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I am having trouble understanding how to compute the connection on a manifold. I see the definition $\nabla_ie_j=\Gamma^k_{ij}e_k$, but I am not sure if this equation is supposed to define $\nabla_i$ or $\Gamma$. If it defines $\Gamma$, then how is $\nabla_i$ defined?

For instance, in his answer to question 270284 on MathOverflow, a user writes the equations \begin{align*} &\nabla_y {\bf e}_x = -{\bf e}_y; \quad \nabla_y {\bf e}_y = {\bf e}_x\\ &\nabla_x {\bf e}_x = \nabla_x {\bf e}_y =0 \end{align*} Once I have these, I can work with them, but I don't understand what exactly they mean. Are the basis vectors ${\bf e}_x$ and ${\bf e}_y$ functions of $x,y$ that are being derivated? How are they related to the usual vectors $\partial_x$ and $\partial_y$?

Can someone show me a choice of basis on the plane and its connection and how things are calculated, explicitly?

thedude
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  • You could take a look at these videos, specifically 6a, b and c. I don't remember exactly what order he did it in, but I seem to recall it made perfect sense at the time. – Arthur May 30 '17 at 12:23
  • In the video the guy defines $\Gamma$ by the equation $\frac{\partial e_i}{\partial x^j}=\Gamma^k_{ij}e_k$ and gives polar coordinates as examples. 1) For coordinate basis such that $e_i=\partial_i$ it seems that $\Gamma^k_{ij}=\Gamma^k_{ji}$ will always hold, but I think this cannot be true. 2) I don't see how to define $\Gamma$ for non-coordinate basis – thedude May 30 '17 at 13:35
  • Chapter 8 does go through the $\Gamma$'s on embedded surfaces (with an explicit calculation for the sphere), maybe that's what you're after. If not, then I can't help you any more. – Arthur May 30 '17 at 14:46
  • Quick note on the equation $\nabla_{i} e_{j} = \Gamma_{ij}^{k} e_{k}$: One might analogously ask, "If $T$ is a linear transformation and $Te_{j} = a_{j}^{i} e_{i}$, does this define $T$ or the matrix $a_{j}^{i}$?" The answer is, neither: The equation defines a connection if the Christoffel symbols are known, and defines the Christoffel symbols if a connection is known. (In case it helps, $e_{x}$ and $e_{y}$ are presumably the standard frame, the ordered pair of vector fields whose values at each point are the standard basis.) – Andrew D. Hwang May 30 '17 at 15:15
  • Nope, chapter 8 did not help either... – thedude May 30 '17 at 15:16
  • @AndrewD.Hwang Yes, but I can describe $T$ geometrically, like saying that it rotates vectors by $\alpha$ degrees, and this will allow me to find the matrix elements. Likewise, is there a way to describe $\nabla$ geometrically and from there find $\Gamma$? – thedude May 31 '17 at 16:02

1 Answers1

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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}\newcommand{\Del}{\nabla}$Let $U \subset \Reals^{n}$ be a non-empty open set, and let $(\Basis_{j})_{j=1}^{n}$ be an arbitrary smooth frame, i.e., an ordered $n$-tuple of smooth vector fields whose values at each point $p$ of $U$ are linearly independent, i.e., are a basis for $T_{p}U \simeq \Reals^{n}$.

A connection $\Del$ in $U$ may be viewed as an axiomization of directional differentiation: If $X$ and $Y$ are smooth vector fields in $U$, then $\Del_{X}Y$, the (covariant) derivative of $Y$ along $X$, is a smooth vector field in $U$ such that for every smooth, real-valued function $f$ in $U$,

  1. $\Del_{fX}Y = f(\Del_{X}Y)$; the value $(\Del_{X}Y)(p)$ depends linearly on, and only on, the value $X(p)$, not on the behavior of $X$ in a neighborhood of $p$.

  2. $\Del_{X}(fY) = (Xf) Y + f(\Del_{X}Y)$; the derivative obeys the Leibniz rule in $Y$.

Writing $X = X^{i} \Basis_{i}$ and $Y = Y^{j} \Basis_{j}$, successive application of the preceding axioms gives \begin{align*} \Del_{X}Y &= \sum_{i=1}^{n} X^{i} \Del_{\Basis_{i}} Y \\ &= \sum_{i=1}^{n} X^{i} \biggl[\sum_{j=1}^{n} \bigl((\Basis_{i} Y^{j}) \Basis_{j} + Y^{j} \Del_{\Basis_{i}} \Basis_{j}\bigr)\biggr] \\ &= \sum_{i,j=1}^{n} \biggl[X^{i} (\Basis_{i} Y^{j}) \Basis_{j} + X^{i} Y^{j} \Del_{\Basis_{i}} \Basis_{j}\biggr]. \end{align*} The derivative $(\Basis_{i}Y^{j})$ of a function along a vector field is determined by the smooth structure. By contrast, the covariant derivatives $\Del_{\Basis_{i}} \Basis_{j}$ are extra structure.

The key idea is this: Since $\Del_{\Basis_{i}} \Basis_{j}$ is a vector field in $U$, it can be expressed uniquely as a linear combination of the frame fields: $$ \Del_{\Basis_{i}} \Basis_{j} = \sum_{k=1}^{n} \Gamma_{ij}^{k} \Basis_{k}. $$

The usual geometric parsing of this equation is: If at the point $p$ the vector field $\Basis_{j}$ is differentiated in the direction of $\Basis_{i}(p)$, the Christoffel symbol $\Gamma_{ij}^{k}$ is the $\Basis_{k}(p)$-component of the derivative.

Equivalently, moving the frame $(\Basis_{j})_{j=1}^{n}$ in the direction of the tangent vector $\Basis_{i}(p)$ infinitesimally deforms the frame by the linear transformation with matrix $A_{j}^{k} = \Gamma_{ij}^{k}$.

When $U$ has a Riemannian metric and the frame field $(\Basis_{j})_{j=1}^{n}$ is orthonormal with respect to the metric, each Christoffel symbol can be interpreted as an angular velocity: $\Gamma_{ij}^{k}$ is the rate at which $\Basis_{j}$ rotates toward $\Basis_{k}$ as the frame is evaluated along a curve through $p$ with velocity $\Basis_{i}$.

Because the covariant derivative $\Del_{X}Y$ is linear over smooth functions in $X$, the map $X \mapsto \Del_{X}Y$ (with $Y$ fixed) may be viewed as a vector-valued $1$-form, and the association $X \mapsto \Del_{X}$ may be viewed as a $1$-form taking values in the bundle of endomorphisms of $TU$.

Locally, such a gadget is represented by a matrix-valued $1$-form $(\theta_{j}^{k})$, the connection form, which satisfies $$ \theta_{j}^{k}(\Basis_{i}) = \Gamma_{ij}^{k}. $$ Barrett O'Neill's Elementary Differential Geometry is an accessible account of this viewpoint. Particularly, he notes the importance of measuring the rate of change of a frame along a curve in terms of the frame itself.

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    Thank you for your answer! I still need to find or work out myself some simple examples to make sure I understand the actual nuts and bolts. I got the book you mentioned, it seems to be just what I need. – thedude May 31 '17 at 22:58