// Workers AI · dad joke modeWhy did Whitney topologies go to therapy? They had boundary issues.

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.

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:

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

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

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:

$W=\bigcup _{k=0}^{\infty }W^{k}.$

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

$\dim \left\{\mathbb {R} ^{k}[x_{1},\ldots ,x_{m}]\right\}=\sum _{i=1}^{k}{\frac {(m+i-1)!}{(m-1)!\cdot i!}}=\left({\frac {(m+k)!}{m!\cdot k!}}-1\right).$

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:

$B_{m,n}^{k}=\bigoplus _{i=1}^{n}\mathbb {R} ^{k}[x_{1},\ldots ,x_{m}],\implies \dim \left\{B_{m,n}^{k}\right\}=n\dim \left\{A_{m}^{k}\right\}=n\left({\frac {(m+k)!}{m!\cdot k!}}-1\right).$

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:

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

Topology

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