Cotangent Vectors and Differential Forms

The cotangent space $T_p^*M$

The cotangent space at $p$ is the dual vector space of $T_pM$:

$$T_p^*M:=\mathrm{Hom}(T_pM,\mathbb{R}).$$

So an element of $T_p^*M$ is a linear map

$$\alpha_p:T_pM\to\mathbb{R}.$$

Such an object is called a covector or one-form at $p$.

In local coordinates, the basis of $T_p^*M$ is

$$dx^1_p,\ldots,dx^n_p.$$

These are defined to be the dual basis to

$$\left.\frac{\partial}{\partial x^1}\right|_p, \ldots, \left.\frac{\partial}{\partial x^n}\right|_p.$$

This means

$$dx^\mu_p\left(\left.\frac{\partial}{\partial x^\nu}\right|_p\right)=\delta^\mu_{\nu}.$$

Why $dx^\mu(\partial/\partial x^\nu)=\delta^\mu_\nu$

First recall that $x^\mu$ is a coordinate function:

$$x^\mu:M\to\mathbb{R}.$$

Its differential is a one-form

$$dx^\mu_p:T_pM\to\mathbb{R}.$$

For any tangent vector $v\in T_pM$, the differential is defined by

$$dx^\mu_p(v)=v[x^\mu].$$

Now take

$$v=\left.\frac{\partial}{\partial x^\nu}\right|_p.$$

Then

$$\begin{align*} dx^\mu_p\left(\left.\frac{\partial}{\partial x^\nu}\right|_p\right) &=\left.\frac{\partial}{\partial x^\nu}\right|_p[x^\mu]\\ &=\frac{\partial x^\mu}{\partial x^\nu}(p)\\ &=\delta^\mu_{\nu}. \end{align*}$$

For example, in $\mathbb{R}^2$ with coordinates $(x,y)$,

$$dx\left(\frac{\partial}{\partial x}\right)=1, \qquad dx\left(\frac{\partial}{\partial y}\right)=0,$$ $$dy\left(\frac{\partial}{\partial x}\right)=0, \qquad dy\left(\frac{\partial}{\partial y}\right)=1.$$

So $dx$ measures the $x$ component of a tangent vector, and $dy$ measures the $y$ component.

If

$$v=v^x\frac{\partial}{\partial x}+v^y\frac{\partial}{\partial y},$$

then

$$dx(v)=v^x, \qquad dy(v)=v^y.$$

Differential forms are not defined only for integration

A common first exposure to differential forms is through line integrals such as

$$\int_C P\,dx+Q\,dy.$$

This can make it look like $dx$ and $dy$ are merely integration symbols. That is not the intrinsic definition.

At a single point $p$, the vector space of alternating $k$-linear maps

$$T_pM\times\cdots\times T_pM\to\mathbb{R}$$

is denoted

$$\Lambda^kT_p^*M.$$

As $p$ varies, these spaces assemble into a bundle

$$\Lambda^kT^*M\to M.$$

A smooth $k$-form is a smooth choice

$$p\mapsto \omega_p\in \Lambda^kT_p^*M.$$

We denote the vector space of smooth $k$-forms by

$$\Omega^k(M).$$

Later, after vector bundles are defined formally, we will write the same statement compactly as

$$\Omega^k(M)=\Gamma(\Lambda^kT^*M),$$

where $\Gamma(E)$ means “smooth sections of the bundle $E$.” Thus symbols like $\Gamma(\Lambda^kT^*M)$ are not new mysterious objects; they are just the space of differential $k$-forms.

Special cases:

• A $0$-form is a function $f:M\to\mathbb{R}$.

• A $1$-form is a covector field:

$$\alpha_p:T_pM\to\mathbb{R}.$$

• A $2$-form is an antisymmetric bilinear map:

$$\omega_p:T_pM\times T_pM\to\mathbb{R}.$$

• A $k$-form eats $k$ tangent vectors and returns a number.

Only after we define forms as geometric objects do we integrate them over curves, surfaces, and manifolds.

One-forms

A one-form on $M$ is written locally as

$$\alpha=\alpha_\mu(x)\,dx^\mu.$$

At each point $p$, it eats a tangent vector

$$v=v^\mu \frac{\partial}{\partial x^\mu}$$

and gives

$$\alpha_p(v)=\alpha_\mu(p)v^\mu.$$

Examples in physics:

• Electromagnetic potential:

$$A=A_\mu dx^\mu.$$

• Berry connection:

$$\mathcal A=\mathcal A_i dk^i.$$

• Gradient of a function:

$$df=\frac{\partial f}{\partial x^\mu}dx^\mu.$$

Two-forms and the wedge product

A two-form is locally written

$$\omega=\frac12\omega_{\mu\nu}\,dx^\mu\wedge dx^\nu,$$

where

$$\omega_{\mu\nu}=-\omega_{\nu\mu}.$$

The wedge product is antisymmetric:

$$dx^\mu\wedge dx^\nu=-dx^\nu\wedge dx^\mu.$$

In particular,

$$dx^\mu\wedge dx^\mu=0.$$

If $\alpha$ is a $p$-form and $\beta$ is a $q$-form, then

$$\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha.$$

Exterior derivative

The exterior derivative

$$d:\Omega^k(M)\to\Omega^{k+1}(M)$$

raises the degree of a form by one.

For a function $f$,

$$df=\frac{\partial f}{\partial x^\mu}dx^\mu.$$

For a one-form $\alpha=\alpha_\mu dx^\mu$,

$$d\alpha=\frac{\partial \alpha_\nu}{\partial x^\mu}dx^\mu\wedge dx^\nu =\frac12(\partial_\mu\alpha_\nu-\partial_\nu\alpha_\mu)dx^\mu\wedge dx^\nu.$$

The most important identity is

$$d^2=0.$$

In electromagnetism, if

$$A=A_\mu dx^\mu,$$

then the field strength is

$$F=dA.$$

In components,

$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$

Integration of forms

Forms are not defined only by integration, but forms are exactly the objects that can be integrated naturally.

Integrating a one-form over a curve

Let $\gamma:[0,1]\to M$ be a curve. Let

$$\alpha=P(x,y)dx+Q(x,y)dy$$

on $\mathbb{R}^2$. The integral over $\gamma(t)=(x(t),y(t))$ is

$$\int_\gamma \alpha=\int_0^1 \alpha_{\gamma(t)}(\dot\gamma(t))\,dt.$$

Since

$$\dot\gamma(t)=\dot x(t)\frac{\partial}{\partial x}+\dot y(t)\frac{\partial}{\partial y},$$

we get

$$\alpha_{\gamma(t)}(\dot\gamma(t))=P(x(t),y(t))\dot x(t)+Q(x(t),y(t))\dot y(t).$$

So

$$\int_\gamma \alpha =\int_0^1\left[P(x(t),y(t))\dot x(t)+Q(x(t),y(t))\dot y(t)\right]dt.$$

This is the familiar line integral.

The general definition of integrating a form

The sentence “a $k$-form can be integrated over a $k$-dimensional object” is not meant to be automatic from the fact that a form is a multilinear dual object. One still has to define the operation of integration. The definition uses three ingredients:

1. an orientation, which tells us what counts as a positive coordinate volume element;

2. pullback, which moves the form to a parameter domain in Euclidean space;

3. the ordinary multiple integral of a function against $du^1\cdots du^k$.

First consider one parametrized $k$-dimensional piece. Let

$$\sigma:U\subset \mathbb{R}^k\to M$$

be a smooth parametrization, and let $\omega\in\Omega^k(M)$. Pull back $\omega$ to $U$:

$$\sigma^*\omega\in\Omega^k(U).$$

Since $U\subset\mathbb{R}^k$ has coordinates $(u^1,\ldots,u^k)$, every $k$-form on $U$ has the form

$$\sigma^*\omega=f(u^1,\ldots,u^k)\,du^1\wedge\cdots\wedge du^k.$$

Then the integral over the parametrized piece is defined by

$$\boxed{ \int_\sigma \omega := \int_U f(u^1,\ldots,u^k)\,du^1\cdots du^k. }$$

Thus the wedge product is not itself the same thing as the measure $du^1\cdots du^k$; rather, once a top-degree form is written as a coefficient times the oriented coordinate form, we integrate the coefficient by ordinary calculus.

For an oriented $k$-dimensional manifold $N$ and a smooth map

$$F:N\to M,$$

one defines

$$\int_N F^*\omega$$

by choosing oriented charts on $N$ and using a partition of unity. More explicitly, if $\{(V_\alpha,\varphi_\alpha)\}$ is an oriented atlas for $N$ and $\{\rho_\alpha\}$ is a smooth partition of unity subordinate to the cover $\{V_\alpha\}$, then

$$\boxed{ \int_N F^*\omega = \sum_\alpha \int_{\varphi_\alpha(V_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha F^*\omega). }$$

Each summand is an ordinary integral of a function on an open subset of $\mathbb{R}^k$. This definition is independent of the oriented atlas and partition of unity. The independence is exactly the change-of-variables theorem from multivariable calculus.

If $N$ is noncompact, the integral is automatically defined for compactly supported forms. For forms without compact support, convergence must be checked. In most physics examples here, the domain is compact, such as $S^1$, $S^2$, or the Brillouin torus $T^2$, or the form has suitable decay.

Integrating a two-form over a surface

Let $\Sigma$ be a two-dimensional surface with local parameters $(u,v)$. A two-form can be integrated over $\Sigma$ by pulling it back to the $(u,v)$ parameter domain and integrating the coefficient of $du\wedge dv$.

This is the conceptual reason pullback matters: it converts forms on the target space into forms on the parameter space where we know how to integrate.

Pullback

Let

$$f:M\to N$$

be a smooth map. Pullback takes differential forms on $N$ and produces differential forms on $M$:

$$f^*:\Omega^k(N)\to\Omega^k(M).$$

For a function $h:N\to\mathbb{R}$,

$$f^*h=h\circ f.$$

For a one-form, use the rule

$$f^*(dy^i)=d(f^i)=\frac{\partial f^i}{\partial x^a}dx^a.$$

Thus, if

$$\alpha=\alpha_i(y)dy^i$$

on $N$, then

$$f^*\alpha=\alpha_i(f(x))\frac{\partial f^i}{\partial x^a}dx^a.$$

Pullback respects the two main operations:

$$f^*(\alpha\wedge\beta)=f^*\alpha\wedge f^*\beta,$$ $$f^*(d\alpha)=d(f^*\alpha).$$

Flows, Pullbacks, Pushforwards, and Lie Derivatives

Lie derivatives combine three earlier ideas: vector fields, flows, and pullback/pushforward. We first give the precise definitions and then allow ourselves the standard shorthand.

Flows of vector fields

Let $X\in\Gamma(TM)$ be a smooth vector field. A flow is a family of maps that moves points according to $X$.

For small enough $t$, $\Phi_t$ is a diffeomorphism onto its image. Where both sides are defined,

$$\Phi_{t+s}=\Phi_t\circ\Phi_s, \qquad \Phi_0=\mathrm{id}_M.$$

If the flow exists for all $t\in\mathbb{R}$ and all $p\in M$, the vector field is called complete. Compact manifolds often make vector fields complete, but compactness is not part of the definition.

In local coordinates,

$$X=X^\mu(x)\partial_\mu,$$

the flow equation becomes the ordinary differential equation

$$\frac{d x^\mu(t)}{dt}=X^\mu(x(t)), \qquad x^\mu(0)=x^\mu(p).$$

This coordinate ODE is a representation of the intrinsic equation $\dot\gamma_p(t)=X_{\gamma_p(t)}$.

Conversely, a one-parameter family of local diffeomorphisms $\Phi_t$ satisfying $\Phi_0=\mathrm{id}_M$ defines a vector field by

$$X_p= \left.\frac{d}{dt}\right|_{t=0}\Phi_t(p) := (c_p)_{*0}\left(\left.\frac{d}{dt}\right|_{0}\right),$$

where $c_p(t)=\Phi_t(p)$. Thus the slogan “differentiate the flow to get the vector field” means “push forward the canonical tangent vector $d/dt$ along the trajectory curve.”

Pullback and pushforward under diffeomorphisms

A diffeomorphism

$$\phi:M\to M$$

acts on functions by pullback:

$$(\phi^*f)(p)=f(\phi(p)).$$

It acts on tangent vectors by pushforward:

$$\phi_{*p}:T_pM\to T_{\phi(p)}M.$$

For a one-form $\alpha$, the pullback is defined by

$$(\phi^*\alpha)_p(v)=\alpha_{\phi(p)}(\phi_{*p}v), \qquad v\in T_pM.$$

For a $k$-form $\omega$,

$$(\phi^*\omega)_p(v_1,\ldots,v_k) = \omega_{\phi(p)}(\phi_{*p}v_1,\ldots,\phi_{*p}v_k).$$

A vector field can also be pulled back by a diffeomorphism, but one must use the inverse pushforward:

$$(\phi^*Y)_p = (\phi^{-1})_{*\phi(p)}\bigl(Y_{\phi(p)}\bigr) \in T_pM.$$

Equivalently,

$$(\phi^*Y)(f)=\phi^*\left(Y\bigl((\phi^{-1})^*f\bigr)\right).$$

This formula is often the most useful one in proofs.

Tensor fields and notation

A type $(r,s)$ tensor at $p$ is an element of

$$(T_pM)^{\otimes r}\otimes (T_p^*M)^{\otimes s}.$$

A smooth tensor field is a smooth assignment of such a tensor to every point. Examples are:

$$\text{functions }(0,0), \quad \text{vector fields }(1,0), \quad \text{one-forms }(0,1), \quad \text{metrics }(0,2).$$

A differential $k$-form is an antisymmetric type $(0,k)$ tensor field. After vector bundles are introduced, we write

$$\Omega^k(M):=\Gamma(\Lambda^kT^*M)$$

for the space of smooth $k$-forms.

General definition of the Lie derivative

Let $X$ be a vector field with local flow $\Phi_t$. If $T$ is an ordinary tensor field on $M$, define

$$\boxed{ \mathcal{L}_XT= \left.\frac{d}{dt}\right|_{t=0}\Phi_t^*T. }$$

For forms this uses the usual pullback. For vector fields and mixed tensors, $\Phi_t^*$ uses inverse pushforwards in the vector slots, as above.

The Lie derivative is canonical: it uses only the smooth structure and the vector field $X$. It does not use a metric or a connection.

Functions

For $f\in C^\infty(M)$,

$$\mathcal{L}_Xf = \left.\frac{d}{dt}\right|_{0} f(\Phi_t(p)) = X_p[f].$$

In coordinates,

$$\mathcal{L}_X f=X^\mu\partial_\mu f.$$

Vector fields and the Lie bracket

For a vector field $Y$,

$$(\Phi_t^*Y)_p=(\Phi_{-t})_{*\Phi_t(p)}Y_{\Phi_t(p)}.$$

The Lie derivative is

$$\mathcal{L}_XY= \left.\frac{d}{dt}\right|_{0}\Phi_t^*Y.$$

We now derive the standard formula

$$\boxed{\mathcal{L}_XY=[X,Y].}$$

Let $f\in C^\infty(M)$. Using the function-action formula for pullback of a vector field,

$$(\Phi_t^*Y)(f) = \Phi_t^*\left(Y(\Phi_{-t}^*f)\right).$$

Differentiate at $t=0$:

$$\begin{align*} (\mathcal{L}_XY)(f) &=\left.\frac{d}{dt}\right|_{0} \Phi_t^*\left(Y(\Phi_{-t}^*f)\right)\\ &=X[Y(f)] + Y\left(\left.\frac{d}{dt}\right|_0 \Phi_{-t}^*f\right)\\ &=X[Y(f)]-Y[X(f)]. \end{align*}$$

Therefore

$$\mathcal{L}_XY=[X,Y], \qquad [X,Y](f):=X(Y(f))-Y(X(f)).$$

In local coordinates,

$$[X,Y] = \left(X^\nu\partial_\nu Y^\mu-Y^\nu\partial_\nu X^\mu\right)\partial_\mu.$$

Differential forms and Cartan’s formula

For a $k$-form $\omega$,

$$\mathcal{L}_X\omega= \left.\frac{d}{dt}\right|_{0}\Phi_t^*\omega.$$

The interior product $\iota_X\omega$ is the $(k-1)$-form obtained by inserting $X$ into the first slot:

$$(\iota_X\omega)(Y_1,\ldots,Y_{k-1}) = \omega(X,Y_1,\ldots,Y_{k-1}).$$

For a one-form $\alpha=\alpha_\mu dx^\mu$,

$$(\mathcal{L}_X\alpha)_\mu = X^\nu\partial_\nu\alpha_\mu+ \alpha_\nu\partial_\mu X^\nu.$$

For a coordinate one-form,

$$\mathcal{L}_X(dx^\mu)=d(X^\mu).$$

What the Lie derivative is not

The Lie derivative is a canonical operation on ordinary tensor fields on $M$. It is not automatically defined for a section of an arbitrary vector bundle $E\to M$. To Lie-differentiate such a section, the flow of $X$ on $M$ must be lifted to a flow on $E$, which is extra structure. A connection, introduced later, is the standard tool for differentiating sections of arbitrary vector bundles.