View source for Integration of differential forms of degree 0 and 1

Next, we consider integration of $0$-forms and $1$-forms -- on [[oriented manifolds]] of these dimensions.

==$0$-manifolds==

Why does orientation matter in calculus? 

Recall from [[calc 1]]:
$$\displaystyle\int_a^b f(x)dx = -\int_b^a f(x)dx.$$
If we rewrite this for oriented $1$-manifolds, we have: $[a,b]$ on the left and $[b,a]=-[a,b]$ on the right.

Simple enough... but we'll start with an even lower level, $0$-manifolds.

Now, $0$-manifolds are points, [[isolated point]]s to be exact. 

How do you "orient" those?

Well, manifolds with orientation (or "oriented manifolds") are manifolds with an extra structure (compare to a [[ring]] being a [[group]] with extra structure).

Specifically: for $0$-manifolds, ''every point is assigned a sign: $+$ or $-$'', or a number $+1$ or $-1$.

Compare: a $0$-manifold and and ''oriented'' $0$-manifolds: 

[[Image:OrientedZeroManifold.png|center|300px]]

Note: Just like any manifold, a [[connected]] $0$-manifold (i.e., a single point) has exactly two possible orientations.

'''Definition:''' Given a $0$-form $\varphi$, the ''integral of the form $\varphi$ over oriented $0$-manifold'' $M=\{p\}$ is:
$$\int_{\{p\}} \varphi = \left\{
\begin{array}{rl} 
\varphi(p) & \colon p {\rm \hspace{3pt} positively \hspace{3pt} oriented} \\
-\varphi(p) & \colon {\rm otherwise}
\end{array}
\right.$$

Why this way? The idea comes from the [[Fundamental Theorem of Calculus]]. 

[[Image:FundamentalTheoremOfCalculus.png|center|300px]]

In the notation we have been using, it states that 
$$\displaystyle\int _{[a,b]} f(x)dx = F(b) - F(a).$$
But now we can also interpret the right hand side as an integral as well!

Indeed:
*$F(b) = \displaystyle\int_{\{b\}}F$, and 
*$F(a) = \displaystyle\int_{\{a\}}F$.

So, the right hand side of the fundamental theorem of calculus is the integral of $F$ over a $0$-manifold of two points, $\{a,b\}$, with $a$ oriented positively and $b$ negatively. It's important to see that these orientations come from the orientation of the 1-manifold $[a,b]$, to be discussed later.

'''Definition:''' Once $\displaystyle\int_M \varphi$ is defined over points, $M = \{p\}$, we extend it over the finite unions, by [[additivity]]: if 
$$M = \{p_1, \ldots, p_s\},$$
then 
$$\displaystyle\int_M \varphi = \displaystyle\sum_{i=1}^s\int_{\{p_i\}}\varphi.$$

'''Definition:''' Given a $0$-manifold $M$, $-M$ is the same manifold with reversed orientation: 
*$M = \{p_1, \ldots, p_s\}$ with signs $\epsilon_1, \ldots, \epsilon_s$ are each $+$ or $-$, 
*$-M = \{p_1, \ldots, p_s\}$ with signs $-\epsilon_1, \ldots, -\epsilon_s$.

'''Theorem:''' $\displaystyle\int_{-M} \varphi = -\displaystyle\int_M \varphi$.

'''Proof:''' Easy. $\blacksquare$

What about the discrete case? 

Recall:
*a discrete manifold = a [[combinatorial manifold]] = a manifold-like [[cubical complex]].

Simple:
*$0$-manifolds $=$ vertices;
*$0$-form $=$ number assigned to each vertex. 
Considering that ''integration is just summation'', there is really no difference...

==Orientations of $1$-manifolds==

Let $M$ be a connected $1$-manifold with boundary. 

It's a curve and there are only two options (why?):

[[Image:1ManifoldWithBoundary.png|center]]

What is the meaning of the orientation of a smooth curve $C$? 

Recall from [[Calculus 1: course|calc 1]], something we know about the [[Riemann integral]]: 

Compare: $\displaystyle\int_a^b$ and $\displaystyle\int_b^a = -\displaystyle\int_a^b$.

This is where it comes from:

[[Image:WhatIsOrientation.png|center]]

In this context, ''orientation is direction'':

[[Image:OrientedCurve.png|center]] 

Here the curve goes ''from $A$ to $B$''. It could also go ''from $B$ to $A$''.  

Simple enough... 

But what if $A=B$? Hmm... 

[[Image:OrientedCurveAEqualsB.png|center]]

Idea: use [[parametrization]]: 
*$p \colon [a,b] \rightarrow C,$
*$p(a)=A$ and $p(b)=B.$
But this isn't enough to establish an [[equivalence relation]] between them to capture orientability...

To formalize this idea, we can use [[tangent]]s. Compare:

[[Image:OrientationsOfCurve.png|center]]

[[Image:OrientationByTangents.png|center]]

Question: But where do tangents come from? 

Answer: the derivative $p'$.

[[Image:OrientationsOnOneManifold.png|center]]

Orange and green have the same orientation, purple opposite. 

How do we tell that two parametrizations have the same orientation? 

One idea: normalize them... 

We can consider $\frac{p'(t)}{||p'(t)||}$ instead of $p'(t)$ for comparison. Then there is no confusion as long as we only use (differentiable) parametrizations with $p'(t) \neq 0$.

So, given two parametrizations $p$,$q$ of $C$: 
*$p \colon [a,b] \rightarrow {\bf R}^N$, $p'(t) \neq 0$, $t \in [a,b]$ and 
*$q \colon [c,d] \rightarrow {\bf R}^N$, $q'(s) \neq 0$, $s \in [c,d]$.

[[Image:ParametrizationsOfACurve.png|center]]

Here $p(t)=x$, $q(s)=x$, then we can assume $p'(t) = \alpha q'(s)$, for some $\alpha(x)$, for every $x$ on $C$. Then
*if $\alpha > 0$, then $p,q$ produce the same orientation; or
*if $\alpha < 0$, the orientations are opposite.

So, all parametrizations are split into two classes. 

Are these, in fact, [[equivalence class]]es? Are there really exactly two of them? This requires proof. '''Exercise.'''

==Integration of $1$-forms over oriented $1$-manifolds==

Suppose $C$ is an oriented smooth $1$-manifold and $\varphi$ is a smooth $1$-form.

Let's define $\displaystyle\int_C \varphi$.

Suppose $C$ is parametrized by $p \colon [a,b] \rightarrow {\bf R}^2$.

Let's review [[line integral]]s ([[Calculus 3: course|calc 3]]) first.

The line integral along $C$ (with respect to this parametrization), is, as an example,
$$\displaystyle\int_C x dx + xydy = \displaystyle\int_a^b p_1(t)p_1'(t)dt + p_1(t)p_2(t)p_2'(t)dt.$$
Observe we use $p=(p_1,p_2)$ and then substitute $x = p_1(t)$ and $y=p_2(t)$.

We want this integral (which could be the [[work]] for example) to be ''independent of parametrization'' (but dependent on orientation). 

We use the definition of $1$-form as a function 
$$\varphi \colon {\bf R}^n \times T{\bf R}^n \rightarrow {\bf R},$$
i.e., a location, a vector at the location $\mapsto$ a number. 

'''Definition.''' The ''integral of a $1$-form over a oriented $1$-manifold $C$'' parametrized by $p \colon [a,b] \rightarrow {\bf R}^n$ is
$$\displaystyle\int_C \varphi = \displaystyle\int_a^b \varphi(p(t),p'(t))dt.$$

The composition $\varphi(p(t),p'(t))$ indeed makes sense since 
$$(p(t),p'(t)) \in C \times TC \subset {\bf R}^n \times T{\bf R}^n.$$

The integral of type $\displaystyle\int_a^b f(x)dx$ is familiar from [[calc 1]]. Let's translate it into the new language.

What are $C$, $p$, $\varphi$ for this integral? 
*$C=[a,b]$, 
*$p(t)=t$, 
*$\varphi = f(x) dx$.

Then 

$\begin{align*}
\displaystyle\int_c \varphi &= \displaystyle\int_a^b \varphi(p,p')dt \\
&= \displaystyle\int_a^b \varphi(t,1)dt \end{align*}$. 

Note: Here, $\{1\}$ is the basis of $T{\bf R}$ 

[[Image:IntegralOf1Manifold.png|center]]

Substitute: 

$\begin{align*} \varphi(t,1) &= (f(x) dx)(t,1) \\
&= f(t)(dx)(1) (dx {\rm \hspace{3pt} evaluated \hspace{3pt} at \hspace{3pt}} 1)\\
&= f(t) \cdot 1 \\
&= f(t),
\end{align*}$ 

so 
$$\displaystyle\int_c \varphi = \displaystyle\int_a^b f(t)dt,$$
as expected.

Recall also that this is the area under the graph of $f$. Therefore, it is independent of parametrization, as requested!

'''Example:''' In dimension $2$, consider, as an example, $C$ is this segment:

[[Image:1ManifoldDimension2.png|center]] 

and
$$\varphi = x dx + dy.$$

Parametrize the curve, 
*$p \colon [0,1] \rightarrow {\bf R}^2$, 
*$p(t) = (t,t)$.
Then
$$\begin{align*}
\varphi(p,p') &= (xdx + dy)((t,t),(1,1)) \\
&= t \cdot (dx)(1,1) + (dy)(1,1) \\
&= t((dx)(1,0) + (dx)(0,1)) + ((dy)(1,0)+(dy)(0,1)) \\
&= t+1
\end{align*}$$

Therefore, 
$$\displaystyle\int_C \varphi = \displaystyle\int_0^1 (t+1) dt = \left. \frac{t^2}{2}+t\right| _0^1 = \frac{3}{2}.$$

'''Exercise:''' Consider
*$\displaystyle\int_C \varphi$, where
*$C = $ upper half circle, and 
*$\varphi = ydx + xdy$. 


What if $C$ is ''not smooth''? It won't work if $p'$ is undefined. 

What if $C$ is ''piecewise'' smooth?

'''Definition:''' Suppose the curve is the union of finitely many smooth curves, 
$$C = C_1 \cup \ldots \cup C_s,$$
each of which is smooth. Then define 
$$\displaystyle\int_C \varphi = \displaystyle\sum_{i=1}^s\int_{C_i} \varphi.$$
(Think of this as the [[additivity]] of the integral.)

Why do we care about this issue? 

''[[Discrete forms]]'' are defined on the grid! 

Indeed, a discrete curve is made of vertical and horizontal pieces. So it's not smooth, but it ''is'' piecewise smooth.

Then, it suffices to define $\displaystyle\int_C \varphi$ for $C$ being vertical edge or horizontal edge (and then use additivity). 

Consider horizontal, 

[[Image:IntegralOfDiscreteForms.png|center]] 

parametrize $p \colon [0,1] \rightarrow {\bf R}^2$ by $p(t) = (t,0)$.

Then,
$$\begin{align*}
\int_C \varphi &= \int_0^1 \varphi(p,p') dt \\
&= \int_0^1 \varphi((t,0),(1,0)) dt \\
&= \int_0^1 m dt \\
&= m,
\end{align*}$$

where $m$ is the value of $\varphi$ on $C$. Same for vertical edges.

Note: What if the length of the grid cell is $h$? Then $\displaystyle\int_C \varphi = mh$.

So, 
$$\displaystyle\int_C \varphi = \displaystyle\sum_{i=1}^s m_i,$$
if $C = $ union of edges $e_1, \ldots, e_s$ with $\varphi(e_i)=m_i$.

Integration is just summation (watch out for orientation though).

==Discrete forms==

Integration allows us to formalize the relation between smooth forms and discrete forms (as [[cochains]]).

For example, $dx$ continuous vs. $dx$ discrete.

What about the definition of a continuous form as  $\varphi \colon {\bf R}^2 \times {\bf R}^2 \rightarrow {\bf R}$ and its [[multi-linearity]]? What does it have to do with cochains? 

Recall, $1$-''cochain'' is $\varphi \colon 1$-cells $\rightarrow {\bf R}$, just any function.

But, $1$-cells can be combined into $1$-[[Chains vs cochains|chains]] $=$ formal [[linear combination]]s of cells: 
$$\displaystyle\sum_{i=1}^n \alpha_i c_i,$$
where $\alpha_i \in {\bf R}$, $c_i$ are $1$-cells. They form $C_1({\bf R}^2)$, a [[vector space]] (assuming ${\bf R}^2$ has the grid).

Then $\varphi$ can be extended to 
$$\varphi \colon C_1({\bf R}^2) \rightarrow {\bf R}$$
by [[linearity]]: 
$$\varphi\left(\displaystyle\sum_{i=1}^n \alpha_i c_i\right) = \displaystyle\sum_{i=1}^n \alpha_i \varphi(c_i).$$ 

That's the meaning of "co" in "cochain". It's a [[linear function]] to ${\bf R}$ on chains (see [[Dual spaces]]).

Let's construct a cochain from a given ''continuous'' differential form, $\omega \in \Omega^1({\bf R}^2)$. It's very simple.

Suppose ${\bf R}^2$ has a grid.

'''Definition.''' The ''cochain corresponding to a given continuous form $\varphi$'' is defined as
$$\varphi(a) = \displaystyle\int_a \omega,$$
where $a$ is a $k$-cell. 

From this point of view,
<center>''integration = evaluation''.</center>

This is a canonical way of discretizing calculus over any grid or mesh.

In particular, all standard operators of [[vector calculus]]: 
*${\rm grad}$, 
*${\rm curl}$, 
*${\rm div}$, 
are automatically discretized as $d$, the [[exterior derivative]], in appropriate dimensions. 

Note: This is also a way of modeling with [[PDEs]].

'''Example:''' Let $\omega = dx$, find the corresponding $\varphi \in C^1({\bf R}^2)$. 

Let $\varphi \colon 1$-cells $\rightarrow {\bf R}$. Pick a cell: 

[[Image:PickACell.png|center]] 

(horizontal and vertical). Then
$$\varphi(a)=\displaystyle\int_a \omega = \displaystyle\int_0^1 \omega(p,p') dt,$$
where $p \colon [0,1] \rightarrow a$ is a parametrization of $a$.

Then $p(t)=(t,0)$, substitute (same as $dx(i)$): 
$$\begin{align*}
\varphi(a) &= \int_0^1 dx((t,0),(1,0)) dt \\
&= \int_0^1 1 dt \\
&= 1.
\end{align*}$$

Next, $p(t)=(0,t)$, substitute (same as $dx(j)$):
$$\begin{align*}
\varphi(b) &= \int_0^1 dx((0,t),(0,1)) dt \\
&= \int_0^1 0 dt \\
&= 0.
\end{align*}$$

Same for all other vertical and horizontal $1$-cells. Then $\varphi$ is: 

[[Image:ExampleOfADiscrete1Cell.png|center]]

All horizontal edges are assigned $1s$ and vertical $0s$. 

It's the opposite for $dy$. This matches our prior definitions.

'''Question:''' What about vice versa? How do you get a continuous form from a cochain? 

'''Answer:''' [[Interpolation]], to be discussed later.



[[category:calculus]]