The h Bar

12:32 PM

@bolbteppa I'm not sure what are you trying to show here. These are just the standard calculations to derive the Christoffel symbols of polar coordinates, no? These aren't affected by the discussion at hand, because you are just dealing with a coordinate system in $\mathbb R^2$ and calculate $\mathbf e^i\cdot\partial_j\mathbf e_k$

It's a simple example of actually computing covariant derivatives, what do you mean they aren't affected by the discussion at hand or "just" doing something as if there's something it's missing

1:07 PM

ok, but I'm not particularly concerned with how to do calculations right now. I know how to do those (well, in simple cases at least). I only care about understanding better the underlying ideas. Which I don't believe to be a simple matter of notational convenience. We are presumably defining a specific mathematical object here, a connection, which then also happens to provide notational/formal advantages.

Here's my current understanding of the matter: we consider a manifold $M$ embedded in an ambient Euclidean space $\mathbb R^n$. Here, we could define a "natural" covariant derivative as …

as per my comment above: I believe we are thinking about this in different ways. You are viewing it as a simple formal tool to do calculations. I know how to do calculations, that was never my issue. My issue was with understanding the underlying mathematical structure (connection) that makes that "notational simplicity" possible. In other words, for me the connection is not a notational trick, but rather a (hopefully) well-defined map $\nabla:\Gamma(TM)\to\Gamma(T^* M\otimes TM)$
(or equivalently, the covariant derivative a map $\nabla_X:\Gamma(TM)\to\Gamma(TM)$ for all vector fields $X\in…

now, this said, we also know that a connection is characterised by the way it acts on a basis. This is direct consequence of its defining axioms. That's one of the reasons I was having troubles making sense of $\nabla'_i\mathbf e_j=0$. But I now understand this because this $\nabla'$ is specifically designed to make this satisfied for this specific basis. It won't hold for other bases, so all's good

at the same time any Levi-Civita connection is tightly tied, via its Christoffel symbols, to the underlying metric. The Christoffel symbols of $\nabla'$, wrt the basis $\mathbf e_i$, vanish: $\Gamma_{ij}^k=0$. Which makes it seem likely that this connection is not Levi-Civita (also, it's probably trivial to see that the Levi-Civita connection in this case would just be the canonical directional derivative, possibly projected on the embedded manifold).
So this makes this notation an interesting case of a fruitful application of a non-Levi-Civita connection.

Or maybe I'm completely wrong lol

1:32 PM

I'm trying to tell you it's just a derivative, that's all it is. We get a new concept not immediately familiar from basic calculus because we're differentiating the basis vectors, we now need to re-write the derivative of a basis vector as a linear combination of the original basis, we call the coefficients 'connection coefficients'. I specifically did the calculation above directly first, then repeating it with connection coefficients, to show how easy it is. Now, when you write $\nabla' \mathbf{V} = (\partial_i V^j) \mathbf{e}_j$ that's wrong, and missing indices, and where did it come fr…

@bolbteppa yes, of course, I meant to write $\nabla'_i\mathbf V$ there. Where it comes from is the definition of covariant derivative with the constraint $\nabla_i'\mathbf e_i=0$. By definition, a covariant derivative must satisfy Leibniz' rule, i.e. $\nabla_X(fY)=X(f) Y+ f\nabla_X Y$. Which becomes $\nabla'_i\mathbf V=(\partial_i V^j)\mathbf e_j$ here.

No, you simply cannot end up with just a partial derivative acting on $V^j$ alone and call it a covariant derivative in general

the notation $\nabla_i V^j$ just doesn't really make sense to me. Well, it makes sense if understood as $\nabla_i V^j\equiv \mathbf e^j(\nabla_i (V^k\mathbf e_k))$, but nothing more than that. A covariant derivative is an object that acts on vector fields, not functions. The components of a vector fields are functions, hence $\nabla_i V^j$ is not, strictly speaking, defined

The notation $\nabla_i V^j$ is extremely standard notation and I've defined it at least five times above

@bolbteppa why not? This is referring to flat Euclidean spaces, so the directional derivative is well defined

sure it's standard. And it's fine to use it, but what it means is that you are taking components of the covariant derivative of the vector field itself. Otherwise, you still haven't told me how you define the covariant derivative itself. Yours is sometimes a map $C^\infty(M,\mathbb R)\to C^\infty(M,\mathbb R)$, and somtimes a map $\Gamma(TM)\to\Gamma(TM)$.

Connection (vector bundle)

1:45 PM

This is becoming incoherent, you want to define a special map $\nabla_i'$ which satisfies $\nabla_i' \mathbf{e}_j = 0$ and then apply it to a vector field $\mathbf{V} = V^j \mathbf{e}_j$ so that you get
$$\nabla_i' \mathbf{V} = \nabla_i' (V^j \mathbf{e}_j) = (\nabla_i' V^j) \mathbf{e}_j + V^j (\nabla_i' \mathbf{e}_j) = (\partial_i V^j) \mathbf{e}_j$$
without explaining why $\nabla_i' V^j$ became just $\partial_i V^j$ and completely ignoring the Christoffel symbol part for no reason, this is simply not a covariant derivative. Further you say a covariant derivative acts on vector fields not f…

In mathematics, and especially differential geometry and gauge theory, a connection on a fiber bundle is a device that defines a notion of parallel transport on the bundle; that is, a way to "connect" or identify fibers over nearby points. The most common case is that of a linear connection on a vector bundle, for which the notion of parallel transport must be linear. A linear connection is equivalently specified by a covariant derivative, an operator that differentiates sections of the bundle along tangent directions in the base manifold, in such a way that parallel sections have derivative zero...

2:06 PM

Compare what I've written above to what's in this section

2:19 PM

@glS After pondering this for a while, Vincent Thacker's answer is right - what the guy in the video uses is not a covariant derivative, it's defined with the wrong sign. Oddly enough, this works out so that his "covariant derivative" is actually the difference between the ambient cov. der. and the surface cov. der., i.e. the second fundamental form or (extrinsic) curvature tensor, but crucially it's not a connection,

it's a map from the tangent to the normal bundle that only makes sense in the embedded context.

so you really shouldn't try to make sense of what's happening there in terms of connections

@bolbteppa I completely agree with what is written in that section of Wikipedia. Where are you seeing them apply the covariant derivative to a function there, precisely? They explicitly write $\nabla_X:\Gamma(E)\to\Gamma(E)$. Here, $\Gamma(E)$ is the space of sections of a vector bundle. In our case, $E=TM$, and $\Gamma(E)$ is the space of vector fields

you might be getting confused for the fact that they write $fs$ where $f$ is a function. But $s$ is a section here, so $fs$ is a section as well. That's because the space of section is a $C^\infty(M)$ module, i.e. sections can be multiplied by functions to give other sectiosn

When they say $\nabla$ satisfies the Leibniz rule, they skip a line they should write $\nabla (f \mathbf{s}) = (\nabla f) \otimes \mathbf{s} + f (\nabla \mathbf{s})$ and then they define $\nabla f = df$ and $\nabla$ acts a different way on $\mathbf{s}$

@bolbteppa except that's not what they are saying, is it? The Leibniz rule is explicitly defined with the differential. If if was $\nabla f$, that would be an abuse of notation, because you are then overloading the symbol $\nabla$. Which is fine.. as long as one understands what the notation is actually saying

@ACuriousMind mh, that's what I'm converging to as well, yes. I was planning to ask another question on math.SE to get the perspective of more people that actually work with the more abstract setting

Derivation (differential algebra)

It's simple, the Leibniz rule just means the product rule, bringing up the Leibniz rule in the context of discussing $\nabla$ acting on $f \mathbf{s}$ means $\nabla (f \mathbf{s})$ expands like in the Leibniz rule, it's incoherent if you say $\nabla$ satisfies the Leibniz rule and then you can't write $\nabla (f \mathbf{s}) = (\nabla f) \otimes \mathbf{s} + f (\nabla \mathbf{s})$

The Leibniz rule far more general than just differentials of functions

In mathematics, a derivation is a function on an algebra which generalizes certain features of the derivative operator. Specifically, given an algebra A over a ring or a field K, a K-derivation is a K-linear map D : A → A that satisfies Leibniz's law: D ( a b ) = a D ( b ) + D ( a ) b . {\displaystyle D(ab)=aD(b)+D(a)b.} More generally, if M is an A-bimodule, a K-linear map D : A → M that satisfies the Leibniz law is also...

cOnnectOrTR 12

2:41 PM

Hi @PM2Ring

3:24 PM

If a 3 manifold has flat level sets, does that mean the global geometry is also flat?

level sets are things a function has, not a manifold

I mean slicing the manifold say at $z=c$ for a constant $c$

and the slices are all flat planes

All 1d manifolds are flat, so slicing a sphere in any way will get you flat slices but a sphere isn't flat

right

should have thought of that

Scientist Smith YT

3:52 PM

@antimony This is in response to the magnetic field question I asked. So the current is a large element in magnetic field strength. Is there a rule or something that would say that I should add say 1.5 volts every 2 or 5 amps? Or say I need 1 ohm per volt and amp or something of that sort?

Slereah

4:09 PM

@ACuriousMind bends your line

:O

Nihar Karve

why are there so many words in all these topological order reviews

where are the equations

Ronald Villiers

4:39 PM

So in QM, matter waves represent a probability distribution of where the matter is right?

as in we don't observe a typical wave passing through a medium, but it works out in predicting all of the quantities and observations

kinda analogous to virtual particles, they're real in the sense that they provide a mechanism to explain the phenomena but not like an observable thing

4:55 PM

Vaguely you could probably say that

Manas Dogra

Hi guys, can you give me an example of a classical mechanical system whose lagrangian depends atmost linearly on velocity...

just trying to find one simple example of a case where constrained hamiltonian dynamics is necessary..

JingleBells

5:18 PM

How do business leaders and other professionals resolve arguments that are based on anecdotal evidence? Let's say I feel that a certain game update will result in positive likes, and my partner thinks the same game update will result in negative likes. We both feel in our guts that our scenario will happen, but neither of us has empirical evidence to show. What do we do?

@ManasDogra how about the example in the first 3 pages here

Manas Dogra

7:23 PM

@bolbteppa Yeah I saw that example in Ashok Das' book on QFT...it's a mechanics example but not linear in velocity...

I am thinking of some potential which depends quadratically on velocity so that it cancels the kinetic energy term...and it may have a linear term too...so the potential must be of the form v^2+v...but can't find a suitable real system with a potential like that!(obviously)

Anyway I have another doubt. Arnold says-"The hamiltonian point of view allows us to solve completely a series of
mechanical problems which do not yield solutions by other means (for
example, the problem of attraction by two stationary centers and the problem
of geodesics on the triaxial ellipsoid)"

Why are these problems not solvable by Lagrangian or Newtonian formulation?
I thought a problem which is solvable in one formulation is solvable in all others(keeping the restrictions in mind).