@0celo7 Around?

Yes, looking for something in Lee

But I am around

I think there's a typo on pg 136.

Lee?

No, G-P.

Shoulda mentioned that.

not in Ted's notes?

Don't see it.

what's the typo?

They divide both sides of an equation by their norms, but there's a plus on the numerator but a minus sign on the norm.

Equation 5 from top to bottom.

Looking

Aha, is this important?

It's a key lemma.

The important fact is that at $t=0$, it's $\vec v/|\vec v|$.

And it's defined for any $t$ small, no matter what sign, right?

OK, that's true.

But it seems to be a typo nonetheless.

Yeah, let Ted know so he can update the list?

I printed his list and stuck it in the front cover.

Guess I'm gonna do that. Thanks for the heads up. Good point about it not being important at $t = 0$

I wrote it on my list

@BalarkaSen Walschap.

200 pages.

Title?

Metric Structures in Differential Geometry.

The treatment is a little nonstandard, but it's very to the point.

You can skip the stuff on homotopy and vector bundles you know

Nice, I'll write down that name. Thanks.

I see it also discusses characteristic classes. That's a topology book!

He represents the cohomolgy classes using curvature tensors.

That's geometry.

Cool stuff, that'd be an interesting perspective to study from.

@0celo7 Sent. I should print the list out and stick it in somewhere too.

@BalarkaSen I want to write a little thing "Why the Levi-Civita connection and why the Riemann tensor"

I think there's a better way of explaining these things than is "standard"

But I'm stuck on page 1

I'd read it if I can understand it.

Why do we want Riemannian metrics in the first place?

Of course, when we have nice things like Morse theory it's clear how useful metrics are, but why do we need them in the first place?

From the little I know about them and have applied them, I think it gives manifolds a sort of rigidity.

@BalarkaSen I'll use lots of theorems on flows, but I'll probably write out all the proofs. I want to go over them again anyway.

From a topological point of view at least. E.g. give a vector bundle a Riemannian metric and you can say things like the unit $S^n$-bundle. That's useful. Or give a manifold a Riemannian metric and you have a specified top dimensional form. Useful.

@0celo7 how many minors does a 3x3 matrix have?

@Obliv I don't know any linear algebra.

@balarka how many minors does a 3x3 matrix have?

@BalarkaSen The top dimensional form in the oriented case, you mean?

@Obliv The only nontrivial bit is to compute the 2x2 minors. Do it by hand.

@0celo7 Right.

I'm going to assume 9

@BalarkaSen The second one is not very obvious, is it?

And the first requires you to know why sphere bundles are useful.

Before reading Bott & Tu I had no idea.

So I'm not really happy with either of those.

@0celo7 Not particularly, I guess not.

My prof said it's so we can measure curve lengths.

That was the original motivation.

@Obliv That is correct.

I guess we might want to ask what the 'speed' of a curve is.

An interesting exercise it to compute the number of k x k minors of an n x n matrix. I have forgotten the exact number, but it's a fun piece of combinatorics.

@BalarkaSen I say Petersen is not introductory because he does not give you the "big view" btw.

He assumes you know RG and doesn't motivate anything

It's more of a reference text.

@0celo7 Well, being topologically biased, I am not satisfied with that. I appreciate the original motivation though.

@BalarkaSen Riemannian geometry is the true generalization of Euclidean geometry.

The basic object of Euclidean geometry is the straight line.

But the existence of geodesics is very far removed from the basic idea of the metric.

Mhm.

Well, one could play fast and loose. It's pretty obvious that to define a curve speed we need a metric.

Then we define the length, and vary the length functional to get the geodesic equation.

It's a nasty computation, but the standard fare in physics books.

Have you heard of geometric group theory, btw? Out of curiosity. I guess you do not like algebra, but it uses loads of ideas from this sort of geometry. I personally find all of that fascinating.

@BalarkaSen I'm assuming that's not the same as Lie theory?

@balarka are there other ways to get the inverse of a matrix besides going through doing the minors, transposing, then dividing by the determinant of the matrix?

No, not really. It's a branch which talks about discrete infinite groups, contrary to the basic object of studies in Lie theory.

@Obliv Mathematica.

@Obliv Yes. Gaussian elimination with block matrix [A | I]

@0celo7 i've never even learned how to use that honestly. @balarka will search that up

I find it much more efficient than Cramer's rule.

@BalarkaSen Oh, what's something interesting from it?

OH that's what my friend was talking about. We were both trying to use matrices to solve a system of linear equations in our physics class and he was talking about row reduction and I was doing this thingy (it's called cramers rule?)

@0celo7 The basic idea is that you take a discrete infinite group, and look at a geometric object associated to it called a Cayley graph which is literally an infinite graph. There is a natural metric which comes on it and which makes it a metric space which admits geodesics. Then one can do all sorts of things with it, e.g., defining curvature etc.

Gromov's original paper was on hyperbolic groups, for which these Cayley graphs are "negatively curved".

Example being free group on 2 generators. Googling will immediately tell you how it' Cayley graph looks like and you can "see" why it's negatively curved.

I can see?

It looks very much like the Poincare disc, doesn't it? As you head towards infinity, the # of vertices increase exponentially, the edges has to shrink exponentially to fit inside a bounded region, etc.

Ah, ok.

Important thing is, look at the triangles. Take three arbitrary vertices, look at the "triangle" formed by the edge-paths connecting them, of minimum length (aka geodesics).

That looks like an extreme version of a triangle in $\Bbb H^2$, heavily deflated. A formalization of this was Gromov's original defn.

@BalarkaSen Ok, I'll take the curve approach. Metrics can give us lengths and angles between curves. Volume too but that's not easy to explain in a short thing.

Right.

@BalarkaSen Hmm.

Are the triangle edges allowed to cross?

Sure.

Then I think I see what you're talking about.

@BalarkaSen Do you consider the question "when is a given Riemannian manifold locally isometric to Euclidean space" to be fairly natural?

Yeah, in that category, I do.

"When are two objects isomorphic/anything close to being isomorphic" is always a natural question to ask.

If you try to answer that question and you know a little PDE, you get a nice equation

$$[\Gamma_i,\Gamma_j]+\partial_i\Gamma_j-\partial_j\Gamma_i=0$$

that's in matrix notation

that's precisely the Riemann tensor.

So the Riemann tensor can be "derived" as the local obstruction to flatness

Interesting, I never heard of this point of view.

@BalarkaSen The literature is dominated by two views:

define $R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$ and that's it.

or the physicist way

somehow parallel transport around a parallelogram is related to curvature (????)

so you compute the commutator of covariant derivatives (????)

and somehow that makes sense to those guys

:P

@BalarkaSen I think that's how Lee explains it.

I don't get it.

They usually try to excuse it somehow

It's indeed clear that parallel transport is path dependent on the sphere

I don't get it either.

I think they are trying to do holonomy someway.

But how that connects to commuting cov. derivatives is not obvious

I will say: there is a way to make that approach formal. But to do it correctly involves the covariant derivative along a map.

And you get the same result: Riemann tensor vanishes iff locally isometric to flat space.

@BalarkaSen Ah!

My approach also gives the zero torsion condition

I don't know torsion.

@BalarkaSen $T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$.

Oh

What does it represent?

Riemannian geometry has 0 torsion.

@BalarkaSen I have no idea. We always throw it away and the only people interested in it are string theorists.

@0celo7 As in, it's zero when $\nabla$ is the Riemannian connection?

Perhaps Sharpe's book on Cartan connections explains it.

@0celo7 lol

@BalarkaSen Yeah.

@BalarkaSen But no one ever explains why we throw it away!

It's equivalent to the condition $\nabla_X\nabla_Yf=\nabla_Y\nabla_Xf$, which seems reasonable.

It's also equivalent to $\Gamma^i_{jk}=\Gamma^i_{kj}$.

Oh right

And you need zero torsion for geodesics to be critical points of the length functional.

Maybe it has something to do with torsion as in diffgeo of curves?

Which measures how 'twisty' the curve is.

@BalarkaSen Maybe...

There's another reason.

If we eliminate torsion, the LC connection is the unique metric connection.

If we allow for torsion, there are lots of metric connections.

(Infinitely many)

weird

@BalarkaSen Another result: given any connection $\nabla$ we can construct another connection $\nabla'$ with zero torsion and the same geodesics.

That's nice

So if we're worried about some geodesics being "lost" by some crazy connection, we don't really have to worry.

It will have the same geodesics as the LC one.

But the LC connection has the benefit that the geodesics actually minimize curve lengths locally.

I think I'll include that proof in my notes

@BalarkaSen I guess it would be nice to compile all of these results.

It'd be nice as a note.

Well, a really long note.

Yeah.

@BalarkaSen Milnor says the torsion should vanish so that $\nabla^2_{X,Y}f$ is symmetric in $X,Y$.

But I don't like the definition of $\nabla^2_{X,Y}$.

$\mathscr D$

pretty.

@BalarkaSen The connection you get that way is actually unique.

Every once in a while I find weird stuff like this