The day this guy has more subs and views than tiktokers will be the day humankind will make a giant leap forward

Such a great video man. Keep it up!

Before seeing this series, I never imagined I could self-learn tensors intuitively. This is what makes Chris' teaching so amazing! Thank you for all your effort!

Small error around 24:50 where you say that for the sphere the geodesics are originally accelerating away from each other and then towards each other. No. With positive curvature, the geodesics are always accelerating towards each other. Thanks for such a well-done series.

Pay this man! I just did. I'm very selective about where my money goes. I can't think of a better investment than Chris's hard work and incredible intrinsic understanding. 🤘

Thank you Chris this is just wonderful. I've bought you a coffee, hope this will help you to produce more videos

A lot of "Heavy Lifting" (hard brain work!) has obviously gone into producing this Superb Video Chris. Thankyou.

I love this series! It's great to see more content!

Thanks so much Chris for this amazing opportunity to learn you are giving us.

This is nothing short of incredible. Thank you eigenchris!

Amazing, unique and clear videos, they should be standard viewing for anyone starting in this field, nothing like them exists.

your videos are treasure, they really help seeing things and what's hidding behind the definitions. Thank you !

Excellent explanation of the tensor notations and the applications for the Riemann curvature tensor

Great job, thank you a lot for sharing this! I love that you don't pretend to know everything (the part with the lie braket), it really reinforces that you don't teach things you don't fully understand, and it makes me trust this videos more than other sources I found! I'll continue watching them even after my exam, this is really fun :]

Oh damn, I didn't realize this was an ongoing series still. Stoked for more!

Fantastic work Chris.

Muchisimas gracias profesor por todos sus videos. Reciba un saludo desde 🇪🇸 España

Keep up ur free education service Ur videos are awesome

Thanks. It is very clear for the explanation. Excellent!

These videos are very helpful for learning the math behind GR. Thanks.

15:20 I think the third term exists so that the curvature tensor doesn't care about the value of u and v anywhere other than the point where it's being evaluated. The lie bracket can be different even if two pairs of vector fields have the same value at some point in the manifold, meaning that if that term was removed here, that would supposedly mean that two pairs of vector fields with the same value at some point could give different linear transformations when put into the curvature tensor. Because of this, applying the riemann curvature tensor to vectors directly wouldn't be as well defined (since each vector can be extended to different vector fields which makes the tensor give different results). I think this would make it not qualify as a tensor field, yet it would still be a coordinate invariant object. A change in coordinates would not change the linear transformations from this modified curvature tensor when two vector fields are put in, but again, this modified curvature tensor asks for vector fields rather than vectors and can't just be defined as being a tensor at each point.

Great explanations ! You are really talent in lecture. Keep going and hope that your chanel will be more successful

YOUR EXPLANATIONS SHOWS THAT YOU REALY UNDERSTAND DIFFRENTIAL GEOMETRY AND GR

Great lecture, thanks for the good explinations

well explained sir,,ur videos making life easy,

Simply wonderful video

Thank you Chris for this incredible work. Here I try to solve the bracket mystery and I hope it will help. I will use "Why Alex Beats Bobbie at Poker" notation in this commentary (with L the parallel transport along [u,v] ) that i hope will help achieve showing the equality with [u,v] not being the null vector . If you consider the points where the covariant derivatives is taken before going to the limit and using the vector field argument it helps leaning towards the correct formula. Let''s note w(A) for instance the vector w at point at the arrival of A and O the Identity at the origin, u_v the vector u at v and Cov(x,y) the covariant derivative along x of y. Let's consider also C' and D' as invert C and D, and d =||[u,v]||. Basically we want to show R(u,v)w= cov(u,cov(v,w)) - cov(v,cov(u,w)) - cov([u,v],w). So applying your line we have Riemann Tensor R = (w(O)-DCBLAw(O)/rs = DC * (C'D' - LBA)w(O) / rs and we have (C'D' - LBA)w(O) = C'D'w(O)-C'w(C)+C'w(C) -w(L)+w(L)-LBAw(O)-LBw(A)+LBw(A)+Lw(B)-Lw(B)+w(L)-w(L) and then it is possible to cancel extra term w(L) which was not possible before. Then we can form the covariant derivatives (C'D' - LBA)w(O)=C'(D'w(O)-w(C)) + (C'w(C)-w(L)) -LB(Aw(O)-w(A))-L(Bw(A)-w(B))-(Lw(B)-w(L)) and we already see that the last term will lead to the missing term cov([u,v],w). Now we get (C'D' - LBA)w(O)=s*C'cov(v,w(C))+r*cov(u_v,w(L))-r*LBcov(u,w(A))-s*Lcov(v_u,w(B))-d*cov([u,v],w(B)) and by rearranging terms we have (C'D' - LBA)w(O) = s*( C'cov(v,w(C)) - Lcov(v_u,w(B)) - r * ( LBcov(u,w(A)) cov(u_v,w(L))) d*cov([u,v],w(B)). Now we see that to take covariant derivatives you have to be at the same place an for example L in C'cov(v,w(C)) - Lcov(v_u,w(B) helps transport the covariant derivative along v to the correct place by filling the gap so you can take the covariant derivative along u of this covariant derivative along v at the same point (arrival of L), and then we understand easily that C'cov(v,w(C)) - Lcov(v_u,w(B) = r* cov(u,cov(v,w)) and LBcov(u,w(A)) cov(u_v,w(L)))=s * cov(v,cov(u,w)) . So you get cov(u,cov(v,w)) - cov(v,cov(u,w)) - (d/rs) cov([u,v],w) and the only missing point is that d must be equivalent to rs when (r,s)->(0,0) so the equality is true, and I have confidence it is true as the decomposition above allows to evaluate all the covariant derivatives in the correct place (indeed the correct Tangent Spaces ) and this cannot be a simple coïncidence.

Thank you so much eigenchris for your work. There is a small typo at 21:29: a basis term e_i is missing in the 4th row. Thank you again!

From your initial explanation I had the same feeling as stated in the textbook that, to define a curved space, the torsion of the connection must be non-zero. It seems like saying that in a curved space the the connection is irrotational.

absolutely great video

Thanks . It is very easy expaination and helpful to us.

Thanks for all your videos. I had a thought about the lie bracket term. Let's think about orthonormal polar cords with the basis vectors representing velocities of the tangent to the flow curves. Now, start at (r1, theta1) and execute the following: move for delta t in radial direction, then move for delta t in theta direction, then move for delta t in minus radial direction and finally move for delta t in minus theta direction. The distance traversed in space will be velocity x time so we will move a distance specified by the basis vectors: (r1, theta 1) to (r1+r basis vector, theta 1) to (r1 + r basis vector, theta 1 + theta basis vector) to (r1 + r basis vector - r basis vector, theta 1 + theta basis vector) to (r1, theta 1 + theta basis vector - theta basis vector). Now note that because we are using an orthonormal basis we haven't executed a closed loop. This is because the theta basis vector is normalised to the speed of 1 so we cover the same distance of arc length when we move + theta basis vector and minus theta basis vector, however to close the loop we have to move a smaller distance in the minus theta direction because the radius we are at, r1, a smaller radius than the radius we were at when we executed the move in the + theta direction, r1 + r basis vector. The Lie bracket is required to close the loop for a non co-ordinate basis, like the orthonormal polar example above. If however, you were using the orthogonal polar basis which is not normalised, then the lie bracket term would go to zero. In conclusion, if you have a coordinate basis you don't need the 3rd term and if you have a non coordinate basis you need the 3 rd term to close the loop. The lie bracket is the closer of quadrilaterals - see MTW, page 250 for geometric derivation.

I LIKE YOUR VIDEOS VERY MUCH - YOU ARE GOOD TEACHER.

I had gone through a lot of videos , how to learn mathematics for general theory of relativity i thought tensors are difficult to understand but now i can assure that your videos changes my opinion.thanks

Chris, first of all, this is some spectacular work! Precise, intuitive and very well explained! Also, I would like to offer a hopefully intuitive way to justify the Lie Bracket term of the Riemann curvature Tensor. Basically, redo the proof but this time calculate w - DCLBAw where L is the operator that lets you transport along the "gap" vector [u,v]. All the calculations remain largely the same, except that we should instead write LB - 1 = (LB - B) + (B - 1). If we do everything carefully, there should be an extra term (BAw - LBAw)/rs in the end, which should be equal to exactly the Lie Bracket term (The difference of transporting all the way to the end of the gap, versus the beginning of it should be exactly equal to the covariant derivative over the gap). This makes sense since we need to "pay" the extra price of having to move along the gap over the (now) pentagon instead of the rectangle. I hope this helps and thank you again for the wonderful work!

Great videos! Thanks! Regarding "flatness," this would seem to be a property of a region of space, rather than a point in space, and this observation might should impact on the presentation somewhere. Hence we could also consider the metric tensor being an unchanging identity matrix over a region, for example.

Just amazing!!

Wonderful video

this is incredible

why this channel is so underrated ??

Brilliant exposition as usual by Chris. Just a question to Chris or anyone who's reading this ? Is there a book where your discussion on holonomy can be found ?

Thank you very much, you are an excellent teacher, I am a Spanish retiree and I do not know English, I am helped by the google translator. I would like to thank you in some way, I try to open the link but it is not possible

Depending on the parameter value the vector field is present over the edges of the loop..... It's value and direction changes at every point on the loop.....

Third term measure that change of W vector if we go through that lie bracket gap, then after measuring that change we are ready to compare those two transports.

excellent.

Hello, this is an excellent resource for learning differential geometry, thanks a lot for your efforts! I have a small comment on the calculation at 21:00. While the first covariant derivative of a scalar is indeed the ordinary derivative, this is now a vector, so the second covariant derivative does involve the Christoffel symbols. This doesn't change the result, however, because of the torsion free property. All the best to you!

15:12-15:25. I think I have an explanation. First off, I'm pretty sure there's an error at 12:47-13:04. The Covariant derivative Nabla(u)(w) represents the deviation of the w field (at (r,0)) from its parallel transported w. This means that for a small step r: w(r,0) ≈ rNabla(u)(w) +Aw, meaning Aw-w ≈ -rNabla(u)(w). A similar mistake can be found with the rest of the differences between parallel transported w's and regular w's. At 14:08, the limit shown at the top is the negative of what it should be, but a similar set of mistakes are made when producing the bottom formula, giving the correct limit. This simply means that when I convert a difference between a parallel transported vector and its normal counterpart, I actually get the negative of the covariant derivative (times a small scalar). We start by adding a fifth linear map which I'll call E, that parallel transports w along [v,u] (the [u,v] vector at 16:02 is pointing the wrong way. You'll see why it starts at u(v) and ends at v(u), making it v(u)-u(v)). Then the Riemann tensor becomes R(u,v)w=lim(r,s->0):(1/rs)(w-DCEBAw). Adding and subtracting (1/rs)(DCBAw) gives lim(r,s->0)(1/rs)(w-DCBAw+DCBAw-DCEBAw). This can be split into lim(r,s->0):(1/rs)(w-DCBAw)+lim(r,s->0):(1/rs)(DCBAw-DCEBAw). The first limit becomes Nabla(u)(Nabla(v)(w))-Nabla(v)(Nabla(u)(w)) as we've already derived, meaning the second limit should give our correction term. We have a limit of lim(r,s->0):(1/rs)(DCBAw-DCEBAw). factor out DC, and substitute BAw for w' (we can first parallel transport w over to BAw, then define a new w' field starting at the location of BAw). This gives lim(r,s->0):(1/rs)DC(w'-Ew'). Now, we know E performs a parallel transport along [v,u]. If you recall, the change of a vector field a, as we travel a tiny amount e along a field b, can be expressed in terms of the covariant derivative. If we move from point p to point q, we get (parallel transported a(p)) + e(Nabla(b)(a))(p) ≈ a(q). This is because the covariant derivative of a vector field along a path measures the field's deviation from parallel transport. We know the direction of the E parallel transport map is along [v,u] (up to a minus sign), but what is the size (the e value) and exact direction? We can figure this out by using a flat-space approximation and keeping track of all the translations we've done. To start, let's find where the start of the path of the E map is: we start at a given starting point *p*. We move r units along *u* giving *p*+r(*u*). We then move along the *v* vector as it is located at *p*+r*u*, but this isn't the same as the v vector at p. This vector is different by a factor of v's derivative with respect to the u direction, giving *v*+r*u*(*v*) as our new v. we walk along this new vector, by s units giving a new position of *p*+r*u*+s*v*+rs*u*(*v*), and we are at the start of the E path. To get to the end of the E path, we walk along v first, then the changed version of u, giving an end position of *p*+s*v*+r*u*+rs*v*(*u*). subtracting final-initial gives a vector of rs[v,u] telling us the approximate size and direction. This size approximation gets better as r and s go to zero. Knowing this, we now know the size and direction of the E path. We also know that w' (at the end of the path) ≈ rsNabla([v,u])(w')+Ew', giving w'-Ew' ≈ rsNabla([v,u])w' = -rsNabla([u,v])(w'). This means our limit for the correction term from before is lim(r,s->0): (1/rs)DC(-rsNabla([u,v])(w'))=lim(r,s->0)DC(-Nabla([u,v])(BAw))=-Nabla([u,v])(w), which is what we see in the video, and on the Wiki page.

Great to have you back Sir.... Respect from pakistan

14:09 Took a while to realize that you meant it became a covariant derivative with respect to u of the preexisting covariant derivative with respect to v operating on w.

Which book for tensor calculus and GR do you suggest as a support for your lectures?

Great series of videos. To determine if a space is flat, can't you also measure circumference and radius of a circle, and check if their ratio is 2 pi?

At 12:20 if follow the rule from Tensor Calculus 21 17:21 the direction of nabla u(w) should be pointed from the end of Aw (parallel transported) toward the end of w actual vector field. But here is vise verse.

Did you check R. Wald's book on general relativity? He gives a sort of a proof for why the lie bracket term vanishes. He argues that the vectors u and v are two linearly independent basis vectors of the local inertial frame and by definition their commutator/lie bracket is equal to zero. I used 'by definition' rather loosely here; in the book he doesn't really give a full fledged proof for why the lie bracket is zero but only provides arguments which are pretty convincing and converting them to a formal proof is more or less trivial. Ironically though, I came here because I found the section on the curvature tensor in the book incomprehensible lol.

Dear Chris when you do the formula for R(u,v)w minute 9:58, usually every formula is done taking the final value minus the initial value, but you are doing the opposite. Is it correct?

The third term at (15.20). Is that because it is a general formula including the situation where there may be torsion so the contribution of non-zero torsion has to be subtracted in the formula for R? See Penrose "The Road to Reality" around page 316.

I am lost a little. How does the limit as r and s go to zero result in zero, when r,s are in deniminator? Also, making w a vector field arbitrarily? How is that done?

Hi Chris! One small formal remark. As written at 13:00, all the numerators inside the four summands (Aw-w), (Bw-w), (C^(-1)w-w) and (D^(-1)w-w) will result into the covariant derivatives of w along u and v with a negative sign in front (-∇w). The same holds also at 14:05 for the repeated covariant derivatives (-∇∇w). However, both steps will produce the same formula for R(u,v)w at the end. :) I'm looking forward to the rest of the video series on relativity. Excellent work!

At 7:25 the A 'linear map that parallel transports w from (0,0) to (r,0)' does not seem to be completely defined, that is you've only specified how it transforms 1 vector.

At 11:35 the introduction of a vector field to allow subtracting the paralleled transported vectors is mindblowing in that it doesn't seem to have any constraints. Can you elaborate on this, or give some accessible reference?

@eigenchris Thanks for this video. However I find your conclusion @ 22:12 a bit misleading. If a space has a metric which in one coordinate system has delta(i,j) coordinates i.e. is the Id., for all points in that space, then the space is flat, correct ?

Wonderful explanation! But would you plz tell me that how can we use matrix i.e. rotation matrix to represent this holonomy or rotation of vector after being parallel transported in a close curve?Is there any analogy here between rotation matrix what we learn in linear algebra class and the rotation of vector in curved space?

hm so you could write down the equations that would prove after moving on two straight paths at a relatively different direction, that it was mathematically impossible for the earth to be flat (though wed have to account for elevation differences by subtracting the normal velocity components continously etc to make such an experiment practicable).

At around 12:59, should the (Aw-w) term be the negative of covariant derivative (similarly for the other 3 terms)?

Thanks. Links for previous videos 17,18,19,20, in the description are not working. The link for video 21 is ok.

Is our definition of intrinsic geometry intrinsic enough? I especially felt this when you were going through the geodesic deviation concept. I can't get out of my head the idea that true inhabitants of a curved space would perceive geodesic deviation as always zero because their sensory organs would be shrinking/expanding as their elementary particles followed geodesics. I know this idea is wrong but I don't know why! 😂

Quick question: when you say linear map around 7:34, are you not implicitly making the assume that r and s have been taken to zero already, as these aren't really linear maps as they move the vector to other coordinates? Is there a way to find the same result without the need for such an assumption, or is it a fine one to make? Cheers :))

Hi Chris. Why are we assuming that a parallelogram can be drawn on a curved surface. Normally this should not be possible. Are we assuming that the curved space is locally flat and therefore this should be possible? Are we also assuming that the manifold is torsion free so when you parallel transport anyone of the basis vectors along the direction of the remaining others the sides of the quadrilateral should close and since parallel transport preserves the magnitude of the vector the resulting quadrilateral shape should be a parallelogram?

when i studied your video 21 i developed the idea that if Lie bracket≠0，such that the loop fails to close,then the space is curved. Not until i saw your video 22,in particular 26:26,when you simply postulated the torsion free. So,is there a version of the Eqn for Geodesic Deviation when you don't use that postulation?what I'm implying is…is it necessary??if it is,why?

is the connection in the definition of Riemann tensor could be any connections, including those are not compatible with the metric?

How do we know there exists such a parallelogram in a curved space? I mean, how do we know vector u here and there are the same or parallel? And if the answer is that the space is locally flat when the parallelogram is infinitesimally small, then I'm having a hard time understanding how we find non-zero curvature in this way, as flat means no curvature, unless the curvature goes "slower" to 0 than the vectors become parallel, mind boggling for me. Thanks to anyone with insights into this, and of course to Eigenchris for the great great videos ! Erez

as we are doing parallel transport of w vector throughout the parallelogram, then why not all the 4 covariant derivative of w are zero (0) at 13:50.

When going about that little parallelogram*, can w be rotated more than 2 pi? If so, how do you tell the difference in curvature between a rotation of say pi/4 and 9 pi/4?

What are the practical applications or potential future applications of all this ? Its very interesting

At right around 6 min re holonomy, you give example pf paralell transport. Is paralell transport only valid with the soldier's arrow is pointing tangent to the curve? If so, then i get why we can't say "What if it points normal to the surface...then it paralell transports perfectly." So in summary, I'm guessing paralell transport is only paralell transport when the spear (tangent vector) is pointing tangent to the surface. This leads me to my next question. If the spear is point normal to the surface and you paralell transport...is that a thing??? Appreciate all your valuable work!

Hi Chris. At 8:38 you say that "We also divide by the product r and s to make the answer independent of the area of the parallelogram". But the product rs is NOT the area of the parallelogram. That has got me confused. Can you explain? Thankyou

Please, make videos on the second and normal fundamental forms to the world sheet of strings in arbitrary background

sir if you would plz share the notes with us, i have downloaded the lecture of you lectures on tensor for beginners

In 4:27, why the Chris. coefficients being zero is equivalent to that the metric looks like Euclidean ? I don't get it. May someone explain it in more details?

13.23 it’s really my first time hearing of Inverse of a Vector.What’s more you alleged its direction is just the opposite of the original….can you explain on that? Sorry if it’s stupid,I took my Linear Algebra five years ago….can’t remember most

Has the mystery of adding the cov derivative in direction of lie bracket being solved?

Please discuss about kahlerian manifold

As far as I understand, if we are parallel transporting the vector, shouldn't the covariant terms be all zero? And Riemann tensor turned out to be, as a result, zero too?

16:29 Shouldnt the third term be automaticly zero if we use levi civita connection, since its torsion free? Its general form of Rieaman tensor so it probably has to accout for other connctions that dont absorb torsion would be my guess, but than again what do i know :D It looks to me like the third term accounts for tangetial change of w along lie bracket ''vector'' , thats basicly what you mantioned as one reason. Best qualitative explenation of relativity a saw , great job.

Can "as straight as possible" be made more mathematically precise? My thinking about parallel transport is as follows: every curve has normal, binormal and tangent vectors which form the so-called Frenet-Serret frame. Parallel transport means that one keeps the arrow at a fixed orientation relative to this frame. Is this right?

This is very advanced math men. I cant understand whats going on but im interested in general theory relativity. Looks challenging like quantum mechanics.

this dude is a genius almost einstein level

I am a bit confused by the argument about why Lie bracket is not linear (at 18:25). Isn’t the derivative a linear operator? Why do you need to use a product rule when multiplying by a scalar? Isn’t this like multiplying by a number, that we can just pull in front of the operator?

Would it be possible to elaborate on the insight on how Riemann came up with the formula ? The absolute top value of all your excellent videos is the deep insight you provide. But for this particular video, I find it hard to swallow that Riemann found the formula of the curvature tensor by right off the bat multiplying linear maps D C C -1 D -1 and proceeding until the final formula was discovered. This quadruple multiplication by linear maps is orders of magnitude beyond simply "squaring the square". Maybe the quadruple part of it can be explained by the 4 sides of the parallelogram but calling in linear maps ? This approach feels more like the standard way math textbooks prove a formula. Sort of like : already knowing the answer, find the most direct way to prove it correct.

In description, you should correct .com instead of /com Can't wait to have time to start watching all your tensor videos, I already studied them but in the old physics way from old books, but...just wanna really understand them in the deepest way possible. Thanks.

Clear explanation. thank for the great lectures. i just watch the lecture 22. There are one question i get stuck . [update:i am still waiting for the solution in question 2. i don't understand because the first step is the same for W(V) and Nablaw V(they are both defined to be the change vector V in direction W) and it is hard to see the difference. in the previous lecture he said the [V,W] is the "ordinary derivative" part but i am not very sure what does he actually mean here. Since two points have different vector space,i am not sure how does W(V) works in a way that is "ordinary" (we know that Nabla w V is a derivative of vector compared with the parallel one in different points such that they are compared in the same point) SO MY MAIN QUESTION IS : HOW DOES [V,W] ACTUALLY WORKS in "ORDINARY WAY"] in the previous lecture,i understand in this way :the main difference is the " connection factor k i j " is not equal to dui duj( ) ,however i dont understand because in the first step they seems to be the same: Nabla w V = w^i * d/du^i ( V^I e^i) = w^j e^j { V^i (ei )} = W(V) it seems this logic fails when e^j is not equal to d/du^i that mean the e^i is not equal to d/du^i under different condition? may be e^i cannot be written in differential operator form or ...?? since d/dx * d/dy = d/dy * d/dx implied they are commutative.. is that the main criteria here? or [V,W] is something that is independent of the connection factor here? i am not sure ....... and i also know how they interpret the [V,W] in different ways(2)other question:what is the different between Nabla v W and V(W) in lecture 21 it seems they are the same thing ?(it is obviously different) what is the "difference" if they are not torsion free however they are expressed in a form: v^i ei ( u^j ej ) = V(U) and nabla v U = v^i d/du^i ( U^j e_j) = V^j e^i (U^j e_j ) however we know they are not the same but there is a extra factor call T[ V,U]... (since you assume d/du^i = ei) in pure algebra,i cannot see the difference between Nabla v W and V(W) however we know the [V,W] is not equal to Nabla v W - Nabla w V = VW-WV in non-torsion free form therefore i just wonder why they are different algebraically thank you

It's an amazing video but I still can not understand why the covariant derivatives of rieman tensor don't go to zero. As I understand W is a vector field and also we want the vectors which are described from this vector field to be parallel transported along the rectangle so W vector field is a field which parallel transports W so it's covariant derivative shouldn't be zero?Also every W field which parallel transports the vectors along the rectangle will have a zero covariant derivative so a zero rieman tensor,but it may be curved because as I understand parallel transport can occur and to curved manifolds like the sphere of video 21. Thank you very much for your time your videos are very helpful!

The third term in the expression $R(X,Y)Z = abla_X abla_Y Z - abla_Y abla_X Z - abla_{[X,Y]} Z$ arises because of the non-commutativity of the Lie bracket of vector fields. In the case where the Lie bracket $[X,Y]$ of the vector fields $X$ and $Y$ is zero, then the third term $ abla_{[X,Y]} Z$ in the definition of the Riemann curvature tensor disappears. This is because in this case, the covariant derivatives $ abla_X abla_Y Z$ and $ abla_Y abla_X Z$ are equal, since the Levi-Civita connection is torsion-free, and the Riemann curvature tensor reduces to: $R(X,Y)Z = abla_X abla_Y Z - abla_Y abla_X Z$ However, when the Lie bracket $[X,Y]$ is non-zero, the two covariant derivatives $ abla_X abla_Y Z$ and $ abla_Y abla_X Z$ no longer coincide, and the difference between them is given by the third term $ abla_{[X,Y]} Z$ in the definition of the Riemann curvature tensor. This term arises because the Lie bracket of the vector fields $X$ and $Y$ measures the failure of the covariant derivative to commute, and it is responsible for the curvature of the manifold. Intuitively, the third term can be interpreted as a correction term that accounts for the non-commutativity of the covariant derivatives. It represents the curvature induced by the non-zero Lie bracket of the vector fields $X$ and $Y$, and it quantifies how much the manifold is curved in the direction of the vector field $[X,Y]$.

9:09 is it R(u,v)w the Riemann Curvature Tensor or is it just R(u,v)??

At 14:27, "D and C become the identity matrix as r and s go to 0." Then how is AB different from DC when r and s go to 0 and not also identity matrices?

9:10 my simpler (but rough) derivation: R(ũ,ṽ)ỹ = ỹ - DCBAỹ = DC(C'D'ỹ - BAỹ) = C'D'ỹ - BAỹ = (1+▽ũ)(1+▽ṽ)ỹ - (1+▽ṽ)(1+▽ũ)ỹ = ▽ũ▽ṽỹ - ▽ṽ▽ũỹ 16:25 my derivation can handle ▽[ũ,ṽ] term. R(ũ,ṽ)ỹ = ỹ - DC(1+▽[ũ,ṽ])BAỹ = C'D'ỹ - (1+▽[ũ,ṽ])BAỹ = C'D'ỹ - BAỹ - ▽[ũ,ṽ]ỹ = ▽ũ▽ṽỹ - ▽ṽ▽ũỹ - ▽[ũ,ṽ]ỹ

Around 15:15 you mention that you find it hard to justify the 3rd term in your expression for the Riemann curvature tensor. Frederic Schuller (a fantastic lecturer) directly proves precisely why you need a term of that form for the torsion tensor. He doesn't explicitly do it for the Riemann tensor (it is assigned as an exercise), but the procedure follows closely what was done for the torsion case. I highly recommend Schuller's lecture series here, great stuff! (Timestamp where he begins the torsion proof here: kzfaq.info/get/bejne/aMuGiriVyK-WmKM.html)

Vectors u and v seem perfect for detecting curvature along their parallelogram in a 2-d plane. But in 4-d, how come you don't need 4 vectors to describe a "hyper-parallelogram"? Seems like you would only determine the curvature of a 2-d surface embedded in a 4-d space. I'm new to this, so forgive my naivete... Thanks!

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24:49 if universe expand with acceleration more than 0 thats mean universe is curve not flat. This General Einstein Theory is like Second Archimedes theorem. This is Archimedes' principle for mass and space. I just wonder if this Differential geometry can bee aplied that same way in Archimedes principle. Question how many stars explode per 1 s in all Universe.

11:46 Hey I actually devise of another theory in lieu of the vector field stuff since I didn’t find that convincing(perhaps you just glossed through it so it’s not very elaborating). I think that since we’re saying r and s both approach zero,such that the area of the parallelogram is infinitesimally small,,then any curved surface under the face of infinitesimally small area simply flattens out to effectively become a flat space. Thus we can just subtract vectors in 11:35 Moreover,if you listen to 8:38,what you alleged had automatically implied that RS gives the area of the parallelogram,but we’re dealing with manifold,so RS really doesn’t give area ain’t it. So had we used the my self-constructed theory,we won’t have all these problems anymore. What do you reckon?