differential-formsdifferential-geometryreference-request
I want to get used to differential forms. Thus I would like to solve a bunch of problems, especially on integration of differential forms.
So I need a collection of problems with answers/solutions, starting from really elementary ones. No theorem proving, just straightforward calculations.
I just want to point out that GA can be used to make covariant multivectors (or differential forms) on $\mathbb R^n$ without forcing a metric onto it. In other words, the distinction between vectors and covectors (or between $\mathbb R^n$ and its dual) can be maintained.
This is done with a pseudo-Euclidean space $\mathbb R^{n,n}$.
Take an orthonormal set of spacelike vectors $\{\sigma_i\}$ (which square to ${^+}1$) and timelike vectors $\{\tau_i\}$ (which square to ${^-}1$). Define null vectors
$$\Big\{\nu_i=\frac{\sigma_i+\tau_i}{\sqrt2}\Big\}$$
$$\Big\{\mu_i=\frac{\sigma_i-\tau_i}{\sqrt2}\Big\};$$
they're null because
$${\nu_i}^2=\frac{{\sigma_i}^2+2\sigma_i\cdot\tau_i+{\tau_i}^2}{2}=\frac{(1)+2(0)+({^-}1)}{2}=0$$
$${\mu_i}^2=\frac{{\sigma_i}^2-2\sigma_i\cdot\tau_i+{\tau_i}^2}{2}=\frac{(1)-2(0)+({^-}1)}{2}=0.$$
More generally,
$$\nu_i\cdot\nu_j=\frac{\sigma_i\cdot\sigma_j+\sigma_i\cdot\tau_j+\tau_i\cdot\sigma_j+\tau_i\cdot\tau_j}{2}=\frac{(\delta_{i,j})+0+0+({^-}\delta_{i,j})}{2}=0$$
and
$$\mu_i\cdot\mu_j=0.$$
So the spaces spanned by $\{\nu_i\}$ or $\{\mu_i\}$ each have degenerate quadratic forms. But the dot product between them is non-degenerate:
$$\nu_i\cdot\mu_i=\frac{\sigma_i\cdot\sigma_i-\sigma_i\cdot\tau_i+\tau_i\cdot\sigma_i-\tau_i\cdot\tau_i}{2}=\frac{(1)-0+0-({^-}1)}{2}=1$$
$$\nu_i\cdot\mu_j=\frac{\sigma_i\cdot\sigma_j-\sigma_i\cdot\tau_j+\tau_i\cdot\sigma_j-\tau_i\cdot\tau_j}{2}=\frac{(\delta_{i,j})-0+0-({^-}\delta_{i,j})}{2}=\delta_{i,j}$$
Of course, we could have just started with the definition that $\mu_i\cdot\nu_j=\delta_{i,j}=\nu_i\cdot\mu_j$, and $\nu_i\cdot\nu_j=0=\mu_i\cdot\mu_j$, instead of going through "spacetime".
The space $V$ will be generated by $\{\nu_i\}$, and its dual $V^*$ by $\{\mu_i=\nu^i\}$. You can take the dot product of something in $V^*$ with something in $V$, which will be a differential 1-form. You can make contravariant multivectors from wedge products of things in $V$, and covariant multivectors from wedge products of things in $V^*$.
You can also take the wedge product of something in $V^*$ with something in $V$.
$$\mu_i\wedge\nu_i=\frac{\sigma_i\wedge\sigma_i+\sigma_i\wedge\tau_i-\tau_i\wedge\sigma_i-\tau_i\wedge\tau_i}{2}=\frac{0+\sigma_i\tau_i-\tau_i\sigma_i-0}{2}=\sigma_i\wedge\tau_i$$
$$\mu_i\wedge\nu_j=\frac{\sigma_i\sigma_j+\sigma_i\tau_j-\tau_i\sigma_j-\tau_i\tau_j}{2},\quad i\neq j$$
What does this mean? ...I suppose it could be a matrix (a mixed variance tensor)!
A matrix can be defined as a bivector:
$$M = \sum_{i,j} M^i\!_j\;\nu_i\wedge\mu_j = \sum_{i,j} M^i\!_j\;\nu_i\wedge\nu^j$$
where each $M^i_j$ is a scalar. Note that $(\nu_i\wedge\mu_j)\neq{^-}(\nu_j\wedge\mu_i)$, so $M$ is not necessarily antisymmetric. The corresponding linear function $f:V\to V$ is (with $\cdot$ the "fat dot product")
$$f(x) = M\cdot x = \frac{Mx-xM}{2}$$
$$= \sum_{i,j} M^i_j(\nu_i\wedge\mu_j)\cdot\sum_k x^k\nu_k$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i\mu_j-\mu_j\nu_i}{2}\cdot\nu_k$$
$$= \sum_{i,j,k} M^i_jx^k\frac{(\nu_i\mu_j)\nu_k-\nu_k(\nu_i\mu_j)-(\mu_j\nu_i)\nu_k+\nu_k(\mu_j\nu_i)}{4}$$
(the $\nu$'s anticommute because their dot product is zero:)
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i\mu_j\nu_k+\nu_i\nu_k\mu_j+\mu_j\nu_k\nu_i+\nu_k\mu_j\nu_i}{4}$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\mu_j\nu_k+\nu_k\mu_j)+(\mu_j\nu_k+\nu_k\mu_j)\nu_i}{4}$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\mu_j\cdot\nu_k)+(\mu_j\cdot\nu_k)\nu_i}{2}$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\delta_{j,k})+(\delta_{j,k})\nu_i}{2}$$
$$= \sum_{i,j,k} M^i_jx^k\big(\delta_{j,k}\nu_i\big)$$
$$= \sum_{i,j} M^i_jx^j\nu_i$$
This agrees with the conventional definition of matrix multiplication.
In fact, it even works for non-square matrices; the above calculations work the same if the $\nu_i$'s on the left in $M$ are basis vectors for a different space. A bonus is that it also works for a non-degenerate quadratic form; the calculations don't rely on ${\mu_i}^2=0$, nor ${\nu_i}^2=0$, but only on $\nu_i$ being orthogonal to $\nu_k$, and $\mu_j$ being reciprocal to $\nu_k$. So you could instead have $\mu_j$ (the right factors in $M$) be in the same space as $\nu_k$ (the generators of $x$), and $\nu_i$ (the left factors in $M$) in a different space. A downside is that it won't map a non-degenerate space to itself.
I admit that this is worse than the standard matrix algebra; the dot product is not invertible, nor associative. Still, it's good to have this connection between the different algebras. And it's interesting to think of a matrix as a bivector that "rotates" a vector through the dual space and back to a different point in the original space (or a new space).
Speaking of matrix transformations, I should discuss the underlying principle for "contra/co variance": that the basis vectors may vary.
We want to be able to take any (invertible) linear transformation of the null space $V$, and expect that the opposite transformation applies to $V^*$. Arbitrary linear transformations of the external $\mathbb R^{n,n}$ will not preserve $V$; the transformed $\nu_i$ may not be null. It suffices to consider transformations that preserve the dot product on $\mathbb R^{n,n}$. One obvious type is the hyperbolic rotation
$$\sigma_1\mapsto\sigma_1\cosh\phi+\tau_1\sinh\phi={\sigma_1}'$$
$$\tau_1\mapsto\sigma_1\sinh\phi+\tau_1\cosh\phi={\tau_1}'$$
$$\sigma_2={\sigma_2}',\quad\sigma_3={\sigma_3}',\quad\cdots$$
$$\tau_2={\tau_2}',\quad\tau_3={\tau_3}',\quad\cdots$$
(or, more compactly, $x\mapsto\exp(-\sigma_1\tau_1\phi/2)x\exp(\sigma_1\tau_1\phi/2)$ ).
The induced transformation of the null vectors is
$${\nu_1}'=\frac{{\sigma_1}'+{\tau_1}'}{\sqrt2}=\exp(\phi)\nu_1$$
$${\mu_1}'=\frac{{\sigma_1}'-{\tau_1}'}{\sqrt2}=\exp(-\phi)\mu_1$$
$${\nu_2}'=\nu_2,\quad{\nu_3}'=\nu_3,\quad\cdots$$
$${\mu_2}'=\mu_2,\quad{\mu_3}'=\mu_3,\quad\cdots$$
The vector $\nu_1$ is multiplied by some positive number $e^\phi$, and the covector $\mu_1$ is divided by the same number. The dot product is still ${\mu_1}'\cdot{\nu_1}'=1$.
You can get a negative multiplier for $\nu_1$ simply by the inversion $\sigma_1\mapsto{^-}\sigma_1,\quad\tau_1\mapsto{^-}\tau_1$; this will also negate $\mu_1$. The result is that you can multiply $\nu_1$ by any non-zero Real number, and $\mu_1$ will be divided by the same number.
Of course, this only varies one basis vector in one direction. You could try to rotate the vectors, but a simple rotation in a $\sigma_i\sigma_j$ plane will mix $V$ and $V^*$ together. This problem is solved by an isoclinic rotation in $\sigma_i\sigma_j$ and $\tau_i\tau_j$, which causes the same rotation in $\nu_i\nu_j$ and $\mu_i\mu_j$ (while keeping them separate).
Combine these stretches, reflections, and rotations, and you can generate any invertible linear transformation on $V$, all while maintaining the degeneracy ${\nu_i}^2=0$ and the duality $\mu_i\cdot\nu_j=\delta_{i,j}$. This shows that $V$ and $V^*$ do have the correct "variance".
See also Hestenes' Tutorial, page 5 ("Quadratic forms vs contractions").
You should check Elementary topology problem textbook, freely available in the English noproof version. An English 2nd ed. with proofs is for sale, but the first ed. (with proofs) is also freely available here (toward the bottom).
Part 2 deals with algebraic topology
Best Answer
You may like Chapter 10 "Differential Forms, Integral Formulae, De-Rham Cohomology" of Mishchenko, Solovyev, Fomenko "Problems in Differential Geometry and Topology" (English translation, Mir Publishers, 1985). I have not seen a newer version of it that may be even better: A.T. Fomenko, A.S. Mischenko, Y.P. Solovyev: Selected problems in differential geometry and topology. Cambridge Scientific Publishers, Cambridge, 2006
Update 1. There is also a very interesting collection of problems by Prof. W.-H. Steeb. See Ch.4 Differential Forms and Applications in "Problems and Solutions in Differential Geometry"
Update 2. A comprehensive set of problems on differential geometry can be found in Analysis and Algebra on Differentiable Manifolds: A Workbook for Students, by P. M. Gadea, J. Munoz Masqué, see Ch.2 "Tensor Fields and Differential Forms".