Think about this. Say you start at the north pole holding a javelin that points horizontally in some direction, and you carry the javelin to the equator, always keeping the javelin pointing "in as same a direction as possible", subject to the constraint that it point horizontally, i.e., tangent to the earth. (The idea is that we're taking "space" to be the 2-dimensional surface of the earth, and the javelin is the "little arrow" or "tangent vector", which must remain tangent to "space".) After marching down to the equator, march 90 degrees around the equator, and then march back up to the north pole, always keeping the javelin pointing horizontally and "in as same a direction as possible".

By the time you get back to the north pole, the javelin is pointing a different direction!

That's because the surface of the earth is curved.

More generally, a space (or spacetime) is "curved" if when we take an arrow and carry it around a loop, doing our best to keep it the same length and pointing in the same direction, it may come back rotated. detail.

Bronis Vidugiris comments: "Unfortunately, from this description I imagine a gyroscope (allowing the arrow to pivot vertically so it is always parallel to surface) attached to the arrow, and the thing winding up pointing in the original direction. :-( "

Why you imagine a gyroscope when I say "javelin" is beyond me.

Bronis adds: "Also I imagine being at/near the north pole, going down to the equator, back up, and being very confused about the forth leg! (I can only go south from the north pole - what do I do _now_?)."

There ain't no fourth leg. Where'd I say there was a fourth leg? True, in my later description of the Riemann tensor I talked about carrying a tangent vector around a little square. But here we are carrying it around a big "triangle". In fact, one can do this parallel translation game for any sort of loop that ends where it started, or even any path.

Bronis asks: "Can you (while retaining this degree of informality) also achieve the precision to avoid the above (presumably unintentional) interpretation of what you actually meant by this?"

Well, I really meant just what I said. Say you were a Roman gladiator up at the north pole (historical accuracy not being my strong point) and you were handed a javelin. "Hold it horizontally, pointing that way," says your commander, pointing at a lump of ice at the horizon. You do so smartly, an exemplar of military precision. It points off to your left. "Now march forwards! Go straight ahead, and never rotate the javelin in the least, under penalty of death! Stop when you reach the equator!" And so on. You march along, never letting the javelin sway or rotate in the least.

Ed Green asks: "I'm sorry, but I have a really obvious question here, so obvious you didn't address it. What do we do with the javelin at the corners? Do we try to hold it "pointing in the same direction" as near as possible, while we turn, or do do we rotate it with us by the angle we turn through?"

Hold the javelin pointing in the same direction even at the corners! The idea is to do your damnedest to never rotate that sucker, even at corners. We want to study how, even when we do our best not to rotate a tangent vector as we carry it around, it can come back rotated due to the curvature of spacetime (or space, or the earth).

Also, since corners are just a limiting case of sharp turns without corners, it would be a bad rule to do something different at corners than at non-corners.

Ed Green adds: "This particular path to the equator and back, walking the outline of an imaginary segment of a chocolate orange, contains the gratuitous symmetry of constant compass direction."

Compass direction!? As you know, any attempt to take coordinates seriously will only confuse us here when studying general relativity, so clearly parallel translation can have nothing to do with something like "constant compass direction". Parallel translation is something you can do on any curved space or spacetime whatsoever, with no reference to any compass or map.

So, for example, if you want to avoid symmetries, you can parallel translate a vector around the coastline of Eurasia. It's just a bit hard to talk about that example using ASCII.

Ed Green comments: "A reasonable person may have some doubt about the well-defined nature of "trying very hard not to rotate the arrow". The point of the exercise is after all, try as hard as we like and the damn thing will have rotated >when we get back! But just how "hard" is "very hard"... ?"

The key thing is the *local* nature of parallel transport. I think I noted this in my first explanation: at *each step of the way* you do your best not to rotate the vector. It's cheating to know your route ahead of time and sneakily diddle with your vector so that it comes out pointing the same way at the end of the journey.

I think you know this and are just trying to cause trouble. :-)

Let's go back to the example I originally gave.

Say our Roman starts at the north pole with his javelin. Say his javelin points directly in front of him as he begins his journey down the meridian to the equator. Assuming the earth is a perfect sphere and he doesn't gratuitously swing his javelin from side to side, he will end up at the equator with the javelin pointing due south. Now he turns 90 degrees --- NOT rotating the javelin, of course!!! --- and walks along the equator due west. The javelin continues to point due south, to his left. He goes 90 degrees along the equator and stops. He does NOT sneakily rotate his vector because he guesses what's coming. He's a Roman soldier, after all, trained to follow orders. He turns 90 degrees again until he's facing north. He does NOT rotate the javelin --- jeez, how many times do I need to say this: he doesn't EVER rotate that javelin --- so it is still facing due south, directly behind him. He now marches up the meridian to the north pole, carrying the javelin pointing directly behind him, not rotating it even a teeny weeny little bit. When he returns to the north pole the javelin is pointing a different direction than when he started. In fact, it's rotated by an angle of 90 degrees from his initial position.

For fun, notice that if we think of the earth as a unit sphere, it's area is 4 pi, and our Roman has just travelled around a region of land having area one eighth of that, hence pi/2. His javelin has rotated by an angle of pi/2! This is no coincidence: on the unit sphere, whenever you go around a simple closed curve enclosing an area A, parallel translation gives a rotation of angle A.

Ed Green comments: "It's like SR... when we speak of length contraction, that means we have tried "very hard" to measure the correct length... but not too hard! If we tried "hard enough", ie, allowing for the Lorentz transfomation, we would measure the correct rest length! And if we tried hard enough here, maybe doing a bit of local surveying, perhaps we *could* triumphantly come back to the same orientation, even in curved space!"

Of course you can always cleverly correct for things, but that's not the point. DON'T cleverly correct for things. In the case of Lorentz contractions, just read what the damn ruler says. In the case of parallel translations, just follow orders like a Roman soldier and don't ever swing that javelin around.

Ed and Bronis remain doubtful. Oz points out: "It would seem to me that (despite all the contortions to the contrary) the legionary is in fact *locally* in a flat spacetime. We can make it as arbitrarily as flat as we like by considering a small enough local area (<<epsilon). In this case there is no problem with local angles, the legionary can head off and maintain at any local angle he likes, and hold the javelin (as long as it's not too long!) at any angle to his path as is required, quite sensibly. Everything he does locally is quite flat, valid and uncomplicated. He has no idea that "globally" his little local area is being rotated by his path though the larger geometry."

Later, Ed sighs: "Thank you, Oz! After letting that thought simmer overnight, that really seems to clear up my CD (cognitive dyspepsia) here."

Ed adds: "I think the *two* infinitesimal distance scales are really the key to thinking about this sensibly."

"On the one hand, we let epsilon get small enough that *something* is effectively constant over the dimensions of our little path, that something presumably being the "curvature" or the Riemann tensor. On the other hand, we then let distances shrink still further, an "infinitesimal of an infinitesimal", in considering the legionary's path, so that space is effectively flat right in front of his face. Then presumably there is no ambiguity, even for drunken wanderings on this scale, in what we mean by "holding the javelin at a constant angle". We piece together all these epsilon-tisimo bits of path to form the path of dimensions or order epsilon."

"I am not sure this view is totally coherent, but it has the strong enough flavor for me of something on its way from speculation into rigor that I think I am about ready to accept it and move on. I assume this insight could be made rigorous."

Yes, it can. Let me try to restate it a little more mathematically, just for fun. Let's take any old n-dimensional space with a Riemannian metric on it. Say we want to parallel transport a vector along a curve. Since this is a "local" process, it suffices to parallel transport it from point P along an itsy-bitsy stretch of the curve, of length epsilon, to the nearby point Q.

So: consider a little patch of spacetime, whose length, width, etc. are all about epsilon, and which contains the curve from P to Q. It's not flat, but it's almost so. Flatten it out, i.e., take a flat metric that closely approximates our original metric in the patch. Now that we're in flat space, parallel transport is unambiguous! So do parallel transport that way.

But wait --- this "flattening out" process was ill-defined: there might
be different flat metrics that approximate our original metric pretty
well in the patch. That's true, but if two people do this flattening
out in different but still reasonable ways, it will only affect the
answer by an amount of order at most espilon^{2}.

This means that if we take a long curve of length L and chop it into L/epsilon pieces of length epsilon, and do the above trick repeatedly, the accumulated error will be at most of order L epsilon, when epsilon is sufficiently small.

So chopping ever finer we can get as good an approximation as we want.

[If any of you ask how the "flattening out" business works I'll challenge you to a duel with javelins at 50 paces. If your manifold is embedded in a metric-preserving way in some higher-dimensional space, this is easy: just project onto the plane tangent to the manifold.]