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# Contents

## 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).

## 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.

###### 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

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

## References

Including the variant of piecewise smooth forms:

Last revised on February 3, 2024 at 18:13:37. See the history of this page for a list of all contributions to it.