How can I understand the fundamental theorem of line integrals in terms of differential forms on manifolds?

Question

Let $f$ be a differentiable function $\mathbb{R}^3\rightarrow\mathbb{R}$. Let $x,y,z$ be the standard coordinates on $\mathbb{R}^3$. Let $r : [0,1]\rightarrow\mathbb{R}^3$ be a smooth parametrized curve, with image $C$.

In the fundamental theorem of line integrals, it is often said that: $$\int_C df = \int_C\nabla f\cdot dr = f(r(1)) - f(r(0))$$

Now, on the far left, I understand "$df$" to be the differential form $$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz\in \Omega^1(\mathbb{R}^3)$$ restricted to $C$. Thus, the left hand side makes sense, as the integral of a differential form.

But what about $\int_C\nabla f\cdot dr$? In what sense is $\nabla f\cdot dr$ a differential form on $C$?

To me, $\nabla f$ is defined to be a vector field on $\mathbb{R}^3$, which via the standard Riemannian metric on $\mathbb{R}^3$ can be identified with a differential 1-form (if it helps) In this case, we probably really want to think of $\nabla f$ as its restriction $\nabla f|_C$ to $C\subset \mathbb{R}^3$.

How should I think of "$dr$"? One thought is to view $dr$ as a vector field on the image $C$ of $r$, namely $dr|_{r(t_0)}$ is the tangent vector at $r(t_0)\in \mathbb{R}^3$ represented by the germ of the parametrized curve $r$ at $t_0$. Ie, it is the tangent vector defined by the derivation: $$dr|_{r(t_0)} = \frac{\partial (x\circ r)}{\partial t}(t_0)\frac{\partial}{\partial x} + \frac{\partial (y\circ r)}{\partial t}(t_0)\frac{\partial}{\partial y} + \frac{\partial (z\circ r)}{\partial t}(t_0)\frac{\partial}{\partial z}$$ on the ring of germs of smooth functions at $r(t_0)$.

If we view $dr$ this way, then the dot product $\nabla f\cdot dr$ makes sense, but unfortunately it's value is a scalar field (ie a function) on $\mathbb{R}^3$, not a differential form. How do we view it as a differential form?

In particular, I would like to understand how exactly is it true that $$"\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz = \nabla f \cdot dr"$$

Answer 1

Most people would probably say that $\int_C \nabla f \cdot dr$ is just shorthand for $$\int_0^1 \nabla f(r(t)) \cdot r'(t) dt,$$ and that the point of writing the integral this way is to make it friendly to those who are learning elementary vector calculus. In the manifold setting, the fundamental theorem can simply be stated as $\int_Cdf = f(r(1))-f(r(0)).$

You can, however, give precise meaning to the dot product version: if we think of $r : [0,1] \to \mathbb R^3$ as a smooth map between manifolds, then its differential is a one-form $dr$ on $[0,1]$ taking values in the pullback bundle $r^*T \mathbb R^3.$ Thus the dot product $\nabla f \cdot dr$ can be interpreted as a metric contraction of the $T\mathbb R^3$ slot of $dr$ with $\nabla f \in T\mathbb R^3$, which leaves you with $\nabla f \cdot dr$ being a (scalar-valued) one-form on $[0,1]$ which can be integrated in the obvious way.

Alternatively, if $C$ is embedded and you want to think about integrals along $C$ instead of along $[0,1],$ you can think of $r : C \to \mathbb R^3$ as the inclusion map, so that $dr$ is one-form on $C$ taking values in $T\mathbb R^3|_C.$ (Explicitly, $dr$ is simply the fibre-wise inclusion of $TC$ in $T\mathbb R^3|_C.$) The expression $\nabla f \cdot dr$ is then the metric contraction of $\nabla f \in T\mathbb R^3$ with $dr \in T^* C \otimes T\mathbb R^3,$ leaving you with a one-form on $C$.

In terms of tensor components we can write this as $$ (\nabla f \cdot dr)_i = g_{ab} (\nabla f)^a (dr)^b_i$$ where $g$ is the Euclidean metric.

Answer 2

To elaborate on Anthony Carapetis's excellent answer, let us take the perspective of our situation being a composition:

$$[0,1]\stackrel{r}{\longrightarrow}\mathbb{R}^3\stackrel{f}{\longrightarrow}\mathbb{R}$$ Let $t$ be the standard coordinate on $[0,1]$ and $x,y,z$ the coordinates on $\mathbb{R}^3$. The tangent bundle $T\mathbb{R}^3$ is the trivial bundle, with standard basis at every point $p\in\mathbb{R}^3$ given by $\left\{\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\}$. The tangent bundle $T[0,1]$ is also trivial, with basis $\frac{\partial}{\partial t}$.

We have three objects:

1. We have $d(f\circ r) = d(r^*f) = r^*df$, a 1-form on $[0,1]$. Given our expression for $df$ in the OP, we have at any $t_0\in [0,1]$: $$d(f\circ r)_{t_0} = \frac{\partial f}{\partial x}(r(t_0))d(x\circ r) + \frac{\partial f}{\partial y}(r(t_0))d(y\circ r) + \frac{\partial f}{\partial z}(r(t_0))d(z\circ r)$$ By definition, we have $d(x\circ r)_{t_0)}(\frac{\partial}{\partial t}) = \frac{\partial(x\circ r)}{\partial t}(t_0)$, and similarly for $d(y\circ r),d(z\circ r)$. Thus, evaluated on the tangent vector $\frac{\partial}{\partial t}$ at $t_0\in[0,1]$, we have $$d(f\circ r)_{t_0}(\frac{\partial}{\partial t}) = \frac{\partial f}{\partial x}(r(t_0))\frac{\partial (x\circ r)}{\partial t}(t_0) + \frac{\partial f}{\partial y}(r(t_0))\frac{\partial (y\circ r)}{\partial t}(t_0) + \frac{\partial f}{\partial z}(r(t_0))\frac{\partial (z\circ r)}{\partial t}(t_0)$$

2. We have the gradient $\nabla f$, which is the vector field on $\mathbb{R}^3$ given at a point $p\in\mathbb{R}^3$ by: $$(\nabla f)_p = \frac{\partial f}{\partial x}(p)\frac{\partial}{\partial x} + \frac{\partial f}{\partial y}(p)\frac{\partial}{\partial y} + \frac{\partial f}{\partial z}(p)\frac{\partial}{\partial z}$$

3. We have the differential $dr$, viewed as the induced map on tangent bundles $T[0,1]\rightarrow T\mathbb{R}^3$. By definition, at $t_0\in[0,1]$, $(dr)_{t_0}(\frac{\partial}{\partial t})$ is the derivation on the ring of $C^\infty$ germs at $r(t_0)$ $C^\infty_{r(t_0)}(\mathbb{R}^3)$ sending $f\in C^\infty_{r(t_0)}(\mathbb{R}^3)$ to $\frac{\partial(f\circ r)}{\partial t}(t_0)$. By the chain rule, we have: $$\frac{\partial(f\circ r)}{\partial t}(t_0) = \frac{\partial f}{\partial x}(r(t_0))\frac{\partial (x\circ r)}{\partial t}(t_0) + \frac{\partial f}{\partial y}(r(t_0))\frac{\partial (y\circ r)}{\partial t}(t_0) + \frac{\partial f}{\partial z}(r(t_0))\frac{\partial (z\circ r)}{\partial t}(t_0)$$ In other words, $$(dr)_{t_0}(\frac{\partial}{\partial t}) = \frac{\partial (x\circ r)}{\partial t}(t_0)\frac{\partial}{\partial x} + \frac{\partial (y\circ r)}{\partial t}(t_0)\frac{\partial}{\partial y} + \frac{\partial (z\circ r)}{\partial t}(t_0)\frac{\partial}{\partial z}$$

Thus, if we view $(\nabla f)\cdot dr$ as the 1-form on $[0,1]$ which at every point $t_0\in[0,1]$ sends $$\frac{\partial}{\partial t}\mapsto (\nabla f)_{r(t_0)}\cdot \left((dr)_{t_0}(\frac{\partial}{\partial t})\right)$$ Then it is clear that we have an equality of 1-forms on $[0,1]$: $$d(f\circ r) = (\nabla f)\cdot dr$$ Thus, we have: $$\int_{[0,1]}(\nabla f)\cdot dr = \int_{[0,1]}d(f\circ r) = f(r(1)) - f(r(0))$$ The second equality is Stoke's theorem, and the combined equalities is the usual statement of the fundamental theorem of line integrals. To connect this with integrals over $C = r([0,1])\subset\mathbb{R}^3$, we note that $\int_C df$ only makes sense if we give $C$ an orientation. In fact, $\int_C df$ is often defined to be $\int_{[0,1]}d(f\circ r)$ for some parametrization $r$ of $C$. The parametrization implicitly gives $C$ an orientation, so this makes sense.