Analysis Theorems

12 Exterior Algebras and Differential Forms

Definition 12.1. If \(M\) is a finite rank free \(R\)-module with an ordered basis \(m_1\dots m_n\), an elememt \(\omega \in \bigwedge ^r(M)\) is in the reduced form if it is written as

\[\sum _{1\leq i_1<\dots < i_r\leq n} a_{i_1\dots i_r}m_{i_1}\wedge \dots \wedge m_{i_r}\]

We will use multi-index notation writing \(\sum _Ia_Im_I\) for unreduced form, and \(\sum _I'a_Im_I\) for reduced form.

Definition 12.2. If \(M\) is a finite rank free \(R\)-module with an ordered basis \(x_1\dots x_n\), the Hodge duality map denoted \(*\) is defined as the linear map sending \(x_I\in \bigwedge ^rM\) to \(\ee (IJ)x_J\in \bigwedge ^{n-r}M\) where \(\ee \) is the sign of the permutation, and \(I,J\) are reduced.

Lemma 12.3. The Hodge dual satisfies \(*(*\omega ) = (-1)^{n(n-r)}\omega \).

Proof. It suffices to show this on \(dx_I\), that \(\ee (IJ)\ee (JI) = (-1)^{n(n-r)}\), but this is obvious by moving entries in \(J\) one at a time across as transpositions. □

Lemma 12.4. If \(\omega ,\sigma \in \bigwedge ^{r},\omega = \sum _I'f_Idx_I, \sigma = \sum _I'g_Idx_I\) then \(\omega \wedge *\sigma = \sum _If_Ig_Idx_1\wedge \dots \wedge dx_n\).

Proof. All the terms in this wedge are \(0\) except the ones where the \(f_I\) and \(g_I\) correspond, in which case the sign is \(\ee (IJ)\ee (IJ) = 1\) □

We will work in the smooth category of \(\cC ^\infty \) functions for simplicity, but if one is careful it is possible to treat \(\cC ^r\) functions. We think of the de Rham complex \(\bigwedge (D)\) of \(D\subset \RR ^n\) as the exterior algebra of the \(\cC ^\infty (D)\)-module of smooth sections of the cotangent bundle \(T^*(D)\), where \(\bigwedge ^r(D)\) is a free module of rank \(\binom n r\). Elements of \(\bigwedge ^r(D)\) are called differential r-forms. Since here \(D\) is in \(\RR ^n\), by choosing an orientation \(x_1\dots x_n\) of \(\RR ^n\), we determine a basis for \(\bigwedge ^1(D)\), \(dx_1\dots dx_n\), namely \(dx_1\) is the section taking every point to the projection onto the \(x_i\) coordinate (recall that we have a canonical identification of the tangent bundle with \(\RR ^n\)).

The de Rham complex is a cochain complex because it has a differential \(d\), the exterior derivative.

Definition 12.5. The exterior derivative is defined by

\[d\bigg (\sum _If_Idx_I\bigg ) = \sum _Idf_Idx_I\]

where \(df\) is the dual of \(f'\) written in the basis of the \(dx\)s, ie \(df = \sum _i\partial _idx_i\)

Note that \(d(x_i) = dx_i\) where \(x_i\) is the \(i^{th}\) coordinate function (indeed this is the meaning of \(dx_i\)). The cohomology of the de Rham complex is called de Rham cohomology. The exterior derivative is characterized by the following properties:

Lemma 12.6. The exterior derivative \(d\) is an \(\RR \)-linear map satisfying \(d(f) = df,(\omega \wedge \sigma ) = d\omega \wedge \sigma + (-1)^r \omega \wedge d\sigma \) and \(dd\omega = 0\) where \(\omega \in \bigwedge ^r(D), \sigma \in \bigwedge ^k(D)\). Moreover it is characterized by these properties.

Proof. \(d\) is clearly linear and satisfies the first property by definition. The second follows from the product rule with the \((-1)^r\) coming from moving the derivative of each \(g\) in the \(\sigma \) to the right place. For the third, it suffices to show that \(ddf = 0\) for a function, but this is true by Theorem 4.10 since \(ddf = \sum _{i,j}\partial _i\partial _jdx_idx_j = 0\). To see that \(d\) is characterized by these properties, note that by induction \(d(dx_I) = 0\), so this allows us to compute \(d(f\wedge dx_I)\) as defined. □

Now when we have a \(\cC ^\infty \) map \(\phi :D \to E\), we would like to be able to pullback differential forms to get a map of cochain complexes, \(\phi ^*:\bigwedge (E)\to \bigwedge ^r(D)\). The pullback will serve as the chain map that turns de Rham cohomology into a functor. To define this, note that we have the pushforward, a map \(d\phi :TD\to TE\), and by taking its dual (transpose) and exterior algebra, this gives a map \(\bigwedge (d\phi )^*:\bigwedge T^*E \to \bigwedge T^*D\). We can define the pullback as the composite in the diagram below:

(-tikz- diagram)

How do we compute the pullback of \(\omega \)? Well first write \(\omega = \sum _If_Idx_I\). The \(f_I\) get sent by \(\phi \) to \(f\circ \phi \), and since \((d\phi )^*\) is \(\cC ^\infty (E)\)-linear on sections, we only need to find out what happens to the \(dx_i\). To do this note that \(d\phi \) sends \((a,v)\mapsto (\phi (a),\phi '(a)v)\), which \(dx_i(\phi (a))\) sends to \(\pi _{x_i}(\phi '(a))v = \phi _i'(a)v\) which is exactly what \(\sum _j\partial _j\phi _idy_j\) does, so \(dx_i \mapsto \sum _j\partial _j\phi _idy_j\), and we can compute the pullback as

\[\phi ^*(fdy_1\wedge \dots \wedge dy_r) = \sum _J(f\circ \phi )\partial _{j_1}\phi _{i_1}\dots \partial _{j_r}\phi _{i_r}dx_{j_1}\wedge \dots \wedge dx_{j_r}\]

which is abbreviated as

\[\phi ^*(f_I(y)dy_I) = f_I(\phi (y))\sum _J\partial _J\phi _Idx_J\]

Lemma 12.7. Pullback is a contravariant functor and is a map of cochain complexes.

Proof. Let’s consider the pullback of a map \(\phi :D\to E\), \(D \subset \RR ^m\), \(E\subset \RR ^n\). The pullback is the composite of maps along the functor \(\bigwedge (d(-))^*\), which is a functor by the chain rule, so it is a functor. By definition pullback is a map of algebras, and so we must check it commutes with the exterior derivative. For any function \(f:E\to \RR \) we have by the chain rule \(\phi ^*(df) = \phi ^*(\sum _i\partial _ifdx_i) = \sum _i(\partial _if \circ \phi )(\sum _j\partial _j\phi _idy_j)\) \(= \sum _j\sum _i(\partial _jf\circ \phi )\partial _j\phi _idy_j = \sum _i\partial _i(f\circ \phi )dy_i= d(f\circ \phi ) = d(\phi ^*(f))\). Now by linearity we only need commutativity on \(\omega = fdx_I\) so

\[\phi ^*(d\omega ) = \phi ^*(df\wedge dx_I) = \phi ^*(df)\wedge \phi ^*(dx_I) = d\phi ^*(f_I)\wedge \phi ^*(dx_I)\]

\[ = d(\phi ^*(f_I)\phi ^*(dx_I)) = d(\phi ^*(\omega ))\]

□

Theorem 12.8 (Poincaré Lemma). The de Rham cohomology of a cell is \(\RR \) in dimension \(0\) and \(0\) elsewhere.

Proof. We assume the cell is a unit cube. That the cohomology is \(\RR \) in dimension \(0\) follows from the fact that the derivative is \(0\) iff the function is constant, and since the cell is connected, there the function must globally be constant. For elsewhere, we consider a closed differential \(r\)-form and show it is exact. To do this we induct on the largest \(k\) so that \(dx_k\) is used in the form. For the base case, if \(\omega =fdx_1\) then \(dg = \omega \) where \(g = \int _0^{x_1}f\). Now to induct on \(k\), it suffices by linearity and induction to consider a \(r\)-form of the sort \(\omega = fdx_I\wedge dx_k\) where \(dx_I\) only use \(dx_1\dots dx_{k-1}\). As it is closed, we have

\[0 = d\omega = \sum _{j=1}^n\partial _jfdx_j\wedge dx_I\wedge x_k \]

so \(\partial _jf\) must be \(0\) for \(j>k\), so \(f\) is constant in those directions, so we define \(h = \int _0^{x_{k}}f\) so that \(\partial _{k}h = f\), and setting \(\omega ' = hdx_I\) and changing \(\omega \) by \((-1)^{r-1}d\sigma \) we are done by induction. □