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I'm self-studying differential geometry and general relativity - so far I know of smooth manifolds, vector fields, smooth curves on manifolds, and basics of connections. I know that a smooth vector field $X$ is parallel w.r.t. a connection $\nabla$ if $\nabla X$ vanishes. My interpretation of this is that $\nabla X(\omega,Y)=0$ for all smooth covector and vector fields $\omega,Y$. i.e. $\nabla_YX=0$ for all smooth vector fields $Y$.

I'm also reading General Relativity by Wald, in which I found a statement: "A space will be curved if and only if some initially parallel geodesics fail to remain parallel."

I tried to search for "parallel curves on manifold" on google, but didn't really find anything concrete on how to identify if two smooth curves on a manifold are parallel. The results are only related to "parallel vector fields", "parallel transport", etc. So I want to ask here: is there a well-defined notion of two smooth curves on a smooth manifold being parallel? If so, what is it? e.g. what makes us say that longitude lines on the Earth are parallel? Or how do we prove that latitude lines are or aren't parallel?

Arctic Char
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Shirish
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    I might be wrong, but it seems to be that this statement is very informal – Didier Feb 26 '23 at 10:51
  • @Didier: Do you mean that there is no well-defined notion of "curves parallel to each other" on a smooth manifold? – Shirish Feb 26 '23 at 15:42
  • Wald is a physicist, not a mathematician. If you want to have a rigorous treatment of the subject, without meaningless sentences like "A space will be curved if and only if some initially parallel geodesics fail to remain parallel", then read O'Neill "Semi-Riemannian Geometry." – Moishe Kohan Feb 26 '23 at 15:58
  • @ShirishKulhari As far as I know, such a notion does not exist. Some might have tried to generalize the notion to (semi-)Riemannian manifolds, but I'm skeptical regarding a potential uniqueness / usefulness of such a notion. – Didier Feb 26 '23 at 16:05
  • @MoisheKohan: Thanks! I'm actually studying something similar to what you suggested ("Semi-Riemannian Geometry" by Newman). But I heard Wald is a mathematically rigorous GR book, plus as a beginner it's tough to know which sentences are meaningless. I suppose I'll have to take some stuff in Wald with a pinch of salt. – Shirish Feb 26 '23 at 16:23
  • Here’s one way I think of this statement: Suppose you have two geodesics and a geodesic segment connecting them that is orthogonal to each geodesic. Then you could define parallel in various obvious ways, and it is easy to see, using Jacobi fields that the geodesics are parallel if and only if they lie in a flat totally geodesic submanifold. – Deane Feb 26 '23 at 20:38
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    @MoisheKohan, I don’t see what is wrong with such a geometrically intuitive informal statement even in an otherwise rigorous math book. The infinitesimal version has a rigorous formulation using Jacobi fields. – Deane Feb 26 '23 at 20:49
  • @Deane: Yes, every professional differential geometer can make sense of it in several different ways. But unloading this undefined notion onto a novice without an explanation is plain wrong. Would you do this when writing a textbook? – Moishe Kohan Feb 26 '23 at 21:46
  • @MoisheKohan, it depends. Ideally, you back up any intuitive statement like this with either a rigorous statement or a reference to another source. But that's not always possible. I don't think it hurts to express thoughts that are left to the reader to wonder about and explore if they are sufficiently intrigued. – Deane Feb 26 '23 at 23:14
  • Another suggestion: This is worth posting on MathOverflow, and there are some people there who might give really interesting answers. – Deane Feb 27 '23 at 01:10
  • @Deane: Thanks! So from what you said, there's a notion of curves being "locally" parallel? So in the domain (the submanifold you talked about) where they're parallel, they still won't converge. Secondly, the notion of parallelism seems to require a metric (since you talked about orthogonality) and seems to be applicable only to geodesics (not general curves)? Finally, I'll follow your advice and try and ask this on MO – Shirish Feb 27 '23 at 06:45
  • @Deane I can think of at least three different ways to define a notion for "two curves are parallel", (using whether they intersect of not, whether the distance between two unit parametrization is constant / bounded, and using Jacobi fields) and it is very not clear to me which one would be useful in that context, if any. – Didier Feb 27 '23 at 08:31
  • @Didier, need it be useful? – Deane Feb 27 '23 at 14:46
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    @Deane Perhaps I should have written meaningful – Didier Feb 27 '23 at 18:23

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This vague statement is merely a comment in the introduction to the chapter on curvature. You just need to keep reading. I recall later on Wald introduces the notion of parallel transport, and the Jacobi equation. The relationship between parallel-transport and curvature, and also Jacobi's equation is also discussed in any Riemannian geometry text, e.g Lee's. On MSE itself, see this and this.

Anyway, since you're so eager, let me briefly outline how Jacobi's equation (not to be confused with the Jacobi identity) gives some nice qualitative insight to the interpretation curvature. The setup is as follows: start with a smooth manifold $M$, a torsion-free connection $\nabla$ on the tangent bundle $TM$, and let $I,J\subset\Bbb{R}$ be open intervals around the origin, and suppose you have a smooth mapping $f:I\times J\to M$ such that for each $s\in J$, $f(\cdot, s)$ is an affinely parametrized geodesic (also called autoparallel since we're being general and not assuming a metric-compatible connection) in $M$. We think of the parameter $s\in J$ as the 'variation parameter', and $t\in I$ as the 'curve parameter' (they're also called the transverse and longitudinal parameters respectively). The mapping $f$ gives rise to two vector fields along $f$, which by common abuse of notation, I shall denote as $T=\frac{\partial f}{\partial t}$ and $S=\frac{\partial f}{\partial s}$ (really I mean $Tf\left(\frac{\partial}{\partial t}\right)$ and $Tf\left(\frac{\partial}{\partial s}\right)$ respectively). It is now a reasonably simple exercise (and also proved in the texts) that the vector field $S$ obeys the following equation along the mapping $f$ (here, torsion-freeness is used): \begin{align} \nabla_T\nabla_TS&=R(T,S)T. \end{align} Here, $R$ is the Riemann curvature endomorphism (i.e the full $(1,3)$ tensor field). This is known as the Jacobi equation, or the geodesic-deviation equation. So, we have the three vector fields

  • The vector field $S$ along $f$ is called in physics texts, the relative geodesic-separation vector field or simply the separation vector field. Heuristically, you'd draw little arrows from one geodesic $f(\cdot, s)$ to a nearby geodesic $f(\cdot,s+\epsilon)$
  • The vector field $\nabla_TS$ along $f$, we call the velocity of the separation field. Roughly, it tells us how the arrows we're drawing change as we move along the geodesics
  • The vector field $\nabla_T\nabla_TS$ along $f$, we call the acceleration of the separation field.

Each of these vector fields tells us how the nearby geodesics (i.e nearby $s$-values) behave as we let their curve parameter $t$ run along. To get a rough pictorial idea, draw a family of geodesics on the sphere, and try to visualize these vector fields. The statement Wald is making is the following:

Theorem.

In order for the Riemann curvature to vanish identically, it is necessary and sufficient that for every possible field of geodesics $f$, the acceleration of the separation field, $\nabla_T\nabla_TS$, vanishes identically.

The necessity is obvious from the Jacobi equation. The sufficiency also follows from the Jacobi equation, with a little more algebraic finesse: the hypothesis implies $R(T,S)T=0$ for all possible $T,S$. In particular, this implies that for all $p\in M$ and all $u,v\in T_pM$, we have $R_p(u,v)u=0$, i.e for all $u,v,w\in T_pM$, we have $R_p(u,v)w=-R_p(w,v)u$. Combining this with the skew-symmetry $R_p(u,v)=-R_p(v,u)$ and the algebraic Bianchi identity (which holds since we have a torsion-free connection) gives (I omit the point $p$ for convenience in notation) \begin{align} 0&=R(u,v)w+R(v,w)u+R(w,u)v\\ &=R(u,v)w+[-R(u,w)v]+R(w,u)v\\ &=R(u,v)w+2R(w,u)v\\ &=R(u,v)w-2R(v,u)w\\ &=3R(u,v)w, \end{align} and hence $R(u,v)w=0$, proving that the Riemann curvature vanishes identically.

So, what this is saying is that if the Riemann curvature is non-zero, then it is possible to find a family of geodesics $f$, such that when $t=0$, the vector field $\nabla_TS$ vanishes 'initially, nearby geodesics are parallel', but that at some later time $t$, $\nabla_TS$ is non-zero 'they fail to remain parallel' (so necessarily $\nabla_T\nabla_TS$ is non-zero). If you want pictures (some good, but some too imprecise), see MTW's tome

peek-a-boo
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  • There's actually alot more we can say regarding Jacobi's equation, both mathematically, and physically (e.g as a stepping stone from Newtonian to Einstein gravity). Lots of cool classical stuff out there, you just have to keep reading further. – peek-a-boo Feb 27 '23 at 07:55
  • Thanks @peek-a-boo. I think I'll read further and get back to your answer when I'm in a better position to understand it. There's a section on Jacobi fields that I'll check out – Shirish Feb 27 '23 at 09:04
  • The notion of geodesic deviation can possibly help to understand Wald’s comment. – Bufo Viridis Feb 28 '23 at 15:46
  • @BufoViridis yes... I talked about it in my answer – peek-a-boo Feb 28 '23 at 22:44