# If differential are covectors, what is the geometric meaning of the exterior derivative of covectors?

I was thinking of the fact that differentials form a contravariant basis, and I understand the interpretation of the line integral:

$$\int_P{f_i\ dx^i}$$

as the aplication of a covector field $${f_idx^i}$$ over a curve $$P$$. But I also had the intuition that integrating over differential forms corresponds to computing the acting of the $$k-$$vector field of the exterior algebra to a domain $$D$$ of integration:

$$\int_D{f_i\ dx^i\wedge dx^I}$$

and so for example, the integral...

$$\int_S{fi\ dx^i\wedge dx^j}$$

I've thought of as the integral of a $$2-$$vector field over a surface $$S$$. I don't know if that intuiton is correct, but thinking otherwise, what would be the geometric (tensor) meaning of $$d(\omega)$$, if $$\omega$$ is a covector? How can be united the "tensor calculus" and "differential form" interpretation?

Edit: My question goes in the dirrection that, knowing that covectors are linear forms, how you arise from the exterior derivative of "a function of the Dual vector space" to the parrallelogram interpretation of $$2-$$forms.

Also, what I searched now is that, may they be like $$2-D$$ covectors or something like that.

It can be understood as an integration of the one form over a tiny parellogram. Elaborately speaking, you get a two form which when you evaluate on two vectors , you get the integration of the one form over the parallelogram spanned by the two vector. (Some scaling factors going on here, but besides the point)

• I made some videos trying to explain this. It is not very rigorous but hopefully you find it interesting Commented Jul 9, 2022 at 3:49
• thanks for your reply. I notice that my question was ambiguos, so I edited it and added that I want to know what is the exterior derivative of a covector (1-form) interpreted as a tensor in tensor calculus. Commented Jul 9, 2022 at 14:08
• I don't seen how my answer fails to answer the question still. The exterior derivative of a one form is a tensor which takes in two vectors and tells you the number you get when you integrate the one form over the loop spanned by the two vectors @DanielMuñoz Commented Jul 9, 2022 at 14:10
• Because covector are not vector, also I said that I understand the interpretation that you told me. But think of this,, how a covector, which is just a function that returns a scalar when applied to a vector, can be transform in the parallelogram, which is the wedge product between 2 vectors? Thats my question. Maybe I shoud add more to the original post Commented Jul 9, 2022 at 14:16
• But it's not a parellogram, my point is that you give two vectors describing a parellogram into the two form(exterior derivative of w), and it gives you the line integral of the one form over it @DanielMuñoz Commented Jul 9, 2022 at 14:20

Given a vector space $$V$$, a covector $$\xi \in V^*$$ is a linear function of vectors in $$V$$. The line integral of a $$1$$-form (i.e., covector field) $$\theta$$ along a parameterized curve is the integral of the value of $$\theta$$ evaluated with the velocity vector, $$\int_C \theta = \int_{t=a}^{t=b} \langle \theta(c(t)),c'(t)\rangle\,dt$$ If $$\theta = df$$, then $$\langle\theta(c(t)),c'(t)\rangle = f'(t)$$ so the fundamental form of calculus implies that $$\int_C df = f(c(b))-f(c(b)).$$

If $$\Xi$$ is an exterior $$2$$-tensor on a $$2$$-dimensional vector space $$V$$, then it can be viewed as defining a unit of measurement for the oriented area of the parallelogram spanned by any two vectors. In other words, if $$\Xi \ne 0$$, then there exist $$v_1, v_2 \in V$$ such that $$\Xi(v_1,v_2) = 1$$. If we now choose units such that the parallelogram spanned by $$v_1$$ and $$v_2$$ has area $$1$$, then given any other pair of vectors $$w_1,w_2 \in V$$, $$\Xi(w_1,w_2)$$ is the oriented area of the parallelogram spanned by $$w_1$$ and $$w_2$$.

Now suppose $$\Theta$$ is a $$2$$-form on a surface $$S$$. Therefore, for each $$x \in S$$, $$\Theta(x)$$ is an exterior $$2$$-tensor. Therefore, it defines a signed area function of parallelograms spanned by two vectors in the tangent space $$T_xS$$.

In particular, if you have a surface parameterized by a coordinate map $$x(u,v)$$, then at each point you have two coordinate tangent vectors $$x_u$$ and $$x_v$$ (which are the velocity vectors of the coordinate curves). You can then evaluate $$\Theta(x(u,v))$$ on the the two coordinate vectors and integrate that, $$\int_S \Theta = \int_D \Theta(x(u,v))(x_u,x_v)\,du\,dv,$$ where $$D$$ is the domain of the coordinate map. Note the similarity of this formula to that of the line integral. For both integrals, the miracle is that the value of the integral is independent of the parameterizations used.

Finally, the exterior derivative is a definition that is forced on you if you want to extend the $$1$$-dimensional fundamental theorem of calculus to a $$2$$-dimensional theorem. Specifically, you want something like this: $$\int_S d\theta = \int_{\partial S} \theta,$$ where the right side is a line integral. There are technical details to work out, but if you start with the special case where $$S$$ is just a rectangle, then the $$1$$-d fundamental theorem of calculus tells you what the formula for $$d\theta$$ has to be for this equation to hold. This theorem of course is usually called Green's Theorem or Stokes' Theorem, but I prefer to call it the $$2$$-dimensional fundamental theorem of calculus.

A covector usually means a linear functional of a vector space $$V$$, that is, an element of $$V^*=Hom(V,\mathbb R),$$ what you are dealing are called the differential forms on a smooth manifold $$M$$, the difference is that a differential form $$\omega$$ is a smooth assignment to each point $$p\in M$$ a covector of the tangent space $$T_pM$$ which is a vector space. Comparing to the family of vector fields, the family of differential forms (denoted by $$\Omega(M)$$) posses additional algebraic structure, that is, the tensor structure of $$C^\infty(M)$$-module which defined pointwise on $$Hom(T_pM,\mathbb R)$$ (the simple multiplication using the multiplicative structure of $$\mathbb R$$), and thus the tensor algebras $$\Omega(X)^{\otimes k}$$ and furthermore the wedge $$C^\infty(M)$$-algebras $$\Omega^k(M):=\bigwedge_{i=1}^k\Omega(X),$$ for $$k\in\mathbb N$$, on which classical algebraic tools can be applied to provide an invariant for the global geometry of a smooth manifold, which turns out to be the de Rham cohomology: The complex $$0\to C^\infty(M)\xrightarrow{d}\Omega(M)\xrightarrow{d^1}\Omega^2(M)\to 0$$where $$d^1$$ is the usual exterior differential operator defined to make $$d^1\circ d=0,$$ generates the exterior $$k$$-differential operators $$d^k:\Omega^k(M)\to\Omega^{k+1}(M)$$ $$\omega_{i_1}\wedge\omega_{i_2}\wedge\cdots\wedge\omega_{i_k}\mapsto \sum_{j=1}^k (-1)^{j-1}\omega_{i_1}\wedge\cdots d^1 \omega_{i_j}\wedge\cdots\wedge\omega_{i_k}$$ for each $$k$$ and gives the de Rham complex $$0\to C^\infty(M)\xrightarrow{d}\Omega(M)\to\cdots \Omega^k(M)\xrightarrow{d^k}\Omega^{k+1}(M)\to\cdots\xrightarrow{d^{n-1}}\Omega^n(M)\to 0,$$ the cohomology sequences, denoted by $$\{H^k_{dR}(M)\}_{k=0}^n$$ encoding certain global property of $$M.$$ Roughly speaking, $$H^1_{dR}(M)$$ measures how often a loop cannot be collapsed into a point, for example, consider a torus in $$\mathbb R^3,$$ there are two such circle loops. As an evidence, for an oriented two dimensional compact manifold, the $$\mathbb R$$-dimension of $$H^1_{dR}(M)$$ is $$2g$$ where $$g$$ is the genus that counts "handles". Similarly $$H^k_{dR}(M)$$ measures how often a $$k$$-submanifold cannot be collapsed into a point, if $$k=n-1,$$ that is a hypersurface.

Even though the construction involves the differentiable structure on $$M,$$ one can show that different differentiable structure gives the same de Rham cohomology, that is, the invariant is topological, in fact it coincides with the sheaf cohomology of constant sheaf $$\mathbb R$$ on $$M$$ regarded as a topological space $$H^k_{dR}(M)\simeq H^k(M,\mathbb R).$$

• Thanks for your answer, but the true is that I don't understand one bit, so I'm study for making sense of it. Do you know free content to study topology? Commented Jul 24, 2022 at 21:57