Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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


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:

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.