Notes on Ehresmann’s connections -Part 1

Ehresmann’s connection

Consider a fibre bundle $\xi=(E,\pi,M)$ with a fibred chart $(x^\mu,v^a),$ where $x^\mu,\quad (\mu=1,\ldots m)$ are local coordinates on the base manifold $M,$ and $v^a\quad (a=1,\ldots n)$ are coordinates along the fibres. If we have two fibred charts, then on their intersection domain we have

$x^{\mu'}=x^{\mu'}(x^\mu),$
$v^{a'}=v^{a'}(x^\mu,v^a).$

The chart $(x^\mu,v^a)$ induces vector fields on the tangent space $TE:$

$\partial_\mu=\frac{\partial}{\partial x^\mu}$
$\partial_a=\frac{\partial}{\partial v^a}$

with the transformation laws:

$\frac{\partial f(x',v')}{\partial x^{\mu'}}=\frac{\partial f(x,v)}{\partial x^\mu}\frac{\partial x^\mu}{\partial x^{\mu'}}+\frac{\partial f(x,v)}{\partial v^a}\frac{\partial v^a}{\partial x^{\mu'}}$
$\frac{\partial f(x'v')}{\partial v^{a'}}=\frac{\partial f(x,v)}{\partial v^a}\frac{\partial v^a}{\partial v^{a'}}.$

Therefore

$\begin{eqnarray*}\partial_{\mu'}&=&\frac{\partial x^\mu}{\partial x^{\mu'}}\,\partial_\mu+\frac{\partial v^a}{\partial x^{\mu'}}\,\partial_a,\\ \partial_{a'}&=&\frac{\partial v^a}{\partial v^{a'}}\,\partial_a,\end{eqnarray*}$

and the inverse

$\begin{eqnarray*}\partial_{\mu}&=&\frac{\partial x^{\mu'}}{\partial x^{\mu}}\,\partial_{\mu'}+\frac{\partial v^{a'}}{\partial x^{\mu}}\,\partial_{a'},\\ \partial_{a}&=&\frac{\partial v^{a'}}{\partial v^{a}}\,\partial_{a'},\end{eqnarray*}$

Vector fields $\partial_a$ span what is called the vertical subspace in $TE.$ This subspace, consisting of vectors tangent to the fibers, is independent of the chart. There are no a priori distinguished `horizontal’ subspaces. The subspaces determined by vector fields $\partial_\mu$ change when the charts change. A preferred distribution of horizontal subspaces is determined by what is called an Ehresmann connection. Once such a connection is given, vector fields $\partial_\mu$ tangent to $M$ can be lifted to horizontal vector fields tangent to $E.$

Remark: In principle we should use a different notation for vector fields $\partial_\mu$ tangent to $M$ and vector fields induced on $E$ by a chart $(x,v).$ Using the same notation for both can lead to misunderstandings when is not paying attention to the context.

Let us denote by $\delta_\mu$ the horizontal lifts induced by an Ehresmann connection on $\xi,$ They can be written uniquely in the form:

$\delta_\mu=\partial_\mu+\Gamma^a_\mu(x,v)\,\partial_a.$

Remark: In Finsler geometry instead of $\Gamma^a_\mu$ one usually uses $N^a_\mu=-\Gamma^a_\mu,$ so that the formula for the horizontal lift reads $\delta_\mu=\frac{\partial}{\partial x^\mu}-N^a_\mu(x,v) \frac{\partial}{\partial v^a}$

The functions $\Gamma^a_\mu(x,v)$ may be called the coefficient functions of the Ehresmann connection with respect to the chart $(x,v).$

If $X=X^\mu\partial_\mu$ is a vector field on $M,$ we denote its horizontal lift by $\delta_X,$ thus

$\delta_X=X^\mu\,\delta_\mu.$

Suppose now we have two charts, $(x,v)$ and $(x'v').$ Corresponding to these two charts we will have two sets of horizontal lifts of the same connection $\delta_\mu$ and $\delta_{\mu'}.$ Since both are supposed to describe the same horizontal distribution, they must be related by an invertible matrix $A^{\mu'}_\mu$ :

$\delta_{\mu'}=A^\mu_{\mu'}\delta_\mu.$

Substituting now the definitions we have

$\partial_{\mu'}+\Gamma^{a'}_{\mu'}\,\partial_{a'}=A^\mu_{\mu'}\left(\partial_\mu+\Gamma^a_\mu\partial_a.\right)$

$\partial_{\mu'}+\Gamma^{a'}_{\mu'}\,\partial_{a'}=A^\mu_{\mu'}\frac{\partial x^{\mu'}}{\partial x^{\mu}}\,\partial_{\mu'}+A^\mu_{\mu'}\frac{\partial v^{a'}}{\partial x^{\mu}}\,\partial_{a'}+A^\mu_{\mu'}\Gamma^a_\mu\frac{\partial v^{a'}}{\partial v^{a}}\,\partial_{a'}$

Comparing the terms we find that

$A^\mu_{\mu'}= \frac{\partial x^\mu}{\partial x^{\mu'}}$

and therefore

$\Gamma^{a'}_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\left(\frac{\partial v^{a'}}{\partial v^a}\Gamma^a_\mu+\frac{\partial v^{a'}}{\partial x^\mu}\right).$

We can distinguish two special classes of transformations. First class consists of changes of coordinates on $M$ without reparametrizing of the fibers. For these transformations $\Gamma^a_\mu$ transforms like one–forms:

$\Gamma^a_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\Gamma^a_\mu.$

The second class consists of transformations of the form $x^{\mu'}=x^\mu,\, v^{a'}=v^a(x,v).$ For these transformations we have

$\Gamma^{a'}_{\mu}=\frac{\partial v^{a'}}{\partial v^a}\Gamma^a_\mu+\frac{\partial v^{a'}}{\partial x^\mu}.$

{\bf Remark}: If $\xi$ is a vector bundle, then the reparametrizations of the fibers compatible with the vector bundle structure are linear and of the form:

$v^{a'}={\Lambda(x)^{a'}}_a\,v^a.$

In this case the last formula reads:

$\Gamma^{a'}_\mu={\Lambda^{a'}}_a\Gamma^a_\mu+\partial_\mu {\Lambda^{a'}}_av^a$

Curvature

Given an Ehresmann connection represented by horizontal vector fields $\delta_\mu$ we can calculate their commutator. As it happens their commutator is a vertical vector field that can be written as

$[\delta_\mu,\delta_\nu]=R_{\mu\nu}^a\,\partial_a,$

where

$R_{\mu\nu}^a=\partial_\mu \Gamma^a_\nu-\partial_\nu \Gamma^a_\mu+\Gamma_\mu^b\,\partial_b\Gamma^a_\nu-\Gamma_\nu^b\,\partial_b\Gamma^a_\mu.$

More generally, for any two vector fields $X=X^\mu\partial_\mu,\,Y=Y^\mu\partial_\mu$ on $M$ one defines

$R_{X,Y}=[\delta_X,\delta_Y]-\delta_{[X,Y]}.$

If $f,g$ are functions on $M,$ then

$R_{fX,gY}(x,v)=f(x)g(x)\,R_{X,Y}(x,v).$

Under coordinate transformations $R_{\mu\nu}$ transforms like a tensor:

$R^a_{\mu'\nu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\nu}{\partial x^{\nu'}}\,R^a_{\mu\nu}.$

Under fiber reparametrization we have:

$R_{\mu\nu}^{a'}=\frac{\partial v^{a'}}{\partial v^a}\,R^a_{\mu\nu}.$

Connection form

It is convenient to code the connection given by a horizontal distribution in one geometrical object. This can be done by introducing the curvature form – a one-form on $E$ with values in the vertical distribution. Given a vector tangent to $E$ at $(x,v)$ we decompose it into a horizontal and vertical part and define the connection form on this vector as the vertical part. It follows that the connection form is the identity map on vertical vectors. In coordinates we can uniquely write the connection form $\omega$ as
$\omega=(dv^a+N_\mu^a\,dx^\mu)\,\partial_a,$
The curvature form vanishes automatically on the horizontal vectors $\delta_\mu,$ therefore it easily follows that $N^a_\mu=-\Gamma^a_\mu.$

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7 Responses to Notes on Ehresmann’s connections -Part 1

grelade says:

2011/01/31 at 21:20

W krzywizna zdefiniowanej jako komutator przy obliczaniu pojawiają mi się jeszcze dwa człony które nie widzę dlaczego powinny się kasować:
$\Gamma^{a}_{\mu}\partial_a \partial_{\nu} - \Gamma^{a}_{\nu}\partial_a \partial_{\mu}$
Można rozjaśnić?

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- grelade says:
  
  2011/01/31 at 21:22
  
  A i jeszcze jest literówka w pierwszym transformation law:
  [img]http://arkadiusz-jadczyk.eu/blog/wp-content/ql-cache/quicklatex.com-5132c752c43f66d1799f89c1d276854e_l2.gif[/img]
  
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arkajad says:

2011/02/01 at 14:28

We have

$[\delta_\mu,\delta_\nu]=[\partial_\mu+\Gamma^a_\mu\partial_a,\partial_\nu+\Gamma^a_\nu\partial_a]$

$=[\delta_\mu,\Gamma^a_\nu\partial_a]+[\Gamma^a_\mu,\partial_\nu]+[\Gamma^a_\mu\partial_a,\Gamma^b_\nu\partial_b]$

We use the following properties of the commutator

$[A,BC]=B[A,C]+[A,B]C$
$[AB,C]=A[B,C}+[A,C]B$

The first term gives:

$[\partial_\mu,\Gamma^a_\nu\partial_a]=[\partial_\mu,\Gamma^a_\nu]\partial_a+\Gamma^a_\nu[\partial_\mu,\partial_a]=[\partial_\mu,\Gamma^a_\nu]\partial_a=\partial_\mu\Gamma^a_\nu\,\partial_a$

The term $[\partial_\mu,\partial_a]$ is zero.

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grelade says:

2011/02/01 at 16:56

Okay, but why your last equality is true? Namely, why
$[\partial_\mu, \Gamma^a_\nu]\partial_a = \partial_\mu \Gamma^a_\nu \partial_a$
? This would mean that term i posted before is 0.

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arkajad says:

2011/02/01 at 17:11

We have two vector fields. Their commutator is also a vector field. Vector fields may be considered as acting on functions. So we act on an arbitrary function f(x,v)

$([\partial_\mu,\Gamma^a_\nu\partial_a]f)(x,v)=(\partial_\mu\Gamma^a_\nu\partial_a\,f)(x,v)-(\Gamma^a_\nu\partial_a\partial_\mu\,f)(x,v)$

Now use Leibniz’s rule when taking derivatives.

$(\partial_\mu\Gamma^a_\nu\partial_a\,f)(x,v)=(\partial_\mu\Gamma^a_\nu\)\partial_a\,f)(x,v)+\Gamma^a_\nu\partial_\mu\partial_a\,f (x,v)$

The second term will cancel with the minus term of the commutator, because partial derivatives commute. What remains is

$(\partial_\mu\Gamma^a_\nu )\partial_a\,f(x,v)$

Therefore

$([\partial_\mu,\Gamma^a_\nu\partial_a]f)(x,v)=(\partial_\mu\Gamma^a_\nu )\partial_a\,f(x,v)$

Now we skip $f(x,v)$

$[\partial_\mu,\Gamma^a_\nu\partial_a]=\partial_\mu\Gamma^a_\nu\)\partial_a$

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grelade says:

2011/02/01 at 17:15

Ok, got it! Thanks

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- arkajad says:
  
  2011/02/01 at 17:25
  
  You are welcome.
  
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