Stokes theorem for the standard simplex

First let $k\ge 2$. Write $\eta ={\sum }_{i=1}^k(-1{)}^{i-1}{f}_{i}d{x}_1\wedge \cdots \wedge \widehat{d{x}_i}\wedge \cdots \wedge d{x}_k\mathrm{.}$ All $f_i$ are smooth near $T_k$. In [F3], every derivative except ${\mathrm{\partial }}_i{f}_i$ wedges a repeated differential and vanishes; moving $d{x}_i$ past $i-1$ factors cancels the coefficient sign. Hence $d\eta =\left({\sum }_{i=1}^k{\mathrm{\partial }}_i{f}_i\right)d{x}_1\wedge \cdots \wedge d{x}_k\mathrm{.}$

Fix $i$. Let $z$ denote the increasing list of all coordinates except $x_i$, let $D_i=\{z\ge 0:\sum z\le 1\}$ and put $b(z)=1-\sum z$. The $x_i$ section of $T_k$ is $[0,b(z)]$ for $z\in D_i$, and it is empty off $D_i$. Coordinate permutation has absolute determinant one by [F6]. The full integrand ${\mathrm{\partial }}_i{f}_i$ is integrable by [F7], and every nonempty section is continuous on its closed interval. Applying [F5] after that permutation and [F4] on sections with $b(z)>0$ gives ${\int }_{T_k}{\mathrm{\partial }}_i{f}_{i}dx={\int }_{D_i}(f_i(z,x_i=b(z))-f_i(z,x_i=0))dz\mathrm{.}$ When $b(z)=0$, both the section integral and the endpoint difference are zero, so the formula holds on these sections too. No exceptional family is discarded.

Parametrize face zero by $y=(x_2,\dots ,x_k)\in T_{k-1}$, with $x_1=1-{\sum }_{j=2}^k{x}_j$. For $i=1$ the omitted differential wedge pulls back to $dy$. For $i>1$, substitute $d{x}_1=-{\sum }_{j=2}^{k}d{x}_j$; only its $-d{x}_i$ term survives, and moving $d{x}_i$ to its increasing position contributes $(-1{)}^{i-2}$. Thus the omitted wedge pulls back to $(-1{)}^{i-1}dy$. The coefficient sign in step 1.1 cancels it, giving ${\delta }_0^{\ast }\eta =\left({\sum }_{i=1}^k{f}_i{\mid }_{\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}0}\right)dy\mathrm{.}$

On face $i\ge 1$, $x_i=0$ and the remaining vertex ordering gives precisely the increasing remaining coordinate basis. All summands of $\eta$ except the one indexed by $i$ pull back to zero, because they contain $d{x}_i$. The signed contribution of this face is therefore $(-1{)}^i\int {\delta }_i^{\ast }\eta =-{\int }_{D_i}{f}_i(z,x_i=0)dz\mathrm{.}$ This matches the lower endpoint term in step 2.1.

For each $i>1$, the map from $y$ to the increasing coordinates $z$ omitting $x_i$ replaces the missing coordinate by $x_1=1-\sum y_j$. Its derivative has determinant $(-1{)}^{i-1}$: expand along the identity rows, leaving the entry $-1$ in the position corresponding to $x_i$. It maps $T_{k-1}$ bijectively onto $D_i$, with inverse obtained by solving $x_i=1-\sum z$. Thus [F6] transforms the upper endpoint integral in step 2.1 into the integral of $f_i$ over the face-zero parametrization, with absolute determinant one. For $i=1$ the map is the identity. Summing over $i$, step 2.2 identifies all upper endpoint terms with $\int {\delta }_0^{\ast }\eta$.

By [F8], sum the identities of step 2.1 and use step 1.1 for the left side, step 3.1 for the lower endpoints and step 3.2 for the upper endpoints. This gives the displayed Stokes identity for $k\ge 2$, with sign $+1$ on face zero and $(-1{)}^i$ on face $i$. All sums are finite.

For $k=1$, $\eta =f$ is a function, and [F3] and [F4] give ${\int }_0^{1}df=f(1)-f(0)$. The face-zero map is the terminal vertex and the face-one map the initial vertex, whose integrals are evaluations by [F1]. This is the same formula. The assertion excludes $k=0$ and does not introduce forms of degree minus one. Zero forms give zero on both sides; all collapsed sections were treated in step 2.1, including intersections of faces. All coordinates, changes and sums are explicit and finite, so the proof uses no choice.