nLab PL de Rham theorem

Idea

The PL de Rham theorem is the variant of the de Rham theorem with the smooth de Rham complex replaced by the PL de Rham complex.

Statement

Proposition

(PL de Rham theorem)

Let $k$ be a field of characteristic zero (such as the rational numbers, real numbers or complex numbers).

Then the evident operation of integration of differential forms over simplices induces a quasi-isomorphism between the PL de Rham complex with coefficients in $k$ and cochain complex for singular cohomology with coefficients in $k$

$$\Omega^\bullet_{PLdR}(X) \underoverset{}{\simeq}{\longrightarrow} C^\bullet(X; k)$$

and hence an isomorphism from PL de Rham cohomology to ordinary cohomology with coefficients in $k$ (such as rational cohomology, real cohomology, complex cohomology):

$$H^\bullet_{PLdR}(X) \underoverset{}{\simeq}{\longrightarrow} H^\bullet(X; k)$$

(for $X$ any topological space).

(Bousfield-Gugenheim 76, Theorem 2.2)

Lemmas

Besides the Poincaré lemma for piecewise polynomial forms, the proof uses:

Lemma

(extension lemma) Given a PL form on the boundary of a simplex, it extends to (hence is the restriction of) a PL form on the full n-simplex.

(e.g. Griffiths & Morgan 2013, Lemma 9.6)

Proof

Consider barycentric coordinates for the given n-simplex

$$\Delta^n \,\equiv\, \left\{ \, (x^0, \cdots, x^n) \,\in\, \mathbb{R}^{n+1} \;\Big\vert\; \textstyle{\sum}_k x^k = 1 \,, \forall_i \, x^i \geq 0 \, \right\}$$

such that the $i$th vertex is the point $v_i \in \Delta^n$ with coordinates

$$x^n(v_i) \;=\; \delta^n_i$$

and the $i$th-face is the subset

$$\sigma_i\big(\Delta^n\big) \;=\; \big\{ \, (x^0, \cdots, x^n) \,\in\, \Delta^n \;\big\vert\; x^i = 0 \, \big\} \,.$$

With these coordinate expressions consider the functions

(1)$$\begin{array}{rcl} \mathllap{ p_i \;\colon\; } \Delta^n \setminus \{v_i\} &\longrightarrow& \sigma_i(\Delta^{n}) \\ (x^k)_{k} &\mapsto& \left( \frac{x^k}{1-x^i} \right)_{k \neq i} \,. \end{array}$$

Observe that given a polynomial form $\alpha$ on $\sigma_i(\Delta^n)$, its pullback along $p_i$ is a form on $\Delta^n \setminus \{v_i\}$ which is polynomial in the variables $\{x^k\}_{k \neq i}$ and in the variable $1/(1-x^i)$. Therefore there is a power $N_i \in \mathbb{N}$ such that

(2)$$\widehat \alpha \;\coloneqq\; (1 - x^i)^{N_i} \cdot p_i^\ast(\alpha)$$

is a differential form on $\Delta^n \setminus \{v_i\}$ such that

1. $\widehat \alpha$ extends to $v_i$ (by zero) to give a polynomial differential form on all of $\Delta^n$;

2. pulled back to the $i$th face, $\widehat \alpha$ coincides with $\alpha$ (there being the pullback of an identity map);

3. if $\alpha$ vanishes on the $j$-face in $\sigma(\Delta^n)$ then $\widehat{\alpha}$ vanishes on the $j$-face of $\Delta^n$ (there being a pullback of the former).

Now to complete the proof, consider a polynomial differential form $\omega$ on the boundary $\partial \Delta^n$. We need to find an extension $\widehat \omega$ to all of $\Delta_n$.

First consider the above construction (2)

$$\widehat \omega_0 \;\coloneqq\; \widehat {\omega \vert_{\sigma_0} }$$

on the restriction of $\omega$ to $\sigma_0(\Delta^n)$ and notice that the difference

$$\omega_1 \;\coloneqq\; \omega - \widehat{\omega}_0 \vert_{\partial \Delta^n}$$

vanishes on $\sigma_0$ and coincides with $\omega$ on all other faces. Therefore consider next the above construction (2)

$$\widehat \omega_2 \;\coloneqq\; \widehat{ \omega_1 \vert_{\sigma_1} }$$

on this difference restricted to $\sigma_1$ and notice that the difference

$$\omega_2 \;\coloneqq\; \omega_1 - \widehat{\omega}_1 \vert_{\partial \Delta^n}$$

vanishes on the union of faces $\sigma_0 \cup \sigma_1$ and coincides with $\omega$ on all remaining faces. Proceeding in this fashion one arrives at

$$\omega \;=\; \underset{ \widehat{\omega} }{ \underbrace{ \textstyle{\sum}_k \widehat{\omega}_k } } \, \Big\vert_{\partial \Delta^n}$$

and hence the term over the brace is an extension as required.

Remark

(extension lemma for piecewise smooth differential forms) The Extension Lemma holds also for other flavors of differential forms over simplices:

In (the proof of) Griffiths & Morgan 2013, Cor. 9.9 this is observed for the case of smooth differential forms: Here the proof of Lemma applies verbatim, except that the multiplication by $(1-x^i)^{N_i}$ in (2) needs to be replaced by multiplication with any bump function which vanishes in a neighbourhood of $x^i = 1$ and is unity for $x^i = 0$.

Moreover, we may observe that this same argument then also applies to differential forms on “extended simplices” (see this Def.)

$$\Delta^n_{ext} \,\equiv\, \left\{ \, (x^0, \cdots, x^n) \,\in\, \mathbb{R}^{n+1} \;\Big\vert\; \textstyle{\sum}_k x^k = 1 \, \right\} \,,$$

where the condition $\forall_i \, x^i \geq 0$ is dropped (which plays no role in the above proof).

This is noteworthy, because it implies, with the discussion at shape via cohesive path ∞-groupoid and using the fundamental theorem of dg-algebraic rational homotopy theory, that that rational space encoded in a Sullivan model/Whitehead $L_\infty$-algebra $\mathfrak{a}$ is equivalently the shape

$$\esh \, \Omega^1_{dR}(-;\mathfrak{a})_{flat}$$

of the smooth set of flat $\mathfrak{a}$-valued differential forms.

Related statements

• PL de Rham complex of smooth manifold is equivalent to de Rham complex

• equivariant PL de Rham theorem

• de Rham theorem

References

• Aldridge Bousfield, Victor Gugenheim, On PL deRham theory and rational homotopy type, Memoirs of the AMS, vol. 179 (1976) (ams:memo-8-179)

Including the variant of piecewise smooth forms:

• Phillip Griffiths, John Morgan, Chapter 9 of: Rational Homotopy Theory and Differential Forms, Progress in Mathematics 16, Birkhauser (2013) [doi:10.1007/978-1-4614-8468-4]