Conservative, Closed, & Exact Covector Fields

Lee Smooth Manifolds

The math is a little sloppy today, sorry, I got really tired while typing this up.

These notes reproduce Lee’s exposition on closed, conservative, exact covector fields. First, some definitions: if $M$ is a smooth manifold (with or without boundary), a smooth covector field $\omega \in {\mathfrak{X}}^{\ast }(M)$ is exact if there exists some $f\in C^{\infty }(M)$ such that $\omega =df$. If $\omega$ is such that ${\int }_{\gamma }\omega =0$ for all piecewise smooth closed curve segments (piecewise smooth $\gamma :[a,b]\to M$ such that $\gamma (a)=\gamma (b)$), $\omega$ is conservative. Note that this is equivalent to saying ${\int }_{\gamma }\omega ={\int }_{{\gamma }^{\prime }}\omega$ for all piecewise smooth curve segments $\gamma$ and ${\gamma }^{\prime }$ with the same start and endpoints.

Theorem 1 (Lee 11.42): Let $M$ be a smooth manifold with or without boundary and $\omega \in {\mathfrak{X}}^{\ast }(M)$. Then, $\omega$ is conservative if and only if it is exact.

Proof) Assume $\omega =df$ for some $f\in C^{\infty }(M)$. For any smooth curve segment $\gamma :[a,b]\to M$, note that ${\gamma }^{\ast }{\omega }_s(d/dt)=\omega (d{\gamma }_s(d/dt))=\omega ({\gamma }^{\prime }(s))=d{f}_{\gamma (s)}{\gamma }^{\prime }(s)=d/dt{|}_s(f\circ \gamma )$. Thus,

$${\int }_{\gamma }\omega ={\int }_a^b{\gamma }^{\ast }{\omega }_s(d/dt)ds={\int }_a^b\frac{d}{dt}{|}_{t=s}(f\circ \gamma (t))ds=f\circ \gamma (b)-f\circ \gamma (a)$$

Concatenating such smooth curve segments to make piecewise smooth closed curve segments, we can clearly see that $\omega$ is conservative. It suffices to show that $\omega$ is exact assuming it is conservative.

Assume $\omega$ is conservative and, for the time being, assume $M$ is connected. We use the following notation: for any $p,q\in M$, ${\int }_p^q\omega ={\int }_{\gamma }\omega$ for some piecewise smooth curve segment $\gamma$ starting at $p$ and ending at $q$; such a $\gamma$ clearly exists for any pair of points, and this is well-defined by conservativeness. Then, fix a point $p_0\in M$, and, for any $p\in M$, define $f:M\to \mathbb{R}$ as:

$$f(p)={\int }_{p_0}^p\omega$$

Clearly, if $\omega$ is exact and $\omega =d\tilde{f}$, $f$ must be a valid potential function as, for any $p$,

$$\tilde{f}(p)-\tilde{f}(p_0)={\int }_{p_0}^p\omega =f(p),\forall p\in M\Rightarrow d{f}_p=d(\tilde{f}-\tilde{f}(p_0){)}_p=d{\tilde{f}}_p={\omega }_p$$

We can thus say with confidence that it suffices to show $f$ is smooth and $df=\omega$ on $M$.

Take any point $p\in M$ and a smooth chart $(U,(x^i))$ centered at $p$ (note $p$ might be a boundary point). In these new coordinates, it suffices to show ${\omega }_p^i={\omega }_p(\partial /\partial x^i)=d{f}_p(\partial /\partial x^i)=\partial f/\partial x^i(p)$ for all $1\le i\le m$. Fixing any arbitrary $i$ (which could be a boundary coordinate), there is clearly some $ϵ>0$ such that the curve $\gamma :[0,ϵ]\to U$ defined by $\gamma (t)=t{e}^i$ is well-defined. For any $s\in [0,ϵ]$, we have that

$$f(\gamma (s))={\int }_{p_0}^{\gamma (s)}\omega ={\int }_{p_0}^p\omega +{\int }_p^{\gamma (s)}\omega =f(p)+{\int }_{\gamma {|}_{[0,s]}}\omega =f(p)+{\int }_0^s{\gamma }^{\ast }{\omega }_v\left(\frac{d}{dt}\right)dv$$ $$=f(p)+{\int }_0^s{\omega }_{\gamma (v)}{\gamma }^{\prime }(v)dv=f(p)+{\int }_0^s{\omega }^i(\gamma (v))dv$$

Therefore,

$$\frac{\partial f}{\partial x^i}(p)={\frac{d}{ds}|}_{s=0}f(\gamma (s))={\frac{d}{ds}|}_{s=0}\left(f(p)+{\int }_0^s{\omega }^i(\gamma (v))dv\right)={\omega }^i(\gamma (0))={\omega }_p^i$$

This simultaneously shows $f$ is smooth and ${\omega }_p^i=\partial f/\partial x^i(p)$, as desired.

A covector field $\omega$ is called closed if, in every smooth chart, $\partial {\omega }_j/\partial x^i=\partial {\omega }_i/\partial x^j$ for every index pair $(i,j)$. The motivation for closedness is that if $\omega =df$ for some smooth $f$, $\omega$ must clearly be closed as partial derivatives of smooth functions can be interchanged. The relationship between a covector field being closed and exact is much more intricate than one might expect; we begin with criteria for evaluating closedness.

Theorem 2 (Lee 11.45): Let $M$ be a smooth manifold with or without boundary and $\omega \in {\mathfrak{X}}^{\ast }(M)$. The following are equivalent. (a) $\omega$ is closed. (b) $\omega$ satisfies $\partial {\omega }_j/\partial x^i=\partial {\omega }_i/\partial x^j,\forall i,j$ in some smooth chart around every point. (c) For any open $U\subset M$ and smooth vector fields $X,Y\in \mathfrak{X}(M)$,

$$X(\omega (Y))-Y(\omega (X))=\omega ([X,Y])$$

Proof) (a) trivially implies (b). We show $(b)\Rightarrow (c)$; it suffices to show (c) holds in some neighborhood of some arbitrary point $p\in U$. Let $(V,(x^i))\subset U$ be a smooth chart containing $p$ in which $\partial {\omega }_j/\partial x^i=\partial {\omega }_i/\partial x^j,\forall i,j$ applies. Using coordinate representations $X=X^i\partial /\partial x^i$, $Y=Y^i\partial /\partial y^i$, and $\omega ={\omega }_{i}d{x}^i$, we have that

$$X(\omega (Y))=X({\omega }_i{Y}^i)={\omega }_{i}X{Y}^i+Y^i{X}^j\frac{\partial {\omega }_i}{\partial x^j}$$

The last term cancels in $X(\omega (Y))-Y(\omega (X))$ by assumption, leaving

$$X(\omega (Y))-Y(\omega (X))={\omega }_{i}X{Y}^i-{\omega }_{i}Y{X}^i={\omega }_i\left(XY\frac{\partial }{\partial x^i}-YX\frac{\partial }{\partial x^i}\right)={\omega }_i(d{x}^i[X,Y])=w([X,Y])$$

To show $(c)\Rightarrow (a)$, take an arbitrary smooth chart $(U,(s^i))$; the vector fields $X=\partial /\partial s^i$ and $Y=\partial /\partial s^j$ commute for any index pair $(i,j)$, implying $\partial {\omega }^j/\partial s^i-\partial {\omega }^i/\partial s^j=X({\omega }^j)-Y({\omega }^i)=\omega ([X,Y])=0$, and thus (a).

A main takeaway of Theorem 2 is that, to show closedness, it suffices to just show closedness in some chart around any point.

Corollary 3 (Lee 11.46): Suppose $F:M\to N$ is a local diffeomorphism. The pullback $F^{\ast }:{\mathfrak{X}}^{\ast }N\to {\mathfrak{X}}^{\ast }M$ takes exact covector fields to exact ones, and closed fields to closed ones.

Proof) The pullback of exact fields are obviously exact as if $\omega =df$, $F^{\ast }(\omega )=F^{\ast }(df)=d(f\circ F)$. Assume $\omega$ is closed; for any $p\in M$, there exists smooth charts $U$ containing $p$ in $M$ and $V$ containing $F(p)$ in $N$ such that the coordinate representation of $F$ is the identity. As $\partial {\omega }_j/\partial x^i=\partial {\omega }_i/\partial x^j,\forall i,j$ in $V$, the same must apply in $U$, and thus every point is in a coordinate chart in which the exchange formula holds. By Theorem 2(b), $F^{\ast }\omega$ is closed.

The relationship between closedness and exactness depends on the topology of the manifold. In some simple cases, these two coincide:

Theorem 4 (Lee 11.49): If $U$ is a star-shaped open subset of ${\mathbb{R}}^n$ or ${\mathbb{H}}^n$, then every closed covector field on $U$ is exact.

Proof) Suppose $U$ is star shaped with respect to a point $c$; as pullbacks of diffeomorphisms preserve closedness and exactness, we can apply a translation such that $c$ is the origin without affecting the core problem. Let $\omega \in {\mathfrak{X}}^{\ast }(U)$ be closed. The proof follows the same flow as Theorem 1, except now we cannot unambiguously define an integral between two points in $U$. Instead, we define a function $f:U\to \mathbb{R}$ as follows: for any $x\in U$, let $\gamma :[0,1]\to U$ be the curve $\gamma (t)=xt$; this is contained in $U$ by assumption. Then, define $f(x)={\int }_{\gamma }\omega$. We must now show that $f$ is smooth and that ${\omega }_j=\partial f/\partial x^j$ for all $1\le j\le n$.

Note that

$$f(x)={\int }_{\gamma }\omega ={\int }_0^1\omega ({\gamma }^{\prime }(t))dt={\int }_0^1{\omega }_i(tx)x^{i}dt$$

As the integrand is smooth, we can legally differentiate under the integral to obtain

$$\frac{\partial f}{\partial x^j}(x)={\int }_0^1\frac{\partial }{\partial x^j}{\omega }_i(tx)x^{i}dt={\int }_0^{1}t\frac{\partial w_i}{\partial x^j}(tx)x^i+{\omega }_j(tx)dt={\int }_0^1\frac{d}{dt}[t{\omega }_j(tx)]dt={\omega }_j(x)$$

Therefore, we conclude that $f$ is smooth and $\omega =df$, as desired.