Tensor Calculus 23: Riemann Curvature Tensor Components and Symmetries

[@eigenchris]

Error at 20:35 : I mixed up the arrows pointing at r and 1/r Fixed an error with one of the symmetries pointed out by AverageKZbinr17. R_abcd = R_cdab. In the original video I put a negative sign in front by accident.

[@eigenchris]

@AverageKZbinr17 I really do appreciate viewers like you catching my mistakes, especially as early on as you did. It saves others a lot of confusion.

[@Snake_In_The_Box]

plz keep going

[@a_llama]

@@eigenchris Hey! Another minor error at 9:20, the second green term should be b then a then c right? Great videos man!

[@frankdimeglio8216]

@@eigenchris UNDERSTANDING WHAT IS THE COMPARATIVE AND BALANCED BODILY/VISUAL EXPERIENCE OF THE MAN WHO IS STANDING ON WHAT IS THE MOON: The BULK DENSITY of the Moon is comparable to that of (volcanic) basaltic LAVAS on the Earth. The energy density of LAVA IS about three times that of water. SO, now, get a good and CLEAR look at what is the ORANGE SUN. The human body is, in fact, about as dense as water ! Look up at the blue sky. Importantly, this is a balanced BODILY/VISUAL EXPERIENCE !! The Moon is ALSO BLUE, AS it is barely visible then; AND the Earth is ALSO BLUE. SO, let's now match up (on balance) the eye, the blue sky, AND the ORANGE "Sun"/LAVA. Notice the basic match in sizes and brightness when considering the SETTING AND fully illuminated Moon AS WELL. (Most of the human body is water, with an average of roughly 60 percent. The average density of the Moon is also 60 percent in comparison with that of the Earth.) Beautiful. Relatively speaking, the terrestrial moon is, in fact, roughly a match (regarding it's size, at 27 percent) WITH the surface land area of the Earth (at 29 percent). In conclusion, in establishing the comparative AND BALANCED BODILY/VISUAL EXPERIENCE of gravity on what is the Moon, we would multiply one HALF times one third in order to get the correct answer. This explains quantum gravity, AS E=MC2 IS F=ma; AS ELECTROMAGNETISM/energy is gravity; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE. Accordingly, the rotation of WHAT IS THE MOON matches it's revolution !!! Great. Consider the relation with the tides. The Moon is ALSO BLUE. E=MC2 IS F=ma ON BALANCE. This NECESSARILY represents, INVOLVES, AND DESCRIBES what is possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity. Gravity IS ELECTROMAGNETISM/energy. In conclusion, I have mathematically proven that the gravity of what is THE MOON is ONE SIXTH in direct comparison with what is THE EARTH/ground; AS gravity/acceleration involves BALANCED inertia/INERTIAL RESISTANCE; AS E=MC2 IS F=ma; AS GRAVITATIONAL force/energy IS proportional to (or BALANCED with/as) inertia/INERTIAL RESISTANCE. Gravity IS ELECTROMAGNETISM/energy. Think QUANTUM GRAVITY !!! Think energy DENSITY. It all CLEARLY makes perfect sense ON BALANCE. The ENERGY DENSITY of LAVA is about THREE TIMES greater than water. There would be no gravity on the blue Earth, AS the EYE then stands in balanced relation WITH what is the BLUE sky !! BALANCED BODILY/VISUAL EXPERIENCE does then dictate that the gravity of WHAT IS THE MOON would be reduced by (and at) one half, thereby bringing the surface gravity of what is THE MOON to BALANCE with that of THE EARTH/ground at one sixth. So, one half (ON BALANCE) of one third is one sixth. The orange "Sun" AND what is THE MOON ARE BALANCED with what is THE EYE. (Think about what is LAVA.) Notice, comparatively, that these forms or manifestations ARE semi-detached in relation to touch/tactical EXPERIENCE !! Moreover, the Moon is TERRESTRIAL in COMPARISON; SO (ON BALANCE) the Moon is ALSO then about one quarter (at 27 percent) the size of EARTH/ground !! The blue Pacific OCEAN takes up about HALF of the view of what is the Earth ON BALANCE !! SO, we WOULD multiply ONE HALF (ON BALANCE) in order to account for WHAT IS THE MOON, this ORANGE "Sun", AND LAVA on balance !! Now, in conclusion, the land surface area of the Earth is 29 percent. This is exactly ON BALANCE WITH BOTH one third AND one fourth, AS ELECTROMAGNETISM/energy is gravity ! E=MC2 IS CLEARLY F=ma ON BALANCE ! Gravity IS ELECTROMAGNETISM/energy. Great. By Frank DiMeglio

[@g04jee]

some miss are good exercises to someone, like me 😂

[@morganhopkins204]

I can't thank you enough for all the hard work you've put into all your videos! These videos have been monumental in helping me achieve my goal of understanding general relativity. These series really mean a lot to me

[@twofiveburn3220]

Eigenchris certainly puts in a lot of hard work. He's helped a lot of people, including myself.

[@NoName-sy4lm]

Absolutely love the way you talk through these concepts in an intuitive way, it makes it a lot easier to visualize what's happening.

[@WilEngl]

After a few attempts, I finally understood the value of tensors thanks to your series of video. Thank you very much.

[@orktv4673]

You've been doing an excellent job at making things intuitive, and I think it would have been a fun exercise to qualitatively check the Riemann curvature tensor (and its symmetries). For the example treated one could look at the u2 basis vector on the equator. Parallel transporting along u1 in this case preserves the angle (you're actually going straight), and transporting along u2 always preserves the angle. Moving first along u2 then u1 therefore yields a completely unchanged vector, but if going along u1 first makes you leave the great circle, and now the path veers to the left. This means that the subsequent transport along u2 introduces a slight change in angle, which results in a difference vector pointing purely in the u1 direction. Close to the poles, the spherical coordinate system effectively becomes the polar one, so the curvature should vanish, in accord with the sin^2(u1) formula.

[@3igen3ggy]

Can't wait till next video. Big shout out from Poland man!

[@Why_Alex_Beats_Bobbie]

Another incredible video! This is the best intro to differential geometry I have every seen! Around the 12:30 mark, I think we need to use the torsion-free property to conclude that the LHS=0. The dot product is indeed a scalar, but after the first partial derivative we are now differentiating wrt a (0,1)-tensor/covector, which means we need to resort to Christoffel symbols this time. Luckily, they are symmetric on the lower indices (tensor-free) so we end up with zero anyway.

[@Why_Alex_Beats_Bobbie]

I guess I should clarify a bit. Let F_ab be the covariant derivative commutator operator (formally: F_ab = Nabla_a Nabla_b - Nabla_b Nabla_a) and Φ be a scalar. In general, it is not true that F_ab(Φ) = 0. Instead, we generally have a torsion term, namely F_ab(Φ) = τ_ab^c Nabla_c Φ where τ_ab^c is the difference of the two corresponding Christoffel symbols with their lower indices swapped (This is due to the fact that when we take the second derivative, we are no longer dealing with a scalar, thus Christoffel symbols are introduced). Of course, this whole point is moot when the torsion is zero, since τ_ab^c will be zero as well. This is only relevant with respect to when these symmetries are applicable. Other than that, this is basically nitpicking and my way to say thanks for this incredible series.

[@SilentFrog-cx7rd]

SOOOO Goooood!!!!

[@djordjianfairpaltrusio4937]

THIS SERIES SO GOOD UPLOAD NEXT VIDEO ALREADY

[@im54d89]

I'm so thankful to universe that this man exists ❤️:-):-)

[@abdullahmudasar9743]

When will you upload Ricci tensor...?Btw enjoyed the series...

[@aliesmaeil1044]

thank you, we are waitting for next video

[@larrydurante6670]

Hi Chris in a17.5 you cleared up a ton for me about the way engineers and mathematicians look at things. Anyway you could please explain the second covariant derivative of a tensor in components like you did for the first in 17.5 please. Theses videos are brilliantly done. I’ll support you with many coffees 😉

[@eigenchris]

I've heard the term "second covariant derivative" used to refer to different things. Which one are you talking about? Just aplying the covariant derivative twice? Or do you mean the definition on wikipedia? en.m.wikipedia.org/wiki/Second_covariant_derivative If it's the second one, I'm don't understand it very well.

[@YossiSirote]

Truly excellent!!! Thank you 🙏🏼

[@person1082]

interestingly the generalized inertia tensor I^{ijkl}=m(r^ir^kg^{jl}-r^ir^lg^{jk}-r^jr^kg^{il}+r^jr^lg^{ik}) has the same symmetries

[@fredrickvanriler7986]

I immensely enjoy your videos! My question is: in the future, will you be doing any tutorials on Quantum Mechanics, specifically, of the operations on the Schrödinger equation?

[@eigenchris]

Thanks. I don't have any plans for covering quantum.

[@ranjitsarkar3126]

@@eigenchris mechanics

[@reihanvojoudi9716]

you're doing a great job here :D thX for these videos

[@TheJuand97]

I really need the Ricci tensor video

[@y0n1n1x]

THANK YOU SO MUCH

[@mtach5509]

YOU ARE VERY GOOD

[@yangyang3824]

Great Lecture! I have a question at 13:17, the fouth line to fifth line of the equation, why e_i can be factored out in this way? Would the order matter?

[@eigenchris]

The order of the dot product vectors doesn't matter. Also, the dot product is bilinear so any scaling constants can be brought out in front.

[@doctormeister]

Would it be possible that for some sort of coordinate system, the lie brackets would not go to zero? If they always go to zero, the 3rd term of the Riemann tensor seems nonsensical (Except for linearity but that would just be a slight of hand then)?

[@jnk594]

so according to 22:55 if I travel from moscow to baghdad then from baghdad to cairo then from cairo to berlin then from berlin to moscow 'i will have a square area less than the square of sine angle between between moscow -baghdad-cairo-moscow right , am I right Dr Chris?

[@JgM-ie5jy]

At [00:50] I was expecting to see the tail-to-tip direction of the difference vector between the initial vector w and the parallel-transported vector w∥ - that this direction be in the opposite direction as the direction shown, i.e. I was expecting that the tail of the difference vector would be at the tip on the initial vector w and that the tip of the difference vector would be at the tip of the parallel-transported vector w∥. So is there an explanation why the difference vector is - apparently - in the opposite direction of the w∥ - w ? Once again, my heartfelt thanks for the wonderful gift of your exemplary work.

[@eigenchris]

I think you're right about that. My mistake.

[@allandavis6116]

Since the curvature of a sphere is everywhere the same, wouldn't you expect R1212 to be a constant?

[@eigenchris]

You're right that the sphere is the same everywhere, but the coordinate system we're using on the sphere is not the same everywhere. The coordinate "grid" squares tend to buldge near the equator and get very skinny near the poles. Because of this, a vector with have smaller u1 components near the equator and larger u1 coordinates at the poles. The Riemann tensor components have some u1 dependence to compensate for this. So the short answer is: the geometry is the same everywhere, but the components/coordinates are not the same everywhere.

[@allandavis6116]

@@eigenchris Curiouser and curiouser. Ok, so R1/212, i.e. R1/212(u1,u2) , or, more specifically non-zero R1/212∙A where A is the area of a small rectangle bounded by (0.0), (r,0),(r,s),(0,s) at (u1, u2), is the U1 coordinate of the difference between U2 and U2 parallel translated around the rectangle. ( ?) And the U2 coordinate of the difference, i.e. R2/212∙A = 0. But parallel translation preserves length (?), so ..... where is the error? ? EDIT: I think the answer is that there is a change in the U2 coordinate but it is a 3rd order effect, i.e. it includes factors like dxdydy, and hence disappears in the limit.

[@ranjitsarkar3126]

3:51 Stop calling these beautiful mathematical expressions ugly

[@ilredeldeserto]

how do we calculate how many the (236) dependent components for space time are? Is there a fast way?

[@eigenchris]

All you can do is look at the symmetry rules and take note when a component goes to zero, or is the same as another component.

[@garytzehaylau9432]

Finish this video.it looks very good and clear. Sorry.i have trouble in lecture 22 when i try to understand the formula DCBA (W) -W/ rs 1.the area of the parallelogram is rs sin(theta) (since it is cross product of two vector) do you need to divide it by theta as well?(because r and s are just two side of the parallelogram but not the base and height) 2.what makes the rate changing of a separation vector constant along a geodesics given a flat plane? thank you.

[@eigenchris]

1. I don't honestly know why there isn't a division by sin(theta). The limiting process only involves taking the lengths of r and s going to zero, so we definitely need to divide by those to "normalize" the area. However the angle theta is constant no matter how big or small the parallelogram is, so it doesn't seem as important when we take the limit. I don't really have a better answer than that. 2. In the flat plane, if I had two geodesics (lines) that are parallel, the rate of change of the separation vector would be zero. If I had two geodesics (lines) that are not parallel, and are at slight angles, the rate of change of the separation vector would be constant.

[@garytzehaylau9432]

@@eigenchris you finally come back!!! math.stackexchange.com/questions/3477028/not-sure-why-the-geodesic-derivation-equation-involved-second-ordinary-derivativ i put this question(about your video25) in the stackexchange and still no response thankk

[@garytzehaylau9432]

@@eigenchris there is trouble when i watch lecture 25. :( miss you

[@DouglasHPlumb]

You need a book with lots of tricky exercises with solutions.

[@silentbubble]

The inconsistency in the definition of covariant derivative between math and physics has been quite frustrating for me as I am trying to learn tensor calculus. My confusion is that the covariant derivative is supposed to treat the basis vectors and the metric as constants, because it carries the connection with it. So the covariant derivatives of the coordinate basis and the metric are supposed to be zero. That is how I learned it before seeing this video. However, the covariant derivatives here act as a "partial derivative" when act on the basis vectors, which gives the connection. I'm not sure how to think about this anymore.

[@eigenchris]

Have you watched any of my videos on the covariant derivative? Videos 17, (17.5), 18, 19, 20 are all on explaining the covariant derivative at different levels of complexity. They are about 2 or 2.5 hours of content so I understand it is a lot, but you can look at them if you want. I would also strongly disagree that the covariant derivative treats the basis vectors as a constant... the entire reason we need the connection coefficients is to keep track of how the basis vectors change. Saying that the covariant derivative of the metric is zero is correct when we use the "metric compatibility" property (see video 20 for more on this).

[@listen_im_physics7559]

at 09:20 u can cut by a single line to easy understand instead of erase them

[@mohd.shadabkhan4477]

Good Video

[@Null_Simplex]

I'm a bit confused. The explanation about parallel transporting a vector w along a parallelgram represented by the vectors u and w is 3 variables, but the Reimann curvature tensor seems to require 4 variables. What does the fourth variable represent geometrically?

[@eigenchris]

Two lower indices are associated with the 2 vectors that make the parallelogram. One lower index is associated with the vector that is transported around the loop. The upper index is associated with an OUTPUT vector (the vector that measures the difference before vs after parallel transport).

[@Null_Simplex]

So after we parallel transport w, the vector we end up with can end up being linearlly indepent of u,v, and w. The top index specifies in which direction you want to measure the error of parallel transported vector from the original, in a similar fashion to a dot product. Am I at all in the right ballpark?

[@eigenchris]

Mostly, yeah. Basically the resulting "error" vector needs to be written as a linear combination of the basis vectors. These basis vectors (lower index) are summed with the Riemann tensor components (with the upper index). The combination of components and basis vectors gives the error vector.

[@tursinbayoteev1841]

Hi Chris, shouldn't it be Rd_bac the whole expression equal to? 4:03

[@tursinbayoteev1841]

liavas.net/courses/math430/files/Surfaces_part3.pdf here it is defined a bit different

[@eigenchris]

​@@tursinbayoteev1841 I think the definition I use is the same one given on page 14 of that PDF (keep in mind I've swapped the lower connection coefficient indexes, when comparing with the formula in the PDF). The formula I use is also shown on wikipedia: en.wikipedia.org/wiki/Riemann_curvature_tensor#Coordinate_expression. It's true that some textbooks use a different convention for the Riemann indexes, and may put the index letters in different positions, but the one that me, wikipedia, and that PDF use is the most common one.

[@empatheticgrinch]

Can you explain why the indices at 4:02 is Rd_cab? Looking at the coordinate independent formula for curvature we started the derivation with, my first impression would have been that it should be Rd_abc instead. I'm aware that the formula in the video is the correct one, just fail to see why it should be that order specifically.

[@eigenchris]

If I understand the question, it's not a math error you're pointing out, but instead you're asking why I used this convention? I'm not sure I have a great reason, other than this is the convention I first came across on wikipedia as well as several other sources: en.wikipedia.org/wiki/Riemann_curvature_tensor#Coordinate_expression. I figured it was better to follow mainstream notation so that viewers could compare to their textbooks. That said, this isn't the only convention and I've seen it written other ways too. Given the symmetries I discuss later in this video, it's arguably "nice" to keep a,b on one side because they can be swapped to get a negative sign, and also to keep c,d on the other side because they can be swapped to get a negative sign as well. This is slightly easier on the eyes when doing derivations that involve a lot of index swapped. But really, there's no good reason for this.

[@empatheticgrinch]

@@eigenchris I'm only seeing your quick reply now, sorry about that. What you said certainly answers a part of my question and thanks for the clarification. However, when I mentioned in my original reply that I expected it to be Rd_abc instead, I was really going by the index order of basis vectors in the formula R(e_a, e_b)e_c [2nd line @ 4:02 ] which has e_a, e_b and then e_c acted on the Riemann tensor. In other words, when we obtain the component formula of the Riemann tensor [bottom line @ 4:02], why is it written with "c" in front of "a" and "b" i.e. Rd_cab? Note that I'm actually referring to the difference between the placement of indices of Rd_cab and that of the starting formula R(e_a, e_b)e_c. Now I skimmed over a couple of texts that follow the same method as yours and all of them invariably present the tensor as Rd_cab without really acknowledging the "switch up" that had taken place, which makes me suspect that there's probably a glaringly trivial reason for this. Btw thanks for taking the time to answer these.

[@eigenchris]

​@@empatheticgrinch Ultimately, there's no good reason. It's just a convention.

[@Panardo777]

Yep just a matter of convention, as we consider R(u,v)w = R(w,u,v) instead of R(u,v)w=R(u,v,w)

[@amithkumar3176]

Can you please make some videos on Linear algebra?

[@eigenchris]

I'd recommend other sources like Khan Academy for linear algebra. There are already a lot of good videos out there online.

[@amithkumar3176]

@@eigenchris Yeah just like Tensors, there is abundant of material out there. But yours was much more easier to start with. So please, consider it a request.

[@eigenchris]

@@amithkumar3176 I'm sorry but I just don't have plans for making a series on that. I'm planning to do general relativity, and making the videos is pretty time-consuming. If you want to ask me any specific questions in comments, I can try to answer. But I don't plan on making videos on it.

[@amithkumar3176]

@@eigenchris Okay. Now I am happy on hearing "general relativity"😅

[@doctormeister]

Why does the Riemann tensor components on the sphere depend on the coordinate u1? The sphere has the same curvature everywhere?

[@eigenchris]

The curvature is the same everywhere (the Ricci scalar is a constant) but the coordinate system does not treat all points equally. The dependence on u1 reflects how the coordinate system looks different at the poles compared to the equator.

[@doctormeister]

@@eigenchris ok but if the Riemann tensor is 0 in that coordinate system(for some u1) doesn’t it imply that it is zero in every coordinate system because it’s a tensor? Thanks for the answer btw!

[@eigenchris]

@@doctormeister ​ @doctormeister That's true, but I think the poles are basically a "coordinate singularity", so the math can't always be trusted there. Note that the Christoffel symbols appear to go to infinity at 21:33. This is because the u2 basis vector isn't defined at the poles due to the coordinate system breaking. The Riemann tensor involves parallel transporting a vector around a small coordinate parallelogram, but at the poles, changing the u2 coordinate doesn't change your location on the sphere. So you end up with a coordinate "triangle" instead of a coordinate parallelogram. If you want to calculate the curvature there, you'd need to change to a new coordinate system. You'll have to keep an eye out for points where coordinate systems "break" in the future. If you ever get to studying black holes, you'll see for Schwarzschild black holes, the curvature appears to go to infinity at the the Schwarzschild radius. But this is the result of a bad coordinate system choice, as opposed to the curvature actually being infinity. You can change coordinates to remove the coordinate singularity and calculate the actual curvature in the new coordinates.

[@doctormeister]

@@eigenchris Thank you!

[@thegreatlazydazz]

9:05, how do the green terms cancel?😅

[@sufyannaeem2436]

What about the next lecture on Ricci curvature? When will you upload??

[@eigenchris]

Probably not for another month or so. There's a number of things I don't understand about the Ricci tensor.

[@Shirsh]

If independent component of riemann tensor is sin^2 u, then at poles riemann tensor vanishes, but the space is not flat on the poles, is it? How do we justify it?

[@eigenchris]

The coordinate system for the sphere "breaks" at the north pole because it has coordinates u1=0 and u2=anything. The same is true for the south pole at u1=pi. Because of this, many calculations will not work for the north and south poles. (For example, the Christoffel symbols for the north pole at u1=0 involve cot(0), which is not defined... as a result the Riemann curvature tensor is not defined.) This doesn't mean these points are different from other points, geometrically-speaking, it just means we picked a coordinate system that can't handle these points. If we changed coordinates to another coordinate system with an "east pole" and "west pole" instead, we could do calculations at the north and south poles just fine.

[@lcchen3095]

if you already assumed Torsion Free,then the Lie Bracket is zero,then the third term ie the term with Lie Bracket at 7:12 is zero. That is to say,you won't have the third term.

[@philamras3732]

If you look at the equation for Torsion, Torsion free doesn't necessarily mean a zero Lie bracket.

[@warrenchu6319]

At 22:39, R _1_212 = (sin (u_1))^2 for the curvature of a sphere where U_1=theta. Since a sphere has only one curvature value everywhere, what is that value of (sin(u_1))^2 and in what way does (sin(u_1))^2 represent the difference between W and its transported result?

[@eigenchris]

R^a_bcd is what happens when you take a parallelogram formed by the c- and d-directions and transport a vector pointing in the b-direction. The resulting number measures the drift in the a-direction. In our case, we're taking a parallelogram formed by u1 and u2 (theta and phi) and transporting vector pointing in the u2 direction, and measuring the drift in the u1 direction. The fact that the u1 coordinate exists in the R^1_212 reflects the fact that the parallelograms change size at the coordinate system's poles vs the equator (whereas they don't change size when we travel along u2). Even if the sphere is totally symmetrical, the coordinate system is not.

[@eigenchris]

@@warrenchu6319 The parallelograms are bigger at the equator, so I think it makes sense that the drift is larger there. The coordinate system basically "breaks" at the poles and tangent vectors aren't well defined there, so I'm not sure the R^1_212=0 result there is very meaningful. I would normally think of R as a pureley mathematical object that doesn't have units attached; however in Einstein's Field Equations in physics the tensor appears to have units of (m^-2), and I'm not sure how to interpret that right now.

[@warrenchu6319]

​@@eigenchris So u_1 is the location of the parallelogram path of W: sin(0)=0 at the pole vs sin(pi/2)=1 at the equator. Is that right?

[@eigenchris]

@@warrenchu6319 Yes, although as I said, the coordinate system breaks at the poles so the idea of a parallelogram doesn't make sense there.

[@warrenchu6319]

@@eigenchris Thank you for your explanations. I understand much better what R means.

[@vatsdimri3675]

I have confusion at 3:16. Are Christoffel symbols scalars?

[@eigenchris]

The components of the Christoffel symbols are individual numbers, so you treat them with partial derivatives. They are not "true scalars" in the sense that they are tensors, though.

[@vatsdimri3675]

@@eigenchris Thanks for clearing that up.

[@cvictor8999]

it seems that a lot of derivations done here rely on fact that Lie bracket goes to zero. does it mean they hold water only under holonomic basis?

[@eigenchris]

I'm assuming a holonomic basis throughout this series.

[@terribiliniandrea4991]

Question about formula at 10:26 R(ea,eb)(r*s). The Riem tensor has 3 slots accepting 3 vectors. Here you are feeding a function (dot product r*s) into the 3. slot. Is this correct ? What is this meaning ?

[@eigenchris]

It is possible to take the covariant derivative of a scalar function, so the 3rd input of R can be a function. The covariant derivative of a function is the same thing as the partial derivative of a function. This is part of the definition I made in Video 20.

[@jackjones5592]

So if the Riemann Tensor is only effectively calculated with three basis vectors, since I can expand each vector into its basis vectors and its components (which I can put out front), and the lie bracket of basis vectors always go to zero, why is the lie bracket term even in the Riemann Tensor anyway? Wouldn‘t it always go to zero for every set of vectors if I just expand those out into basis vectors with components?

[@eigenchris]

The covariant derivative is not linear for the input vector field. The Lie bracket is also not linear (see video #21). In both cases, we get "extra terms" when we pull vector components out in front. It just so happens that the extra terms from the covariant derivatives and the extra terms for the lie bracket cancel out perfectly in the Riemann tensor formula. I show in the previous video (#22) that the "pulling vector components out in front" of the Riemann tensor doesn't work unless the Lie bracket term is there, and you can see the "cancelling" of the extra terms.

[@jackjones5592]

Oh yeah I forgot about that. But then the lie bracket term just feels sort of like a „linear maker“ of the Riemann Tensor; so I just need it to put the components out front but it still vanishes at the end right?

[@eigenchris]

Yeah, unfortunately I have no better explanation for the Lie Bracket term, other than saying it makes the Riemann Tensor linear. The lie bracket term will not always vanish, but it removes the parts of the other 2 terms that "stop" R from being linear.

[@ilredeldeserto]

Is there an error with the second green part indices order in 9:26 ? It would be c b a, rather it's b c a

[@eigenchris]

Yup. My bad, sorry.

[@monsurrahman7663]

Sir, Please upload the lecture on ricci curvature

[@eigenchris]

I'm working in it. There are still gaps in my understanding.

[@AkshayKumar-nr7vb]

What is multilinear map and n-form?

[@eigenchris]

A multilinear map is a function with several inputs, and each individual input has the linearity property. An n-form is a multilinear form is just a multilinear map where the output is a scalar (instead of a vector or other tensor).

[@wjrasmussen666]

The computer voice is better than most on this video.

[@deepbayes6808]

What happened to Ricci tensor video? Can't see it anymore

[@eigenchris]

It should be back. Found a mistake and re-uploaded it.

[@abdallaobeidat5608]

since R is a multilinear map, that implies the Lie bracket is zero. so, what is the point of last term in R?

[@eigenchris]

The Lie bracket is only zero if the vector inputs are basis vectors. If the input are other types of vector fields, the Lie bracket may not be zero. See Tensor Calculus 21 for examples.

[@abdallaobeidat5608]

@@eigenchris I do understand this, but as R is a multilinear map, which means we can always factor the vector in terms of basis vectors. in return, the R could be always written in terms of basis vectors! which means that the last term is always zero! I am right?

[@eigenchris]

​@@abdallaobeidat5608 You need to be a bit careful about how you state this. If you haven't watched Tensor Calc 22, I would make sure you do, particularly around the 18-minute mark. R is a multilinear map, but the 3 terms in R are not multilinear maps individually... they only become a multilinear map when you add/subtract them together as seen in the formula. In particular, the Lie bracket is not a multilinear map. If you factor out the coefficients and leave only the basis vectors, you'll get extra terms which prevent it from being a true multilinear map. It just so happens that when the 3 terms are put together, all the extra terms from the Lie bracket and so on cancel out exactly, so that the total (Riemann tensor) is a multilinear map.

[@abdallaobeidat5608]

@@eigenchris in fact I did watched all the videos, but I will watch video 22 again, thanks alot

[@abdallaobeidat5608]

@@eigenchris I got it, thanks a lot, you are really an amazing lecturer, and explain things very well

[@iiiiii-w8h]

22:48 Ah... so satisfying!

[@cademosley4886]

11 was a racehorse, and 22 was 12. 1111 horse race, and 22112.

[@commentquiva]

I've been calculating all the symmetry terms for fun. R1212 and R1221 I get sin(u)^2 and -sin(u)^2. However, for R2121 and R2112 I get 1 and -1. I'm not sure what I'm doing wrong...

[@eigenchris]

Can you share your calculation?

[@commentquiva]

@@eigenchris My bad... I forgot the metric tensor to lower the indices...

[@mohamedshalaby8253]

I can not find video 24 on Ricci tensor .

[@eigenchris]

I'm still working on it.

[@sufyannaeem2436]

how much time you would take for Ricci....

[@eigenchris]

It will take probably another 2 weeks.

[@chenlecong9938]

A blunder at 17:10,the first subscript of BI and 12-s should’ve been superscripted

[@ranjitsarkar3126]

I just can't get it that why someone will hate math