The h Bar

Danu

21:00

@ChrisWhite Consider $F:M\to N$, then $dF_p: T_pM \to T_pN$ is its differential

David Z

@Danu both for differentials? why?

Danu

@DavidZ No, \dif for differentials and \d for exterior derivatives

David Z

ah, okay

Danu

(which, I guess, are often confused for differentials by many)

DanielSank

@Danu What does "differential" mean?

ACuriousMind

21:01

@DanielSank In differential geometry, the differential is a well-defined notion

DanielSank

Oh?

Danu

@ACuriousMind Exactly

@DanielSank What I just wrote down. THe induced map on tangent spaces by a map between manifolds

David Z

@Danu so am I, but then I usually wind up doing this anyway eventually and it saves a ton of effort

Danu

@DavidZ Okay, okay, okay.

DanielSank

@Danu With that definition what on earth does $\int f(x) dx$ mean?

Danu

21:02

@DanielSank Heh, chat hating on you again?

DanielSank

^ that.

Thomas Klimpel

@ACuriousMind Wow, once again I'm impressed. (Because I'm not so far yet...) I'm curious whether you will manage to find a good explanation for the tensor product based on the projective Hilbert space alone.

user54412

@Danu Ok. Haven't used that construction since I took a topology class. I guess I've never mixed it with exterior differentiation.

Danu

@DanielSank Well, the integral over the 1-form $\mathrm{d} x$ along the path defined by the composition of $f$ with some curve $\gamma: \mathbb{R}\to M$?

DanielSank

Yeah, but now you're taking $dx$ to be a form. What does it mean if $d$ is a differential?

Danu

21:04

@DanielSank It's not.

One integrates over forms

Is at least the impression that I have.

user54412

^ agreed. The d's in integrals are exterior derivatives as far as I'm concerned

DanielSank

So again, what does "differential" mean?

ACuriousMind

@Danu Which is correct

Danu

Not something one integrates over lol

DanielSank

facepalm

ok

:D

Danu

21:05

The word differential as physicists use it is totally ill-defined. For mathematicians, it's well-defined

It's just an induced map on tangent spaces

ACuriousMind

@DanielSank Well, the map between manifolds tells you where points go. The differential $df$ tells you where the vectors attached to the points go.

Danu

The really perplexing thing is that 'our' terminology actually works

0celo7

why do people like using \mathrm d's in derivatives?

DanielSank

It's not perplexing.

I recommend Munkres's book :)

0celo7

what's wrong with the italic one?

Danu

21:06

@DanielSank It is when viewed from a purely modern perspective

DanielSank

?

Danu

Of course, historically, it's no coincidence

@DanielSank It is highly nontrivial to me, at least.

user54412

@0celo7 Personally, because it's an operator, not a variable? I'm not sure how consistent I am with that rule though

ACuriousMind

@0celo7 The italic letters usually denote variables in LaTeX's math mode. But the $\mathrm{d}$ is not a variable, it is an operator, and operators are conventionally typeset upright

Danu

@0celo7 Basically, one must distinguish the exterior derivative and the differential

DanielSank

21:07

I have manifold. I have coordinate chart. Coordinate chart has Jacobian. Jacobian has determinant. $\mathrm{d}x$ is determinant of Jocabian. Bam.

Not surprising that it works!

Danu

@DanielSank Ehm... what about arbitrary forms?

user54412

I follow Sean Carroll's notation almost without exception. \mathrm{d} is the exterior derivative for things like $\mathrm{d}x$, but ds^2 has an italic d to indicate it's not the exterior derivative of anything

DanielSank

@Danu Define arbitrary forms, plz.

@ChrisWhite: That's the opposite of what a lot of math books do.

But given their convention for spherical coordinates, one only assumes that mathematicians are just trying to be ornery.

Danu

@DanielSank The wikipedia definition is adequate

A smooth section of the $k$th exterior power of the cotangent bundle of the manifold

DanielSank

@Danu I have absolutely no idea how that fits into the idea of notation used in integrals. This is a limit of my background.

ACuriousMind

21:12

@DanielSank The notation in the integral as you commonly write them is historical, and not really related to forms

Danu

@DanielSank Hmm okay. Well, it's really not trivial to me.

DanielSank

Also, there's no Wikipedia page for "arbitrary form".

Danu

Differential form

In the mathematical fields of differential geometry and tensor calculus, differential forms are an approach to multivariable calculus that is independent of coordinates. Differential forms provide a unified approach to defining integrands over curves, surfaces, volumes, and higher-dimensional manifolds. The modern notion of differential forms was pioneered by Élie Cartan. It has many applications, especially in geometry, topology and physics. For instance, the expression f(x) dx from one-variable calculus is called a 1-form, and can be integrated over an interval [a,b] in the domain of f : and...

DanielSank

@ACuriousMind Indeed, but actually if you understand forms the notation is actually fine.

@Danu: Oh, I thought "arbitrary forms" were a special thing. You just mean any ol' form.

ACuriousMind

@DanielSank Yes, because we defined the form notation to match it, I think ;)

DanielSank

21:13

@ACuriousMind: Indeed.

Danu

@ACuriousMind For instance, do you know how to rigorously define integration over noncompactly supported forms?

DanielSank

@Danu: Sure.

Take a limit, as usual.

Danu

Can you show me rigorously?

ACuriousMind

@DanielSank Is it certain that every non-compact domain of definition can be expressed as the limit of compact ones?

DanielSank

@ACuriousMind No!

no no no

Ack

What you do is you make a sequence of integrands, not a sequence of the domain!!!

user54412

21:15

@Danu What's your rigorous definition of integration in general?

Danu

@ChrisWhite That's part of the question...

DanielSank

Each point in the sequence of integrands (forms, functions, it matters not) has larger and larger support.

Therein you find a well defined limit.

Danu

@DanielSank I'm not sure I buy this

DanielSank

That's standard technique in analysis.

Danu

Compare the following statement (same wiki page, part on integration)

DanielSank

21:16

You always do it that way.

Danu

There is in general no meaningful way to integrate k-forms over subsets for k < n because there is no consistent way to orient k-dimensional subsets; geometrically, a k-dimensional subset can be turned around in place, reversing any orientation but yielding the same subset. Compare the Gram determinant of a set of k vectors in an n-dimensional space, which, unlike the determinant of n vectors, is always positive, corresponding to a squared number.

I am unconvinced that it works that easily on manifolds, @DanielSank

DanielSank

It's the only thing that works.

Read Munkres's book!

It's sooooo good.

ACuriousMind

@Danu You don't integrate over subsets, you integrate over submanifolds, no?

DanielSank

@Danu: en.wikipedia.org/wiki/Partition_of_unity

user54412

Also, is this the kind of question that goes away when one assumes the manifold is paracompact or similar?

Danu

21:17

@ACuriousMind I'm not sure what we (as physicists) are doing: Are we always integrating over submanifolds?

ACuriousMind

There is an interesting factoid that integration is sometimes "just" a cohomology operation. But it seems to be lacking any good treatment, even the nlab is clueless.

Danu

@DanielSank I know about partitions of unity. Do you have a more concrete reference to an actual proof or construction (possibly in Munkres' book: I have it)?

ACuriousMind

@Danu I think that, yes, we do only integrate over submanifolds

DanielSank

I'm 95% sure it's in that book.

@Danu: The analysis one, not the topology one.

Danu

Of course

DanielSank

21:19

Don't hate me if I'm wrong. It's been ten years since I read that book.

Danu

Yeah, sorry, I think you're wrong

THat book doesn't cover anything but $\mathbb{R}^n$

last chapter is on general manifolds.

DanielSank

Still, the best (standard) way to handle integration on noncompact domains is to form a sequence of integrands which have larger and larger (but compact) support.

Danu

It's not obvious to me that this is generally possible

DanielSank

That way, the tools for compact supports just run through, and all you have to do is show that your sequence of integrands has as a limit the thing you actually wanted to integrate.

user54412

@Danu And what about R^n doesn't instantly apply to general manifolds?

ACuriousMind

21:21

@DanielSank My initial objection still holds - for this, you need that the non-compact domain is a limit of compact domains.

DanielSank

@Danu: Are you working with manifolds which are not locally mappable to $R^n$?

@ACuriousMind: I'm pretty sure the point of doing it this way is that you don't need that.

Danu

@DanielSank They are always locally homeomorphic, by definition. But there are significant complications (in my experience)

Thomas Klimpel

@ACuriousMind My second question relates to your comment "But it is not my style" with respect to my presentation "trafo_vektorfelder.pdf". Do you still remember what you meant by "style" here? Did this just refer to the mathematical practice to write a long text expecting that no-one will ever want to read it, and then compressing the interesting results into a short presentation?

Danu

@ChrisWhite Many things

E.g. nontrivial topological things

ACuriousMind

@DanielSank But what you are doing is taking comp. supported function, whose support becomes larger and larger. You only can say that this approaches the integration over the non-compact domain if these functions approach a function defined on the whole domain

user54412

21:24

@Danu Not local things. And I feel the existence of a partition of unity usually lets you do what you want over extended regions.

DanielSank

@ACuriousMind: Yes. That is usually much easier than dealing with sequences of domains.

In fact, I've been a bit sloppy with the construction. You actually use a partition of unity.

Danu

@ChrisWhite Hmm... 'usually' is a very tricky word :P I'm very cautious about these things (mostly because I don't understand them very well myself)

Thomas Klimpel

Or did it refer to the fact that the topics concentrated on objects which one still can imagine easily, if one really wants to. Because I wanted to illustrate the whole thing with images, this was a natural restriction of the topics I could discuss...

Danu

@ChrisWhite For instance, all closed forms are exact in $\mathbb{R}^n$ but this is not true in general

Anyways, @ACuriousMind, maybe you have some reference for these type of things?

ACuriousMind

@ThomasKlimpel I think it was something like that, but I am not sure anymore. I am certain that I found nothing wrong about it, though. (And I totally understand writing up things regardless of whether people will read them ;) )

Danu

21:27

In my course we did cover Stokes, but it seems insufficient for what we physicists do

ACuriousMind

@Danu Hahaha...I've already said today that I'm abysmal at references

DanielSank

@Danu Yeah, but we're talking about local things.

Thomas Klimpel

The reason I ask this is because some of your answers indicate that you don't care much about exponential blowup. But objects which exhibit exponential blowup are naturally also objects which are hard to imagine in the mind.

ACuriousMind

@Danu Why would Stokes be insufficient?

DanielSank

The business about cohomology classes doesn't matter here.

Thomas Klimpel

21:28

The tensor product is a prime example leading to such a blowup

user54412

@Danu Good point. That's a truly topological thing, whereas my intuition for integration is that it's nonlocal geometry so to speak.

Danu

@ACuriousMind It has these caveats about compactness etc

at least the version I've seen in my textbooks (and course)

ACuriousMind

@ThomasKlimpel I'm not sure what you mean by exponential blowup here. That the equations you have to write down gain an annoying number of terms?

Danu

@ACuriousMind lol

@ACuriousMind Not even a set of lecture notes? (I can read German like a native!)

@DanielSank When we are integrating over the entire manifold, or a submanifold, are you 100% sure we can just work in a chart?

DanielSank

@Danu: That's the point of a partition of unity.

Danu

21:31

That doesn't just make all possible problems evaporate

DanielSank

Ah, are you worried about how you make a partition of unity on a non-compact manifold?

Thomas Klimpel

If you take the tensor product of 10 particles, you have at least 2^10 degrees of freedom. In differential geometry, you have to ask yourself whether you want to work with alternating differential forms. The have a beautiful theory, but they also have this exponential number of degrees of freedom, which makes them very hard to imagine in the mind.

Danu

@DanielSank I'm not worried about anything specific yet, because I don't know what the (most) general construction would be like. I am mostly looking for a good reference

ACuriousMind

@Danu The only restriction I know on Stokes' theorem off the top of my head is that it needs a compactly supported form. In physics, these issues quite often go away by just compactifying $\mathbb{R}^n$ to $S^n$, I think. (which works because we almost always have a fall-off at infinity, and hence a well-defined value at infinity)

@ThomasKlimpel I see. Yes, indeed, in that sense I do not care for the exponential blowup - I much prefer the clarity of the theory over such "practical" considerations

DanielSank

@Danu: I really think Munkres covers this.

Danu

21:34

@ACuriousMind Hmm, okay. I've heard an argument like that before. Where the heck would I be able to read about things like "compactifying $\mathbb{R}^n\Leftrightarrow $ fall-off at infinity"?

DanielSank

I know he doesn't go to abstract manifolds until the last chapter, but there's a good reason for that: all theory goes through verbatim via charts.

He definitely talks about noncompact domains.

He doesn't even write the proof of Stokes for general manifolds because by his own words he doesn't have to, since it's the same proof.

Danu

@DanielSank I'm looking at it right now. Let me see if he covers my question.

DanielSank

@Danu Hahaha, how come when ACuriousMind says it, then it's ok?

That idea, that compactifying $R^n$ is equivalent to the integrand falling off at infinite, is called "regularization" a lot of math contexts.

Danu

@DanielSank You never mentioned anything about compactifying $\mathbb{R}^n$ though

DanielSank

facepalm

Danu

21:37

^that's a little condescending

DanielSank

You do this when you prove how Fourier transforms work, for example, It's done in every analysis book, but they don't always outright say it.

Didn't mean it to be condescending. I meant it like "I need to learn to communicate better".

Danu

It comes across as facepalming at the other person's inability to understand you.

DanielSank

Sorry.

Didn't mean that.

ACuriousMind

@Danu Well, you always can compactify $\mathbb{R}^n$. But to extend a function on it to a (continuous) function on the compactification, it needs a limit at infinity - else the extension is not continuous. I'm not even sure where I got that idea from.

DanielSank

IMHO it is always the job of the explainer to be clear, no exceptions.

Danu

21:38

Also, I'm seeing some cautionary statements in Munkres

DanielSank

Oh?

Danu

"We have indicated how Stokes' theorem and the deRham groups generalize
to abstract differentiable manifolds. Now we consider some of the other topics
we have treated. Surprisingly, many of these do not generalize as readily.
Consider for instance the notions of the volume of a manifold M, and of
the integral $\int_M f dV$ of a scalar function over M with respect to volume.
These notions do not generalize to abstract differentiable mainifolds. "

which is exactly what I was talking about

DanielSank

Does he say something about non-orientability?

What are the reasons for the caution?

Danu

I'm reading on, I'll report back :P

ACuriousMind

@Danu That, strictly speaking, already fails for $\mathbb{R}^n$, right? (Because you cannot integrate $f(x) = 1$ over $\mathbb{R}^n$)

Danu

21:42

@ACuriousMind I'm not sure exactly what Munkres means to say cannot be generalized

DanielSank

@ACuriousMind Right. In that case the limit of the sequence of integrals over $f_n$ where each $f_n$ is e.g. a top hat function with larger support, doesn't converge.

Danu

Okay, after that section I quoted he goes on to talk about Riemannian geometry and that one can define an integral over such things on a compact Riemannian manifold (one crucially needs the extra structure of the inner product, I guess)

No mention of how to generalize this to non-compact manifolds

so my question kind of remains

ACuriousMind

@Danu Well, the Riemannian metric defines a volume form (and automatically orients the manifold). Munkres most probably means that there is no natural choice of the volume form on an arbitrary manifold.

(And indeed no possible choice on the non-orientable ones)

DanielSank

Can the question be framed as "under what conditions can I not naively integrate things on a manifold?"

Danu

I guess we just stick to hoping the things we try to do are 'nice' enough

@DanielSank ^ sorta

ACuriousMind

21:47

It is perhaps worthwhile to recall that Riemann integration is also not defined on non-compact sets - the "improper" integrals including $\pm\infty$ are also only defined by a limit as Daniel wants to construct it

So really, integration is a notion that lives on compact sets.

DanielSank

^ YES!

ACuriousMind

(The Lebesgue integration doesn't help either, non-compact sets have infinite/undefined Lebesgue measure, or zero measure, if they're lower in dimension)

DanielSank

crowd boos Lebesgue integration

Danu

I understand that, but I'm not sure how straightforwardly the usual limit thing extends

DanielSank

@ACuriousMind: Yeah, in my analysis course we handled these issues with Lebesgue integration by forming sequences of integrals on sets of finite measure.

@Danu: Once you map to coordinate charts it generalizes directly, AFAIK.

@ACuriousMind: What I mean is, you make a sequence of functions which converges to the thing you want to integreate.

Each point in the sequence has support of finite measure.

(Hence the partition of unity)

In fact, we defined integration on sets of nonfinite measure as the limit of such a sequence.

ACuriousMind

21:52

Yes. I too believe this generalizes, because the whole structure of manifolds seems to be meant to generalize this.

The usual theorems and definitions that Danu cited that always have some kind of compactness in them make me wary, though

DanielSank

There is an issue with switching the order of two limits, and you have to be quite careful (and clever) to make that work.

That's where regularization of the integrals come in.

@ACuriousMind Again, in my experience on defines the cases with noncompact support as a limit.

You have to do this even in complex analysis when you deal with contours that go to infinity.

Or, as you pointed out, with integration on $R$.

ACuriousMind

@DanielSank Yes, but it would be natural to do this directly after these compact statements in the books

That they don't do it makes me wary, as I said

DanielSank

@ACuriousMind Indeed.

ACuriousMind

But, perhaps, the books are simply written by people who never had much desire to integrate stuff over non-compact sets :P

DanielSank

@ACuriousMind Or they just don't want their book to be a zillion pages long.

I think the reason Munkres stops where he does is that he doesn't want to write several more chapters about non-orientable manifolds, etc.

Danu

21:57

@DanielSank I'm pretty sure that won't stop a mathematician from including a significant result.

DanielSank

@Danu Did you find the bit in Munkres about improper integrals?

Danu

@DanielSank As can be read in his intro, his book is not actually meant to be a book on differential geometry: It's like a intro, so it only does $\mathbb{R}^n$, and also only analysis, of course.

DanielSank

Right, it was written for a course.

Danu

...but not for graduate level differential geometry

DanielSank

I can't think of how integration over a non-compact domain could introduce any more complications than what one has on e.g. $R$.

Precisely because of the fact that manifolds are by definition locally equivalent to $R^n$.

$R^n$, sorry.

Danu

22:00

@DanielSank Like I said before, I'm precisely worried about the parts where general manifolds do not precisely conform to the intuition from $\mathbb{R}^n$

DanielSank

Every trick I know to handle improper integrals on $R^n$ works exactly the same way on a manifold.

@Danu You can always write an integral over a manifold as a sum of integrals over $R^n$.

Danu

@DanielSank Certainly not: $\mathbb{H}^n$ (half-space) at best :P

DanielSank

Sure, just multiply your integrand by zero in the right places.

That's the usual trick: put the complexity in the function, not the domain.

ACuriousMind

Silly question: Does anyone know a non-compact smooth manifold without boundary that is not diffeomorphic to $\mathbb{R}^n$?

Danu

@DanielSank Manifolds with boundary are not actually like $\mathbb{R}^n$, but like $\mathbb{H}^n$

@ACuriousMind You're in the wrong chat :D (maybe I am, too, but I'm afraid of getting slaughtered over at the math chat)

DanielSank

22:04

@ACuriousMind Yeah, pluck a hole out.

@Danu True.

You can still write integrals over those manifolds as integrals over $R^n$.

Danu

@DanielSank Doesn't $\mathbb{R}^n\setminus \{0\}$ have boundary?

DanielSank

@Danu: I don't think so

ACuriousMind

@Danu No, its boundary is $0$, which is not part of it

Danu

ok

DanielSank

@SabreTooth: Octopus sandwich today?

Danu

22:06

@DanielSank You mean like splitting off the boundary and integrating over $\mathbb{R}^{n-1}$ on there?

ACuriousMind

@Danu: Doesn't your whole problem go away if you consider that you can smoothly embed the manifold into some $\mathbb{R}^n$? Then your notion of integration is just coming from $\mathbb{R}^n$, and everything works out just fine.

Why do I keep writing mathrm for mathbb?

DanielSank

@Danu That's not what I had in mind.

Danu

@ACuriousMind I don't think that this should work, since we can also embed non-orientable manifolds in $\mathbb{R}^n$

DanielSank

I just meant that you can map to $R^n$ and define the function as zero on any point which is on the wrong side of the boundary.

ACuriousMind

@Danu Hmmmmmmmm...damn.

Okay, let me rephrase my silly question then: Does anyone know a smooth non-compact manifold without boundary of dimension $n$ that is not diffeomorphic to a subset of $\mathbb{R}^n$?

DanielSank

22:10

@ACuriousMind I think that's impossible by definition.

Danu

@ACuriousMind Lol. I'm always afraid of using these "but then some other unreasonable case would also work" type arguments but it seems reasonable?

@ACuriousMind Coming up with counterexamples is always hard :P

ACuriousMind

@DanielSank How? For compact ones, the circle is an easy example

DanielSank

@ACuriousMind but you said noncompact.

ACuriousMind

Yeah, but I don't see how the non-compactness would make it impossible.

I mean, intuitively I believe it is impossible, but I don't really see why

@Danu Yeah. But I want to understand what the problems of integration over arbitary non-compact manifolds could be, but I keep just thinking of $\mathbb{R}^n$.

DanielSank

@ACuriousMind: I'm thinking of a rigorous answer, but I'm held up by manifolds with nontrivial topology.

Sabre Tooth

22:15

@DanielSank ?

DanielSank

@SabreTooth Yes?

@ACuriousMind: Imagine an infinitely long iron chain.

but where the links are welded together.

ACuriousMind

@DanielSank Countably or uncountably infinite? :P

DanielSank

@ACuriousMind: Haha.

countably

In fact, consider a finite chain.

One link is a torus, a compact manifold.

This is equivalent to a square with opposite edges identified, yes?

ACuriousMind

@DanielSank Yes

DanielSank

Two links is two toruses glued together.

tori?

ACuriousMind

22:25

Tori, and...glued where/along what?

DanielSank

Imagine unfolding a single torus into a square.

ACuriousMind

Are you trying to construct a Riemannian surface of infinite genus?

DanielSank

Now cut out a circle in the middle of the square.

Now do this for a second square, and then finally identify the edges of the holes in the two squares.

ACuriousMind

@DanielSank Ah, it's the connected sum!

DanielSank

This makes a two-link chain.

@ACuriousMind Precisely.

Now consider a generic non-compact manifold.

I think you can isolate the topologically nontrivial parts and identify them as "tori" glued onto your manifold.

The point is that these things are compact.

So I think you can decompose your manifold into a sum of parts that are like $R^n$ (i.e. topologically boring) and other compact parts.

Sabre Tooth

22:29

what are you talking about here:

http://chat.stackexchange.com/transcript/message/20077808#20077808

ACuriousMind

@DanielSank That works in 2D (every smooth orientable compact 2D manifold is the connected sum of tori), but fails in higher dimensions, AFAIK

DanielSank

@SabreTooth: It was a joke. You used to tease me about having octopus for lunch since you know I like/respect octopuses.

@ACuriousMind Ok, this is the part I was hoping you can help with.

Is it not the case that the topologically nontrivial bits in higher dimension are compact?

(I think I want to ask this as a Math.SE question now)

ACuriousMind

@DanielSank I think the problem lies in making "topologically non-trivial bits" precise

@DanielSank Heh. If you don't ask, I will ;)

DanielSank

@ACuriousMind: I'm not about to do it until I think/research a bit. If you wanna ask just ping me with a link, plz.

ACuriousMind

Ooooh, I just remembered an easy counterexample:

Exotic R4

In mathematics, an exotic R4 is a differentiable manifold that is homeomorphic but not diffeomorphic to the Euclidean space R4. The first examples were found in 1982 by Michael Freedman and others, by using the contrast between Freedman's theorems about topological 4-manifolds, and Simon Donaldson's theorems about smooth 4-manifolds. There is a continuum of non-diffeomorphic differentiable structures of R4, as was shown first by Clifford Taubes. Prior to this construction, non-diffeomorphic smooth structures on spheres — exotic spheres — were already known to exist, although the question of the...

DanielSank

22:34

Bah! My mathbrain is rusty.

ACuriousMind

The problem is that the topological sum decomposition is not always compatible with the smooth structure

@DanielSank Well, "easy" counterexample means here that it is well-known, not that it is easy to imagine or find :D

DanielSank

@ACuriousMind The thing is, I'm not even sure this matters for Danu's question, but if the topologically interesting bits break down into compact pieces then I think Danu can stop worrying.

It's sufficient, but I think not necessary.

@ACuriousMind Truth.

Danu

@ACuriousMind I read that as "Erotic" lol

DanielSank

hahahaha

Starring: Max Wedge and Long Division.

ACuriousMind

@Danu Well, if you ever find an erotic manifold, you've done math for too long :D

@DanielSank Rule 34. It always holds.

Danu

22:40

math.uchicago.edu/~mduchin/womp/manf.pdf CLOSE ENOUGH

(look at the first quote)

Sabre Tooth

@DanielSank that's clear now

ACuriousMind

@Danu lol

user54412

Have any of you heard of Edward Frenkel?

user54412

mathematician who's starred in some.... nsfw... films (involving math)

Danu

That's... hilarious

So... it starts of as something like a classy porn... and then switches to just plain weird? :D

Thomas Klimpel

23:05

@ACuriousMind From the summary of the long text: "Das zentrale Element des Textes ist die Untersuchung der Richtungsableitung und ihrer Verallgemeinerungen. Es geht aber auch um drei sehr unterschiedliche Beweistechniken. Diese sind die Verwendung von Testkurven, die Manipulation von Integralen mit Testfunktionen, und die Begradigung eines Vektorfeldes."

English translation: "A central element of the text is the study of the directional derivative and its generalizations. But a goal is also to present three very different proof methods. These are the application of test curves, the manipulation of integrals with test functions, and the canonical straightening of a vector field."

Danu

Where's the 'canonical'? :P Or do you guys not employ an equivalent word in German?

Thomas Klimpel

These are basically techniques to reduce questions and computations in arbitrary dimensions to one dimensional questions and computations.

@Danu You are right, the 'canonical' is missing in the Germain summary. But it is of course the "kanonische Begradigung"...

ACuriousMind

@ThomasKlimpel It's not missing, the proper German term is just Begradigung.

At least, Google shows no hits for "kanonische Begradigung"

Danu

Kanonische sounds crappy compared to canonical, so I can't blame you guys for leaving it out.

Also, canonical sounds kinda pompous anyhow

user54412

@Danu I always hated "grand canonical ensemble" for that reason. All those pompous words conveying absolutely no information about the term.

Thomas Klimpel

23:19

I would have to grab a copy of V.I. Arnold to verify. But I think V.I. Arnold didn't use a canonical straightening, because it would have increased the space dimension by one. I used the canonical straightening as a tool to exploit transformation invariance for simplifying computations.

One goal of the text was to show the importance and usefulness of transformation invariance, without diving into all the technicalities of differentiable manifold theory.

Danu

@ChrisWhite So, so true.

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