流形上的分析 Calculus on Manifold 笔记

1. Coordinate Charts and Local Diffeomorphisms.
   By definition of a ${\mathrm{𝒞︀}}^1$ $𝑘$-dimensional manifold, there exists an open set $𝑈\subset {ℝ}^{𝑘}$ and a ${\mathrm{𝒞︀}}^1$ coordinate chart $𝜑:𝑈\to 𝑀$. This map $𝜑$ is a homeomorphism onto its image, and its inverse ${𝜑}^{-1}$ is defined on $𝜑(𝑈)\subset 𝑀$.

2. Differential is an Isomorphism.
   For each point $𝑝\in 𝑈$, the differential (a.k.a. Jacobian matrix) of $𝜑$ at $𝑝$, denoted

   $$\mathrm{d}𝜑(𝑝):{ℝ}^{𝑘}⟶{\mathrm{T}}_{𝜑(𝑝)}𝑀,$$

   is a linear map from ${ℝ}^{𝑘}$ into the tangent space of $𝑀$ at $𝜑(𝑝)$. Since $𝜑$ is a ${\mathrm{𝒞︀}}^1$ coordinate chart, its differential is a linear isomorphism at each point $𝑝$. In other words, $\mathrm{d}𝜑(𝑝)$ is invertible.

3. Dimension Count.
   Because $\mathrm{d}𝜑(𝑝)$ is an isomorphism, it maps a $𝑘$-dimensional space ${ℝ}^{𝑘}$ onto the tangent space ${\mathrm{T}}_{𝜑(𝑝)}𝑀$. Hence
   

   $$dim{\mathrm{T}}_{𝜑(𝑝)}𝑀=dim{ℝ}^{𝑘}=𝑘$$

   Renaming $𝜑(𝑝)$ as $𝑥\in 𝑀$ completes the argument.

Since the point $𝑥$ was arbitrary, this shows that every tangent space ${\mathrm{T}}_{𝑥}𝑀$ has dimension $𝑘$, which matches the manifold dimension.

Q.E.D.

Suppose $𝑐=\sum {𝑎}_{𝑖}{𝑐}_{𝑖}$ where ${𝑐}_{𝑖}$ is singular $𝑘$ cube. Then we have

and

Thus, it is suffice to show

for standrad $𝑘$ cube ${𝐼}^{𝑘}$. Likewise, we may simplify the $𝜔$ into following type $𝑘-1$-form

Notice that

Also, we may conclude that

Compare these eqs, it is clearly that

Q.E.D.

Following the definition, we know that

As we know ${𝑐}^{♯}𝜔=𝑓\mathrm{d}{𝑥}_1\cdots \mathrm{d}{𝑥}_{𝑘}$, we have

Because $det{𝐽}_{𝑝}\ge 0$, we may apply change of variable formula here and conclude that

Q.E.D.

Existence

Since ${𝜑}_{♯(𝑎)}:{\mathrm{T}}_{𝑎}{ℝ}^{𝑘}\to {\mathrm{T}}_{𝜑(𝑎)}(𝑀)$ is a linear isomorphism and $𝐹(𝜑(𝑎))\in {\mathrm{T}}_{𝜑(𝑎)}𝑀$, there exists a unique vector $𝜉\in {\mathrm{T}}_{𝑎}{ℝ}^{𝑘}$ such that

We define $𝐺(𝑎):=𝜉$. Because ${𝜑}_{♯(𝑎)}$ is an isomorphism, this choice of $𝜉$ is unique. Thus, we can define a vector field $𝐺:𝑊\to 𝑇({ℝ}^{𝑘})$ by repeating this construction at each point $𝑎\in 𝑊$.

Uniqueness

Suppose there exists another vector field $\widetilde{𝐺}$ on $𝑊$ such that

Then define $𝐻(𝑎):=𝐺(𝑎)-\widetilde{𝐺}(𝑎)$. We have

Since ${𝜑}_{♯(𝑎)}$ is injective, it follows that $𝐻(𝑎)=0$ for all $𝑎$, and thus $𝐺(𝑎)=\widetilde{𝐺}(𝑎)$. Therefore, $𝐺$ is unique.

Independence of Coordinate Charts

Let $𝜓:𝑉\to 𝑀$ be another coordinate chart such that for some smooth map $𝛾:𝑊\to 𝑉$, we have

Let $𝐻$ be the vector field corresponding to $𝐹$ in coordinates via $𝜓$, i.e.,

Then, we compute:

Since both ${𝜑}_{♯(𝑎)}^{-1}$ and ${𝜓}_{♯(𝛾(𝑎))}$ are smooth such that they preserve differentiability, it follows that $𝐺\in {\mathrm{𝒞︀}}^{𝑟}$ if and only if $𝐻\in {\mathrm{𝒞︀}}^{𝑟}$.

Q.E.D.

Consider two coordinate charts

On the overlap of their images, we can write:

which implies that for a differential form $𝜔$ defined on $𝑀$, the pullbacks satisfy:

Since pullbacks along smooth maps preserve differentiability, and $𝜓\circ {𝜑}^{-1}$ is a smooth transition map between coordinate charts, it follows that ${𝜓}^{♯}𝜔$ and ${𝜑}^{♯}𝜔$ have the same differentiability class.

Q.E.D.

Let $𝜔$ be a $𝑝$-form on $𝑀$. We wish to define a $(𝑝+1)$-form $\mathrm{d}𝜔$ on $𝑀$ by using local coordinate charts.

Local Definition via a Coordinate Chart

Consider a coordinate chart $𝜑:𝑊\to 𝑀$ such that $𝑎\in 𝜑(𝑊)$. We define

where $𝐯=({𝐯}_1,\dots ,{𝐯}_{𝑝+1})$ is an ordered $(𝑝+1)$-tuple of tangent vectors at $𝑎$, and

Here, ${𝜑}^{♯}𝜔$ denotes the pullback of $𝜔$ to $𝑊\subset {ℝ}^{𝑘}$, and $\mathrm{d}\left({𝜑}^{♯}𝜔\right)$ is the exterior derivative of ${𝜑}^{♯}𝜔$ in the standard Euclidean space ${ℝ}^{𝑘}$.

Clearly, if this procedure is well-defined (i.e., independent of the chosen chart) and yields a unique $(𝑝+1)$-form on $𝑀$, then we have defined $\mathrm{d}𝜔$ correctly in global coordinates.

Independence of the Coordinate Chart

Let ${\mathrm{d}}^{𝜑}𝜔$ denote the $(𝑝+1)$-form arising from the above construction using the chart $𝜑$. Suppose we choose another chart $𝜓:𝑉\to 𝑀$ with overlapping image, and let ${\mathrm{d}}^{𝜓}𝜔$ be the corresponding definition on $𝜓(𝑉)$.

Define the transition map

which is smooth on the overlap of $𝜑(𝑊)$ and $𝜓(𝑉)$. Then for $𝑥$ in the overlap, we have

Thus, ${\mathrm{d}}^{𝜑}𝜔={\mathrm{d}}^{𝜓}𝜔$ on the overlap region. Since every point of $𝑀$ is contained in at least one coordinate patch, and any two such patches overlap in a region where the definitions agree, it follows that ${\mathrm{d}}^{𝜑}𝜔$ is independent of the chart $𝜑$. This justifies the notation $\mathrm{d}𝜔$.

Uniqueness of the $(𝑝+1)$-Form

Suppose there is another $(𝑝+1)$-form $𝜂$ on $𝑀$ such that for every coordinate chart $𝜑$,

Then, for each $𝜑$,

If the pullback of $𝜂-\mathrm{d}𝜔$ is zero in every coordinate system, it means $𝜂-\mathrm{d}𝜔$ vanishes on each coordinate patch. Hence, by the sheaf property (or the formary fact that if a form vanishes in each coordinate system, it must vanish globally),

Consequently, $𝜂=\mathrm{d}𝜔$. This shows that our construction yields a unique $(𝑝+1)$-form on $𝑀$.

Q.E.D.

Orientation in ${ℝ}^{𝑘}$

Recall that an orientation ${𝜇}_{𝜑(𝑎)}$ at a point $𝜑(𝑎)\in {ℝ}^{𝑘}$ can be identified with an ordered orthonormal basis $[{𝑣}_1,\dots ,{𝑣}_{𝑘}]$. In particular, there exists a unique volume form $𝜔$ up to scalar such that:

1. $𝜔({𝑣}_1,\dots ,{𝑣}_{𝑘})=1$ whenever $[{𝑣}_1,\dots ,{𝑣}_{𝑘}]$ is the positively oriented orthonormal basis corresponding to ${𝜇}_{𝜑(𝑎)}$.
   
2. $𝜔(𝐿𝐯)=det(𝐿)𝜔(𝐯)$ for every linear transformation $𝐿$ and every $𝑘$-tuple of vectors $𝐯=({𝑣}_1,\dots ,{𝑣}_{𝑘})$.

Determinant and Orientation

From this setup, for a coordinate chart $𝜑:𝑊\to {ℝ}^{𝑘}$ and standard basis ${𝑒}_1,\dots ,{𝑒}_{𝑘}$ in ${ℝ}^{𝑘}$, we have:

By the defining property of $𝜔$,

Hence, a coordinate chart $𝜑$ preserves orientation if and only if $det(𝐷𝜑(𝑎))>0$ for all $𝑎$. If $det(𝐷𝜑(𝑎))<0$ for some $𝑎$, then $𝜑$ reverses the orientation at that point.

Composition with an Orientation-Reversing Chart

Finally, suppose $𝜑$ is an orientation-reversing chart. Let $𝑇$ be another smooth map (often a linear transformation in examples). Then, for $𝑥$ in the domain of $𝑇$, we compute the determinant of the composition:

In situations where $det(\mathrm{d}{𝜑}_{𝑇(𝑥)})$ is negative and $det(𝑇)$ is also negative , their product can be nonnegative, illustrating how the combined map may preserve, reverse, or degenerate the orientation.

Q.E.D.

One may ask that if (1) and (2) can be satisfied at a same point $𝑎$. The answer is no.

Suppose, for the sake of contradiction, that there exists a point $𝑎$ such that both conditions (1) and (2) are satisfied in neighborhoods of $𝑎$. Thus, we have:

1. A diffeomorphism ${ℎ}_1:{𝑈}_1\to {𝑉}_1$ satisfying condition (1) on an open neighborhood ${𝑈}_1$ of $𝑎$.
   
2. A diffeomorphism ${ℎ}_2:{𝑈}_2\to {𝑉}_2$ satisfying condition (2) on an open neighborhood ${𝑈}_2$ of $𝑎$.

Since $𝑎\in {𝑈}_1\cap {𝑈}_2$, we may replace ${𝑈}_1$ and ${𝑈}_2$ with smaller open sets (if necessary) so that

is an open neighborhood of $𝑎$ on which ${ℎ}_1$ and ${ℎ}_2$ remain diffeomorphisms. Denote

Consider the map

Since ${ℎ}_1$ and ${ℎ}_2$ are diffeomorphisms, $𝑓$ is itself a ${\mathrm{𝒞︀}}^{𝑟}$-diffeomorphism and therefore a homeomorphism onto its image.

As

• By condition (1), ${𝑉}_{{\left(1\right)}^{\prime }}$ locally resembles ${ℝ}^{𝑘}\times \{0{\}}^{𝑛-𝑘}$ in a way that is ${\mathrm{𝒞︀}}^{𝑟}$-diffeomorphic.
  
• By condition (2), ${𝑉}_{{\left(2\right)}^{\prime }}$ locally resembles ${ℍ}^{𝑘}\times \{0{\}}^{𝑛-𝑘}$ (where ${ℍ}^{𝑘}$ is a half-space in ${ℝ}^{𝑘}$) in a way that is ${\mathrm{𝒞︀}}^{𝑟}$-diffeomorphic.

under $𝑓$, the set

(an open neighborhood in ${ℝ}^{𝑘}$) must map to

which is a set open in ${ℍ}^{𝑘}$ (in its subspace topology) but not open in the standard topology of ${ℝ}^{𝑘}$. Since a homeomorphism preserves topological openness, this situation is impossible. The image of an open set in ${ℝ}^{𝑘}$ under $𝑓$ would have to remain open in ${ℝ}^{𝑘}$, contradicting the fact that ${ℍ}^{𝑘}$ is not open in ${ℝ}^{𝑘}$.

Thus, our assumption that conditions (1) and (2) both hold at $𝑎$ leads to a topological contradiction. Therefore, no single point $𝑎$ in the manifold can satisfy both conditions simultaneously.

Q.E.D.

Setup and Transition Map

Since $𝑓$ and $𝑔$ are diffeomorphisms onto their images in $𝑀$, define the local transition map

whose domain is ${𝑓}^{-1}(\mathrm{I}\mathrm{m}(𝑔))$ and codomain is ${𝑔}^{-1}(\mathrm{I}\mathrm{m}(𝑓))$. By construction, $𝛾$ is a ${\mathrm{𝒞︀}}^{𝑟}$ diffeomorphism between open subsets of ${ℝ}^{𝑘}$, and

Half‐Space Condition

Both $𝑓$ and $𝑔$ are half‐space charts, meaning:

• $𝑓(\{{𝑡}_{𝑘}=0\})\subset 𝜕𝑀$ and $𝑓(\{{𝑡}_{𝑘}>0\})\subset 𝑀\setminus 𝜕𝑀$.
  
• $𝑔(\{{𝑠}_{𝑘}=0\})\subset 𝜕𝑀$ and $𝑔(\{{𝑠}_{𝑘}>0\})\subset 𝑀\setminus 𝜕𝑀$.

Because $𝛾={𝑔}^{-1}\circ 𝑓$, it follows that if $({𝑡}_1,\dots ,{𝑡}_{𝑘})$ lies in $\{{𝑡}_{𝑘}=0\}$, then $𝑓({𝑡}_1,\dots ,{𝑡}_{𝑘})\in 𝜕𝑀$. Consequently,

Similarly, $𝛾$ preserves the interior condition $\{{𝑡}_{𝑘}>0\}\mapsto \{{𝑠}_{𝑘}>0\}$. Altogether,

Tangent Cones and Their Images

Denote by ${𝐻}_0^{𝑘}\subset {\mathrm{T}}_0{ℝ}^{𝑘}$ the set of all velocity vectors ${𝛼}^{\prime }(0)$ of curves $𝛼(𝑡)\subset \{{𝑡}_{𝑘}\ge 0\}$ with $𝛼(0)=0$. Equivalently, ${𝐻}_0^{𝑘}$ is the tangent cone of the closed half‐space $\{{𝑡}_{𝑘}\ge 0\}$ at the origin. Since

any curve $𝛼$ in $\{{𝑡}_{𝑘}\ge 0\}$ has its image $𝛾\circ 𝛼$ in $\{{𝑠}_{𝑘}\ge 0\}$. Taking derivatives at $0$ shows:

so $𝐷𝛾(0)({𝐻}_0^{𝑘})\subset {𝐻}_0^{𝑘}$. By applying the same argument to ${𝛾}^{-1}$ (which is also a diffeo), we deduce

Hence $𝐷𝛾(0)$ is a linear isomorphism from ${𝐻}_0^{𝑘}$ onto itself.

Consistency of the Pushed‐Forward Subspaces

From $𝛾={𝑔}^{-1}\circ 𝑓$, we write

But ${𝛾}_{♯}({𝐻}_0^{𝑘})$ is exactly ${𝐻}_0^{𝑘}$ again, so

Thus both charts $𝑓$ and $𝑔$ push forward the same $𝑘$-dimensional cone in ${\mathrm{T}}_{𝑥}{ℝ}^{𝑛}$, which reflects the tangent‐plus‐normal decomposition at $𝑥$.

Uniqueness of the Normal Direction

Within this common $𝑘$-dimensional space, the direction of $\frac{𝜕}{𝜕{𝑡}_{𝑘}}$ is precisely the one normal to $𝜕𝑀$. Concretely, the subspace spanned by $\{\frac{𝜕}{𝜕{𝑡}_1},\dots ,\frac{𝜕}{𝜕{𝑡}_{𝑘-1}}\}$ projects onto ${\mathrm{T}}_{𝑥}(𝜕𝑀)$, leaving the line spanned by $\frac{𝜕}{𝜕{𝑡}_{𝑘}}$ as the normal complement. Since both charts determine the same pushforward space, they also determine the same normal line. By convention, we take the positively oriented side to define the outward normal direction.

Therefore, the outward normal vector at $𝑥$ (up to positive scaling) given by $𝑓$ agrees with that given by $𝑔$. This shows that the outward normal on $𝜕𝑀$ is independent of the choice of half‐space boundary chart, provided those charts align on the same side of $𝜕𝑀$.

Q.E.D.

Let ${𝜔}_{𝑥}$ be the oriented volume form on ${\mathrm{T}}_{𝑥}𝑀$ for each $𝑥\in 𝜕𝑀$. Denote by $𝑛(𝑥)$ the outward normal vector at $𝑥$. We define a $𝑛-1$-form ${𝑤}_{𝑥}$ on the tangent space ${\mathrm{T}}_{𝑥}(𝜕𝑀)$ by

Since ${𝜔}_{𝑥}$ is a non-degenerate $𝑛$-form on ${\mathrm{T}}_{𝑥}𝑀$ and $𝑛(𝑥)$ is linearly independent of ${\mathrm{T}}_{𝑥}(𝜕𝑀)$, it follows immediately that ${𝑤}_{𝑥}$ is non-degenerate on ${\mathrm{T}}_{𝑥}(𝜕𝑀)$. Hence ${𝑤}_{𝑥}$ is a valid volume form on ${\mathrm{T}}_{𝑥}(𝜕𝑀)$, and it defines an orientation at each $𝑥\in 𝜕𝑀$.

Consequently, an ordered basis $[{𝑣}_1,\dots ,{𝑣}_{𝑛-1}]$ of ${\mathrm{T}}_{𝑥}(𝜕𝑀)$ is positively oriented (i.e., lies in $(𝜕𝜇{)}_{𝑥}$) if and only if the extended basis $[𝑛(𝑥),{𝑣}_1,\dots ,{𝑣}_{𝑛-1}]$ is positively oriented in ${𝑇}_{𝑥}𝑀$ (i.e., lies in ${𝜇}_{𝑥}$). Symbolically,

Thus, if $𝑀$ is orientable with orientation $𝜇$, the boundary $𝜕𝑀$ inherits a natural orientation $𝜕𝜇$, determined by adjoining the outward normal to oriented bases of $𝜕𝑀$.

Q.E.D.

We start with the half‐space

with its usual orientation given by the standard volume form

Its boundary is

which is naturally identified with ${ℝ}^{𝑘-1}$ (with the usual orientation $\mathrm{d}{𝑥}^1\wedge \cdots \wedge \mathrm{d}{𝑥}^{𝑘-1}$).

The induced orientation on $𝜕{ℍ}^{𝑘}$ is defined by using the outward unit normal to ${ℍ}^{𝑘}$. Since the interior of ${ℍ}^{𝑘}$ is $\{{𝑥}^{𝑘}>0\}$, the outward normal is the vector pointing in the direction where ${𝑥}^{𝑘}$ decreases. In standard coordinates this is

The induced $(𝑘-1)$-form is given by contracting the $𝑘$-form $𝜇$ with $𝑛$:

Since

we now obtain

This shows that the induced orientation on the boundary is given by the $(𝑘-1)$-form

But the “usual orientation” on ${ℝ}^{𝑘-1}$ is $\mathrm{d}{𝑥}^1\wedge \cdots \wedge \mathrm{d}{𝑥}^{𝑘-1}$. Hence, the induced orientation on $𝜕{𝐻}^{𝑘}$ is $(-1{)}^{𝑘}$ times the usual orientation. This factor $(-1{)}^{𝑘}$ comes exactly from the contraction with the outward normal $-{𝑒}_{𝑘}$ and reflects the combinatorial sign arising from removing the $\mathrm{d}{𝑥}^{𝑘}$ factor from the volume form.

Q.E.D.

TODO

1. Computing the Normal Vector
   A local normal vector can be obtained by taking the cross product of the partial derivatives

   $$\frac{𝜕𝜑}{𝜕𝜃}(𝜃,𝑡)\text{and}\frac{𝜕𝜑}{𝜕𝑡}(𝜃,𝑡).$$

   At the line $𝑡=0$, these derivatives simplify. Denote

   $$𝐧(𝜃,0)=\frac{𝜕𝜑}{𝜕𝜃}(𝜃,0)\times \frac{𝜕𝜑}{𝜕𝑡}(𝜃,0).$$

2. Values at $𝜃=0$ and $𝜃=2𝜋$
   By direct (though somewhat tedious) computation, one obtains:

   $$𝐧(0,0)=(0,0,-\frac{1}{2}),𝐧(2𝜋,0)=(0,0,+\frac{1}{2}).$$

   Meanwhile, the base points in the strip coincide:

   $$𝜑(0,0)=𝜑(2𝜋,0).$$

   That is, the same physical point on the Möbius strip has two different normal vectors if we demand continuity of the normal across this identification.

3. Contradiction
   If $𝑀$ were an orientable surface in ${ℝ}^3$, one could choose a continuous unit normal vector field $𝐧(𝑥)$ at every point $𝑥\in 𝑀$. But the Möbius strip is nonorientable, so no such globally continuous choice exists. Concretely, the mismatch $𝐧(0,0)\ne 𝐧(2𝜋,0)$ while $𝜑(0,0)=𝜑(2𝜋,0)$ contradicts the requirement of continuity for a globally well‐defined normal.

Q.E.D.

Let ${𝜉}_1,{𝜉}_2$ be the orientation-preserving charts corresponding to ${𝑐}_1,{𝑐}_2$. Thus,

Since ${𝜉}_2$ is a local diffeomorphism onto its image, we can write

for some function $𝑓:{𝜉}_2^{-1}(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(𝜔))\to ℝ$.

Step 1: Define the transition map

Set

By hypothesis, $𝑇$ is a local diffeomorphism between open sets in ${ℝ}^{𝑘}$. In particular, on $[0,1{]}^{𝑘}\subset {ℝ}^{𝑘}$, it restricts to a map whose image remains in ${𝜉}_2^{-1}(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(𝜔))$.

Step 2: Pull back $𝜔$ via ${𝜉}_1$

We compute ${\int }_{{𝑐}_1}𝜔$ by expressing it as an integral in coordinates:

Since ${𝜉}_1={𝜉}_2\circ ({𝜉}_2^{-1}\circ {𝜉}_1)={𝜉}_2\circ 𝑇$, this becomes

Thus,

Step 3: Change of variables using $𝑇$

Recall that for a smooth map $𝑇:𝑈\to 𝑉\subset {ℝ}^{𝑘}$ and a $𝑘$-form $𝑓\mathrm{d}{𝑥}_1\wedge \cdots \wedge \mathrm{d}{𝑥}_{𝑘}$ on $𝑉$, the pullback is given by

Hence,

By standard change-of-variable arguments in ${ℝ}^{𝑘}$,

But $𝑇([0,1{]}^{𝑘})\subset {𝜉}_2^{-1}(\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(𝜔))$, so continuing,

Putting everything together, we obtain

Because $𝜔$ vanishes outside ${𝑐}_1([0,1{]}^{𝑘})\cap {𝑐}_2([0,1{]}^{𝑘})$, all of the above integrals make sense and indeed coincide.

Q.E.D

TODO

Suppose we have a $𝑘$–cube $𝑐$ that is mapped into $𝑀$ and that $𝑐$ is orientation preserving with respect to the given orientation on $𝑀$. Only its face

lies in $𝜕𝑀$.

• When we pull back the orientation from $𝑀$ to the cube by $𝑐$, the standard orientation on ${ℝ}^{𝑘}$ is given by $\mathrm{d}{𝑥}^1\wedge \cdots \wedge \mathrm{d}{𝑥}^{𝑘}$.
• The induced orientation on the boundary (via contraction with the outward normal) is, as computed above in Remark 1, $(-1{)}^{𝑘}\mathrm{d}{𝑥}^1\wedge \cdots \wedge \mathrm{d}{𝑥}^{𝑘-1}$.

Thus, if $𝑐$ is originally orientation preserving, then the face ${𝑐}_{(𝑘,0)}$ inherits the induced orientation that is $(-1{)}^{𝑘}$ times the “usual” orientation of a $(𝑘-1)$–cube.

Q.E.D

Suppose $𝜔$ is a $𝑘$-form on $𝑀$ such that $𝜔$ vanishes outside the image of a singular $𝑘$-cube $𝑐:[0,1{]}^{𝑘}\to 𝑀$. By continuity, $𝜔$ also vanishes on the interior boundary faces of $[0,1{]}^{𝑘}$, except possibly the face on which $𝑐$ meets $𝜕𝑀$. Concretely, if we write

then $𝜔$ vanishes on all faces $𝑐(𝑖,𝛼)$ except the one that intersects $𝜕𝑀$, which we label $𝑐(𝑘,0)$ without loss of generality. Hence,

Moreover, by the choice of orientation on $𝑀$ and how $𝑐$ meets $𝜕𝑀$, the integral of $𝜔$ over $𝑐(𝑘,0)$ relates to the integral of $𝜔$ on $𝜕𝑀$. Specifically,