@DanielSank Ok, think about this. The abstract index labels the "slot" of a tensor, right?

So what on Earth does the $a$ on $\partial_a$ label?

Now you could have two indices, $\partial_{\mu a}$.

Or not, that doesn't make sense either :P

@Danu A diffeo can just be $C^1$, right?

@0celo7 Sure

@Danu So why does the pushforward, viewed as a map of tangent bundles, have to be differentiable? It's already given in local coordinates by the Jacobian, so you'd need to be $C^2$ for the pushforward to be differentiable.

@0celo7 Right.

@0celo7 Dunno. Never thought about it. Probably it's how something changes as the argument labeled a changes.

@DanielSank Uh, what?

You asked what the a labels.

It labels what you're differentiating with respect to.

How is that confusing?

What direction are you differentiating in?

@0celo7 Where are you getting this from?

@Danu It's an exercise in Arnold, why?

I mean, the differnetial is a linear map

@Danu Yes

No, idk

I feel like I'm not getting your statement

The differential is just the Jacobian! No reason for it to be differentiable!

No, the differential can be expressed in coordinates as

@0celo7 None. It's like a gradient of a function. Gradient-ing a function gives you a tensor which eats vectors to give you a directional derivative.

dammit

Right, but the definition of "differentiable map" is that, when expressed in coordinates, its partial derivatives are defined.

@0celo7 That's a stupid definition.

@0celo7: Have you examined Arnold's definition of a diffeomorphism before asking random people what they think a diffeomorphism is?

So if $f:M\to N$ then $f_*:TM\to TN$ so "$f_*$ is a differentiable map" means it is a differentiable map of local bundle coordinates

@ACuriousMind This should hold for all maps, I just said diffeo for no good reason

At least, he doesn't specify if this only holds for a certain class of maps

I can at least tell you that the differential of a linear map is just the map itself---did you mean a statement like that?

@DanielSank Uh, what do you propose?

@0celo7 I dunno, but it seems odd to define stuff like this in terms of properties of representations (i.e. coordinates).

@DanielSank I see what you're saying.

@DanielSank Dude, tell me about it. These coordinate proofs take forever.

Lee's/Straumann's are so much more elegant.

@Danu Hm? What do you mean?

@0celo7 What do you mean, "what do you mean"? I mean exactly what I said, it's not ambiguous :P

@Danu Linear map from what to what?

Like $f:M\to N$?

How do you define "linearity" there?

$d F$ is a linear map

@Danu Considering the differential as a map $TM\to TN$ it makes indeed no sense to say "it is a linear map". What is linear are the individual maps $T_x M\to T_{f(x)} N$.

@ACuriousMind Sure, that's true

unless

as a map $\Omega(M)\to C^{\infty}(M)$

hurr durr

almost made it pointwise again :P

How do you want to consider a map $TM\to TN$ as a map $\Omega(M)\to C^\infty(M)$? And what exactly is $\Omega(M)$ supposed to be, the space of forms on $M$?

@Danu It's amazing how much of the technical discussion in this chat is math.

@DanielSank Physics is just too non-technical

@Danu Facepalm

@DanielSank Knew that one was coming

@ACuriousMind I was talking about $dF$ and yes, then it wouldn't be regarded as a map $TM\to TN$

I was just being facetious---show some humor (oh wait your kind ain't got none of that) :D

(and yes $\Omega(M)$ are the forms)

@DanielSank tensor indices can be viewed are labels on function arguments? As $T^{ijk}_w=T(\theta^i,..., e_w)$

?

back

@Danu that was the previous exercise

I'm asking why $dF$ is differentiable

because I think I have shown it isn't

@dmckee can you delete the "Me and MAR" chat room?

@NeuroFuzzy huh?

@dmckee why did you put a bounty on that electron question? I didn't understand your reasoning...

@NeuroFuzzy no, in the following sense: $\omega_a$ is an object $\omega$ that can "eat" another object if it has an upper index $a$

and that object is precisely a vector $v^a$ (in this case)

@TanMath Because of physics.stackexchange.com/questions/215758/… .

so the abstract $\omega(v)$ is then $\omega_a v^a$

Today's question is a duplicate of the earlier one, but the OP is justifiably unhappy with the answers on the original. I want that second condition to not happen again.

the confusing bit is that if $i$ is a coordinate label, this is also $\omega_i v^i$

@0celo7 I'm pretty sure Arnold is working in the smooth category.

(making your question obsolete)

@Danu pretty sure because otherwise I'm right and it's not necessarily differentiable?

It's obvious from his definitions

where

I just found your exercise and checked out its page

The definition of differentiable map

he never mentions degrees of differentiability

@TanMath It will go away on it's own if you stop using it. I don't want to be in the habit of dealing with individual chat rooms unless an exceptional circumstance develops. That's what mods are: human exception handlers.

it doesn't say smooth

This cannot happen unless he's in the smooth case

differentiable means at least $C^1$

smooth means smooth

Take it or leave it :P

just so we're clear, if the map is $C^1$, then I'm right and $df$ isn't differentiable?

Exercise: Find a $C^1$ diffeomorphism with non-differentiable differential :P

Once again, I'm not your personal tutor

doesn't have to be a diff!

I said that for no good reason

@Danu maybe you should be

@0celo7 If you'd pay me...

@0celo7 why are you reading Arnold, Hawking/Ellis, string theory books etc if you can't factor quadratic polynomials?