The standard simplices $\Delta^n \subset \{\mathbf{x}\in\mathbb{R}^{n+1}\mid x\_0 + \ldots + x\_n =1 \} =: \mathbb{A}^n$ carry two natural sorts of smooth differential forms:


1. Those differential forms on the interior of $\Delta^n$ that extend smoothly to a neighbourhood of $\Delta^n$ in $\mathbb{A}^n$ (this definition is used for instance by Dupont in his book *Curvature and Characteristic Classes*). This is implicitly using Whitney's extension theorem, I think.
2. Consider $\Delta^n$ as a diffeological subspace of $\mathbb{A}^n$, where a function $\phi\colon \mathbb{R}^k \to \Delta^n$ is a plot if and only if the composite $\mathbb{R}^k \to \Delta^n \hookrightarrow \mathbb{A}^n$ is smooth. A differential form $\omega$ on $\Delta^n$ then consists of the data of a differential form $\phi^\*\omega$ for each plot $\phi$ with a compatibility condition when a plot factors through another via a smooth map between Euclidean spaces.


The first of these is more of a "maps out" viewpoint, and probably corresponds to a natural smooth space structure defined via smooth real-valued functions. The second is a "maps in" viewpoint. Note that the $D$-topology arising from the diffeology in 2. above is the standard topology on the simplex. We then get a cochain complex of differential forms of each type, as exterior differentiation can be defined in the more-or-less obvious way in each case.



> 
> My question is: how do these relate? Is one a subcomplex of the other? Or are they quasi-isomorphic, via a third cochain complex?
> 
> 
> 


The motivation is that Dupont's simplicial differential forms on semisimplicial manifolds $X\_\bullet$ look like they should be differential forms on the fat geometric realisation considered as a diffeological space, since a simplicial differential form is more or less descent data for the sheaf of differential forms and the 'cover' $\coprod\_{n\geq 0} \Delta^n\times X\_n \to ||X\_\bullet||$, assuming *the first definition* above. However, if his differential forms on $\Delta^n$ (or more precisely on $\Delta^n\times X\_n$) aren't diffeological differential forms, they don't give a form on $||X\_\bullet||$ as a diffeological space. I guess all we really need is a map of cochain complexes from the first to the second given above.



Differential forms on standard simplices via Whitney extension vs diffeological structure
The two chain complexes are isomorphic for any $n≥0$.


Fix some $n≥0$, $k≥0$ and consider $k$-forms on the $n$-simplex.
I will use the notations $Ω\_e^k$ and $Ω\_d^k$ for forms of type 1 and 2 respectively.
I also use the notation $Δ\_d^n$ for the diffeological $n$-simplex
so that $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} \def\d{\,{\rm d}} Ω\_d^k=\Hom(Δ\_d^n,Ω^k)$.


First, we have a canonical map $ι\colon Ω\_e^k→Ω\_d^k$ that
sends $ω∈Ω\_e^n$ to the map $Δ\_d^n→Ω^k$
that sends a smooth map $φ\colon S→Δ\_d^n$ to the $k$-form $φ^\*ω∈Ω^k(S)$.


Secondly, the map $ι$ is injective,
which follows from the following two observations.
By the Yoneda lemma the restriction of $ι(ω)\colon Δ\_d^n→Ω^k$ to the sheaf represented by the open interior of $Δ\_d^n$ equals the restriction of $ω$ to the open interior of $Δ^n$.
Furthermore, any two forms in $Ω\_e^k$ whose restrictions to the open interior of $Δ^n$ coincide must be equal (by continuity).


Thirdly, the map $ι$ is surjective.
This can be shown using a two-step construction:
in the first step we construct a certain element $ω$ of $Ω\_e^k$
starting from some given element $ψ$ of $Ω\_d^k$
and in the second step we show that $ι(ω)=ψ$.


First, let's briefly examine the simplest nontrivial case $d=n=1$.
We have a form $ψ∈Ω\_d^1(Δ\_d^1)=Ω\_d^1([0,1])$,
and we already know its restriction $f(x) \d x$ to the open interval $(0,1)$.
In particular, $f$ is a smooth function on $(0,1)$.
Pulling back $ψ$ along the plot $\R→\R$ ($x↦x^2$)
yields some 1-form $g(x) \d x$, where for any $x≠0$ we have $g(x)=2xf(x^2)$
and $g$ is a smooth function on $(-1,1)$.
Since $g$ is odd, we have $g(0)=0$,
so the even function $h$ given by $h(x)=g(x)/(2x)$ is smooth on $(-1,1)$
and we have $h(x)=f(x^2)$ for all $x≠0$.
Set $f(0)=h(0)$.
We claim $f$ is smooth on $(-1,1)$.
Indeed, $h'(x)=2xf'(x^2)$ for $x≠0$,
and since $h'$ is odd, we have $h'(0)=0$
and $h'(x)/(2x)$ is a smooth function on $(-1,1)$
whose value at $x=0$ is $f'(0)$.
We repeat this step by induction, proving $f$ has derivatives of all orders at 0.


The general case is nothing else than a multivariable paremetrized
version of the above argument.
More precisely, we argue as follows.
For the first step, suppose we are given some $ψ∈Ω\_d^k$.
Given some point $x∈Δ^n$, we would like to define $ω(x)$.
Denote by $d≥0$ the codimension of the stratum of $Δ^n$ that contains $x$.
Parametrize the simplex $Δ^n$ using $n$ coordinates
in such a way that $x\_1=⋯=x\_d=0$
and the other $n-d$ coordinates of $x$ are strictly positive,
so that the points $y$ of the stratum containing $x$ are defined by the relations $y\_i≥0$ for $1≤i≤d$.


Decompose $$ω=∑\_I g\_I ∏\_{i∈I} \d x\_i$$ into its individual components with respect to this coordinate system.
Pick one such component $g\_I ∏\_{i∈I} \d x\_i$; we would like to define $g\_I$ as a smooth function on some open neighborhood $U$ of the given stratum of $Δ^n$, which is parametrized by the remaining $n-d$ coordinates $x\_{d+1}$, …, $x\_n$.
Consider the smooth map $U→Δ^n$ that (in the coordinates introduced above)
sends $x\_i↦x\_i^2$ for $1≤i≤d$ and $x\_i↦x\_i$ for $i>d$.
This defines a morphism $τ\colon U→Δ\_d^n$, so we have a form $τ^\*ψ∈Ω^k(U)$.
Now take the coefficient $h$ of $τ^\*ψ$ before $∏\_{i∈I} dx\_i$.
Take the partial derivative of $h$ with respect to all coordinates $x\_i$
such that $i∈I$ and $i≤d$.
Divide the resulting function by $2^{\#(I∩\{1,…,d\})}$.
This is the function $g\_I$.
This argument also proves that the resulting $k$-form $ω$ is smooth.


For the second step, we have to show that $ι(ω)=ψ$.
Reusing the notation of the previous paragraph,
consider some arbitrary plot $φ\colon S→Δ\_d^n$ such that $φ(s)=x$ for some $s∈S$.
The first $d$ coordinates of $φ$ must be nonnegative in a neighborhood $U$ of $s$.
Since $φ$ is differentiable, the first derivatives of these $d$ coordinates must vanish at the point $s$.
Thus, taking the square root of each of the first $d$ coordinates produces a smooth map $ε\colon U→\R^n$ such that $φ=τε$, where the map $τ$ was defined in the previous paragraph.
Now $φ^\*ι(ω) = ε^\* τ^\* ι(ω) = ε^\* τ^\*ψ = φ^\*ψ$,
where $τ^\*ι(ω)=τ^\*ψ$ by definition of $ω$.


