Tensor Calculus 21: Lie Bracket, Flow, Torsion Tensor (contains error; see pinned comment)

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[@eigenchris]

ERROR IN VIDEO: Looking back on this 2 years later, I actually made a semi-serious error in this video that I didn't realize. At 5:46 and 6:15, I should not have set these 2nd-order partial derivatives to zero (∂x∂y = 0 and ∂y∂x = 0). Instead I should have let these two terms cancel each other in the Lie bracket subtraction (∂x∂y - ∂y∂x = 0). The standard derivative without a connection doesn't let us say that ∂x∂y = 0. On the other hand, with the Covariant derivative explicitly contains a way to take the derivatives of basis vectors using the Christoffel symbols/connection coefficients. This is what makes the Lie Bracket different than the Covariant derivatives. I'm sorry for the confusion. I did the best I could with the understanding I had at the time.

[@miloszivanovic2908]

If ii8i

[@Mikey-mike]

Thanks for the addenda. Your series and vids are awesome and so are you. Thanks.

[@abdullahamr8028]

Thank you for your big effort Can you explain the meaning of vector field in a separate lesson?

[@abrahamalebachew4133]

Hi Chris. Thank you for these enlightening videos you made on tensorial concepts. Just a little confusion about the correction you gave: so, what do the justifications @6:30 and @6:45 you explained imply? Thank you.

[@Panardo777]

thank you so so much ! understanding that normal derivative commutator of basis vectors being 0 ( as the parallelogram closes naturally as you shown in a coordinate system) doesnt imply covariant derivative commutator is 0 unless there is no torsion caused be me some headache this week but thank to your fantastic video finaly did it !

[@BorisNVM]

When I was in 5th semester at the university, I was seeing your videos and i was understanding nothing. Now, after I completed my 4 years of Physics I can say "I undestard you, please don't stop.". Thanks for the student friendly form of express yourself

[@chikitronrx0]

Thank you. My favorite KZbin series is this one, tensor calculus.

[@chimetimepaprika]

Dude, it's so nice.

[@jonasdaverio9369]

I'm yet happier that this series is going on ! I was very pleased to see a new video on your channel when you lanched your correcting code series, but I wanted to see the next part of your tensor calculus series, and here is it! Thank you so much

[@diegocastillo6470]

WOW... Just wow... Amazingly clear video, very intuitive, love your take on the subject and your approach. You are the kind of teacher that very clearly have passion for the subject, and that passion becomes infectious. Thank you for bringing this quality of classes to youtube. Thank you thank you thank you.

[@metarestephanois3262]

Best and most comprehensive explanation on KZbin. It is very difficult to get started in differential geometry as there are several different definitions for the same object and notations. But this video series is thankfully very clear.

[@alvarolouzi]

You’re the best, man. I already had a comprehension about this topic, but now I feeling that all subjects are linked together. Your vids give me a deep understanding about it. For that, thank you

[@haroldhawaiki]

thank you very much. Superb series with astounding knowledges, in tensors and linear codes. you're a Great Professor.

[@eugeniobrelles2240]

awesome videos, I'm going through my second playthrough and everything is falling into place glad to see you are still making videos on the topic thanks

[@imaginingPhysics]

21:50 This does make intuitive sense. Cristoffel symbol Gamma^k_ij measures, by definition, the change of i:th basis vector in the direction of the j:th basis vector. Then, if we want a "closed rectangle", it should be the same as the change of e_j in the direction of e_i, that is, Gamma^k_ji.

[@michaelcheung6290]

You are making me to fulfil my lifetime goal - GR

[@mythic3187]

Welcome back Bro, great to see another tensor calc video. These lie brackets really remind me of Green's Theorem. But I do remember that's greens has to be a closed path loop..

[@gguevaramu]

Hi Chris, Like the other followers I am glad to see another tensor video. Even when you say that you don't know some topics you are a good teacher. Please keep on your videos are very helpful.

[@Panardo777]

If you want a no torsion version of the connection of your curved understanding then this the place to be !! God bless Eigenchris !

[@eugenesong2128]

This is a great and wonderful explanation. I cannot thank you enough. Please make us more videos.

[@jaimeafarah7445]

In a Riemannian manifold the torsion tensor is always zero. The Levi-Civita connection used in Riemannian geometry is the only torsion free connection that preserves the metric.

[@signorellil]

Hurrah! Tensor calculus is back!

[@tony_narchuk]

And then I tried to read one of the differential geometry books and understood that you're the best explainer of math. I really cannot comprehend why math in books are so dry and boring, as like you not read but think and think all the time without any enjoyment

[@thankall]

One thing confuses me: if Lie Bracket [u, v] is defined as u(v)-v(u) as slide at 11:44, then at 18:00, where are u(v) and v(u), which are not shown on the slide?

[@reinerwilhelms-tricarico344]

I'd been pondering what that bizarre parallel transport with using the Levi Civita connection means for the sphere: The example you show is a rotation on a fixed altitude along the longitudinal line. The "boring" connection lets that vector pointing southwards continue to point southwards. But the fancy parallel transport is actually also quite natural: Imagine you are sitting on that point that moves eastwards and you look in Southern direction - Where the sun is at noon. Then, when the Earth rotates further, you're finding the sun further and further in the West, and that arrow still points in the direction where the sun is (projected on the ground). So apparently you need to rotate just like with what the Levi-Civita connection delivers. I hope i'm not completely off. A further application of this might be the computation of Coriolis forces. Whow! Now I want to see a movie about fluid flow on the surface of earth.

[@andrewandrus3296]

Your explanations are so clear. Thank you SO much!!!

[@vampireegg]

Hello Eigenchris, greetings. I am confused about the difference between u(v) and nabla_u(v). In both cases we take di(ej) (I mean curly d here) which should expand to connection coefficients Gamma_ij^k e_k, shouldn’t it? Thanks in advance.

[@danlii]

Yeah, I have also the same question. When expanding u(v) one gets the derivative di(ej) which when dealing with the covariant derivative is written as Gamma^k_ij ek. The question is why in the case of the Lie brackets it is not like that? I’ve been searching in books and the internet but I haven’t found a convincing explanation

[@erikstephens6370]

Something interesting to note is that in extrinsic geometry, the (Levi-Civita) covariant derivative of v in direction w, where v and w are vector fields tangent to whatever surface you are working with, is the result of projecting w(v) onto the tangent surface. This means that Nabla(w)(v)=projection of w(v) onto the surface. When you take the Lie bracket of v and w, all of the second derivatives of the position vector vanish, making the projection of [v,w] equal to [v,w]. This results in Nabla(w)(v)-Nabla(v)(w)=[v,w]. This gives a way to compare the Levi-Civita covariant derivative (Nabla(w)(v)) and the directional derivative (w(v)) without worrying about orthogonal components (to the surface) and second derivatives, which cannot be computed without embedding the surface within a higher dimensional flat background space (which we want to avoid when dealing with intrinsic geometry, using an arbitrary metric tensor). For anyone wondering why we don't just state Nabla(w)(v)=w(v), it might be because it isn't true when w(v) has a normal component to the surface and v(w) has some second derivatives which cannot be found without an embedding in a higher dimensional space (making the problem an extrinsic one). Using [v,w] resolves both issues.

[@hljav787]

Men, looks like you are the only one who gets the essence

[@pomeron490]

Hello Chris, Thank you so much for the clarification. Everything is now clear. May I suggest that you make the same clarification in TC20 under pinned comment, if you consider it appropriate. Cheers.

[@pomeron490]

Hello Chris, Thank you so much for your response. Much appreciated. (1) Please do correct me if I am wrong, after all it is long time since I delved into mathematics. The grey arrows and the black arrow in the diagram at 19.28 show how the black arrow comes about.Not that their sum is zero. So, as the diagram now stands there is still a torsion tensor. Correct? (2) I think we both agreed on torsion free is the cause (re: my last sentence in my previous message). Mathematically, torsion free is expressed as having the Lie bracket =0, that was what I meant by "the Lie bracket being torsion free". Apologies for the clumsy way I have expressed it. The crux of what I wanted to say is: instead of wriing T= ... (which puts the emphasis on T), it may be better to say [u,v] =... (which puts the emaphasis on the connection which is tosion free, and the consequnces on the r.h.s of the equation). Cheers.

[@eigenchris]

Maybe you could look at the diagram at 18:00. The blue arrow is the result of the torsion tensor after we plug in the 2 vector fields "u" and "v". In the diagram at 19:28, the blue arrow is zero, so this means the output of the Torsion tensor is zero. If the blue output vector is zero for all possible inputs, the conclusion we draw is that the Torsion tensor T must be zero. When you say "Mathematically, torsion free is expressed as having the Lie bracket =0", this is not correct. The formula for the Torsion tensor is given at the bottom of the screen at 18:05. The formula for computing the Torsion tensor output does involve the Lie bracket, but the Lie bracket is only part of the formula... we still need to compute the covariant derivatives as well and check if the combination of the 3 terms is zero (geometrically, these 3 terms correspond to 3 arrows--2 grey and 1 black--which combine to zero). At 21:21, you'll see that the components of the Torsion tensor only depend on the Connection coefficients... there's no Lie bracket in that calculation. The Lie bracket is only there to cancel out some other terms in the original formula. I'm putting the emphasis on "T" on purpose because "T" is the Torsion tensor. I'm a bit confused why you want the emphasis to be on [u,v].

[@jkli6031]

The Lie bracket is also called the Lie derivatives. where, let X, Y be elements in some Lie algebra TG, then the lie derivatives of Y w.r.t. X is denote by L_x(Y) equals XYAB where A is X inverse and B is Y inverse.

[@longsarith8106]

thanks teacher for all your videos ........you're really good teacher for me.

[@pomeron490]

Hello Chris, It was exactly 50 years ago that I took a course in GR. Having left the academia for a long long time, I have had the good fortune to stumble upon your excellent video series on Relativity and Tensor Calculus. Thank you truly for all the hard work you have done. There are two matters that trouble me a bit: (1) your diagram at 19.28 - when the parallel transported vectors are closed and the torsion tensor=0, the Lie bracket vector should have vanished too, correct? (2) I think it is a bit misleading to say at 19.07 that when T=0, the difference between the covariant derivative vectors = the separation between the vector field vectors. Two reasons for this comment: (1) you are putting the emphais on T as the cause, rather than the Lie bracket being torsion free. (2) you are relating two differnces each of which is in a different vector space, this may be a bit difficult to comprehand immediately. Perhaps, would it be better to say - when we have torsion free in vector space, then the quantities on the other side of the equation living in the tangent vector space also =0, meaning the the parallet transported vectors close properly, which in turn means the torsion tensor =0. Am I missing something? Thanks again for your excellent videos.

[@eigenchris]

Hey, I'm glad you find the videos helpful! (1) At 19:28, the Lie Bracket is the black arrow at the upper-right. The Torsion would normally be the arrow linking the dashed-purple and dashed-green arrows (or the the base of the two grey arrows). So here the Lie Bracket is non-zero, but the Torsion (the combination of the black arrow and two grey arrows) is zero. (2) I am putting emphasis on T being zero because being torsion-free is indeed the cause. The concept of being "Torsion-free" is a property of the connection (gamma coefficients), and it shouldn't depend on any particular vector field or Lie Bracket. I'm not sure I follow your reasoning here. Also, for the issue of relating differences in separate vectors spaces, I think when you take the limit in the derivatives, you are basically dealing with vectors that are all in the same vector space, tangent to the lower-left point on the diagram. Separating the vectors out is just to help visualization.

[@ObsessiveClarity]

For anyone who is having a hard time understanding what the hell is happening conceptually: Don’t think of u and v as derivatives. Just think of them as vectors fields (i.e. plain old vector-valued functions). Plain and simple. Then you can take the exterior derivative of u to get a vector-valued 1-form du, and you can take the exterior derivative of v to get a vector-valued 1-form dv. These could be rewritten as the Jacobian of u and the Jacobian of v. Then the Lie Bracket [u,v] is just dv(u) - du(v) which is conceptually very clear, and involves no 2nd derivatives nor product rules, but you get the same result as he did in this video. PS: See Keenan Crane’s video on vector valued differential forms if u don’t know what those are.

[@chrismaudsley127]

I am a little confused re the difference between “normal derivatives” and covariant derivatives. I thought a previous video said the only difference was that the normal component was dropped from the covariant derivative. This was not mentioned re Lie bracket discussion. Great videos.

[@eigenchris]

I tried to explain in the previous videos that there are several different definitions of covariant derivative. When we're talking about the covariant derivative on a 2D surface living in 3D space, then the covariant derivative is the ordinary derivative with the normal component subtraced. Later in video 20 I tried to explain there's a more "abstract" version of the covariant derivative that works in abstract curved spaces ("Riemannian Manifolds"). This covariant derivative is inspired by the above definition on 2D surfaces, but it's ultimately just a set of abstract properties (for example, the covariant derivative is defined to obey the product rule). This abstract definition is the one I'm using here. Does that make sense?

[@yamansanghavi]

This is an excellent lecture. Thank you very much.

[@harrysvensson2610]

Does this mean that you can close all fields by just subtracting the Lie Bracket?

[@ht4792]

Substracting Lie Bracket from what?

[@harrysvensson2610]

​@@ht4792 Well, as someone who is not knowledgeable in the area. I would imagine that you subtract it from the equation you used to calculate the Lie Bracket from. In other words, you just add an extra term to the equation (but negative) that closes it. Maybe it makes no sense or becomes an iterative solution, who knows. I have no clue.

[@ht4792]

​@@harrysvensson2610 I'm also still learning it. (If I understood what you said correctly.) If we perform the said operation, then it just becomes Lie Bracket - Lie Bracket and becomes zero which does not mean closing the field.

[@ryam4632]

Another masterful presentation! I have sent my support through co-fi.

[@chenlecong9938]

17:13 “we can measure how much these vectors deviate from (those vectors that are) parallel transport by looking at the Cov Der….” I suppose that’s because we know in the least that those vectors that are parallely transported,their Cov Der are zero(by definition)

[@thehockyebox4517]

In previous videos, you said that the connection coeficients aren’t tensors because when converting to a new basis, you get an extra term in addition with the (1,2)-tensor conversion rule, but a difference of connection coeficients is a tensor. Do these extra terms just cancel out? btw, thanks for these amazing videos.

[@eigenchris]

I believe they cancel, because the "extra" term in the transformation law us a 2nd partial derivative with respect to the lower indexes on the connection coefficients. Since you can swap the order of differentiation for partial derivatives, the terms cancel whem we subtract the connection coefficients.

[@tinkeringengr]

Thanks for the amazing video series!

[@huonghuongnuquy7272]

Hello, thank you very much for your videos. Your explications are very clear and you are very talent in lecture.

[@rvoinescu]

This series is great!

[@xinyuli6603]

Hi Chris! Thanks a lot for the video! It is the best diff geometry video I've ever seen. I think I am still a bit confused on two concepts. i.e., what is the difference between u(v) and \Nabla_u(v). I know the second one is to compute covariant derivative but I think both of them are doing 'direction derivative of u along u'? Looking back from the covariant derivative, if we cannot easily move one vector v from one point to another since the two tangent space are not related yet (without a given connection), how can we define u(v) as directional derivative? Sorry my question is a bit wordy but I hope it is understandable since it really confused me on these two concepts........Thanks in advance for the explanation.

[@vitrums]

17:18 It might be due to my short memory, but I really struggle to recall this particular geometrical sense of covariant derivative presented as this grey vector. Looking at the image we state that covariant derivative equals to the grey difference vector between (1) the value of v vector field at the point matching the tip of u and (2) the vector gained via the parallel transport of v across u to the same point. Again, it might be something I somehow totally missed from the previous videos or maybe I simply forgot, but at this moment for me it lowkey just fell from the sky. I would appreciate any directions pointing to the explanation of such geometry.

[@reinerwilhelms-tricarico344]

I'm a little confused here: When you write the Lie bracket [a,b] do you mean the covariant differentiation V_a (b) - V_b(a) (V stands for nabla operator) or the simple derivatives. i.e., without the Christoffel symbols? - Which would drop out when you're using covariant differentiation. Mmh I guess I answered it myself. I always believed the point of the Lie bracket is precisely to avoid these connection coefficients, but I'm probably wrong.

[@TenzinLundrup]

Questions: (1) Am I right that the choice of connection is arbitrary? If so, are final results (for example the Ricci tensor) independent of the connection? (2) I now understand what the Lie bracket [u, v] means in terms of just writing it out. However I am unsure how to write out Del_[u, v] (the last term in the defn. of the Riemann curvature). I'll give it a try though. (3) Both v(u) and Del_v u mean a change of u along v, however the latter involves parallel transport (one can write it in terms of the limit of a ratio of changes. How does one write v(u) in terms of a limit? Thank-you for the accessible videos!

[@nahblue]

The L-C connection seems to be the unique connection that has some certain properties (mentioned in previous videos), like torsion free and metric compatibility.

[@eskandable]

Best youtuber!

[@angrytreespainting]

I subbed to this awesome channel! ;-)

[@erikstephens6370]

17:21, shouldn't the covariant derivatives be the grey vectors, not the second green and purple ones? The black arrows on the diagram point to the purple and green ones. The top purple one represents the actual value of v's vector field, at the tip of u. I think you even say that the covariant derivative is the deviation of v from it's parallel transported version, which would mean that at the tip of u: v=(parallel transported v)+(Covariant derivative of v along u). This would make the covariant derivatives equal to the grey vectors, but your expressions for the covariant derivatives have arrows pointing to the purple and green ones.

[@patrikengas6479]

at 23:08, why are the transported vectors not paralell to the original ones here? is there like a curvature that causes the slight change in direction like with the transportation on the sphere? Or is it just like the vector field earlier in the vid?

[@kimchi_taco]

20:30 sir, how different between didj and nabla_di dj? I thought didj is Christopher ijk dk... 😢

[@MrEzystreet]

Great video, thank you very much for all the detailed explanations.

[@tommaullin1197]

Hey @eigenchris, First off, I want to say thank you for this incredible series! This has been immensely invaluable to my understanding of tensors and the course is incredibly clear/well laid out. I just had a quick question about the proofs appearing around 5:24 . I understand the steps as you have laid them out by interpreting ex and ey as derivative vectors and performing the expansion. I was just wondering could you also interpret v(u) in some way using the tensor product/dot products of v and u and perform the proof using methods from tensors from beginners? Perhaps I have forgotten/missed something but I feel a little unclear on what applying v to u would mean purely in terms of what we learnt in the tensors for beginners series. I think part of my confusion is that I work with matrix representations a lot and am struggling to think what v(u) would look like in matrix notation. I have a clear intuition for when its a covector applied to a vector but not so much a vector applied to a vector.

[@eigenchris]

So, in my Tensors for Beginners series, I don't talk about calculus at all. I do talk about covectors acting on vectors, such as alpha(v) or beta(u)... this makes sense because a covector is a function. In Tensors for Beginners, the idea of a vector acting on a vector v(u) doesn't make much sense. In Tensor calculus, things change, because we re-interpret vectors to be derivative operators. Now a vector can act on just about any field. This includes a scalar field (a normal scalar-valued function)... in this case v(f) gives us the derivative of f in the direction of v. We can also apply v to other vector fields like u, which is v(u). I don't know of a simple way to write this operation out in "matrix" form since it's not a dot-product... it's an actual derivative operation where we need to do things like product rule, and matrix notation might not deal well with that.

[@muhammedustaomeroglu3451]

@@eigenchris I have the same question. I think I get what you say. However, I could not get what does it mean to vectors acting on vectors in physical (or geometric) point of view. Yes we can think partial derivatives as base vectors but when I try to represent a physical vector field partial derivatives become meaningless to me. For instance let E be electric field vector and B be magnetic field vector. if I write E(B) it does not mean anything to me. Is there a more intuitive way of thinking that you can suggest because obviously I am missing somethings.

[@behnamparsaeifard3883]

Thank you very much! I didn't understand the difference between coverant derivative of v along u and u (v). To me they both seem the same. We can also define cristoffel symbols in the definition of u (v) and get back coverant derivative. What's the difference?

[@eigenchris]

They are the same in the special case of the connection having the "torsion-free" properties. The main example of a covariant derivative is the Levi-Civita connection, which is indeed Torsion-free. But there are other connections that are not torsion-free. I actually haven't given any examples of a connection with torsion in this series, unfortunately. You could make one up just by setting Gamma^k_ij to something other than Gamma^k_ji.

[@behnamparsaeifard3883]

@@eigenchris is /partial_i /partial_j equal to cristoffel symbol^k_{I,} /partial_k ?

[@michaellewis7861]

7:19 A small step on what parameter? Is there an invariant notion of interval of movement here.

[@chenlecong9938]

21:20 can you provide the intuition as to why you alleged that bracket (the difference between Chris Symbol) to be the component for the Torsion Tensor? That said,when we discussed the metric tensor in like the last two videos of "Tensor for Biginner",and subsequently in the beginning of this series when we discussed general tensor as Covector-Vector mixture,we also simply alleged that "the component of the metric tensor is the dot product of the unit basis vectors"..... So really is there regulation as to how we define the Component of General Tensor??

[@John-lf3xf]

Is the Lie Bracket defined only for linearly independent vector fields? (d/dx)Y-(d/dy)X where X is a vector field defined to be only in direction x and Y is a vector field defined to be in only direction y. Example: Y(x,y)= X(x,y)= In this case, actually for any kind of constant vector fields like this, this Lie Bracket would be zero wouldn’t it? The most basic it would have to be is at the very least linear. Y(x,y)= X(x,y)= In this case, it is easy to see that the Lie Bracket defined on all points (x,y) in R^2 is . The latter here is the same as if I were to factor out the inside components, so the basis vector then has nothing to do with the differentiation. Are we differentiating vectors here or are we differentiating vector fields? Problem in Video: The derivative of u w.r.t. the direction v is 0. The derivative of v w.r.t. the direction u is 1. What is the meaning of this? Is this quantity 0 if and only if it is the case that traveling down dimension 1 and then up the dimension 2 results in the same path being traversed as traveling dimension 2 first and then dimension 1 first. Say if we travel for the same amount of time in each of the dimensions. Travel t in dimension 1 and then t2 in dimension 2 as opposed to in the other order.

[@carsonyanningli3301]

Hi Chris, is it possible for you to make a video on the difference between Lie and covariant derivatives? There's a lot of questions on the internet. I don't think they give any intuitive explanation.

[@eigenchris]

Other than saying "the covariant derivative has connection coefficients", i'm not sure what else to say. what is confusing you?

[@carsonyanningli3301]

@@eigenchris I watched your previous videos and fully understand that the covariant derivative essentially describes the extrinsic total change of a tensor without counting the change due to the curvature of the manifold in which it lives. For example, if you draw a straight line and then warp the paper however you like, the tangent vector to the line will always have zero covariant derivative. I am looking for a similar intuitive graphical understanding of the Lie Derivative. What character does a tensor field have when its Lie derivative is zero along certain integral curves? In this video, you talked about the special case of vector, ie. a (1,0) tensor. Then the Lie Derivative takes the form of a Lie Bracket. There are two things in this video that I do not agree with.

[@carsonyanningli3301]

@@eigenchris At 5:24, you interpret d_y d_x as d_y acting on d_x, d_y(d_x) or d_y of d_x, the change of the x-basis vector in the y-direction and is therefore equal to zero. By similar argument, you have d_r d_theta = d_theta d_r = 1/r d_theta at 13:17. But I would simply interpret this as a mixed partial derivative, ie. successive partial differentiation on an incoming test function. Because U and V are both vector operators acting on some test function f. So your interpretation can be written as (UV)f, while mine is U( V(f) ). This sounds very nit-picking especially considering that since partial derivatives commute, both interpretations will eventually arrive at the final expression of the Lie bracket: [U, V]^i = U^j d_j V^i - V^j d_j U^i. But there IS a difference in Quantum Mechanics. Because the canonical commutation relation says [x, p] = ih_bar, while p = -ih_bar d_x. This only works if you use [x, p]f = x( p(f) ) - p( x(f) ). Basically, when I first learned operator commutators, it was interpreted as successive applications of the two operators to some test function in different orders. So I think it is simpler to keep that consistent interpretation of the math notation, whether in QM or GR. Let me know if I am wrong.

[@carsonyanningli3301]

@@eigenchris Sorry I keep getting distracted on this subject and forgot to talk about the 2nd thing I don't agree with, which I think is less pedantic and more important. I don't think the Lie Bracket tells us if the two flow curves close properly as stated at 10:39. The flow curves of two vector fields will always intercept into a closed shape as long as they are not parallel. For example, in the two examples at 10:39, the flow curves are obviously just horizontal and vertical lines. So they will always intercept into rectangular grids. The problem with the left example is that, in each rectangular grid, the two opposite purple sides represent different increments of the flow curve's parameter. I think that should be the correct symptom of a nonzero Lie Bracket. So you won't be able to tell apart a nonzero Lie bracket just from flow curves. Just like we cannot tell the velocity simply by looking at the trajectory of the motion. The same trajectory could be traced out very quickly or slowly. We must label the time parameter along the trajectory to know that information. So for two vector fields with zero Lie bracket, each pair of opposite sides of the area enclosed by their integral curves must represent the same increment of the respective parameter. This is obviously required for coordinate grids, since the parameters are just the coordinate values themselves. That's why partial derivatives always commute.

[@kimchi_taco]

20:25 can we swap derivative order? I think it requires torsion free….

[@robertprince1900]

I noticed that too! It's the same error he pinned earlier in the video but does it here too. Certainly the partials of the coordinate basis made as an embedding will commute, because they commute in the euclidian space, but then what is the point if it's all torsion free to begin with? I guess with the intrinsic view, 20:25 is wrong?

[@aliesmaeil1044]

thanks , its an amazing series ,what is the best book to study this kind of thing

[@chikitronrx0]

Spiegel: Vector analysis from the Schaum outlines. At least as an introduction, in the Tensor Calculus chapter.

[@debendragurung3033]

im confused by the product rule on 4:35 . I have known about product rule on calculus , about taking derivative of products of two functions , most of them being functions of the same variables like x. I am quite sure we are not directly migrating that concept here. Is the proof too advanced and more rigorous. Also i kind of cant grasp the idea of v(u) as measuring change of u along v and vice versa. Cant quite wrap my head around it. Any insights would be helpful. Thanks

[@JohnJoss1]

Just a slight confusion. At 1:38 you say the VELOCITY is given by v = 1e_x +xe_y, but at 7:45 you have u = 1e_x and v = xe_y . So that's a little confusing (maybe use big V for the total velocity?) . The other thing is I thought that u and v were coordinates rather than velocities anyway? For example in Video 12, 4:53 you use u and v as coordinates and convert them to XYZ on the surface of a sphere? In 12:09, this video, you say that coordinate lines are just the flow curves along basis vectors. So coordinates or velocities please? Thank you Chris.

[@eigenchris]

I wasn't intending for that first example at 1:38 to be connected to the others. My bad if that's confusing. I ultimately have to re-use the same letter for different things at various times, since there are only so many letters I can use. If something is a vector, I always try to write it with an arrow on top.

[@safkanderik7217]

i dont understand the difference between lie brackets opertion and and normal covariant dervitive

[@eigenchris]

The Lie Bracket doesn't involve the Christoffel Symbols. The Lie Bracket and Covariant Derivative end up being the same if the connection is "torsion-free".

[@safkanderik7217]

Can you make a video about an example on Lie Brackets in polar cordinates? It would be nice of you. thanks in advanced.

[@eigenchris]

@@safkanderik7217 I think I show an example of the Lie Bracket in polar coordinates in this video? I'd rather move on with the relativity videos I'm making. Is something unclear about the Lie Bracket in polar coordinates?

[@tusharjain6562]

How can you subtract vectors at different points? Don't they have to be parallel transported to the same tangent plane in order for the subtraction to make sense?

[@eigenchris]

Can you point to a specific part of the video? In some cases my answer is that the vectors are in the space tangent space, and so you don't need to parallel transform. In some other cases I might say that you would follow the flow curves given my the vector fields to form closed shapes. I think I'm going to have to re-edit and re-upload this video because it contains mistakes and some parts that aren't clear.

[@tusharjain6562]

@@eigenchris At 18:02 the separation vectors that you use to represent lie bracket and the torsion tensor. Also the formulas that follow make it appear that they are actually being subtracted. Perhaps some more steps that can justify that the formula makes mathematical sense? Also from the animation it appears that u is being transported along v in the tangent plane instead of the surface because it sits at the tip of the v vector, whereas I believe it should transported along the curve along which the vector field v is defined?

[@eigenchris]

@@tusharjain6562 I'll probably have to look into this again as I said. I think the reasoning I use works in flat space but not curved space. I'll aim to make a new video with more correct reasoning before the end of the year.

[@tusharjain6562]

@@eigenchris I see, thanks!

[@waynechau9884]

At 13:14, I am a little confused when computing ∂_r(∂_θ). Here, do we need to apply the product rule to ∂_r(-r sinθ ∂_x + r cosθ ∂_y) giving us -sinθ[∂_r(r) ∂_x + r ∂_r(∂_x)] + cosθ[∂_r(r) ∂_y + r ∂_r(∂_y)]? I tried continuing down this path and obtained -sinθ ∂_x + cosθ ∂_y - r sinθ [∂_r(∂_x)] + r cosθ [∂_r(∂_y)]. I then expanded ∂_x and ∂_y in the r and θ basis and found the last two terms to be -1/r ∂_θ + ∂_r(∂_θ). Since -1/r ∂_θ = -sinθ ∂_x + cosθ ∂_y, we have only ∂_r(∂_θ) remaining. But this is just getting right back to where we started. I cannot see why ∂_r(∂_θ) = -1/r ∂_θ.

[@yianniskalogeris1023]

You are amazing!

[@Sharikkhursheed]

plz upload further work on tensor analysis...

[@tulliolevicivita4443]

5:37 v(u) is not a vector because vector operators are first derivatives and v(u) is a second one

[@eigenchris]

I assume it's possible to write 2nd derivatives as linear combinations of 1st derivatives.

[@garytzehaylau9432]

excuse me,i get stuck in 10:32 in this video your explanation seems to make sense to me in 10:32 which describe how the "velocity change" in different path might give different displacement. but i am not sure why e^y is equal to the "gap" here i guess it might be some factors multiply e^y but not exactly e^y itself. since it is the difference between two "total derivative" ( e.g d/du^i (V) =U(V))which is the difference between two "change" of velocity vector along the lines/curve my assumption is : the gap here is the displacement as you said it will be longer if you have faster velocity i think the lines described the "displacement" rather than the "velocity",i am not sure why it describe the displacement which is the "gap" here(which is the difference between two "difference of total derivative in direction direction" itself? (2) in 21:50 you said "torsion tensor"... how do you know it is a tensor as you subtract two C-symbol(they are not tensor as you said before -connection aare not tensor..) together and obtaining this Torsion tensor (3) how can i borrow the coffee to support your videos.. thank you

[@eigenchris]

1) In this particular case, you can imagine the vector fields bwing like conveyor belts, where bigger arrows mean faster conveyor belts. If you imagine riding along U for one second, we move to the right 1m. If we move along V for one second, we move 1m if we are along the Y axis, and 2m if we are father along, 1m away from the Y axis. In this case the difference in the paths is exactly 1m in the Y direction (in other words, one ey vector). If this rule applies in the xy directions, I believe it should apply in all directions, since derivatives are linear. 2) It is a tensor because the result is an invariant vector that can be built from a basis.... you can also check using the transformation rule for Christoffel Symbols given in video 17.5 that the result of the subtraction transforms like a tensor. 3) A link to my donations page is here: ko-fi.com/eigenchris

[@garytzehaylau9432]

@@eigenchris thank! very clear explanation. (1)Still one question left in previous lesson ( the reason that a_ dot = apha implied da/du^i dot___ = d alpha/du^i ___ and i dont understand the logic behind it) (updated 8/1/2019: have solved) (2)other question:what is the different between Nabla v W and V(W) 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 in algrebra form(i mean in algebra but not in "intuitive sense") do you assume d/du^i is not equal to ei here? ( i just rearrange the stuff in the post such that it become more clear)

[@PedroCruz-ku3qi]

Hi @eigenchris Maybe my question is too silly but here I go: At 20:02 you expanded the covariant derivatives and the partial derivatives in components, but what causes me confusion is, aren't they the same? why in the partial derivatives there are no connection coefficients and we have the term ∂i ∂j ? I understand we can expand the expression ∂i ∂j in terms of the Christoffel symbols. I appreciate your videos, great job!

[@eigenchris]

You're not the first to ask this. I actually made a semi-serious error in this video that I didn't realize at the time... but at 5:46 and 6:15, I should not have set these 2nd-order partial derivatives to zero... instead I should have let them cancel each other in the Lie bracket subtraction. The standard derivative without a connection doesn't let us say that ∂x∂y = 0. Instead we get ∂x∂y - ∂y∂x = 0. On the other hand, with the Covariant derivative, this explicitly contains a way to take the derivatives of basis vectors using the Christoffel symbols/connection coefficients. This is what makes the Lie Bracket different than the Covariant derivatives.

[@PedroCruz-ku3qi]

@@eigenchris I see, it makes sense, thank you very much!

[@pomeron490]

Hello Chris, Thank you for your patience and clarification. I had a temporary mental aberration when I said "So, ...there is still a torsion tensor ". What I should have said is "...there is still a Lie bracket term ". I got the idea "Mathematically, torsion free is expressed as having the Lie bracket =0" from Tensor Calculus 20 The Abstract Covariant Derivative 10.45 to 11.21. I am a bit confused now that you say it is not correct. Am I missing something?

[@eigenchris]

Yeah, I can see you getting confused there. I didn't fully understand the torsion tensor or the torsion-free property when I made TC 20, so I "weaselesd" out of it using the special case where the two vector fields are basis vector fields. In this case the lie bracket does go to zero since d^2/dxdy - d^2/dydx = 0. However with more general vector fields this simplification doesn't happen and you need the lie braclet term to stick around. I'm regret not having explained this better but my understanding was limitted back when I made these videos. I hope that clears it up.

[@Ohmashaal]

Anyone knows a derivation of spin connection that's explained as simple as this?

[@Simon-ed6zc]

Hiya, sorry for bothering you again, but I think all the different derivatives get mixed up in my head. At 20:55 you make a difference between the ordinary derivative and the covariant derivative. At 19:22 you used the symbol of the covariant derivative for u(v) and v(u), right? Is that just notation abuse or is u(v) = cov_u(v)? When I (probably wrongly) apply u(v) and then cov_u(v) = cov_{u1d1 + u2d2}(v) = u1*cov_d1(v) + u2*cov_d2(v), I get the same final expression for the derivative of the basis vector. Do you have any idea where that could have gone wrong? And what exactly do you mean with the ordinary derivative? Around 5:40, I believe, you derive the jth basis vector by the ith basis vector. Could one not expand that in terms of christoffel symbols? Again, sorry for all these questions. I jsut really want to understand TC and later GR and your videos are just the best step-by-step lectures I have found. All the best!

[@eigenchris]

The line at 19:22 is not the definition of the Lie bracket. The definition is [u,v] = u(v) - v(u). So when I write [u,v] = ∇_u(v) - ∇_v(u), this is not the definition of the Lie bracket... it's a property satisfied by a special type of connection called a "torsion-free connection". (When I say "connection", I'm talking about a choice of Christoffel symbols/gammas.) Some connections are torsion-free and so this property holds, and other connections are not torsion free and so this property fails. For the purposes of GR, we assume the connection is torsion-free, so we accept this property as being true 100% of the time. There are some more fringe/theoretical modifications of GR that allow for connections that are not torsion-free, but you don't need to worry about those when you are first studying GR. The u(v) notation at 5:40 should not be expanded in terms of the Christoffel symbols... maybe check my pinned comment for more info. Let me know if this answers your question.

[@Simon-ed6zc]

​@@eigenchris Ahh, so in the 19:22 case what we really have us that u(v) - v(u) = ∇_u(v) - ∇_v(u), correct? Now I am still not sure if I understand the difference between ∇_u(v) and u(v) If you would bear with me for a wee longer, I would much appreciate it. The left hand side are "ordinary" partial derivatives (u^i ∂_i)(v^j ∂_j), while the right hand side are covariant derivatives: ∇_u(v) = ∇_{u^i ∂/∂_i} (v^j ∂_j) = u^i ∇_{∂/∂_i} (v^j ∂_j) # ∇_{a*t + b* w} v = a*∇_t (v)+ b*∇_ w (v) = u^i ∂/∂_i (v^j ∂_j) = u^i * [ ∂_i(v^j) ∂_j +v^j* (∂_i∂_j) ] # expand (∂_i∂_j) as christoffel symbols "Gij^k" = u^i * [ ∂_i(v^j) ∂_j +v^j* Gij^k ∂_k] = u^i ∂_i(v^k) ∂_k + u^i * v^j* Gij^k ∂_k = {u^i ∂_i(v^k)+ u^i * v^j* Gij^k ] ∂_k u(v) = (u^i ∂_i)(v^j ∂_j) = u^i * [ ∂_i(v^j) ∂_j +v^j* (∂_i∂_j) ] # same expression as above, however something must be different (?) I think the mistake I make might be that I expand ∇_u(v) = ∇_{u^i ∂/∂_i}(v^j ∂_j). This leads to the same expression (∂_i∂_j) in the ordinary derivative and I get confused why one can be expanded as christoffel symbols and the other cannot. I am not sure how to do it differently, though. . If I remember correctly, the covariant derivative is the ordianry derivative with the normal component subtracted, correct? So ∇_u(v) = u(v) - n, but I don't think this helps me understand why I can expand one in terms of christoffel symbols and not the other. Sorry, for the long and confused sounding post. But I really want to understand this... All the best, and thank you a lot for your videos and your help so far! Edit. Just saw your pinned comment. Its interesting that the covariant derivative contains a way of takign the derivative of the basis vectors and the ordinary doesn't. I was always under the impression that (∂_i∂_j) could always be expanded in terms of christoffel symbols. Maybe thats also a point of confusion... I always thought the christoffel symbol gives you the component of the kth direction if you derive the jth basis vector in the ith direction. And if you have a sum over all i and j oone could say u^i v^j (∂_i∂_j) = u^iv^j Γ_ij^k e_k. Edit2: (sorry :/) At 23:19 you have a huge diagram. You have the vectors u and v as thick lines, you have their parallel transported versions as dashed lines. The grey arrows are the covariant derivatives, correct? And the second pair of thick lines connecting to the original vectors are the vectors of the field, which you reach by U(v) and v(u), correct? Then it would make sense, that their tips are connected by the lie bracket.

[@eigenchris]

You're expressions are correct. I think I've had a bad habit in some of my videos of writing the covariant derivative using partial derivative notation, because I really didn't see the difference between them at the time. I'd suggest writing "u^i ∇_{∂/∂_i} (v^j ∂_j)" as u^i [ ∂/∂_i(v^j) + ∇_{∂/∂_i}(∂_j)], using the nabla symbol to help remind you that it's the covariant derivative. Whereas using the ordinary derivative just gives you the expression "∂_i∂_j" instead. This is one of the more frustrating misunderstandings I've made.

[@Simon-ed6zc]

@@eigenchris I see! Thanks a lot! I think you derived the covariant derivative from pretty much the same expression you used for u(v), so that was a bit confusing. To summarize, the covariant derivative inherently takes care of basis-vector derivatives through the christoffel symbols, which do not appear in simple, ordinary partial derivatives (how would you calculate them there? Just "brute force"?). The covariant derivative gives you a way of connecting the tangent spaces on a surface at two different points, while u(v) tells you how v changes if you vary it in the direction of u (or do both do that?). I greatly appreciate these videos! I always wanted to learn this but my PhD took me in a completely different direction. So this is a complete pleasure project for me and I could have never afforded the time to do all of this myself. Thanks a lot.

[@rasraster]

@eigenchris - I'm mystified about how you got the formula at 18:15. When I look at the right hand side, it seems to me like that equals (v-u). Unless maybe the label nabla sub u of v (and nabla sub v of u) are referring to the GREY vectors... but if so, why?

[@eigenchris]

The covariant derivative vectors are represented by the grey vectors. The "dashed" vectors are what parallel transport of u and v would look like. The "solid" vectors are what u and v ACTUALLY look like. The grey vectors are how much the ACTUAL u and v (solid) deviate away from the parallel transported u and v (dashed), so it's just the difference between the dashed and solid vectors.

[@firdausalfaraby6356]

I have a question. Here, you said that swapping the indices of the Christoffel symbols is possible due to the connection being torsion free and that torsion free is a property of the connection. But in video 19 you argued that we can swap the indices because we can swap the order of partial derivatives, which feels like a property of the space or coordinates instead of the connection. Then with metric compatibility we can derive LC connection without torsion free. So, what am I missing here? Are the conditions where we can swap partial derivatives the same as torsion free somehow?

[@eigenchris]

So I did my best to explain this, but I don't know if I did a very good job. After videos 17-19 and starting in video 20, there's a change in perspective of what the covariant derivative means. In video 17, the covariant derivative is built from the concrete situation of needing to define a derivative that works in any basis in flat space. In video 18, the covariant derivative is built from the concrete need of needing to define a derivative that works for a small creature living on a 2D surface embedded in 3D space. These definitions are built "from the ground up" using concepts that are already familiar to us like surfaces and partial derivatives. Starting in video 20, there's a change in perspective where try to invent a definition of the covariant derivative "from the top down"... a definition that is based on 4 properties/axioms without reference to anything else. It turns out that in video 20, the property of "torsion free"/"swapping lower Christoffel symbol indices" isn't something we can prove from the 4 properties in the definition. "Torsion-Free" is a property that some connections have, and others don't. As seen with the "Boring Connection" in video 20, we can come up with covariant derivatives that don't result in the standard "parallel" transport rules that we're used to.

[@rahmatkhan3982]

sir your videos are very helpful

[@yousufnazir8141]

Excellent explanation

[@operatorenabla8398]

I'm curious about one thing: At the times Einstein was "mathematicyzing" general relativity, was today's notation already in use? I don't mean the symbols themselves, but the theory: was it in the form we know it today for mathematicians? Or was it being discovered in the same years of Einstein's general relativity?

[@eigenchris]

I'm not really a historian, so take what I say with some caution, but I think the tensor calculus Einstein used was conveniently developed a decade or two before he needed it for general relativity. Levi-Citiva published a 60-page work on tensor calculus in 1900 called "Méthodes de calcul différentiel absolu et leurs applications", or in English, "Methods of the absolute differential calculus and their applications". I think Einstein only started considering this type of math around 1907 or later. Wikipedia links to the original French paper here, at the first bullet point under "articles": en.wikipedia.org/wiki/Tullio_Levi-Civita#Works I haven't read it fully but here are some observations from a quick look: I don't think Einstein notation had been invented yet, so you can see summation symbols everywhere. Also the Christoffel symbols didn't use the capital Gamma Γ we use today. Instead they used a weird notation using a tall { } with the 3 index letters inside (see section 1.5 for an example). There's something like the Lie bracket defined in section 2.3 but it looks different. The metric seems to be written in various places with the letter "a" instead of our modern "g". I think the "dot product" notation would have been pretty new at the time and I'm not sure if it was well-known. I don't see it used in the paper anywhere. I also don't see the idea of basis vectors anywhere. Again I'm not sure what the state of linear algebra was at the time, and whether the author would have viewed a vector as just a list of components. Personally, tensor calculus was incomprehensible to be without using basis vectors. Basis vectors are half the story, with components being the other half.

[@operatorenabla8398]

@@eigenchris One thing I'm sure of is that at the times Einstein was a student, there was an exam called "determinants" so a theory of linear systems existed and was probably complete (after all Cramer lived in the 17th century). If I have to guess, I would say that linear algebra the way we know it today was developed after or contemporary to the Bourbaky school, but again this is just a guess. Anyway, I think it is remarkable that Einstein studied contemporary mathematics instead of limiting himself to what was meant to be learned by physicists. After all, many of the contemporaries that criticized General Relativity didn't know about differential geometry (and that's probably the main part of the reason they criticized Einstein, besides other possible human reasons). I mean, it would be remarkable today if a physicist is so devoted that he spends time in learning new mathematics, think about a century ago where you couldn't simply go on internet and look for articles or books.

[@eigenchris]

@@operatorenabla8398 I think Einstein had help. The excerpt below says that when Einstein had the idea to use non-euclidean geometry, he consulted his mathematician friend Grossmann for help, and learning about the Riemann and Ricci tensors from him: hsm.stackexchange.com/questions/5188/how-was-einstein-led-to-make-a-contact-with-differential-geometry-for-his-theory In general, I think Einstein had a lot of help from other people getting the mathematics of special and general relativity fleshed out. I think it was largely Minkowski who was the first one to draw spacetime diagrams, light hones, and hyperbolas related to Lorentz transformations (and also Lorentz transformations existed before Einstein published relativity).

[@jongminlee5634]

Thanks for very useful video!

[@J.P.Nery.N.]

Simply amazing!

[@domenicobianchi8]

really appreciate all your work. which textbook or sources you referred to in order to produce this tensor calculus serie?

[@eigenchris]

For most of my videos I use a lot of sources in parallel, so it's not just 1 source. Also for my older videos I was less diligent at keeping track of sources, so if there's nothing listed in the video or description, I've forgotten which sources I used.

[@domenicobianchi8]

I understand, the long indicial derivations are so extreme i tought you had to follow one precise book in order to produce them. so, right now, would you suggest any particular text or notes (about ricci's tensor and gr mainly) as a follow up to your videos? i'm asking you cause i think u really are able to discerne a good reference

[@toaj868]

Why exactly is the term torsion used here?

[@silentbubble]

Thank you very much for this series! This video is fantastically helpful for me to visualize and finally understand what torsion really is. But I have questions. The Levi-Civita connection is defined in terms of the metric and its derivatives and thus it is always torsion free. However, for nonzero torsion, does it mean that the connection has extra terms that are unrelated to the metric? What is the formula for a general connection with nonzero torsion? Thanks.

[@ht4792]

Having a formula for connection coefficients means, it can be uniquely defined. But, when it's uniquely defined we get the Levi-Civita connection. So, you can define connection coefficients as you like in a general case.

[@HansTube1]

If a christoffelsymbol is not a tensor, how come a torsion tensor T is a tensor? I ask this because the torsion tensor is made out of 2 christoffelsymbols.

[@eigenchris]

The Torsion tensor can be written as a linear combination of a in a basis, but the Christoffel symbols on their own cannot be written as a linear combination of a basis. The Torison formula is written using a special combination of Christoffel symbols where the final result is a tensor.

[@AkshayKumar-nr7vb]

Please make videos on lie derivatives and killing vector field

[@gguevaramu]

Hi Chris. I had some misconceptions that finally could resolve and I removed my former questions. But here in minute 17:58 I see a problem with the vector [u,v]. If it represents a difference between u(v) - u(v), then these two vectors should be considered with their tails together and that is not happening in your vector diagram. I have some days trying to find a similar graph in other texts but not successes jet.

[@eigenchris]

I got the visualization from: math.stackexchange.com/questions/465603/are-there-simple-examples-of-riemannian-manifolds-with-zero-curvature-and-nonzer You're right that we can't subtract vectors unless their tails are at the same point (the same "tangent space"). The diagram I made is "lying" somewhat, since the u(v) and v(u) vectors should really have their tails at the lower-left corner of the diagram, as they are derivatives taken at that point. If the space is flat, we can move the vectors around anyplace we want, and create the diagram shown in the video. However in curved space, we can't do this, and so all the derivative vectors should really be drawn with tails at the lower-left corner, since this is the tangent space they live in.

[@gguevaramu]

​@@eigenchris Thank you, Chris. If you check the diagram maybe you also will feel a little bit uncomfortable with what is said. Most of the books say that [u,v] = u [ v ] - v [ u ], but in the diagram showing the Torsion tensor not null, you can see that it is not the case because both covariant derivatives do not have the same origin. I think the diagram is saying that [ u , v ] = { u(P) + v(Q) } - { v(P) + u(R) } and it is acceptable. But also is said that Torsion tensor is defined using Lie bracket but In the diagram, Lie bracket is not what was said. In the end, in the Torsion tensor definition, we have an expression with 2 new elements not define jet T(u,v) and [u, v ] and because of that I feel I can not calculate T(u,v) Maybe I am losing something

[@axiomateorema6355]

Hi Gerardo, You are right. The diagram is not correct. The tails of all four vectors lie in the same point because they are obtained by differentiation of vectors u and v at their intersection and origen point. As a matter of fact you dont need to paralel transport any vector at the end of the other vector. You only must obtain the covariant derivative at the origen point and make it equal to zero.

[@gguevaramu]

@@axiomateorema6355 Thanks. I am still trying to understand this diagram in the correct way

[@azziahmed4721]

Thank's, but in 20:32 we have (...) - (..) - (...) + (...)

[@tursinbayoteev1841]

Hi Chris, the direction of the covariant derivatives doesn't matter at 17:21? By definition it is pointed to the end of parallel transported vector.

[@eigenchris]

I don't quite understand what you mean. nabla_u v = 0 says we are parallel transporting the vectoe v in the direction of the u vector. Why do you say the direction doesn't matter?

[@tursinbayoteev1841]

@@eigenchris The result of covariant derivative is a vector itself. I mean what is the direction of resulting nabla_u (v)? v is not parallel transported one, the actual field.

[@tursinbayoteev1841]

It is directed from the end of dashed lines toward the end of solid ones. How do you choose this direction?

[@j.k.sharma3669]

Very nice video

[@miguelaphan58]

beatiful job !!!!

[@deepbayes6808]

Here (20:00) by ordinary derivatives you mean covariant derivative with levi-civita connection? (I understand the two coincide in flat spaces, but this geometric example is not requoring flatness I suppose?)

[@eigenchris]

By "ordinary derivatives" I just mean "partial derivatives". There's no connection coefficients involved.

[@deepbayes6808]

@@eigenchris now I understand, thanks.

[@jibeshbeura6453]

Sir please make vedios on general relativity.

[@eigenchris]

If anyone wants to see an "in-progress" version of some of my early general relativity videos, you can see them here: twitter.com/eigenchris/status/1229681192502362114?s=20 Feel free to leave comments/feedback.

[@sufyannaeem2121]

components of the torsion tensor are zero for torsion free property only? As christoffel symbols are symmetric in lower indices so shouldn't it be zero for every condition???

[@eigenchris]

The lower indices of the connection coefficients being symmetric is not something you can prove. You can only state it. There are connections that are not torsion free.

[@temp8420]

Excellent

[@allandavis6116]

At 19:40 we have that the property of being torsion free depends only on the connection, so, is the Levi-Civita connection torsion free? And, why are we interested in connections other than the Levi-Civita connection? Also, when vector v acts as a derivative operator on vector field w, won't v(w) in general have 'normal' components outside of the tangent space?

[@eigenchris]

Levi-Civita connection is torsion-free by definition. (It is defined as a torsion-free connection with the metric compatibility property.) For basic general relativity, the Leci-Civita connection is the only one we care about. Many GR courses don't even bother saying "Levi-Civita" connection because they are assuming it's the connection being used. I have never stuided any other connections so I can't tell you when the would be useful. I should have said we were working with intrinsic geometry in this video. Normal components never show up so you can ignore them.

[@allandavis6116]

@@eigenchris It looks to me like your definition of u(v) is too complicated. From wiki .....en.wikipedia.org/wiki/Lie_derivative ..... The Lie bracket of X and Y at p is given in local coordinates by the formula {\displaystyle {\mathcal {L}}_{X}Y(p)=[X,Y](p)=\partial _{X}Y(p)-\partial _{Y}X(p),}{\displaystyle {\mathcal {L}}_{X}Y(p)=[X,Y](p)=\partial _{X}Y(p)-\partial _{Y}X(p),} where {\displaystyle \partial _{X}}{\displaystyle \partial _{X}} and {\displaystyle \partial _{Y}}{\displaystyle \partial _{Y}} denote the operations of taking the directional derivatives with respect to X and Y, respectively. Here we are treating a vector in n-dimensional space as an n-tuple, so that its directional derivative is simply the tuple consisting of the directional derivatives of its coordinates.

[@tursinbayoteev1841]

Hi Chris, when the change of a vector along another is not used Г? Is Г only for covariant derivative?

[@eigenchris]

Yes, the Г are only used in the covariant derivative. The Lie Bracket (also called "Lie Derivative", which you can look up online) doesn't require a connection in its definition.

[@tursinbayoteev1841]

@@eigenchris , Thank you! The change of a vector along the other (in definition of Lie bracket) is the same as covariant derivative of a vector in the direction of the other? Or they are same only for torsion free connections?

[@takomamadashvili360]

Thank u!!!!

[@kamaellewis1545]

What are the remaining planned topics of the series

[@eigenchris]

I only have 2 more videos planned: Riemann Curvature Tensor amd Ricci Curvature Tensor. I will likely do some amount of relativity afterwards, but I haven't planned that out yet.

[@gguevaramu]

@@eigenchrisAll math you have developed including those 2 more videos have the final reason in relativity. It will be great to see how to read Einstein's equation in general relativity. You have a very intuitive way to explain complicated topics.

[@IntegralMoon]

Now I’m a little confused. Isn’t ∂i∂j = ∂(e_j)/∂x^i, which would make this whole torsion tensor zero for any case wouldn’t it? Is there something I missed?

[@IntegralMoon]

Sorry I mean at 20:35

[@eigenchris]

That derivative would be zero for cartesian coordinates, where the basis vectors are constant. In another basis, the derivative could be non-zero.

[@IntegralMoon]

Sorry, I perhaps haven’t been too clear in asking this, I realise this would be zero for Cartesian coordinates, but I can’t avoid making it zero in general, so I must be missing something. I wrote it out in more detail over at math exchange math.stackexchange.com/questions/3397449/torsion-tensor-always-zero but I must be missing something. The comments certainly indicate that I am :P

[@Thatcher_Lai]

Might be a dumb question, but can the torsion tensor be non-zero if the space is flat?

[@eigenchris]

Torsion is a property of the connection chosen, not the curvature of the space. In this series, and general relativity, we only ever use a one particular connection called the "Levi-Civita connection", which is torsionless by definition. But it's possible to choose another connection which has torsion. You can do this in any space, including flat space. I think I talk about this in later videos.

[@Thatcher_Lai]

@@eigenchris Thanks, that makes sense now. Just wanted to say thank you so much for your videos. In my opinion, your videos on tensor calculus and general relativity are by far the most clearly and intuitively explained on KZbin. You deserve so many more subscribers!

[@monsurrahman7663]

Sir I have a question,,in 17.07 time in this video we said "accutual vector in the vector field" there i got confused,,, The accutual vector is a result of the parallel transport of the vector which you had already mentioned to be zero

[@monsurrahman7663]

In 16.53 time I understand because of the parallel transport the covariant derivative is zero,, but later in this video in 17.22 you used the same covariant derivative as difference between the actual vector and parallel transportered vector.. there i am suffering

[@eigenchris]

The "dashed" arrows represent vectors that would be parallel transported versions of u and v (zero covariant derivative). The "solid" arrows represent the actual vector field, which can be any vectors we like (no necessarily parallel transported).

[@tursinbayoteev1841]

@@eigenchris , Can we imagine the solid vectors as "continuation" of flow curves for each vector?

[@abhishekaggarwal2712]

Finally!!

[@muhammedustaomeroglu3451]

I could not get what does it mean that vectors acting on vectors in physical (or geometric) point of view. Yes we can think partial derivatives as base vectors but when I try to represent a physical vector field partial derivatives become meaningless to me. For instance, let E be electric field vector and B be magnetic field vector. if I write E(B) it does not mean anything to me. Is there a more intuitive way of thinking that you can suggest because obviously I am missing things.

[@eigenchris]

The vectors that is behaving like a derivative is like the derivative's "direction". And the vector being acted on is like a vector field. So U(V) is the rate of change of the vector field V in the direction of U.

[@muhammedustaomeroglu3451]

@@eigenchris so it is just an extra property that we add to the vectors, that has nothing to do with physics (it is something like notational device) if it is useful to find change of B as we travel along E we use it " E(B)/||E|| ". I thought that it might have more meaning that I miss but it does not have. Thank you.

[@jacobvandijk6525]

Your videos are priceless, Chris! But I will give the cup of coffee to someone who probably needs it more than you. Although I haven't got a clue what your profession is, I think you're doing fine ;-)

[@eigenchris]

That's fair. I don't need the money. I spend a lot of time making these though, so i figure I would give people to chancr to donate if they want.

[@jacobvandijk6525]

@@eigenchris No problem with that. I just told ya why I don't donate. The time you put in to make these videos is well reflected in their quality. I think everyone watching agrees on that. I know at least one resident of the Netherlands who is very happy with them ;-)