018).
3. Sandholm, W. H. Population games and evolutionary dynamics. (MIT press, 2010).
4. Lee, J. M. Introduction to Smooth Manifolds. (Springer-Verlag New York, 2012).

Answer 1

The decomposition you're looking for is referred to as the Hodge decomposition, and for forms over a compact oriented Riemannian manifold one has $\Omega^k(M)=\mathrm{im} d_{k-1}\oplus \mathrm{im} \delta_{k+1}\oplus \mathcal{H}^k$ where $\mathcal{H}^k$ is the space of harmonic $k$-forms.

In the non-compact case things are somewhat more subtle, since even though $d_k$ and $\delta_{k+1}$ are formally adjoint, they need not truely be adjoint in a more meaningful (functional analytic) sense.

Consider $U=(-1,1)\times (-1,1)\subset \mathbb{R}^2$. There is a diffeomorphism $\phi:U\to \mathbb{R}^2$ by $(x,y)\mapsto (\tan(\pi x/2, \tan(\pi y/2))$. Pulling back the standard Riemannian metric on $\mathbb{R}^2$ makes $\phi:U\to \mathbb{R}^2$ an orientation preserving isometry. The codifferential on $\Omega^2(\mathbb{R}^2)$ is $\delta=\star^{-1}d\star$, so if $\omega=fdx\wedge dy$ then $\delta\omega=\star^{-1}(\partial_x fdx+\partial_y f d_y)=\partial_xf dy-\partial_yfdx$. We have $d\delta\omega=(\partial_x^2f+\partial_y^2f)dx\wedge dy$, so if $f$ is chosen to be non-constant harmonic (i.e. the real part of an entire holomorphic function), $d\delta\omega=0$ and by the Poincare lemma, $\delta\omega=dg$ for some nonconstant $g\in C^\infty(\mathbb{R}^2)$. This of course means that $\mathrm{im\delta}\cap \mathrm{im d}\neq 0$ on $\mathbb{R}^2$. Since $U$ is isometric to $\mathbb{R}^2$, one sees that $\delta\phi^*\omega=d\phi^*g$, so the statement need not be true for sets which are bounded in $\mathbb{R}^n$.

As I have suggested, things are somewhat nicer in the compact case. Given $M$ a compact oriented Riemannian manifold without boundary, for $\omega\in \Omega^k(M)$ and $\alpha\in \Omega^{k+2}$ we have \begin{align}(d\omega,\delta\alpha)&=\int_Md\omega\wedge d\star\alpha\\ &=\int_M d(\omega\wedge d\star\alpha)-(-1)^{n-k-1}\omega\wedge dd\star \alpha\\ &=\int_M d(\omega\wedge d\star\alpha)\\ &=0\end{align} The final equality coming from Stokes' theorem. This means that $\mathrm{im}d_{k-1}$ is perpendicular to $\mathrm{im}\delta_{k+1}$ with respect to the inner product structure on $\Omega^k(M)$ induced by the metric. If the manifold has $H^1(M,\mathbb{R})=0$, the Hodge decomposition and DeRham's theorem tell us that $\Omega^1(M)=\mathrm{im} d_{k-1}\oplus \mathrm{im}\delta_{k+1}$.

EDIT: One has $\star^2\sigma=(-1)^{n(n-k)}\sigma$ so $\star^{-1}=\pm\star$, when restricted to $\Omega^k$ for fixed $k$, consistent with the convention.