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Difference between revisions of "Parallel transport"

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# Let $E=\R^n\smallsetminus\{0\}$ be the punctured Euclidean space, $B=\mathbb S^{n-1}$ the standard unit sphere and $\pi$ the radial projection $\pi(x)=\|x\|^{-1}\cdot x$. This is a topological bundle with the fiber $F=(0,+\infty)\simeq\R^1$.
 
# Let $E=\R^n\smallsetminus\{0\}$ be the punctured Euclidean space, $B=\mathbb S^{n-1}$ the standard unit sphere and $\pi$ the radial projection $\pi(x)=\|x\|^{-1}\cdot x$. This is a topological bundle with the fiber $F=(0,+\infty)\simeq\R^1$.
 
# Let $E=\mathbb S^{n-1}$ as above, $B=\R P^{n-1}$ the real [[projective space]] (all lines in $\R^n$ passing through the origin) and $\pi$ the map taking a point $x$ on the sphere into the line $\ell_x$ passing through $x$. The preimage $\pi^{-1}(\ell)$ consists of two antipodal points $x$ and $-x\in\mathbb S^{n-1}$, thus $F$ is a discrete two-point set $\mathbb Z_2=\{-1,1\}$. This is a topological bundle, which cannot be trivial: indeed, if it were, then the total space $\mathbb S^{n-1}$ would consist of two connected components, while it is  connected.
 
# Let $E=\mathbb S^{n-1}$ as above, $B=\R P^{n-1}$ the real [[projective space]] (all lines in $\R^n$ passing through the origin) and $\pi$ the map taking a point $x$ on the sphere into the line $\ell_x$ passing through $x$. The preimage $\pi^{-1}(\ell)$ consists of two antipodal points $x$ and $-x\in\mathbb S^{n-1}$, thus $F$ is a discrete two-point set $\mathbb Z_2=\{-1,1\}$. This is a topological bundle, which cannot be trivial: indeed, if it were, then the total space $\mathbb S^{n-1}$ would consist of two connected components, while it is  connected.
# More generally, let $\pi:M^m\to N^n$ be a differentiable map between two smooth (connected) compact manifolds of dimensions $m\ge n$, which has the maximal rank (equal to $n$) everywhere. One can show then, using the implicit function theorem and partition of unity, that $\pi$ is a topological bundle with a fiber $F$ which itself is a smooth compact manifold.  
+
# More generally, let $\pi:M^m\to N^n$ be a differentiable map between two smooth (connected) compact manifolds of dimensions $m\ge n$, which has the maximal rank (equal to $n$) everywhere. One can show then, using the implicit function theorem and partition of unity, that $\pi$ is a topological bundle with a fiber $F$ which itself is a smooth compact manifold<ref>This statement is also known as the Ehresmann theorem, see Ehresmann, C., ''Les connexions infinitésimales dans un espace fibré différentiable'', Colloque de Topologie, Bruxelles (1950), 29-55. The compactness assumption can be relaxed by the requirement that the map $\pi$ is [[proper map|proper]]. </ref>
  
 
===Cocycle of a bundle===
 
===Cocycle of a bundle===

Revision as of 12:17, 7 May 2012

A very flexible construction aimed to represent a family of similar objects (fibres or fibers, depending on the preferred spelling) which are parametrized by the index set which itself has an additional structure (topological space, smooth manifold etc.).

The most known examples are the tangent and cotangent bundles of a smooth manifold. The coverings are also a special particular form of a topological bundle (with discrete fibers).

Formal definition of a topological bundle

Let $\pi:E\to B$ be a continuous map between topological spaces, called the total space[1] and the base, and $F$ yet another topological space called fiber, such that the preimage $F_b=\pi^{-1}(b)\subset E$ of every point of the base is homeomorphic to $X$. The latter condition means that $E$ is the disjoint union of "fibers", $E=\bigsqcup_{b\in B} F_b$ homeomorphic to each other.

The map $\pi$ is called fibration[2] of $E$ over $B$, if the above representation is locally trivial: any point of the base admits an open neighborhood $U$ such that the restriction of $\pi$ on the preimage $\pi^{-1}(U)$ is topologically equivalent to the Cartesian projection $\pi_2$ of the product $F\times U$ on the second component: $\pi_2(v,b)=b$. Formally this means that there exists a homeomorhism $H_U=H:\pi^{-1}(U)\to F\times U$ such that $\pi=\pi_2\circ H$.

Examples

  1. The trivial bundle $E=F\times B$, $\pi=\pi_2: F\times B\to B$, $(v,b)\mapsto b$. In this case all trivializing homeomorphisms are globally defined on the entire total space (as the identity map).
  2. Let $E=\R^n\smallsetminus\{0\}$ be the punctured Euclidean space, $B=\mathbb S^{n-1}$ the standard unit sphere and $\pi$ the radial projection $\pi(x)=\|x\|^{-1}\cdot x$. This is a topological bundle with the fiber $F=(0,+\infty)\simeq\R^1$.
  3. Let $E=\mathbb S^{n-1}$ as above, $B=\R P^{n-1}$ the real projective space (all lines in $\R^n$ passing through the origin) and $\pi$ the map taking a point $x$ on the sphere into the line $\ell_x$ passing through $x$. The preimage $\pi^{-1}(\ell)$ consists of two antipodal points $x$ and $-x\in\mathbb S^{n-1}$, thus $F$ is a discrete two-point set $\mathbb Z_2=\{-1,1\}$. This is a topological bundle, which cannot be trivial: indeed, if it were, then the total space $\mathbb S^{n-1}$ would consist of two connected components, while it is connected.
  4. More generally, let $\pi:M^m\to N^n$ be a differentiable map between two smooth (connected) compact manifolds of dimensions $m\ge n$, which has the maximal rank (equal to $n$) everywhere. One can show then, using the implicit function theorem and partition of unity, that $\pi$ is a topological bundle with a fiber $F$ which itself is a smooth compact manifold[3]

Cocycle of a bundle

On a nonvoid overlapping $U_{\alpha\beta}=U_\alpha\cap U_\beta$ of two different trivializing charts $U_\alpha$ and $U_\beta$ two homeomorphisms $H_\alpha,H_\beta: \pi^{-1}(U_{\alpha\beta})\to F\times U_{\alpha\beta}$ are defined. Since both $H_\alpha$ and $H_\beta$ conjugate $\pi$ with the Cartesian projection on $U_{\alpha\beta}$, they map each fiber $F_b=\pi^{-1}(b)$ into the same space $F\times\{b\}$. The composition $H_\alpha\circ H_\beta^{-1}$ takes then the form $$ H_\alpha\circ H_\beta^{-1}:(v,b)\mapsto (H_{\alpha\beta}(x,v),x),\qquad H_{\alpha\beta}(\cdot,x)\in\operatorname{Homeo}(F) $$ with the homeomorphisms $H_{\alpha\beta}(\cdot, x)$ continuously depending on $x\in U_{\alpha\beta}$. The collection of these "homeomorphism-valued" functions defined in the intersections $U_{\alpha\beta}$ is called the cocycle associated with a given trivialization of the bundle $\pi$ (or simply the cocycle of the bundle. They homeomorphisms $\{H_{\alpha\beta}\}$ satisfy the following identities, obvious from their construction: $$ H_{\alpha\beta}\circ H_{\beta\alpha}=\operatorname{id},\qquad H_{\alpha\beta}\circ H_{\beta\gamma}\circ H_{\gamma\alpha}=\operatorname{id}, $$ the second being true on every nonvoid triple intersection $U_{\alpha\beta\gamma}=U_\alpha\cap U_\beta\cap U_\gamma$.

Vector bundles and other additional structures on the fibers

The general construction of bundle easily allows various additional structures, both on the base space and (more importantly) on the fibers. By far the most important special case is that of vector bundles.

To define a vector bundle, one has in addition to the principal definition assume the following:

  1. The fiber $F$ is a vector space[4], and
  2. The trivializing homeomorphisms must respect the linear structure of the fibers.

The second assumption means that rather than being arbitrary homeomorphisms, the maps $\{H_{\alpha\beta}\}$ forming the bundle cocycle, must be linear invertible of each "standard fiber" $F\times \{b\}$; if the fiber is identified with the canonical $n$-space $\Bbbk^n$ (over $\Bbbk=\R$ or $\Bbbk=\C$), then the cocycle will consist of invertible continuous matrix-functios $M_{\alpha\beta}:U_{\alpha\beta}\to\operatorname{GL}(n,\Bbbk)$, so that $H_{\alpha\beta}(v,x)=(M_{\alpha\beta}(x)\, v, x)$, $v\in\Bbbk^n$. The cocycle identities become then the identites relating the values of these matrix-valued functions, $$ M_{\alpha\beta}(x)\cdot M_{\beta\alpha}(x)\equiv E,\qquad M_{\alpha\beta}(x)\cdot M_{\beta\gamma}(x)\cdot M_{\gamma\alpha}(x)\equiv E, $$ where $E$ is the $n\times n$-identical matrix.

For vector bundles all linear constructions become well defined on fibers.

Following the way, one may define vector bundles with extra algebraic structures on the fibers. For instance, if the cocycle defining the bundle, consists of orthogonal matrices, $M_{\alpha\beta}:U_{\alpha\beta}\to\operatorname{SO}(n,\R)$, then the fibers of the bundle naturally acquire the structure of Euclidean spaces. Other natural examples are bundles whose fibers have the Hermitian structure (the cocycle should consist of unitary matrix functions then) or symplectic spaces (with canonical cocycle matrices preserving the symplectic structure).

Tangent and cotangent bundle of a smooth manifold

If $M$ is a smooth manifold with the atlas of coordinate charts $\{U_\alpha\}$ and the maps $h_\alpha:U_\alpha\to\R^m$, then the differentials of these maps $\rd h_\alpha$ allow to identify the tangent space $T_a M$ at $a\in U_{\alpha}$ with $\R^m$ and the union $\bigsqcup_{a\in U_\alpha}T_a M$ with $\R^m\times U_\alpha$ (we write the tangent vector first). For a point $a\in U_{\alpha\beta}$ there are two identifications which differ by the Jacobian matrix of the transition map $h_{\alpha\beta}=h_\alpha\circ h_\beta^{-1}$. This shows that the tangent bundle $TM$ is indeed a vector bundle in the sense of the above definition.

The cotangent bundle is also trivialized by every atlas $\{h_\alpha:U_\alpha\to\R^m\}$ on $M$, yet in this case the direction of arrows should be reverted[5]: the cotangent space $T_a^*M$ is identified with $\R^n$ by the linear map $(\rd h_\alpha^*)$, thus the corresponding cocycle will consist of the transposed inverse Jacobian matrices.

Equivalence of cocycles

The trivializing maps defining the structure of a bundle (vector or topological) are by no means unique, even if the covering domains $U_\alpha$ remain the same. E.g., one can replace the the collection of maps $\{H_{\alpha}\}$ trivializing a vector bundle, by another collection $\{H'_{\alpha}\}$, post-composing them with the maps $F\times U_\alpha\to F\times U_\alpha$, $(v,x)\mapsto (C_\alpha(x)\,v, x)$ with invertible continuous matrix functions $C_\alpha:U_\alpha\to\operatorname{GL}(n,\Bbbk)$. The corresponding matrix cocycle $\{M_{\alpha\beta}\}$ will be replaced then by the new matrix cocycle $\{M'_{\alpha\beta}(x)\}$, $$ M'_{\alpha\beta}(x)=C_\alpha(x)M_{\alpha\beta}(x)C_\beta^{-1}(x),\qquad x\in U_{\alpha\beta}. \tag{CE} $$ Two matrix cocycles related by these identities, are called equivalent and clearly define the same bundle.

Example. The trivial cocycle $\{M_{\alpha\beta}(x)\}=\{E\}$ which consists of identity matrices, corresponds to the trival bundle $F\times B$: the trivializing maps agree with each other on the intersections and hence define the global trivializing map $H:E\to F\times B$. A cocycle equal to the trivial one in the sense of (CE) is called solvable: its solution is a collection of invertible matrix functions $C_\alpha:U_\alpha\to\operatorname{GL}(n,\Bbbk)$ such that on the overlapping of the domains $U_{\alpha\beta}=U_\alpha\cap U_\beta$ the identities $$ M_{\alpha\beta}(x)=C_{\alpha}^{-1}(x)C_\beta(x),\qquad \forall\alpha,\beta,\ x\in U_{\alpha}\cap U_\beta. $$ Thus solvability of cocycle is an analytic equivalent of the topological triviality of the bundle.



  1. Also names fibre space or fibered space are used.
  2. Also the terms bundle or fiber bundle are used.
  3. This statement is also known as the Ehresmann theorem, see Ehresmann, C., Les connexions infinitésimales dans un espace fibré différentiable, Colloque de Topologie, Bruxelles (1950), 29-55. The compactness assumption can be relaxed by the requirement that the map $\pi$ is proper.
  4. The fiber $F$ should be equipped with some topology, but often it is finite-dimensional, $F\simeq\R^n$ or $F\simeq\C^n$, thus leaving only the default option.
  5. Covectors form a covariant rather than contravariant tensor of rank $1$.

How to Cite This Entry:
Parallel transport. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Parallel_transport&oldid=26172