21. Connection¶

This notebook is part of the Introduction to manifolds in SageMath by Andrzej Chrzeszczyk (Jan Kochanowski University of Kielce, Poland).

Roughly speaking, connection $∇$ on a smooth manifold $M$, is a rule to calculate the directional derivatives of the vector fields on $M$. If $X$ and $Y$ are two vector fields, $∇_X Y$ denotes the vector field whose value at each point $p ∈ M$ is equal to the directional derivative of $Y$ in the direction of $X_p$ .

Euclidean connection in $U\subset R^n$¶

Let $U\subset R^n$ be an open subset and $\ \ (x^1,\ldots,x^n)\ \ $ define Cartesian coordinates on $U$. For $X,Y\in \mathfrak{X}(U),$ $ \ X=X^i\frac{\partial}{\partial x^i},\ Y=Y^i\frac{\partial}{\partial x^i}\ $ define $\ D_XY\in \mathfrak{X}(U)\ $ by

$$D_XY=X(Y^j)\frac{\partial}{\partial x^j}=X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial}{\partial x^j}.$$

Thus, each component of the vector field $D_XY$ is the usual directional derivative in $R^n$ of $Y^j$ in the direction of $X$.

The map $$(X,Y)\to D_XY$$

is $R$-linear with respect to $Y$ since

$$X(\alpha Y_1^j+\beta Y_2^j)=\alpha X(Y_1^j)+\beta X(Y_2^j),\quad \text{for } \alpha,\beta\in R.$$

It is $C^\infty(U)$ -linear in $X$ since

$$(f_1X_1+f_2X_2)(Y^j)= f_1X_1(Y^j)+f_2X_2(Y^j), \ \text{for } \ f_1,f_2\in C^\infty(U).$$

It satisfies the Leibniz rule since

$D_XY(fY)=X(fY^j)\frac{\partial}{\partial x^j}= (Xf)Y^j\frac{\partial}{\partial x^j}+fX(Y^j)\frac{\partial}{\partial x^j} =(Xf)Y+fD_XY.$

Affine connection in a manifold¶

The above observations motivate the following definition.

Let $M$ be a smooth manifold. A connection on $M$ assigns to each $X ∈ \mathfrak{X}(M)$ an operator $∇_X$ from $\mathfrak{X}(M)$ into itself, such that for all $X, Y, Z ∈ \mathfrak{X}(M), a, b ∈ R$ and $f ∈ C^∞ (M)$,

\begin{eqnarray} \tag{21.1} &∇_X (aY + bZ) = a∇_X Y + b∇_X Z,\\ \tag{21.2} &∇_X ( f Y) = f ∇_X Y + (X f )Y,\\ \tag{21.3} &∇_{aX+bY} Z = a∇_X Z + b∇_Y Z,\\ \tag{21.4} &∇_{f X} Y = f ∇_X Y. \end{eqnarray}

$∇_X Y$ is called the covariant derivative of $Y$ with respect to $X$.

If $\ (x^1 , x^2 , . . . , x^n )\ $ is a coordinate system in some neighborhood $U$ of $M$, and $\ X = X^i \frac{∂}{∂ x^i},\ \ Y = Y^j \frac{∂}{∂ x^j},\ $ then from (21.3),(21.4) it follows

$$\displaystyle \nabla_{X^i\frac{∂}{∂ x^i}}Y=\displaystyle \nabla_{(X^1\frac{∂}{∂ x^1}+\ldots+X^n\frac{∂}{∂ x^n})}Y \stackrel{\text{(3)}}{=}\displaystyle \nabla_{X^1\frac{∂}{∂ x^1}}Y+\ldots+\nabla_{X^n\frac{∂}{∂ x^n}}Y\\ \stackrel{\text{(4)}}{=}\displaystyle X^1\nabla_{\frac{∂}{∂ x^1}}Y+\ldots+X^n\nabla_{\frac{∂}{∂ x^n}}Y =X^i\nabla_{\frac{∂}{∂ x^i}}Y. $$

From (21.1),(21.2) we have $$ \nabla_{\frac{∂}{∂ x^i}}Y= \nabla_{\frac{∂}{∂ x^i}}\big(Y^j\frac{∂}{∂ x^j}\big) =\nabla_{\frac{∂}{∂ x^i}}\big(Y^1\frac{∂}{∂ x^1}+\ldots Y^n\frac{∂}{∂ x^n}\big)\\ \stackrel{\text{(1)}}{=}\displaystyle \nabla_{\frac{∂}{∂ x^i}}\big(Y^1\frac{∂}{∂ x^1}\big)+\ldots+ \nabla_{\frac{∂}{∂ x^i}}\big(Y^n\frac{∂}{∂ x^n}\big)$$ $$ \stackrel{\text{(2)}}{=} Y^1\nabla_{\frac{∂}{∂ x^i}}\big(\frac{∂}{∂ x^1}\big)+ \big(\frac{∂}{∂ x^i}Y^1\big)\frac{∂}{∂ x^1}+ \ldots+ Y^n\nabla_{\frac{∂}{∂ x^i}}\big(\frac{∂}{∂ x^n}\big)+ \big(\frac{∂}{∂ x^i}Y^n\big)\frac{∂}{∂ x^n}\\ =Y^j\nabla_{\frac{∂}{∂ x^i}}\frac{∂}{∂ x^j}+ \big(\frac{∂}{∂ x^i}Y^j\big)\frac{∂}{∂ x^j} =\big(\frac{∂}{∂ x^i}Y^j\big)\frac{∂}{∂ x^j} +Y^j\nabla_{\frac{∂}{∂ x^i}}\frac{∂}{∂ x^j}. $$

Thus (21.1)-(21.4) give us

$$ ∇_X Y=X^i\big[\big(\frac{∂}{∂ x^i}Y^j\big)\frac{∂}{∂ x^j} +Y^j\nabla_{\frac{∂}{∂ x^i}}\frac{∂}{∂ x^j}\big]. $$

Since $\nabla_{\frac{∂}{∂ x^i}}\frac{∂}{∂ x^j}$ is a smooth vector field, it is a linear combination of the vector fields $\{\frac{∂}{∂ x^k}\}_{k=1}^n$, so there is a set of $n^3$ smooth functions $\displaystyle\Gamma^k_{ji}$ on $U$, such that

\begin{equation} \nabla_{\frac{∂}{∂ x^i}}\frac{∂}{∂ x^j}=\Gamma^k_{ji}\frac{∂}{∂ x^k}. \tag{21.5} \end{equation}

Functions $\displaystyle\Gamma^k_{ji}$ which are called Christoffel symbols define the connection in the coordinate system $(x^1,\ldots,x^n)$ since

\begin{equation} \begin{matrix} \displaystyle ∇_X Y=X^i\big[\big(\frac{∂}{∂ x^i}Y^j\big)\frac{∂}{∂ x^j} +Y^j\Gamma^k_{ji}\frac{∂}{∂ x^k}\big]\\ \displaystyle =X^i\big(\frac{∂}{∂ x^i}Y^k+ \Gamma^k_{ji}Y^j\big)\frac{∂}{∂ x^k}. \end{matrix} \tag{21.6} \end{equation}

Comparing (21.6) with the definition of the Euclidean connection we see that all Christoffel symbols for the Euclidean connection vanish.

Observe that if in the definition of the Euclidean connection, we put $X=\frac{∂}{∂ x^i}$ and $Y=\frac{∂}{∂ x^j}$, then the components of $Y$ in Cartesian coordinates are $(0,\ldots,1,\ldots,0),\ $ ($1$ on $j$-th position) and $\displaystyle D_{\frac{∂}{∂ x^i}}\frac{∂}{∂ x^j}= \frac{∂}{∂ x^i}(0)\frac{∂}{∂ x^1}+\ldots+ \frac{∂}{∂ x^i}(1)\frac{∂}{∂ x^j}+\ldots+ \frac{∂}{∂ x^i}(0)\frac{∂}{∂ x^n}=0. $

Example 21.1

Consider the two-dimensional half-plane $y>0$ with connection coefficients defined by $\ \ \Gamma^1_{12}=\Gamma^1_{21}=\Gamma^2_{22}=-\frac{1}{y}\ \ $ and $\ \ \Gamma^2_{11}=\frac{1}{y}.$

As we can see, only nonzero Christoffel symbols are displayed.

We can force displaying all symbols using only_nonzero=False switch.

The coordinate_labels=False switch prevents using coordinate labels instead of indices.

Covariant derivative along a curve¶

If $\gamma:I\to M$ is a smooth curve in a smooth manifold $M$ and $Y$ is a smooth vector field defined on the image of $\gamma$, then the covariant derivative of $Y$ along $\ \gamma$ is the vector field $∇_{\gamma'} Y$ on $\gamma$ defined by

\begin{equation} \displaystyle (∇_{\gamma'} Y)_{\gamma_t } = ∇_{\gamma'_t} Y\quad \text{for } t ∈ I. \tag{21.7} \end{equation}

($\gamma'_t$ denotes the tangent vector to $\gamma$ at $t$, defined in (7.6))

Recall that in local coordinates the tangent vector $\gamma'$ can be represented by (cf. (8.10))

$$ \gamma'_{t_0}=\frac{d(x^i\circ \gamma)}{dt}\Big|_{t_0}\frac{\partial}{\partial x^i}\Big|_{\gamma(t_0)}, $$

furthermore $$\frac{d(x^i\circ\gamma)}{dt}\Big|_{t}\frac{\partial Y^k}{\partial x^i}\Big|_{\gamma(t)}= \frac{\partial Y^k}{\partial x^i}\Big|_{\gamma(t)} \frac{d(x^i\circ\gamma)}{dt}\Big|_{t} =\frac{d(Y^k\circ\gamma)}{dt}\Big|_t,$$ and consequently \begin{equation} \begin{matrix} \displaystyle ∇_{\gamma'(t)} Y =\frac{d(x^i\circ\gamma)}{dt}\Big|_{t} \Big(\frac{\partial Y^k}{\partial x^i}+\Gamma^k_{ji}Y^j\Big)\Big|_{\gamma(t)}\frac{\partial}{\partial x^k}\Big|_{\gamma(t)}\\ \displaystyle =\Big(\frac{d(Y^k\circ\gamma)}{dt}+\frac{d(x^i\circ\gamma)}{dt}(\Gamma^k_{ji}Y^j)\circ\gamma\Big)\Big|_{t}\frac{\partial}{\partial x^k}\Big|_{\gamma(t)}. \end{matrix} \tag{21.8} \end{equation}

Vector field parallel along a curve¶

A vector field $Y$ is parallel along the curve $\gamma$ if $∇_{\gamma'}Y = 0.$

$Y$ is parallel along $\gamma$ if and only if its components with respect to the basis $\{\frac{∂}{∂ x^i} \}_{i=1}^n$ satisfy the system of linear (with respect to $Y^k$) differential equations

\begin{equation} \frac{d(Y^k\circ\gamma)}{dt}+\frac{d(x^i\circ\gamma)}{dt}(\Gamma^k_{ji}\circ\gamma)(Y^j\circ\gamma)=0. \tag{21.9} \end{equation}

Often we use a simplified form \begin{equation} \frac{dY^k}{dt}+\frac{dx^i}{dt}\Gamma^k_{ji}Y^j=0, \tag{21.9a} \end{equation}

where all functions are restricted to the image of $\gamma$.

For example if $n=2$ we obtain

\begin{equation} \begin{matrix} \frac{dY^1}{dt}+\frac{dx^1}{dt}\Gamma_{1,1}^1Y^1 +\frac{dx^2}{dt}\Gamma_{1,2}^1Y^1 +\frac{dx^1}{dt}\Gamma_{2,1}^1Y^2 +\frac{dx^2}{dt}\Gamma_{2,2}^1Y^2=0,\\ \frac{dY^2}{dt}+\frac{dx^1}{dt}\Gamma_{1,1}^2Y^1 +\frac{dx^2}{dt}\Gamma_{1,2}^2Y^1 +\frac{dx^1}{dt}\Gamma_{2,1}^2Y^2 +\frac{dx^2}{dt}\Gamma_{2,2}^2Y^2=0. \end{matrix} \tag{21.9b} \end{equation}

The map $P_{t,t_0} :T_{\gamma(t_0)}M\to T_{\gamma(t)}M$, defined by

$$P_{t,t_0}(Y_0)=Y(\gamma(t)),$$

where $Y$ is parallel along $\gamma$ and $Y(\gamma(t_0)) = Y_0$ , is called parallel transport along of $\gamma$ from $\gamma(t_0)$ to $\gamma(t)$.

If $R^2$ is endowed with the connection $Γ_{ij}^k = 0$ for $i, j, k = 1, 2$, then the system (21.9b) takes the form $$ \frac{dY^1}{dt}=0,\\ \frac{dY^2}{dt}=0, $$ i.e., the vector field $Y$ is parallel along $\gamma$ iff $Y$ is constant.

Example 21.2

Consider (again) the connection $\nabla$ on the upper halfplane $U = \{(x, y)\ | \ y > 0\}$ with Christoffel symbols

$$ \begin{matrix} Γ^1_{11}=0,\quad Γ^2_{11}=1/y,\\ Γ^1_{12}= Γ^1_{21}= −1/y,\\ Γ^2_{12}= Γ^2_{21}= 0,\\ Γ^1_{22}= 0,\quad Γ^2_{22}= −1/y. \end{matrix} $$

Let us check that along the curve $x=t,\ y=1$, the vector field $\ Y=-\cos t \frac{\partial}{\partial x}+\sin t\frac{\partial}{\partial y}\ $ is parallel.

The system (21.9b) reduces to

\begin{equation} \begin{matrix} \frac{dY^1}{dt}+\frac{dx}{dt}\cdot 0\cdot Y^1 +\frac{dy}{dt}(-\frac{1}{y})Y^1 +\frac{dx}{dt}(-\frac{1}{y})Y^2 +\frac{dy}{dt}\cdot 0\cdot Y^2=0,\\ \frac{dY^2}{dt}+\frac{dx}{dt}(\frac{1}{y})Y^1 +\frac{dy}{dt}\cdot 0\cdot Y^1 +\frac{dx}{dt}\cdot 0\cdot Y^2 +\frac{dy}{dt}(-\frac{1}{y})Y^2=0. \end{matrix} \tag{21.9c} \end{equation}

Since $\frac{dx}{dt}=1,\ y=1, \ \frac{dy}{dt}=0$ we obtain \begin{equation} \begin{matrix} \frac{dY^1}{dt}-\frac{1}{y}Y^2=\frac{dY^1}{dt}-Y^2=0,\\ \frac{dY^2}{dt}+\frac{1}{y}Y^1=\frac{dY^2}{dt}+Y^1=0. \end{matrix} \tag{21.9d} \end{equation}

If we put $Y^1=-\cos t, \ Y^2=\sin t,$ then both equations are fulfilled.

Example 21.3

Use Sympy to prove that the vector field $Y$ from the previous example is parallel along the curve from that example.

Example 21.4

Solve the system (21.9d) with respect to $Y^1,Y^2\ $ with initial conditions $\ Y^1(0)=-1,\ Y^2(0)=0.$

Example 21.5

Let us show graphically that the vector field $Y$ from the previous example realizes the parallel transport of the vector $Y_0$ with components $(-1,0)$ at $(0,1)$ to the vector $Y_1$ with the same components at $(2\pi,1)$ along the curve $x=t,\ y=1,\ t\in [0,2\pi].$

Example 21.6

Now take the same $\nabla$ on the upper halfplane but consider the vertical curve $\gamma:\ x=0,\ y=t+1, \ t\in [0,1]$ and the vector field $Y=(t+1)\frac{\partial}{\partial x}.$

First let us check that $Y$ is parallel along $\gamma$.

System (21.9c) takes the form

\begin{equation} \begin{matrix} \frac{dY^1}{dt}-\frac{1}{y}Y^1=0,\\ \frac{dY^2}{dt}-\frac{1}{y}Y^2=0. \end{matrix} \tag{21.9e} \end{equation}

Now solve the system (21.9e) (we have restarted the Kernel).

Example 21.7

Show graphically the vector field $Y$ from the previous example along the curve $x=0,\ y=t+1.$

Example 21.8

Consider the plane $R^2$ with Christoffel symbols $$Γ^1_{11}= Γ^1_{22}=\frac{4u}{1+u^2+4v^2},\\ Γ^2_{11}= Γ^2_{22}=\frac{4v}{1+u^2+4v^2}, $$ and the remaining symbols equal to 0.

Assume that the curve $\gamma$ is defined by $\ x=t,\ y=t,\ t\in (0,1)$.

System (21.9b) takes the form

$$ \frac{dY^1}{dt}+\frac{4t}{1+8t^2}(Y^1+Y^2)=0,\\ \frac{dY^2}{dt}+\frac{4t}{1+8t^2}(Y^1+Y^2)=0. $$

and can be solved with Sympy (with initial conditions $Y^1(0)=1,\ Y^2(0)=0$).

Example 21.9

Using the data from the previous example show the parallel transport of the tangent vector at $(0,0)$ with components $(1,0)$ along the curve $x=t, \ y=t, \ t\in [0,1].$

Remark. In many examples, the exact solutions to ODE systems are not accessible.

Fortunately, the same figure as in previous example can be obtained, solving the system numerically.

Example 21.10

Use numerical tools to obtain the figure from the previous example.

First we check that the connection coefficients from the previous example correspond to an elliptic paraboloid.

Now we can use the method

eparaboloid.parallel_translation_numerical(curve, t, v0, tinterval).

In the above command

curve =[x(t),y(t)] can be replaced for example by [s,s],

t by s,

v0 by [1.0,1.0],

and tinterval =[t_0,t_1,number_of_subintervals] by [0.0, 1.0, 9].

Use the command eparaboloid.parallel_translation_numerical? to obtain more information.

Let us note that the values of the compositions $\Gamma_{ij}^k\circ\gamma$ in the above calculations are cached, and can be used for example in symbolic computations:

Remark. In previous examples we have used linear functions to define $\gamma$. If the numerical approach is used we can try more complicated curves.

Example 21.11

Let us investigate the parallel transport from the previous example but along the curve $x=t,\ y=t^2, \ t\in [0,1]$.

Remark. This time the symbols $\Gamma_{ij}^k\circ\gamma$ are more complicated:

Example 21.12

Show how the equations (21.9b) of the parallel vector field along a curve can be obtained in SageMath Manifolds for a concrete parametric surface in and concrete curve.

If the equations (21.9b) of the parallel transport system are written in the form $$ \frac{dY^1}{dt}=F^1(Y^1,Y^2),\ \ \frac{dY^1}{dt}=F^1(Y^1,Y^2),$$

then the right hand sides can be obtained as follows (we use the symmetry of connection coefficients for parametrized surfaces in $R^3$).

Thus the right hand sides of equations for parallel transport on the surface $\ \ z=u^2+v^2\ \ $ along the curve $\ u=t,\ v=t^2,\ $ are:

Example 21.13

Make a three-dimensional illustration for parallel transport on the surface $\ \ z=u^2+v^2\ \ $ along the curves $\ u=t,\ v=t,\ $ and $\ u=t,\ v=t^2.\ $

Since the connection from the last example corresponds to the parametrized surface $z=u^2+v^2$ we are able to make a 3d illustration of the parallel transport.

First the case of the curve $x=t, y=t$: