Elie Cartan
1. Differential Forms and the Wedge Product
Antisymmetric tensors are very important. There are a very special class of antisymmetric covariant tensors called differential forms. They are objects you have no doubt seen before. In fact, one of the simplest cases if \(f’(x) dx\). That might give you a clue to what we do to them - integrate! We will get to that in a moment. First, what is a differential form? A \(k\)-form is a \((0,k)\) tensor that is totally antisymmetric: swap any two arguments and it changes sign.
\[ \omega(\ldots, X, \ldots, Y, \ldots) = -\omega(\ldots, Y, \ldots, X, \ldots) \tag{1} \]
In components, \(\omega_{\mu_1 \ldots \mu_k}\) flips sign under exchange of any two indices. That's a strong condition, and several properties immediately follow. If two indices are equal, antisymmetry says the component equals its own negative, so it's zero. Only components with all indices distinct can be nonzero. And once you know the component for one ordering, every reordering is determined by the sign of the permutation. So a \(k\)-form is fixed by its values on increasing index sequences \(\mu_1 < \mu_2 < \cdots < \mu_k\), and the number of those is
\[ \binom{n}{k} \tag{2} \]
Compare with a general \((0,k)\) tensor, which has \(n^k\) components. In four dimensions a general \((0,3)\) tensor has 64 components; a 3-form has 4. Write the counts out for \(n = 3\):
\[ 1, \quad 3, \quad 3, \quad 1 \tag{3} \]
and for \(n = 4\):
\[ 1, \quad 4, \quad 6, \quad 4, \quad 1 \tag{4} \]
Pascal's triangle, which is what \(\binom{n}{k}\) always gives. The total across all \(k\) is \(2^n\). Two things. First, there are no \(k\)-forms for \(k > n\), because you can't choose \(k\) distinct indices out of \(n\). Second, look at the middle of the \(n = 4\) row: a 2-form in four dimensions has exactly six independent components. Three electric and three magnetic! The electromagnetic field tensor \(F_{\mu\nu}\) is antisymmetric, and that is why \(\mathbf{E}\) and \(\mathbf{B}\) have three components each. The \(n=3\) row explains a different coincidence. There, 1-forms and 2-forms both have three components. That is why you can get away with treating the curl of a vector field as another vector field. It's really a 2-form, and in any other dimension this does NOT work.
1.1 Building a product
We want a way to multiply forms. The tensor product \(\alpha \otimes \beta\) won't do, because it isn't antisymmetric, so the result isn't a form. The fix is to antisymmetrise the product by hand. For two one-forms, define
\[ (\alpha \wedge \beta)(X, Y) = \alpha(X)\beta(Y) - \alpha(Y)\beta(X) \tag{5} \]
Swapping \(X\) and \(Y\) gives \(\alpha(Y)\beta(X) - \alpha(X)\beta(Y)\), which is minus the original. So \(\alpha \wedge \beta\) is a 2-form. In components, feeding basis vectors,
\[ (\alpha \wedge \beta)_{\mu\nu} = \alpha_\mu \beta_\nu - \alpha_\nu \beta_\mu \tag{6} \]
which is a \(2\times2\) determinant… That’s the Jacobian! Two properties follow. Swapping the forms rather than the arguments,
\[ \alpha \wedge \beta = -\beta \wedge \alpha \tag{7} \]
and setting \(\beta = \alpha\),
\[ \alpha \wedge \alpha = 0 \tag{8} \]
For higher degrees the definition extends by requiring bilinearity and associativity, which works nicely once you know it on basis one-forms. So a \(k\)-form has the basis
\[ dx^{\mu_1} \wedge dx^{\mu_2} \wedge \cdots \wedge dx^{\mu_k}, \qquad \mu_1 < \mu_2 < \cdots < \mu_k \tag{9} \]
with the ordering condition avoiding double-counting, since any other order is \(\pm\) one of these.
1.2 The sign rule
The general commutation rule. For a \(p\)-form \(\alpha\) and a \(q\)-form \(\beta\),
\[ \alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha \tag{10} \]
The reason is bookkeeping. Write each side in terms of basis one-forms. To move \(\beta\)'s block of \(q\) one-forms in front of \(\alpha\)'s block of \(p\), each of the \(q\) must pass each of the \(p\), costing a sign each time. That's \(pq\) transpositions. Now the issue. Everyone remembers \(\alpha \wedge \alpha = 0\), but setting \(\beta = \alpha\) above gives \(\alpha \wedge \alpha = (-1)^{p^2} \alpha \wedge \alpha\), and \((-1)^{p^2} = (-1)^p\). So the argument only forces \(\alpha \wedge \alpha = 0\) when \(p\) is odd. For even-degree forms the wedge with itself can be nonzero. In four dimensions take
\[ \omega = dx^1 \wedge dx^2 + dx^3 \wedge dx^4 \tag{11} \]
and wedge it with itself. The terms \(dx^1 \wedge dx^2 \wedge dx^1 \wedge dx^2\) and \(dx^3 \wedge dx^4 \wedge dx^3 \wedge dx^4\) vanish, since each repeats an index. The cross terms survive, and moving a 2-form past a 2-form needs \((-1)^{2 \cdot 2} = +1\), so they add rather than cancel:
\[ \omega \wedge \omega = 2 \, dx^1 \wedge dx^2 \wedge dx^3 \wedge dx^4 \tag{12} \]
Not zero. If you set \(\omega = F\), the electromagnetic 2-form, then \(F \wedge F\) is proportional to \(\mathbf{E} \cdot \mathbf{B}\). This is a very interesting object - the term that shows up in discussions of the strong CP problem and instantons. That’s for another post though.
1.3 The top of the tower
The case \(k = n\) is special. There, \(\binom{n}{n} = 1\), so the space of \(n\)-forms is one-dimensional: every \(n\)-form is a function times
\[ dx^1 \wedge dx^2 \wedge \cdots \wedge dx^n \tag{13} \]
That single fact is actually why determinants exist. Change coordinates, and the new top form must be some multiple of the old one. The multiple is
\[ d\tilde{x}^1 \wedge \cdots \wedge d\tilde{x}^n = \det\left( \frac{\partial \tilde{x}^\mu}{\partial x^\nu} \right) dx^1 \wedge \cdots \wedge dx^n \tag{14} \]
which is the Jacobian.
1.4 An invariant volume
Top forms transform, so no particular one is canonical. But with a metric we can build one that is. From last post, under a change of chart the metric components satisfy \(\tilde{g} = J^T g J\) as matrices, where \(J^\mu{}_\nu = \partial x^\mu / \partial \tilde{x}^\nu\). Taking determinants,
\[ \det \tilde{g} = (\det J)^2 \det g, \qquad \sqrt{|\det \tilde{g}|} = |\det J| \sqrt{|\det g|} \tag{15} \]
Meanwhile the top form picks up \(1/\det J\). Multiply the two together and the Jacobians cancel, leaving only a sign:
\[ \sqrt{|\det \tilde{g}|} \, d\tilde{x}^1 \wedge \cdots \wedge d\tilde{x}^n = \operatorname{sign}(\det J) \sqrt{|\det g|} \, dx^1 \wedge \cdots \wedge dx^n \tag{16} \]
So the volume form
\[ \operatorname{vol} = \sqrt{|\det g|} \, dx^1 \wedge \cdots \wedge dx^n \tag{17} \]
is chart-independent, provided you restrict to coordinate changes with \(\det J > 0\). Making that restriction consistently across the whole manifold is a choice of orientation. Not every manifold has this property though - a Möbius strip doesn't, which is what "non-orientable" means. Let's use it. Take the sphere with the metric derived last post:
\[ g_{\mu\nu} = \frac{4}{(1+u^2+v^2)^2} \begin{pmatrix} 1 & 0 \\\\ 0 & 1 \end{pmatrix} \tag{18} \]
Its determinant is \(16/(1+s)^4\) with \(s = u^2+v^2\), so \(\sqrt{\det g} = 4/(1+s)^2\) and
\[ \operatorname{vol} = \frac{4}{(1+u^2+v^2)^2} \, du \wedge dv \tag{19} \]
Integrate it over the whole chart. In polar coordinates \(u = r\cos\phi\), \(v = r\sin\phi\),
\[ \int \operatorname{vol} = \int_0^{2\pi} \!\! \int_0^\infty \frac{4r}{(1+r^2)^2} \, dr \, d\phi = 8\pi \left[ \frac{-1}{2(1+r^2)} \right]_0^\infty = 4\pi \tag{20} \]
The area of the unit sphere. We integrated over the entire infinite plane and got a finite answer, because the chart stretches the tiny region near the north pole out to infinity while the metric factor shrinks correspondingly. The pole missing from the chart contributes nothing, being a single point.
1.5 Why these are the things you integrate
Whenever you integrate in ordinary calculus and change variables, a Jacobian appears. An \(n\)-form carries it automatically, because that factor is how top forms transform. Integration of \(n\)-forms over \(n\)-dimensional regions is invariant by construction. That follows downward. A 1-form is what you integrate along a curve, which is what line integrals of work are. A 2-form is what you integrate over a surface, which is flux. The degree of the form is the dimension of the thing you integrate it over. So now we have an algebra: forms of each degree, a product raising degree, a sign rule, and a top-level volume form. What we don't have is a derivative. We defined \(df\) for a function, taking a 0-form to a 1-form. So we extend the idea of the derivative!