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#41 Xerxes

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Posted 7 March 2017 - 07:14 PM

The tangent space is an n-dimensional vector space spanned by the partials.

Yes. In fact these are called differentiable operators, and are closely related to the directional derivative. They are also the closest we can get, in an arbitrary manifold, to the notion of a directed line segment that is used to define vectors in Euclidean space.

Anyway, recall I wrote the property of linear independence for these bad boys as \sum\nolimits_{j=1}^n \frac{\partial}{\partial x^j}x^k = \begin{cases}1 \quad j=k\\0\quad j \ne k \end{cases}

Yes the K. delta is a tensor - its called a "numerical tensor", rather special case.

Anyway, from the above, the following is immediate...

If I accept these differential operators as a basis for T_mM then I can write an arbitrary tangent vector as v=\sum\nolimits_{j=1}^n \alpha^j \frac{\partial}{\partial x^j} so that

v(x^j) = \alpha^j which is unique to this vector.

Anyway.....

Suppose the point m \in M and space C_m^{\infty} of all smooth functions  M \to \mathbb{R} at m.

Recall I defined the tangent space at m as space of mappings T_mM:C_m^{\infty} \to \mathbb{R} so that v(f) \in \mathbb{R}

For the mapping f:M \to \mathbb{R} I now define the differential df:T_mM \to \mathbb{R}. This is sometimes called the pushforward - see my post http://www.sciencefo...ng-and-dualing/

I insist on a numerical identity df(v)= v(f) for some f \in C_m^{\infty} and and any v \in T_mM

To see why we care, let me replace the arbitrary function f by the coordinate functions x^j so that dx^j(v)=v(x^j)

I now replace the vector v \in T_mM by the basis vectors \frac{\partial}{\partial x^k} so that

dx^j(\frac{\partial}{\partial x^k})=\frac{\partial}{\partial x^k}(x^j)

So we know that the RHS is \frac{\partial}{\partial x^k}(x^j)= \delta ^j_k, so that the LHS implies that dx^j and \frac{\partial}{\partial x^k} are linearly independent.

But since the basis for T_mM is already complete, we have to say that the dx^j are a basis for another but related vector space space.

This is called the dual space and is written T^*_mM.

Note the existence of the dual space is thus a mathematical inevitability, not a mere whim

PS Note this is not a unique situation in mathematics. Consider the space of eigenvectors - the eigenspace - obtained by the action of an operator on a vector space.
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#42 wtf

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Posted 7 March 2017 - 10:35 PM

You're two posts ahead of me FYI. I haven't worked through the earlier one yet. Been a little busy with other things.


Note the existence of the dual space is thus a mathematical inevitability, not a mere whim


I'm thinking that you are intending this remark as a response to my questions about why dual spaces creep into tensor products, but I don't think you are understanding my question then. Of course I understand what dual spaces are. But in the algebraic definition of tensor products, duals NEVER show up; while in diffGeo/physics discussions, they ALWAYS show up. That's the gap I'm trying to bridge. Apparently no algebraist has ever set foot in the same room as a differential geometer, else there would be a clear and simple explanation of this expositional mismatch somewhere.

I hope to get through the earlier post today or tomorrow or the day after.
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#43 wtf

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Posted 8 March 2017 - 06:33 AM

Ok now that I'm going through this I'm completely confused by where all this is taking place. We don't know how to take derivatives on a manifold yet but your notation is assuming that we can.
 

First the boring bit - notation. One says that a function is of class C^0 if it is continuous. One says it is of class C^1 if it is differentiable to order 1.


Picky refinement, my understanding is that a C^1 function has a continuous derivative. There are functions with derivatives that fail to be continuous on one or more (even infinitely many) points. http://math.stackexc...e/292380#292380
 

One says it is of class C^{\infty} if it is differentiable to all imaginable orders


More pickiness, this is a trivial point but of course you mean for all positive integer orders. Then it's no longer a function of someone's imagination. I was thinking fractional derivatives, who knows what else.
 

in which case one says it is a "smooth function". I denote the space of all Real C^{\infty} functions at the point m \in M by C^{\infty}_m


Ok here is an expositional problem that confuses me. This is not pickiness, I'm genuinely confused. We've been letting M stand for a manifold. But we don't know how to differentiate a function on a manifold. In fact you said that the charts are only homeomorphisms, so for all we know our manifold M is so full of corners it can't be differentiated at all. In order to get past this point I have to either assume we've defined differentiability on a manifold somehow, or else that we're working in \mathbb R^n. I hope you will clarify this point.

 

So recall from elementary calculus that, given a C^1 function f:\mathbb{R} \to \mathbb{R} with a \in \mathbb{R} then \frac{df}{da} is a Real number.


Little point of notational confusion. I'd believe \frac{df}{dx}\biggr\rvert_{x=a} or \frac{df}{dx}(a) but I'm not sure about your notation. Is that a typo or a standard notation?
 

Recall also that this can be interpreted as the slope of the tangent to the curve f(a) vs a.


Yes.
 

Using this I make the following definition:

For any point m \in U \subsetneq M with coordinates (functions) x^1,x^2,....,x^n then I say a tangent vector at the point m \in U \subsetneq M is an object that maps C^{\infty}_m \to \mathbb{R}


Now you see I have the M problem in spades. I see you talking about tangent vectors to a point on a manifold but I have no idea how to define differentiability on a manifold. Rather than look it up I thought I'd just ask.

Of course if we're in \mathbb R^n this is clear.

This is still an interesting point of view even if I imagine that we are talking about Euclidean space and not manifolds. We're fixing a point and letting the functions vary. If we are in single-variable calculus, we can let x = 1 for example, and then \frac{df}{dx}(1) : C^\infty_1 \to \mathbb R is a function that inputs x^2 and outputs 2, inputs x^3 and outputs 3, inputs e^x and outputs e, and so forth.

You see I'm still bothered by your notation. Did you really want me to write \frac{df}{d1} as you indicated earlier? I have a hard time believing that but I'll wait for your verdict.

It's clear to me that by the linearity of the derivative, \frac{df}{dx}(1) is a linear functional on C^\infty_1. But the domain is the real numbers, not some arbitrary one-dimensional manifold that I don't know how to take derivatives on. For one thing don't we need an algebraic and metric structure of some sort so that we can add and subtract vectors and take limits?

So I do sort of see where you're going with this. But I'm totally confused about how we lift the differentiable structure of \mathbb R^n to M.
 

so that, for any f \in C^{\infty}_m and since m = \{x^1,x^2,...,x^n\} we may write v=\frac{\partial}{\partial x^1}f + \frac{\partial}{\partial x^2}f+....+\frac{\partial}{\partial x^n}f.

Or more succinctly as v= \sum\nolimits^n_{j=1} \frac{\partial}{\partial x^j}f \in \mathbb{R}.


Ok I believe this. Or maybe not. First, you are using those set brackets again and I do not for the life of me see how that can make any sense. There's no order to sets so how do you know which coordinate function goes with which coordinate? Secondly of course there is the manifold problem again, I don't know how to define a differentiable function on a manifold.

Now if I forget manifolds and pretend we're in \mathbb R^n then I suppose we could define the functional v=\frac{\partial}{\partial x^1}(m)f + \frac{\partial}{\partial x^2}(m)f+....+\frac{\partial}{\partial x^n}(m)f. I would almost believe this notation as I have written it.

This particular functional is defined at the point m. However I see that you've left that part out and you're defining this functional for all points? But then it's not defined correctly. I don't know what is the input to the functional.

Can you clarify please?

Well like I say it's more or less clear what you're thinking but I'm lost n the points I've indicated.

ps -- Ah ... slight glimmer ... since m itself has coordinates, we can break up the partials as acting on each coordinate separately, and we'll end up with some Kronecker-fu leading to the rest of your exposition. Is that the right intuition?

I'll push on.

(Later edit) ...

I can see a way to define differentiability.

If M is a manifold and U \subset M is an open set, and if \varphi : U \to \mathbb R^n is a chart, and f : U \to \mathbb R is a function, then we would naturally look at f \varphi^{-1} : \varphi(U) \to \mathbb R.

If f \varphi^{-1} is smooth then (since \varphi(U) \subset \mathbb R^n) we can take the partials with respect to the coordinate functions and then I think the rest of your notation works.

Is that right?

Edited by wtf, 8 March 2017 - 07:24 AM.

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#44 Xerxes

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Posted 10 March 2017 - 11:18 PM

I'm completely confused by where all this is taking place. We don't know how to take derivatives on a manifold yet but your notation is assuming that we can.


We can. This is because of the continuous isomorphism (homeomorphism) U \simeq R^n. Or if you prefer, our manifold is locally indistinguishable from an open subset of R^n
 
 

Picky refinement, my understanding is that a C^1 function has a continuous derivative.

Yes, but a C^0 function is by definition a continuous function, and C^1 subsumes C^0. As I said
 
 

I'm genuinely confused. We've been letting M stand for a manifold. But we don't know how to differentiate a function on a manifold.


Yes we do - see above

for all we know our manifold M is so full of corners it can't be differentiated at all.

If it is of class C^{\infty} all functions (including coordinate functions) are continuous - no corners!

In order to get past this point I have to either assume we've defined differentiability on a manifold somehow, or else that we're working in \mathbb R^n.

Roughly speaking we are working in R^n, or something that "looks very like it", namely the open subset of M where the homeomorphism U \simeq R^n holds. D

 
 

Little point of notational confusion. I'd believe \frac{df}{dx}\biggr\rvert_{x=a} or \frac{df}{dx}(a) but I'm not sure about your notation. Is that a typo or a standard notation?

Its standard (see below)
 
 

You see I'm still bothered by your notation. Did you really want me to write \frac{df}{d1} as you indicated earlier? I have a hard time believing that but I'll wait for your verdict.

It's clear to me that by the linearity of the derivative, \frac{df}{dx}(1) is a linear functional on C^\infty_1.

I'm afraid I cannot parse this.

Look, suppose that f(x)=y. Then I can write \frac{dy}{dx}=\frac{d(f(x)}{dx}. But the "x" in the "numerator" MUST be the same as the "x" in the deminator, so I introduce no ambiguity by wring \frac{df}{dx}. This is standard
 

you are using those set brackets again and I do not for the life of me see how that can make any sense. There's no order to sets so how do you know which coordinate function goes with which coordinate?

The superscripts in x^1,x^2,....,x^n are just tracking indices - they do not imply a natural order. I may have x=x^1,\,y=x^2,\,z=x^3 or equally I may have x=x^2,\,y=x^3,\,z=x^1. It doesn't matter

Now if I forget manifolds and pretend we're in \mathbb R^n then I suppose we could define the functional v=\frac{\partial}{\partial x^1}(m)f + \frac{\partial}{\partial x^2}(m)f+....+\frac{\partial}{\partial x^n}(m)f. I would almost believe this notation as I have written it.

Well, you need to be careful. If I write, say, \frac{d}{dx}(m) I really mean \frac{d(m)}{dx}, and this not what you meant. What you write has no meaning.In terms of notation you could, if you wanted to specify a point of application you could write \frac{df}{dx}|_m \in U
 

This particular functional is defined at the point m. However I see that you've left that part out and you're defining this functional for all points? But then it's not defined correctly. I don't know what is the input to the functional.

The input for any functional is, by definition, a vector. The output is a Real number. What you wrote (sorry, I lost it in transcription) is not a functional.

In my last post I gave you 2 functionals - df and dx^j. Please check that they are mappings fron a vector space to the Real numbers


 

ps -- Ah ... slight glimmer ... since m itself has coordinates, we can break up the partials as acting on each coordinate separately, and we'll end up with some Kronecker-fu leading to the rest of your exposition. Is that the right intuition?

Oh yes. Good.

 

If M is a manifold and U \subset M is an open set, and if \varphi : U \to \mathbb R^n is a chart, and f : U \to \mathbb R is a function, then we would naturally look at f \varphi^{-1} : \varphi(U) \to \mathbb R.

If f \varphi^{-1} is smooth then (since \varphi(U) \subset \mathbb R^n) we can take the partials with respect to the coordinate functions and then I think the rest of your notation works.

Is that right?

Sort of, but your reasoning escapes me. If on LHS of the above you mean f(\varphi^{-1}):\varphi(U) \to \mathbb{R} or f\circ \varphi^{-1}:\varphi(U) \to \mathbb{R} (they mean the same) and since (\varphi^{-1} \circ \varphi)U= U then how does your composite unction differ from f:U \to \mathbb{R} (which I gave as a definition)?
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#45 Xerxes

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Posted 18 March 2017 - 07:05 PM

So, in spite of a sudden lack of interest, I will continue talking to myself, as I hate loose ends.

Recall I gave you in post#27 that, for set open sets in U \cap U' we will have the coordinate transformations x'^j=x'^j(x^k). Notice I am here treating the x'^j as functions, and the x^k as arguments

Suppose some point m \in U \cap U' and a vector space T_mM defined over this point.

Recall also I said in post#41 that for any v \in T_mM that v(x^j)=\alpha^j which are called the components of v= \alpha^j \frac{\partial}{\partial x^j}.

Likewise I must have that v=\alpha'^k \frac{\partial}{\partial x'^k}. We may assume these are equal, since our vector v is a Real Thing


So that \alpha^j=v(x^j) and \alpha'^k=v(x'^k), we must have that \alpha'^k= \alpha^j\frac{\partial x'^k}{\partial x^j}.

This is the transformation law for the components of a tangent vector, also known (by virtue of the above) as a type (1,0) tensor.

It is no work at at to extract the transformation laws for higher rank tensors, and very little to extract those for type (0,n) tensors.

PS I do wish that members would not ask questions where either they they are not equipped to understand the answers, or have no real interest in the subject they raise
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#46 wtf

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Posted 18 March 2017 - 10:13 PM

No lack of interest. I'm working through your posts. I've been busy with other things and you're four posts ahead of me now but I intend to catch up.

However you're wrong about differentiability. If I map the graph of the Weirstrass function to the reals by vertical projection, I have a homeomorphism but no possible differentiable structure on the graph because the graph has no derivative at any point. I'll get busy on my next post (which I've drafted but not yet cleaned up) and elaborate on this point.

https://en.wikipedia...strass_function

Well never mind I'll just put this bit up here.
 
IMG_0986_800.jpg

Now the point is that if the map f \varphi^{-1} : \mathbb R^n \to \mathbb R happens to be differentiable (or smooth, etc.) then we say that f is differentiable. Also we need the transition maps to be smooth as well. We talked about them a while back. You can confirm all this in volume one of Spivak's DiffGeo book. I'll add that working through your posts has enabled me to make sense of parts of Spivak; and reading parts of Spivak has enabled me to make sense of your posts. So I am making progress and finding this valuable.

You need to define differentiability this way. Mere homeomorphism is not enough, surely you agree with this point but perhaps forgot? Plenty of continuous functions aren't differentiable. Remember that almost all continuous functions are just like the Weirstrass function, differentiable nowhere.

Likewise your definition of C^1 is wrong, you need the function to be continuously differentiable and not just differentiable. There are functions that are differentiable but whose derivative is not continuous, and such functions are not C^1. It is of course my curse in life that my ability to be picky and precise exceeds my ability to understand math, and I'm right about these two points despite being ignorant of differential geometry.

I will see if I can focus some attention this week on catching up with your last four posts.


"PS I do wish that members would not ask questions where either they they are not equipped to understand the answers, or have no real interest in the subject they raise ..."

Sorry was that for me? I'm paddling as fast as I can. If it's for someone else, personally I welcome any and all posts. This isn't the Royal Society and I'm sure I for one would benefit from trying to understand and respond to any questions about this material at any level.

Edited by wtf, 18 March 2017 - 10:39 PM.

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#47 StringJunky

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Posted 19 March 2017 - 03:23 AM

PS I do wish that members would not ask questions where either they they are not equipped to understand the answers, or have no real interest in the subject they raise

Would you like me to find someone to dust off your ivory tower? :) You are a clever guy but your intellectual aloofness and the implicit self-aggrandising I get from the quoted  post, does you no favours. 

 

One doesn't know that one might not understand the answer until one asks. Even though one may not understand completely, it may add a useful piece or two in the jigsaw puzzle for them. At the very least, it gives a person an indication how far they've got to go learning before they can understand and can put signposts in the road ahead for them. Besides, an answer may not prove useful to the questioner but will to someone else that is capable, now or in the future; It's never wasted. 

 

I read all your posts in this thread and don't have a clue about most of it but they give me a sense of the scale of what is necessary to be learnt in order  to understand this subject. It sets the stage for me, if not the details just yet. With increasing exposure, one becomes familiar with the unfamiliar.


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#48 Xerxes

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Posted 20 March 2017 - 09:29 PM

However you're wrong about differentiability. If I map the graph of the Weirstrass function to the reals by vertical projection, I have a homeomorphism but no possible differentiable structure on the graph because the graph has no derivative at any point.

Yes, but at no pint did I assert that a continuous function needs to be differentiable. Rather I asserted the converse - a differentiable function must be continuous.
 

Likewise your definition of C^1 is wrong, you need the function to be continuously differentiable and not just differentiable.

Maybe I did not make myself clear. I said that the C^k property for a function "subsumes" the C^0 property. If we attach the obvious meaning to the C in C^k we will say that a C^0 function is continuous to order zero, a C^1 function is continuous to order one..... a C^k function is continuous to order k

I am sorry if my language was not sufficiently clear.
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#49 wtf

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Posted 26 March 2017 - 04:27 AM

PS I do wish that members would not ask questions where either they they are not equipped to understand the answers or have no real interest in the subject they raise


I take this to heart and plead guilty. As my philosophy prof once said: The spirit is willing but the flesh is weak. I have the math skills but my interest is drifting. The good news is that your posts enabled me to read parts of Spivak (*) and reading Spivak enabled me to understand parts of your posts. Learning has taken place and this has been valuable. You've moved me from point A to point B and I am appreciative.

I have not given up. I'm going far more slowly than I thought I would. I'll post specific questions if I have any. For the record you have no obligation to post anything. I regret encouraging any expectations that have led to disappointment. No one is more disappointed than me.


(*) Michael Spivak, A Comprehensive Introduction to Differential Geometry, Volume I, Third Edition. PDF here.


Now, all that said ... I have four specific comments, all peripheral to the main line of your exposition. Regarding the main line of your exposition, I pretty much understand all of it, but not well enough to turn it around and say something meaningful in response. The concepts are in my head but can't yet get back out. You should not be discouraged by that. Your words are making a difference.

Question 1) Definition of differentiable structure on U

You wrote:
 

Yes, but at no pint did I assert that a continuous function needs to be differentiable. Rather I asserted the converse - a differentiable function must be continuous.


First I stipulate that this issue is unimportant and if we never reach agreement on it, I'm fine with that.

However this remark was in response to my pointing out that you need the map f \varphi^{-1} : \varphi(U) \to \mathbb R to be differentiable order to define the differentiability of f. It's the only possible thing that can make sense. And yes of course by f \varphi^{-1} I mean f \circ \varphi^{-1}, sorry if that wasn't clear earlier.

For whatever reason you seem to have forgotten this. It's true that we think of U as having a differentiable structure. But we have to define it as I've indicated. I verified this in Spivak. Homeomorphism can't be enough because there's no differentiability on an arbitrary manifold till we induce it.

Your not agreeing with this puzzles me. And your specific response about differentiable implying continuous doesn't apply to that at all.

As I say no matter on this issue but wanted to register my puzzlement.

* Question 2) Definition of C^1

In response to this issue you wrote:
 

Maybe I did not make myself clear. I said that the C^k property for a function "subsumes" the C^0 property. If we attach the obvious meaning to the C in C^k we will say that a C^0 function is continuous to order zero, a C^1 function is continuous to order one..... a C^k function is continuous to order k

I am sorry if my language was not sufficiently clear.


I apologize but you are still not clear. What does subsume mean? You can't mean subset, because the inclusions go the other way. If a function is n-times continuously differentiable then it's certainly n-1-times. So C^n \subset C^{n-1}. So subsume doesn't mean subset.

Of course it does mean that a C^n function is conntinuous. Differentiable functions are continuous, we all agree on that (is this what you were saying earlier?) So you are saying that a C^n function must be continous. Agreed, of course. That's "subsumed."

However this seems to be missing the point. The point is that there exists a differentiable function whose derivative is not continuous.

Therefore it's not good enough to say that C^1 is all the differentiable functions. It's all the differentiable functions whose derivative is continuous. There's no way I can fit "subsumes" into this.

Again like I say, trivial point, not important, we can move on. But I wanted to be as clear as I could about my own understanding, since like any beginner I must be picky.

* Question 3) The notation \frac{df}{da}

Earlier you wrote:
 

So recall from elementary calculus that, given a C^1 function f:\mathbb{R} \to \mathbb{R} with a \in \mathbb{R} then \frac{df}{da} is a Real number.


I have never seen this notation. a is a constant. I asked about this earlier and did not understand your response. If a = \pi would you write \frac{df}{d\pi}? I would write \frac{df(a)}{dx} or \frac{df}{dx}(a), which you seem to think are radically different. Or even \frac{df}{dx} \bigg\rvert_{x = a}. I'm confused on this minor point of notation.

* Question 4) The real thing I want to know

After glancing through Spivak I realized that I am never going to know much about differential geometry. Perhaps looking at Spivak was a mistake :-)

I'm trying to refocus my search for the clue or explanation "like I'm 5" that will relate tensors in engineering, differential geometry, and relativity, to what I know about the tensor product of modules over a commutative ring in abstract algebra.

What I seek, which perhaps may not be possible, is the 21 words or less -- or these days, 140 characters or less -- explanations of:

- How a tensor describes the stresses on a bolt on a bridge; and

- How a tensor describes the gravitational forces on a photon passing a massive body; and

- Why some components of these tensors are vectors in a vector space; and why others are covectors (aka functionals) in the dual space.

And I want this short and sweet so that I can understand it. Like I say, maybe an impossible dream. No royal road to tensors.

Ok that is everything I know tonight.
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#50 Xerxes

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Posted 26 March 2017 - 05:14 PM

Ha! So I am fired, in the nicest possible way! *wink*

Do not feel bad, wtf. Differential geometry is a hard subject, as you would see if you had all 5 volumes of Michael Spivak's work.

I do not pretend to have his depth of knowledge - I merely took a college course. Moreover his reputation as a teacher is extremely high, whereas mine is ....... (do NOT insert comment here!)

Regarding applications, all I can say is that I am neither an engineer nor a physicist, so as far as bridge bolts etc. you would need to ask somebody else.

On the other hand, it is not possible to study differential geometry without at some point encountering tensor fields, especially metric fields and the curvature fields that arise from them. These are the principle objects of interest in the General Theory of Gravitation.

If I offered to give guidance on this subject, it would be strictly as an outsider, an amateur.
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#51 wtf

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Posted 27 March 2017 - 11:11 PM

Ha! So I am fired, in the nicest possible way! *wink*


Not at all. I acknowledge letting my enthusiasm exceed my concentration, discipline, and ability to make this a priority. Your expectations were reasonable and your disappointment perfectly understandable.

I simply want to get us both out of that loop. No expectations. I do want to get back to your posts. I'm stuck on one piece of notation as I'll get to.

I seek only to go slowly. I've always been interested in this material. Just no expectations about my pace, which might be arbitrarily small but not zero.
 

Do not feel bad, wtf. Differential geometry is a hard subject, as you would see if you had all 5 volumes of Michael Spivak's work.


What struck me was the depth of the detailed examples and calculations he did. He has fifty pages of explicit calculations of various manifolds before he defines the differentiable structure. So you and I are flying way way above the territory.

The good news is I can undertand things in there. The bad news is I will never have the storehouse of examples he gives. I'm not sure where you're suppose to learn those.
 

I do not pretend to have his depth of knowledge - I merely took a college course. Moreover his reputation as a teacher is extremely high, whereas mine is ....... (do NOT insert comment here!)


Writing clear exposition is challenging for everyone.
 

Regarding applications, all I can say is that I am neither an engineer nor a physicist, so as far as bridge bolts etc. you would need to ask somebody else.


I was thinking some of the engineering-oriented people here probably know. Studiot perhaps.
 

On the other hand, it is not possible to study differential geometry without at some point encountering tensor fields, especially metric fields and the curvature fields that arise from them. These are the principle objects of interest in the General Theory of Gravitation.


I can definitely imagine associating a tensor at each point of a manifold, by analogy with a vector field. I know what vector fields are.
 

If I offered to give guidance on this subject, it would be strictly as an outsider, an amateur.


Fine by me. I'm really curious about the \frac{df}{da} notation. That's the exact spot I got stuck. Is that a typo or a feature? Would you write \frac{df}{d\pi}?

ps -- I am finding this article most enlightening. https://en.wikipedia...ance_of_vectors. I'm going to work through it. It explains why some components of the tensor are vectors and others covectors. It depends on which way the coordinates of a vector transform under a change of basis. Contravariant and covariant. This is a big piece of the puzzle.

Edited by wtf, 28 March 2017 - 12:14 AM.

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#52 Xerxes

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Posted 31 March 2017 - 05:14 PM

I'm really curious about the \frac{df}{da} notation. That's the exact spot I got stuck. Is that a typo or a feature?

Neither. If I say for some arbitrary a \in \mathbb{R} that f(a) \in \mathbb{R} I am entitled to ask how the image varies as the argument varies. In other words, a is a variable. For reasons I gave in another post, I can write this as \frac{df}{da} it being understood this is unambiguous shorthand for \frac{df(a)}{a}.

You seem to believe that x is the only possible label I can attach to an arbitrary Real number - it's not

Would you write \frac{df}{d\pi}?

Of course not. \pi is a constant. You cannot differentiate with respect to a constant
 

ps -- I am finding this article most enlightening. https://en.wikipedia...ance_of_vectors. I'm going to work through it. It explains why some components of the tensor are vectors and others covectors. It depends on which way the coordinates of a vector transform under a change of basis. Contravariant and covariant. This is a big piece of the puzzle.

Interesting - I don't like it one bit. But let's not go there......
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