Covector Fields, Pullbacks, & Basic Line Integrals

Lee Smooth Manifolds

Recall that if $M$ is a smooth manifold (with or without boundary) and $f\in C^{\infty }(M)$, the differential $d{f}_p\in T_p^{\ast }M$ is the covector such that, for any $v\in T_{p}M$, $d{f}_{p}v=vf$. Furthermore, for any linear map between vector spaces $A:V\to W$, the dual $A^{\ast }:W^{\ast }\to V^{\ast }$ satisfies

$$(A^{\ast }\omega )v=\omega (Av),\forall v\in V,\omega \in W^{\ast }$$

Given a smooth map $F:M\to N$ between manifolds (with or without boundary) and covector field $\omega$ on $N$, the pullback covector field $F^{\ast }\omega$ is such that $F^{\ast }{\omega }_p=d{F}_p^{\ast }({\omega }_{F(p)})$. We begin with some elementary computations using these objects.

$df$ in Coordinates Let $(U,x^i)\subset M$ be a smooth chart and $p\in U$. Note that $d{x}_p^i$ is the covector such that $d{x}_p^i(\partial /\partial x^j{|}_p)={\delta }_i^j$ ($\delta$ is the Kronecker delta). Then, for any $f\in C^{\infty }(M)$ and $v\in T_{p}M$, with $v$ represented as $v=v^i\partial /\partial x^i{|}_p$,

$$d{f}_{p}v=vf=v^i\frac{\partial f}{\partial x^i}(p)\Rightarrow d{f}_p=\frac{\partial f}{\partial x^i}(p)d{x}^i$$

Thus, $d{f}_p$ has a coordinate representation analogous to the gradient in calculus. It resides in the covector space rather than the vector space, however, as the gradient changes “covariantly” with respect to coordinates while vectors change “contravariantly,” and thus a coordinate-agnostic representation must be in the form of covectors.

Pullback Fields Let $F:M\to N$ be a smooth map between manifolds (with or without boundary), and suppose $u$ is a continuous real-valued function on $N$, and $\omega$ is a (continuous) covector field on $N$. Then

$$F^{\ast }(u\omega )=(u\circ F)F^{\ast }\omega$$

This is a one-line computation as follows: for any $p\in M$, $F^{\ast }(u\omega {)}_p=d{F}_p^{\ast }(u(F(p)){\omega }_{F(p)})=u(F(p))d{F}_p^{\ast }({\omega }_{F(p)})=u(F(p))F^{\ast }(w{)}_p$. Thus, $F^{\ast }(u\omega )=(u\circ F)F^{\ast }\omega$. Furthermore, if $u\in C^{\infty }(N)$, for any $p\in M$ and $v\in T_{p}M$, $(F^{\ast }du{)}_{p}v=(d{F}_p^{\ast })(du{)}_{F(p)}v=(du{)}_{F(p)}(d{F}_{p}v)=d(u\circ F{)}_{p}v\Rightarrow d(u\circ F)=F^{\ast }du$.

We use this to compute a coordinate representation of the pullback. Let $F:M\to N$ be smooth and $\omega$ be a (continuous) covector field on $N$. Given $p\in M$ and a smooth chart $(V,(y^i))$ defined in a neighborhood of $F(p)$, on the neighborhood $F^{-1}(V)$ of $p$ we have that:

$$F^{\ast }\omega =F^{\ast }({\omega }_{i}d{y}^i)=({\omega }_i\circ F)F^{\ast }d{y}^i=({\omega }_i\circ F)d(y^i\circ F)$$

Thus, pullback computations are (fortunately) straightforward.

We now review the base theory for line integrals. Recall that, if $\omega \in {\mathfrak{X}}^{\ast }([a,b])$ with coordinate representation ${\omega }_t=f(t)dt$, we define ${\int }_{[a,b]}\omega ={\int }_a^{b}f(t)dt$This is diffeomorphism-invariant. In particular, if there is an increasing diffeomorphism $\phi :[c,d]\to [a,b]$, then the pullback is $({\phi }^{\ast }\omega {)}_s=f(\phi (s)){\phi }^{\prime }(s)ds$, and

$${\int }_{[c,d]}{\phi }^{\ast }\omega ={\int }_c^{d}f(\phi (s)){\phi }^{\prime }(s)ds={\int }_a^{b}f(t)dt={\int }_{[a,b]}\omega$$

If $\gamma :[a,b]\to M$ is a smooth segment, then ${\gamma }^{\ast }{\omega }_t(d/dt)={\omega }_{\gamma (t)}({\gamma }^{\prime }(t))$. Thus,

$${\int }_{\gamma }\omega ={\int }_a^b{\gamma }^{\ast }{\omega }_t(d/dt)dt={\int }_a^b{\omega }_{\gamma (t)}({\gamma }^{\prime }(t))dt$$

The same identity can be shown for piecewise-smooth curve sections analogously. The fundamental theorem of line integrals follows, as if $f\in C^{\infty }(M)$ is smooth and $\gamma :[a,b]\to M$ is a smooth curve segment, ${\int }_{\gamma }df={\int }_a^{b}d{f}_{\gamma (t)}({\gamma }^{\prime }(t))dt={\int }_a^b(f\circ \gamma {)}^{\prime }(t)dt=f(\gamma (b))-f(\gamma (a))$the same trivially follows for piecewise smooth curve sections.