192
IX – Multivariate Differential and Integral Calculus
ago, laboriously reach the result. In fact, Stokes’ formula is one the hardest
to prove rigorously in an elementary manner; and even using the highest level
of mathematics, no one knows how to exactly define the right category for
“ surfaces ” (of arbitrary dimension) to apply it to; the problem is including
integration domains whose “ edges ” are sufficiently regular for integration to
be possible on it, while including sufficiently general singularities so as not
to exclude important practical or theoretical cases. For example, the border
of a polyhedra is not a “ smooth ” surface ; it has sharp corners and edges.
Excluding such a simple example from the application domain of Stokes’ formula would run counter to common sense, but proving it in a sufficiently
general framework so as to include polyhedrons (in other words, “ simplicial
complexes ” from algebraic topology) poses substantial difficulties since at the
same time the case of perfectly smooth surfaces has to be covered. Physicists
reply that the surface of a tetrahedron is in reality made of four perfectly
smooth triangles and that the edges do not count in the integration, or that
“ corners may be cut ” without significantly changing the result. True, but
they obviously do not have to prove it. It is for good reason that this problem is at the origin of the Bourbaki group; in the 1930s, when its program
was to write a usable treatise for university teaching, a mathematician of
the level of Andre Weil asked Henri Cartan if he knew a good method for
proving the formula, it being agreed that all mathematicians – I used to do
in in 1947 – have always been able to present the type of “ proof ” that satisfy
physicists.
(i) The exterior derivative as an infinitesimal integral. Let us again consider a differential form p i (x)h
i = ω(x; h) of degree 1 and class C
1 on an
open subset G of a Cartesian space E. Take a point x ∈ G, and fix two
vectors h and k and, for given s, t ∈ R, consider the plane parallelogram
P (x, sh, tk) = P , i.e. the set of points of the form x + uh + vk, where u
(resp. v) vary between 0 and s (resp. t) ; assume s and t to be sufficiently
small so that P can be contained in G. When we follow the sides of this
parallelogram in the direction indicated in fig. 2, the border of P is transformed into a closed integration path written ∂P , the boundary of P . Let us
calculate the integral I(x; sh, tk) of ω along this path. On the side connecting
x to x + sh, the parametric representation u → x + uh can be used, which
gives a contribution equal to
s
0
ω(x + uh; h)du .
Contributions from the other sides are calculated in a similar fashion.
I(x; sh, tk) is thus seen to be equal to
IX – Multivariate Differential and Integral Calculus
ago, laboriously reach the result. In fact, Stokes’ formula is one the hardest
to prove rigorously in an elementary manner; and even using the highest level
of mathematics, no one knows how to exactly define the right category for
“ surfaces ” (of arbitrary dimension) to apply it to; the problem is including
integration domains whose “ edges ” are sufficiently regular for integration to
be possible on it, while including sufficiently general singularities so as not
to exclude important practical or theoretical cases. For example, the border
of a polyhedra is not a “ smooth ” surface ; it has sharp corners and edges.
Excluding such a simple example from the application domain of Stokes’ formula would run counter to common sense, but proving it in a sufficiently
general framework so as to include polyhedrons (in other words, “ simplicial
complexes ” from algebraic topology) poses substantial difficulties since at the
same time the case of perfectly smooth surfaces has to be covered. Physicists
reply that the surface of a tetrahedron is in reality made of four perfectly
smooth triangles and that the edges do not count in the integration, or that
“ corners may be cut ” without significantly changing the result. True, but
they obviously do not have to prove it. It is for good reason that this problem is at the origin of the Bourbaki group; in the 1930s, when its program
was to write a usable treatise for university teaching, a mathematician of
the level of Andre Weil asked Henri Cartan if he knew a good method for
proving the formula, it being agreed that all mathematicians – I used to do
in in 1947 – have always been able to present the type of “ proof ” that satisfy
physicists.
(i) The exterior derivative as an infinitesimal integral. Let us again consider a differential form p i (x)h
i = ω(x; h) of degree 1 and class C
1 on an
open subset G of a Cartesian space E. Take a point x ∈ G, and fix two
vectors h and k and, for given s, t ∈ R, consider the plane parallelogram
P (x, sh, tk) = P , i.e. the set of points of the form x + uh + vk, where u
(resp. v) vary between 0 and s (resp. t) ; assume s and t to be sufficiently
small so that P can be contained in G. When we follow the sides of this
parallelogram in the direction indicated in fig. 2, the border of P is transformed into a closed integration path written ∂P , the boundary of P . Let us
calculate the integral I(x; sh, tk) of ω along this path. On the side connecting
x to x + sh, the parametric representation u → x + uh can be used, which
gives a contribution equal to
s
0
ω(x + uh; h)du .
Contributions from the other sides are calculated in a similar fashion.
I(x; sh, tk) is thus seen to be equal to