# Transport of structure

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In mathematics, transport of structure is the definition of a new structure on an object by reference to another object on which a similar structure already exists. 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 V and W are vector spaces, and if $\phi \colon V\to W$ is an isomorphism, and if $(\cdot ,\cdot )$ is an inner product on $W$ , then we can define an inner product $[\cdot ,\cdot ]$ on V by

$[v_{1},v_{2}]=(\phi (v_{1}),\phi (v_{2}))$ .

Although the equation makes sense even when $\phi$ is not an isomorphism, it only defines an inner product on V when $\phi$ is, since otherwise it will cause $[\cdot ,\cdot ]$ to be degenerate. The idea is that $\phi$ allows us to consider V and W as "the same" vector space, and if we follow this analogy, we can transport an inner product from one to the other.

A more involved example comes from differential topology, in which we have the notion of a smooth manifold. If M is such a manifold, and if X is any topological space which is homeomorphic to M, we can consider X as a smooth manifold as well. That is, let $\phi \colon X\to M$ be a homeomorphism; we must define coordinate charts on X, which we will do by "pulling back" coordinate charts on M through $\phi$ . Recall that a coordinate chart on $M$ is an open set U together with an injective map

$c\colon U\to \mathbb {R} ^{n}$ for some n; to get such a chart on X, we let

$U'=\phi ^{-1}(U)$ and $c'=c\circ \phi$ .

Furthermore, it is required that the charts cover M, we must check that the transported charts cover X, which follows immediately from the fact that $\phi$ is a bijection. Finally, since M is a smooth manifold, we have that if U and V, with their maps

$c\colon U\to \mathbb {R} ^{n}$ and $d\colon V\to \mathbb {R} ^{n}$ ,

are two charts on M, then the composition, the "transition map"

$d\circ c^{-1}\colon c(U\cap V)\to \mathbb {R} ^{n}$ (a self-map of $\mathbb {R} ^{n}$ )

is smooth. We must check this for our transported charts on X. We have

$\phi ^{-1}(U)\cap \phi ^{-1}(V)=\phi ^{-1}(U\cap V)$ ,

and therefore

$c'(U'\cap V')=(c\circ \phi )(\phi ^{-1}(U\cap V))=c(U\cap V)$ , and
$d'\circ (c')^{-1}=(d\circ \phi )\circ (c\circ \phi )^{-1}=d\circ (\phi \circ \phi ^{-1})\circ c^{-1}=d\circ c^{-1}$ .

Therefore the transition map for $U'$ and $V'$ is the same as that for U and V, hence smooth. Therefore X is a smooth manifold via transport of structure.

Although the second example involved considerably more checking, the principle was the same, and any experienced mathematician would have no difficulty performing the necessary verifications. Therefore when such an operation is indicated, it is invoked merely as "transport of structure" and the details left to the reader, if desired.

The second example also illustrates why "transport of structure" is not always desirable. Namely, we can take M to be the plane, and we can take X to be an infinite one-sided cone. By "flattening" the cone we achieve a homeomorphism of X and M, and therefore the structure of a smooth manifold on X, but the cone is not "naturally" a smooth manifold. That is, we can consider X 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 by definition not diffeomorphic) to $S^{7}$ , the 7-dimensional sphere in 8-space. Thus, transport of structure is most productive when there exists a canonical isomorphism between the two objects.