Differential Forms

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Dual space

Let $E$ be a finite dimensional vector space.

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Cotangent space

Let $M$ be a manifold. For $p \in M$, we define the cotangent space at $p$ to be the dual of the tangent space, i.e. \(T_p^* M = (T_p M)^*\) Elements of the cotangent space are called cotangent vectors or covectors.

For $F \in C^\infty(M, N)$, the cotangent map is the dual of the tangent map: \(\begin{align*} T_p^* F : T_{F(p)}^* N &\to T_p^* M \\ g &\to (T_p F)^* g \end{align*}\)

Differential of $f$ at $p$

For $f \in C^\infty(M)$, we define the differential of $f$ at $p$ to be the element $(df)_p \in T_p^M$ defined by $(df)_p (v) = v(f)$ for $v \in T_p M$.

Differential commutes with pullback (primitive)

For $F \in C^\infty(M, N)$ and $g \in C^\infty(N)$, we have \(d(F^* g)\big\vert_p = T^* F ((dg)_{F(p)})\)

1-forms

A 1-form on $M$ is a map \(\begin{align*} \alpha : \X(M) &\to C^\infty(M) \\ X &\mapsto \alpha(X) = \angles{\alpha, X} \end{align*}\) This is $C^\infty(M)$-linear, i.e. if $f, g \in C^\infty(M)$ and $X, Y \in \X(M)$, then \(\alpha(fX + gY) = f\alpha(X) + g\alpha(Y)\) The vector space of 1-forms on $M$ is denoted $\Omega^1(M)$.

1-forms and cotangent vectors

For any $\alpha \in \Omega^1(M)$ and $p \in M$, there exists $\alpha_p \in T_p^*(M)$ so that for all $X \in \X(M)$, \(\alpha(X)_p = \alpha_p(X_p)\) Proof.

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Exterior derivative of a 1-form

If $f \in C^\infty(M)$, the exterior derivative of $f$ is the 1-form $df \in \Omega^1(M)$ defined by $\angles{df, X} = X(f)$.

• if $\alpha \in \Omega^1(M)$ and $U$ is an open subset of $M$, the restriction of $\alpha$ to $U$ is a 1-form on $U$, and $\left(a \big\vert_U\right)_p = a_p$ for $p \in U$.
• if $U \subseteq \R^m$ is open and $\alpha \in \Omega^1(U)$, then $\ds \alpha = \sum_i \alpha_i dx^i$ where $\ds \alpha_i = \alpha\left(\frac{\partial}{\partial x^i}\right)$