Transport of structure

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Short description: Property of structural isomorphism

In mathematics, particularly in universal algebra and category theory, transport of structure refers to the process whereby a mathematical object acquires a new structure and its canonical definitions, as a result of being isomorphic to (or otherwise identified with) another object with a pre-existing structure.[1] Definitions by transport of structure are regarded as canonical.

Since mathematical structures are often defined in reference to an underlying space, many examples of transport of structure involve spaces and mappings between them. For example, if [math]\displaystyle{ V }[/math] and [math]\displaystyle{ W }[/math] are vector spaces with [math]\displaystyle{ (\cdot,\cdot) }[/math] being an inner product on [math]\displaystyle{ W }[/math], such that there is an isomorphism [math]\displaystyle{ \phi }[/math] from [math]\displaystyle{ V }[/math] to [math]\displaystyle{ W }[/math], then one can define an inner product [math]\displaystyle{ [\cdot, \cdot] }[/math] on [math]\displaystyle{ V }[/math] by the following rule:

[math]\displaystyle{ [v_1, v_2] = (\phi(v_1), \phi(v_2)) }[/math]

Although the equation makes sense even when [math]\displaystyle{ \phi }[/math] is not an isomorphism, it only defines an inner product on [math]\displaystyle{ V }[/math] when [math]\displaystyle{ \phi }[/math] is, since otherwise it will cause [math]\displaystyle{ [\cdot,\cdot] }[/math] to be degenerate. The idea is that [math]\displaystyle{ \phi }[/math] allows one to consider [math]\displaystyle{ V }[/math] and [math]\displaystyle{ W }[/math] as "the same" vector space, and by following this analogy, then one can transport an inner product from one space to the other.

A more elaborated example comes from differential topology, in which the notion of smooth manifold is involved: if [math]\displaystyle{ M }[/math] is such a manifold, and if [math]\displaystyle{ X }[/math] is any topological space which is homeomorphic to [math]\displaystyle{ M }[/math], then one can consider [math]\displaystyle{ X }[/math] as a smooth manifold as well. That is, given a homeomorphism [math]\displaystyle{ \phi \colon X \to M }[/math], one can define coordinate charts on [math]\displaystyle{ X }[/math] by "pulling back" coordinate charts on [math]\displaystyle{ M }[/math] through [math]\displaystyle{ \phi }[/math]. Recall that a coordinate chart on [math]\displaystyle{ M }[/math] is an open set [math]\displaystyle{ U }[/math] together with an injective map

[math]\displaystyle{ c \colon U \to \mathbb{R}^n }[/math]

for some natural number [math]\displaystyle{ n }[/math]; to get such a chart on [math]\displaystyle{ X }[/math], one uses the following rules:

[math]\displaystyle{ U' = \phi^{-1}(U) }[/math] and [math]\displaystyle{ c' = c \circ \phi }[/math].

Furthermore, it is required that the charts cover [math]\displaystyle{ M }[/math] (the fact that the transported charts cover [math]\displaystyle{ X }[/math] follows immediately from the fact that [math]\displaystyle{ \phi }[/math] is a bijection). Since [math]\displaystyle{ M }[/math] is a smooth manifold, if U and V, with their maps [math]\displaystyle{ c \colon U \to \mathbb{R}^n }[/math] and [math]\displaystyle{ d \colon V \to \mathbb{R}^n }[/math], are two charts on [math]\displaystyle{ M }[/math], then the composition, the "transition map"

[math]\displaystyle{ d \circ c^{-1} \colon c(U \cap V) \to \mathbb{R}^n }[/math] (a self-map of [math]\displaystyle{ \mathbb{R}^n }[/math])

is smooth. To verify this for the transported charts on [math]\displaystyle{ X }[/math], notice that

[math]\displaystyle{ \phi^{-1}(U) \cap \phi^{-1}(V) = \phi^{-1}(U \cap V) }[/math],

and therefore

[math]\displaystyle{ c'(U' \cap V') = (c \circ \phi)(\phi^{-1}(U \cap V)) = c(U \cap V) }[/math], and
[math]\displaystyle{ d' \circ (c')^{-1} = (d \circ \phi) \circ (c \circ \phi)^{-1} = d \circ (\phi \circ \phi^{-1}) \circ c^{-1} = d \circ c^{-1} }[/math].

Thus the transition map for [math]\displaystyle{ U' }[/math] and [math]\displaystyle{ V' }[/math] is the same as that for [math]\displaystyle{ U }[/math] and [math]\displaystyle{ V }[/math], hence smooth. That is, [math]\displaystyle{ X }[/math] is a smooth manifold via transport of structure. This is a special case of transport of structures in general.[2]

The second example also illustrates why "transport of structure" is not always desirable. Namely, one can take [math]\displaystyle{ M }[/math] to be the plane, and [math]\displaystyle{ X }[/math] to be an infinite one-sided cone. By "flattening" the cone, a homeomorphism of [math]\displaystyle{ X }[/math] and [math]\displaystyle{ M }[/math] can be obtained, and therefore the structure of a smooth manifold on [math]\displaystyle{ X }[/math], but the cone is not "naturally" a smooth manifold. That is, one can consider [math]\displaystyle{ X }[/math] as a subspace of 3-space, in which context it is not smooth at the cone point.

A more surprising example is that of exotic spheres, discovered by Milnor, which states that there are exactly 28 smooth manifolds which are homeomorphic but not diffeomorphic to [math]\displaystyle{ S^7 }[/math], the 7-dimensional sphere in 8-space. Thus, transport of structure is most productive when there exists a canonical isomorphism between the two objects.

See also

References

  1. Holm, Henrik (2015). "A Note on Transport of Algebraic Structures". Theory and Applications of Categories 30 (34): 1121–1131. http://www.tac.mta.ca/tac/volumes/30/34/30-34.pdf. 
  2. Bourbaki, Nicolas (1968), Elements of mathematics: Theory of sets, Hermann (original), Addison-Wesley (translation) , Chapter IV, Section 5 "Isomorphism and transport of structures".