Differential structures: the pivotal concept of tangent vector spaces - Lec 09 - Frederic Schuller

This is from a series of lectures - "Lectures on the Geometric Anatomy of Theoretical Physics" delivered by Dr.Frederic P Schuller

[@keyyyla]

It is absolutely fascinating how Prof. Schuller produces everything on the blackboard by memorizing/understanding. This is by far not standard compared to my professors. Much respect for this excellent lecture.

[@Hannibill]

happens when this is your LIFE

[@EvilStalkerBE]

Absolutely agree. That was the first thing I noticed about his lectures.

[@Bignic2008]

He certainly does seem to know / remember a lot without looking at any references. However there are times when the camera doesn’t follow him or when an edit has been made, so maybe he is human like all of us and looks at notes occasionally.

[@rishabhkaushik1020]

@@Bignic2008 I think the parts are edited out where he erases the blackboard.

[@franciscoaguero9028]

He is very intelligent

[@KipIngram]

Ok, so I've been working through these, and I'd just like to take a minute to thank Dr. Schuller and his institution for making these materials available on KZbin. I'm in my late 50s, and finished my education when I was 29 years old. Ever since then I've made a fairly regular effort to improve my knowledge. Things got a lot easier when it became so possible to find good materials online. Anyway, the whole business of abstract algebra has just always been an obstacle for me, and I had his notion that it was a tough hurdle to clear. But this course has been so plain and clear, and I now feel like I've got a "starting point" level of knowledge about this part of mathematics. Plenty more to learn, of course, but this is a great start.

[@Almentoe]

I came at this from the opposite end to you, I know all the abstract stuff; but he is able to connect the "abstract nonsense" together into a cohesive picture. I left university with my brain as a jumble of abstraction, I feel that Dr. Schuller is combing through this mess so I can think clearly again.

[@xinqi7733]

Great lecture! Everything is very well-explained. Thank you so much for teaching us such a fascinating course, Prof. Schuller!

[@thomaslo9605]

Great lecture!!! I see the passion and humour as well. Absolutely fantastic insight! Thank you!

[@aaronTNGDS9]

It's too bad we can't clone Fred because he gives such a good exposition of a a tricky subject with the right amount amount of energy, pace, and enthusiasm. The German accent helps also.

[@LordVysh]

Really nice to see the formation of the Jacobian and covariant basis from the ground up.

[@tim-701cca]

0:30 tangent space 38:47 algebra 42:50 derivation 58:49 dim TpM=dim M 1:16:50 span 1:27:15 linear indep.

[@kockarthur7976]

There is a slight error in the definition of derivation (46:53). He modified his definition to allow derivations to take you from A->B, but these algebras must be the same because in the Leibniz rule the blobs on the right have to be between elements from the same algebra!

[@redaabakhti768]

a derivation Is actually from a ring A to an algebra B over the ring A that is additive and satisfying the leibniz rule (obviously A must be a ring with unity and B must be a unital algebra in wich the external operation or scaling multiplication : A x B -> B works both on the left and on the right a bialgebra I think ) also some people actually speak of germs of curves and germs of function to only capture the local behavior of the curve or C infinity M function at neighborhoods of p but its too much distraction from the main idea for a beguinner

[@josegarcia-cuerva7863]

In order to view a vector at p as a derivation, you do not take the algebra of all smooth functions on M, but rather the algebra of smooth functions defined on some open set containing p. To be more precise you take a quotient by declaring equivalent functions that coincide on an open neighborhood of p.

[@brandonwillnecker8060]

If we make C infinity M into an algebra and make R into a (C infinity M)-bimodule then X is a derivation from C infinity M to R because the . operation on the RHS of X(f.g)=X(f).g +f.X(g) will be the left and right scalar multiplication from the (C infinity M)-bimodule R.

[@rewtnode]

Wow, this was a quite amazing lecture. It was rather difficult in some parts. I wonder though if this is truly going beyond the “classic” idea where tangent spaces are pictured as little flat pieces of paper stuck to a curved surface. It was said that one should try to make no reference to some kind of embedding of a compact manifold into a higher dimensional space. But de facto with all those charts it’s still the same story, the difference being that we avoid to imagine those local charts being stuck to a surface that represents the manifold, that is, embedding it in d+1 dimensions. Their intrinsic dimensionality is still d, the same as the intrinsic dimensionality of the manifold. I find it hard to not think of an embedding of the whole construction into a d+1 dimensional space, where then these tangent spaces are hyperplanes, even if the mathematical apparatus presented here makes no use of that. It’s true; for the physical universe we can’t just make up yet another higher dimensional space in which the universe is embedded. It doesn’t have obvious physical reality then. But the same is true for all those charts: nice imaginations and a lot of technical stuff to get right. I guess I have to watch this again. Also, the stuff with derivation, that fat blob operator etc, was a bit of rather empty abstraction. Hope this will get clearer when he really gets to Lie groups.

[@nono-bt8gy]

The result on derivation is actually stronger than the one shown in this lecture. Because ANY local derivation on C\infty(M) at p is a tangent vector, not just the one defined against a curve (ie, the set of derivation at p against a curve is actually equal to the set of deviations at p). See spivak, introduction to differential geometry volume one, chapter 3 for instance.

[@klinexxx1871]

I think that in the definition of the tangent space at p, we should take into the account an equivalence relation of the equal derivatives at p (we are dealing with a quotient structure here). [Maurin, K. - Analiza: Tom 2. Warsaw: PWN]

[@llrecova]

Thanks for sharing. Brilliant.

[@millerfour2071]

11:52, 30:35, 40:46, 45:36 (haven’t constructed a h=fg to show closure under multiplication), 47:56, 50:57, 53:58, 56:10, 1:03:45, 1:11:15, 1:29:55, 1:31:53, 1:41:23 (position are not vectors, no linear map between two Cartesian charts with different origins), 1:43:09

[@possiblepilotdeviation5791]

The cover to the batteries on his wireless microphone transmitter is open.

[@electronselectron2321]

What bothers me though in the definition of T_p(M) is the redundancy: shouldn't we consider equivalence classes of smooth curves passing through p which coincide in a open around p rather than a tangent vector for each smoth curve since it is the same for the whole classe?

[@theflaggeddragon9472]

You can indeed create that equivalence class but it isn't strictly necessary since T_p M as a set is just that, a set. So this means that duplicates go away {1,2,2,3} = {1,2,3}. So if you imagine the set of tangent vectors, consider two curves gamma:R --> M and delta:R --> M that have the same derivative at p. Well if you don't create an equivalence class, these vectors are the same and become one vector (as they would in the set T_p M), and as such, you don't really need an equivalence class.

[@md2perpe]

Given a smooth curve γ the tangent vector is well-defined. The problem is the other way around; given a tangent vector, the smooth curve is not well-defined. The calculations in the video, more precisely the proofs that ⊕ and ⊗ are well-defined, show that the choice of smooth curve for a given tangent vector doesn't matter. Although not said explicitely, we in fact have an equivalence class of smooth curves. In the following paper, the tangent space is actually defined as an equivalence class: www.wisdom.weizmann.ac.il/~yakov/Geometry/3.pdf

[@mappingtheshit]

I know in Moscow they construct tangent spaces from equivalence classes

[@bouquet_of_circles4484]

You should look at the definition of tangent space that uses germs, it can be found at the end of Chapter 3 of Lee's Introduction to Smooth Manifolds.

[@jms547]

At 27:09 he says x(p) is a map from R -> R^d, but is it? By definition the chart map x:U -> R^d, so is there some hidden gamma somewhere to take us from R -> U first?

[@cartmansuperstar]

x(p) = x(p)* lambda^0

[@mbivert]

In case the other answer isn't clear enough, x(p) can be understood as the constant function (λ ↦ x(p)), which takes a real argument λ, discards it, and systematically returns the value x(p), which is in R^d. Like saying that the symbol "3" implicitly corresponds to the function (λ ↦ 3). The two following notations are mixed on the chalkboard, causing some further confusion: σ'(λ) := x⁻¹ ⚬ ( (x ⚬ γ)(λ) + (x⚬ δ )(λ) - x(p)λ⁰) = x⁻¹ ⚬ ( (x ⚬ γ)(λ) + (x⚬ δ )(λ) - x(p)) σ' := x⁻¹ ⚬ ( (x ⚬ γ) + (x⚬ δ) - (λ ↦ x(p))) Where in the second expression, the +/- are on { R ⟶ R^d }, while they are on R^d in the first.

[@aeroscience9834]

Your definition of a derivation from A to B is wrong I think. Since Df is in B and g is in A, how would you blob them in B? Or even in A?

[@vikramsundara5065]

This is very true. They are elements in different spaces with different blobs. It should be defined on the same algebra.

[@frederickdelpozo6976]

I have a question on the proof that the "partials" form a basis for TpM. For linear operators A and B with equal domain D the equality A = B holds iff for every x in D we have Ax = Bx. By choosing the coordinate projections x^{b} however you have only proved that the vectors are independent on the subset U generated by the coordinate projections. So I was wondering, shouldnt we still need to proove that the x^{b}'s generate all smooth functions on M, which in fact is wrong as the C_{infinity}(M) v-space is infinite dimensional and there are only finitely many x^{b}'s. Maybe I have misunderstood something in the proof, so I would be very grateful for clarification! :)

[@frederickdelpozo6976]

nevermind, I have found my own mistake! :D

[@WLVBS]

At 1:34:50 there's a proof linear independence of the partials (to prove that they are a basis of TpM). Prof. Schuller wants to show that: for a linear combination of the partial/partial_x^a operators to be the operator 0 (as a map from the manifold to the real numbers), then all the coefficients must be 0 (as real numbers). For this he says he will prove it by letting the operators act on an arbitrary function f, but then chooses particular functions x^b. At 1:34:50 he says "in particular, it must be 0 acting on particular functions". For this to be an actual proof, there's a step missing which says that if this is true for x^b then it is true for any f, which I don't see as being obvious from anything said previously. My question is: is it true that if one shows that the partial operators acting on x^b are "linearly independent" then this implies that they are so for any f? Is this somehow implying that the f is a linear combination of the x^b? I'm still digesting all the content from this and the previous lectures, which is very abstract for me... But (intuitively and totally informally) I kind of relate this to a Taylor expansion. The x^b are functions which are "linear in one direction and constant in all others", and the f could be Taylor expanded into a constant term + x^b terms + higher order terms. So it is true that the f is a "linear combination of the x^b" in the sense that the partial derivatives only care for those "linear terms". But this totally informal reasoning is not enough proof for the above statement (especially because we are dealing with C^infinity and not C^omega so Taylor expansion is not always possible).

[@kyubey3166]

Hello. A little too late answer but here it goes. He is assuming that the linear combination of the partial/partial_x^a terms (which is itself an operator) is zero. If an operator (in this case from C infinity M to the reals) is zero, this means that no matter which function we let it act on, it will always return zero (a number). This is particularly true for the component functions of the chart, from which the rest of the proof follows. In short, we are not trying to prove that the linear combination acting on any function is zero. Instead, we are assuming that to be true, so we can choose wisely our functions and show that the coefficients are zero. Hope this helped.

[@lukasmoudry9973]

@@kyubey3166 I am similarly confused. I understand it like "we assume the combination is zero. If we act on components of chart function, then we get coefficient to be zero meaning the components are linearly independent". I can't see why then it follows that it would work for any function...

[@balendubhooshan9391]

Really an insightful lecture...

[@taylankargin7474]

There is a problem in the definition of derivations for functions f:A->B where A and B are two different algebras (kzbin.info/www/bejne/i4GqoIt4m9JpgNU). The "blob" operation on the algebra B cannot be exploited in the Leibniz rule "blob" operation on B applied an element from A and an element from B (f "blob_B" Dg and Df "blob_B" g). The derivations, therefore, can be defined for the functions of form f:A->A, A being an algebra.

[@olehvinichenko5571]

can someone explain if this also valid for a special relativity, and space-time indices of four-vectors? what is manifold in this case? i am particularly confused with formalism of vectors in diffenetial geometry and partial derivatives in cft and qft and sr. jumping back and forth in books is very upsetting. differential geometric vectors and derivative in positions ( slots ) is much more precise, than four-vectors t, x, y, z derivtives, though they are clearly understood

[@alexheaton2]

At about 48 minutes, you talk about how a tangent vector is an example of a derivation. You write that the tangent vector X takes smooth functions on M to the vector space R. But then the way you have it acting on the product of two functions it looks like the result is actually another smooth function on M. Do we evaluate those functions at the point p? Or is the second algebra actually smooth functions on M again, rather than R?

[@electronselectron2321]

You need always to bear in mind that when talking about a directional derivative X its always actually X_p as the professor did define.

[@aeroscience9834]

Alex Heaton his definition of a derivation from A to B was inconsistent, since on the right hand side he was trying to multiply elements of A with elements of B. This is where the error is coming from

[@juliangarcia3416]

What is a smooth function from M to R? Is considering M and R to be two topological spaces? And then considering the map f to be differentiable?

[@JacobGaiter]

These are functions f, such that f compose with theta^-1, the chart transition map, which creates a function from R^m to R is infinitely differentiable, we have to define the manifold by saying it has an atlas, a set of coordinate patches, and you can think of these coordinate patches as identifying points on the manifold, making it intrinsically look like a subset of R^m, the topological space. These transition maps allow us to define differentiability intrinsically i.e. without reference to an ambient space that the manifold is embedded in.

[@tim-701cca]

It can be viewed as smooth map between smooth mfds M and N=R, and the atlas of N is just (N, identity function), and the chart transition map is smooth. The definition can be found in Lect7.

[@klgamit]

at 27:04 - x(p) is technically U->R^d not R->R^d , right? Just a technicality I don't think it changes the algebra in any way

[@amolvaidya4740]

Yea, I was confused by that as well. A bit after he writes that down he says it's just the constant function, so I think there's sort of an implied x(p)=x(s(q)) where q is in R and s(q)=p, which makes his R->R statement, and his statement about a constant function make sense. That's the way I'm conceptualizing it anyway.

[@RalphDratman]

Lecturers should take a five- or ten-minute break after about 45 minutes of working at the blackboard. This improves their (and the class's) staying power over a long lecture. If the total lecturing time goes on for more than 1.5 hour, they should take another break.

[@UnforsakenXII]

Good thing this is on video so I can always walk away for a bit and have a chat with a friend or something : ) Nice seeing you here again Ralph, you are everywhere.

[@billf7585]

Coffee makes all things possible.

[@RalphDratman]

@@billf7585 True, but people's minds work better with a bit of relaxation from time to time.

[@abrlim5597]

at 1:30:37, why is it crucial to change the range from the subset of R to R?

[@cartmansuperstar]

it´s not, i guess. He just wants to make x^b comparable to f, by which he is about to explain, what differentiability for a function, defined on M, means. Also it can´t be wrong, because, when mapping to a subset of R, you are still mapping to R....i guess.

[@rounak5106]

Why does he say @ 40:53 that the "product" for the algebra can't be defined for 4 dimension, what restricts it? Is it somehow related to Hurwitz theorem?

[@JacobGaiter]

It's because you can't uniquely define an orthogonal subspace of R^4 from 2 vectors. You'll end up with a hyper plane, meaning that the cross product cannot be cannonically chosen, since you'll end up with two linearly independent vectors and hence the produce is not well defined.

[@studywm_pls]

mi suster said he bost tutor every!

[@RohitMishra-yx2ie]

at 37:22 how does the proof for s multiplication goes through?? multiplying the curve by a number does not make the curve pass through p at parameter value zero??

[@ccosm4587]

Multiply the parameter by a scalar. It's a "sped up" version of the curve.

[@rohitmishra6351]

c cosm yes I got it.

[@ernestomamedaliev4253]

Im not sure if you are able to consider it as a proof at 1:35:31, since you are specifying the functions the operator is acting on. I mean, the point is that "if they are zero in general, they will be zero for this election; but, if they are zero for this election, will they be zero in general?". Of course, if it can be proven that the set of all chart projections x^b construct, by lineal combination, any function f, the proof given is 100% correct. So, my question is, is this last sentence correct?

[@richerzd]

That's incorrect; a linear combination of the x^b will be linear, but a general f won't be. So his argument goes like this: - We want to show that "if λ^a ∂/∂x^a = 0, then for any a, λ^a = 0". - Assume that λ^a ∂/∂x^a = 0. - Then λ^a ∂/∂x^a f = 0 for any f. - In particular, for any b, if we set f = x^b, then we get λ^b = 0 (from his derivation). - Therefore, for any b, λ^b = 0, which is what we wanted to prove. So actually, there's no need to "express" f as a combination of x^b.

[@ericgamliel8500]

Is the "precise intuition" (12:50) for the directional derivative correct? No, i don't think so. The function Gamma takes a parameter, say 'p', and maps to the manifold. Then 'f' takes Gamma and maps back to R. We are presumably taking the derivative of f w.r.t. p in the directional derivative. Okay, then how is this the "velocity" of Gamma? It is not. This explanation seems independent of the choice of 'f'. Doesn't sound correct. It seems more like this is the change of f along the curve Gamma

[@garinb5013]

To talk about the change in the function with respect to the curve you need to talk about how quickly this curve is changing, in other words its velocity. He is not saying that the directional derivative is the "tangent vector" or "velocity" of gamma, but that the linear map as a whole is the directional derivative. The X(gamma,P) operator is defined such that it works the way we want it, i.e. it takes the directional derivative. Intuitively, this is the vector that is tangent to the curve, or its velocity at that point.

[@martinzika7370]

I think you misread "The morphism f \mapsto (f \circ \gamma)'(0) is the directional derivative along \gamma." as " (f \circ \gamma)'(0) is the directional derivative along gamma."

[@olehvinichenko5571]

i still cannot wrap my head around why derivatives are not ill-defined, when doing inverse map? i understand why adding curves can be terribly ill-defined. ( from 29:38 )

[@adimurthy5576]

Who proposed pivotal concept...

[@reinerwilhelms-tricarico344]

50:00 Memo: A Lie algebra is an algebra where the blob is a smiley bracket box with two dot eyes and a comma nose. Clear? 😂 And with some more blobbing you get immediately the Leibniz rule … wow man that’s deep man. 🧐

[@joefuentes2977]

Big boy math!

[@jamesalexander5455]

Can someone direct me to Schuller's definition of $\partial_a$ ?

[@OnTheThirdDay]

It is just \partial / \partial x_a, because he is dealing with a function from \R^d to \R.

[@danideboe]

It is the derivative w.r.t. the a-th entry since f maps from R^d to R

[@dientesfelices416]

22:51 ¿🇪🇦?

[@cocoaaa2680]

1:39:00

[@cocoaaa2680]

38:34 algebra and derivations

[@Algebrodadio]

Who gave this thumbs down?

[@MohammedMohammed-xv8ud]

i am at 1:06:04

[@mappingtheshit]

good boy