# Where are the covectors in a line integral?

I'm trying to connect the idea of covectors to integration.

From what I understand, given a basis $\{e_1, \dots, e_n\}$ of a vector space $V$, there exists a dual basis of covectors $\{f^1, \dots, f^n\}$ of the dual space $V^*$ such that $\langle f^i, e_j\rangle = {\delta^i}_j$.

A line integral in $\Bbb R^n$ has the form $$\int_\gamma \vec f \cdot d\vec r = \int_\gamma f_{x^1}dx^1 + f_{x^2}dx^2 + \cdots + f_{x^n}dx^n$$

Where are the covectors in the above expression? Does $dx^i$ really mean $\langle f^i, x^ie_i\rangle$? That doesn't seem right, because isn't $dx^i$ supposed to be some kind of derivative (the exterior derivative)?

Vector calculus is full of abbreviations and shortcuts that let you focus on what you're actually trying to compute, but when you examine those shortcuts with scrutiny, you might realize that there's a little more going on than you realized.

This problem is a good example. First, what does it mean to take a line integral of a vector field?

$$\int_C f \cdot dr$$

If the only tool you have to evaluate this is a (1d) Riemannian integral, this notation by itself doesn't tell you how to evaluate the integral. It's not in the form of a Riemannian integral.

The solution is to parameterize the curve $C$ by some parameter $t$ over some interval $I$. Then the expression becomes something you can integrate:

$$\int_I (f \circ C)(t) \cdot C'(t) \, dt$$

where $C(t)$ traces out the curve. Though this expression involves vectors, it is something that Riemannian integration can actually deal with.

Now wait a minute. This expression involves a dot product. What if we want to integrate something and don't have a dot product?

Answer: we can still integrate forms, or covector fields. Let $\omega$ be a covector field. Then the Riemannian integral

$$\int_I \omega_{C(t)}(C'(t)) \, dt$$

is a sensible expression. It involves a covector field being used as a map from vector fields to scalar fields. No problem. This is a sensible way to integrate something without reference to a metric, and if you do happen to have a metric, and want to integrate a vector field, you can find the natural corresponding covector field (e.g. by lowering indices and using the covector basis that you described instead) and integrate that.

But there's still something unsatisfying: the paramterization using $t$, which is still arbitrary. You could paramterize the curve using another parameter $\tau$ instead, and with a function $\Gamma(\tau)$, and a corresponding interval $\iota$, and you should get the same answer. There is a coordinate freedom to describe the curve.

The traditional answer to this problem is, well, to erase all reference to the parameterization.

The tangent vector $C'(t)$? Get rid of it. $dt$? Get rid of it. Just write the integral as

$$\int_C \omega$$

And then, write $\mathrm dt$ (not to be confused with $dt$) to mean the basis covector associated with $C'(t)$, such that $\mathrm dt(C'(t)) = 1$, so you can in principle write the projection of $\omega$ onto the curve $C$ in terms of $\mathrm dt$.

And yes, that $\mathrm dt$ is what would appear when you use the exterior derivative to differentiate using a coordinate system that is aligned to the curve.

In summary:

• Integrals with vector fields are often avoided in higher mathematics because they can be written in terms of integrals of covector fields instead, and those integrals are more generally applicable.
• The traditional notation for integrals of covector fields (integrals of differential forms, to use the more common terminology) suppresses the tangent vector field and paramterization by convention.
• Covetor fields may be expressed as a linear combination of basis covector fields, associated with the coordinates of the space. These are confusingly written similar to differentials in an integral ($\mathrm dx^i$ vs. $dx^i$, or even identically when the meaning is clear in context), but they do not mean the same thing.
• Since traditional differential forms notation suppresses the paramterization, it may be taken for granted that the parameterization uses the coordinates, and that the form to be integrated should be written in terms of those basis covector fields.

That is, this vector integral takes parameterization as implicit:

$$\int A \cdot dr = \int A \cdot e_1 \, dx^1 + \int A \cdot e_x \, dx^2$$

With explicit parameterization, it would be something like

$$\int A \cdot e_1 f^1 \cdot C'(t) \, dt + \int A \cdot e_2 f^2 \cdot C'(t) \, dt$$

In differential forms, it might look like (with implicit parameterization and tangent vector)

$$\int \alpha = \int \alpha_1 \mathrm dx^1 + \alpha_2 \mathrm dx^2$$

and with explicit parameterization and tangent vector

$$\int \alpha_1 \mathrm dx^1(C'(t)) + \alpha_2 \mathrm dx^2(C'(t)) \, dt$$

• Wow. That was a great answer. Thanks! – user275673 Oct 1 '15 at 13:42

The term to use here would be differential form. The operator $dx^i$ is a differential form. Specifically, it is a 1-form. It lives in the cotangent space, which is the dual space of the tangent space. The tangent space is where the tangent vectors live, and here it is equal to $Span\{x_1, ..., x_n\}$. The operator $dx^i$ acts on this tangent space via $<dx^i, x_j> = \delta_{i, j}$, where $\delta_{i, j}$ is the Kroeneker delta function.

• This is not right. What are the $x_i$? – Ted Shifrin Oct 1 '15 at 2:28

$dx^i$ is a covector field. It eats vector fields and spits out functions.

The way you integrate a covector field $\omega$ on $M$ along a parameterized path $\gamma: [0,1] \to M$ is by partitioning $[0,1]$ into subintervals $[x_i, x_{i+1}]$ and summing $\sum_1^n \omega(\gamma'(x_i)) (x_{i+1}-x_i)$, then taking a limit. This is what line integrals do.