Whitney topologies

In mathematics, and especially differential topology, functional analysis and singularity theory, the Whitney topologies are a countably infinite family of topologies defined on the set of smooth mappings between two smooth manifolds. They are named after the American mathematician Hassler Whitney.

Construction

Let M and N be two real, smooth manifolds. Furthermore, let C^∞(M,N) denote the space of smooth mappings between M and N. The notation C^∞ means that the mappings are infinitely differentiable, i.e. partial derivatives of all orders exist and are continuous.^[1]

Whitney C^k-topology

For some integer k ≥ 0, let J^k(M,N) denote the k-jet space of mappings between M and N. The jet space can be endowed with a smooth structure (i.e. a structure as a C^∞ manifold) which make it into a topological space. This topology is used to define a topology on C^∞(M,N).

For a fixed integer k ≥ 0 consider an open subset U ⊂ J^k(M,N), and denote by S^k(U) the following:

$S^k(U) = \{ f \in C^{\infty}(M,N) : (J^kf)(M) \subseteq U \} .$

The sets S^k(U) form a basis for the Whitney C^k-topology on C^∞(M,N).^[2]

Whitney C^∞-topology

For each choice of k ≥ 0, the Whitney C^k-topology gives a topology for C^∞(M,N); in other words the Whitney C^k-topology tells us which subsets of C^∞(M,N) are open sets. Let us denote by W^k the set of open subsets of C^∞(M,N) with respect to the Whitney C^k-topology. Then the Whitney C^∞-topology is defined to be the topology whose basis is given by W, where:^[2]

$W = \bigcup_{k=0}^{\infty} W^k .$

Dimensionality

Notice that C^∞(M,N) has infinite dimension, whereas J^k(M,N) has finite dimension. In fact, J^k(M,N) is a real, finite-dimensional manifold. To see this, let ℝ^k[x₁,…,x_m] denote the space of polynomials, with real coefficients, in m variables of order at most k and with zero as the constant term. This is a real vector space with dimension

$\dim\left\{\R^k[x_1,\ldots,x_m]\right\} = \sum_{i=1}^k \frac{(m+i-1)!}{(m-1)! \cdot i!} = \left( \frac{(m+k)!}{m!\cdot k!} - 1 \right) .$

Writing a = dim{ℝ^k[x₁,…,x_m]} then, by the standard theory of vector spaces ℝ^k[x₁,…,x_m] ≅ ℝ^a, and so is a real, finite-dimensional manifold. Next, define:

$B_{m,n}^k = \bigoplus_{i=1}^n \R^k[x_1,\ldots,x_m], \implies \dim\left\{B_{m,n}^k\right\} = n \dim \left\{ A_m^k \right\} = n \left( \frac{(m+k)!}{m!\cdot k!} - 1 \right) .$

Using b to denote the dimension B^k_m,n, we see that B^k_m,n ≅ ℝ^b, and so is a real, finite-dimensional manifold.

In fact, if M and N have dimension m and n respectively then:^[3]

$\dim\!\left\{J^k(M,N)\right\} = m + n + \dim \!\left\{B_{n,m}^k\right\} = m + n\left( \frac{(m+k)!}{m!\cdot k!}\right).$

Topology

Consider the surjective mapping from the space of smooth maps between smooth manifolds and the k-jet space:

$\pi^k : C^{\infty}(M,N) \twoheadrightarrow J^k(M,N) \ \mbox{where} \ \pi^k(f) = (j^kf)(M) .$

In the Whitney C^k-topology the open sets in C^∞(M,N) are, by definition, the preimages of open sets in J^k(M,N). It follows that the map π^k between C^∞(M,N) given the Whitney C^k-topology and J^k(M,N) given the Euclidean topology is continuous.

Given the Whitney C^∞-topology, the space C^∞(M,N) is a Baire space, i.e. every residual set is dense.^[4]

References

1. Golubitsky, M.; Guillemin, V. (1974), Stable Mappings and Their Singularities, Springer, p. 1, ISBN 0-387-90072-1
2. Golubitsky & Guillemin (1974), p. 42.
3. Golubitsky & Guillemin (1974), p. 40.
4. Golubitsky & Guillemin (1974), p. 44.