# Line integral and integration of differential forms

The definition of integral of a $k$-form $\omega$ over a parametrized manifold $Y_\alpha$ is $\int_{Y_\alpha}\omega = \int_A\alpha^*\omega$ where $\alpha\colon A \to R^n$.

Let $\gamma:(a, b) \to R^3$. $C_\gamma$ is a curve in $R^3$ and the line integral over that curve is written as $\int_{C_\gamma}Pdx + Qdy + Rdz$ where $P$, $Q$ and $R$ are component function of $\gamma$. By applying the definition of integration of a $k$-form over a manifold we should get the formula $$\int_a^bP(\gamma(t)) \frac{d\gamma_1}{dt} + Q(\gamma(t)) \frac{d\gamma_2}{dt} + R(\gamma(t)) \frac{d\gamma_3}{dt}dt.$$ However, I seem to be missing something. Here's my computation: $$\int_{C_\gamma}Pdx + Qdy + Rdz = \int_a^bP(\gamma(t))dx(\gamma(t))(\alpha_*(t; v)) + ... \\= \int_a^bP(\gamma(t))dx(\gamma(t))(\gamma(t);D\gamma(t) v) + ...$$

where $v$ is some vector, in this case just a scalar, since it has only one component. Obviously, the disappearance of $v$ would give the expected result, but I can't see why it should disappear. I know I'm missing something quite simple, but could someone point that out for me, cause I'm feeling quite helpless right now...

For reference: I'm using the book "Analysis on manifolds" by Munkres.

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I don't see where your vector $v$ comes from. I've written out a derivation of the integral below, where I make no use of such a vector.

Let's take it from the beginning. I've taken some notational freedom, just tell me if something is unclear. $C_{\gamma}$ is a smooth $1$-dimensional submanifold of $\mathbb{R}^3$ with a global chart $\gamma:(a,b)\rightarrow \mathbb{R}^3$, $\gamma = (\gamma_1,\gamma_2,\gamma_3)$, and $F=(F_1,F_2,F_3)\in C^{\infty}(\mathbb{R}^3,\mathbb{R}^3)$. Then we have a 1-form $F\text{d}x = F_1\text{d}x_1 + F_2\text{d}x_2 + F_3\text{d}x_3$ and a straightforward calculation gives

$\gamma^*(F\text{d}x)(t)= \sum_i \gamma^*(F_i)\wedge \gamma^*(\text{d}x_i)$

$= \sum_i \gamma^*(F_i)\text{d} \gamma^*(x_i)$

$=\left[(F_1\circ \gamma)(t)\gamma_1^{\prime}(t)+(F_2\circ \gamma)(t)\gamma_2^{\prime}(t)+(F_3\circ \gamma)(t)\gamma_3^{\prime}(t)\right]\text{d}t$

$=\langle (F\circ \gamma)(t),\gamma^{\prime}(t)\rangle\text{d}t$.

So your integral becomes $$\int_{C_{\gamma}} F_1\text{d}x_1 + F_2\text{d}x_2 + F_3\text{d}x_3 = \int_{(a,b)}\langle (F\circ \gamma)(t),\gamma^{\prime}(t)\rangle\text{d}t$$

EDIT:

My definition of $\gamma^*$ when $\omega=\sum_i f_i \text{d}x_i$ is a 1-form, is $\gamma^*(\omega)(x)(v) = \omega(\gamma(x))(D_x \gamma (v))$ where $D_x\gamma (v) = \sum_{i=1}^3 \frac{\partial \gamma}{\partial x_i}(x) v^i$. Here is a link to the wikipedia article: http://en.wikipedia.org/wiki/Pullback_%28differential_geometry%29#Pullback_of_differential_forms

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Could you write down the definition of $\gamma^*$ that you're using here? –  Ormi Nov 30 '12 at 21:52
I added it at the bottom. –  Espen Nielsen Nov 30 '12 at 22:09