What is a Tensor 7: Rank of a TSP

[@Flutterstorm]

This series is one of the best I've ever seen when it comes to explaining such complex subjects

[@PunmasterSTP]

Rank of a TSP? More like "Really great information for me!" Thanks again for sharing all of your knowledge.

[@joelhaggis5054]

Title text here

[@nathanboettcher7421]

Title text here.

[@juanmanuelpedrosa53]

@@nathanboettcher7421 TITLE TEXT HERE

[@robertwilsoniii2048]

I really resonate with your way of wanting to present bare bones material and explain exactly why everything works without any "tricks." You are 100% correct. I hate it when books play games and beat around bushes for whatever rationalization the author, or worse, instructor of a class, might have as an excuse or reason for playing such games. Great job man! You're awesome!

[@XylyXylyX]

Robert Wilson III Thank you. And thank you for helping to make the presentation better.

[@stephenweber33]

16:40 Schaums Outline series of Tensor Calculus back in 2002 was exactly like that. It jumped right to here to start as you state. Just claiming a mixed tensor and some covariant and contravariant without all the rigor that you have held to through six earlier videos. I have been following along so I know I really knew this basic stuff, but I got into major trouble learning it. I worked at SLAC as a control room operator in 2002 , filling the conference room white boards with all these tensor notations as I went through that book in my off hours. Thank you for your contribution to Science . I look forward to many hours listening to your talks.

[@davidhand9721]

Boom. Wires connected. Lightbulb. You finally got me to understand tensors.

[@XylyXylyX]

Wow. Glad to hear it :). It is a mess of a subject, IMO.

[@levels1937]

Thank you so much once again .Please write a book if already have not!.This is truly priceless.

[@XylyXylyX]

Ha ha, actually I have written a book, but it is a touch advanced. It is not about elementary GR, it is about orbits around black holes. I hope to get to that topic here someday.

[@__donez__]

What's your book called?

[@XylyXylyX]

+Donez Horton-Bailey The tentative title is "Why Does it Move: Elliptic Integrals, Elliptic Functions, and the Geodesics of the Schwartzschild Spacetime." It is not generally available because I have not distributed it, but I plan to eventually. It is basically complete, but it is hard to claim that it is "finished". Kinda like a painting....you can always make it better. It is somewhere between intermediate and advanced and relatively esoteric as far as GR texts go. If you would like an advance copy and promise to keep it to yourself let me know and I will send you one. THe price you pay is that you must tell me where I can improve the exposition and report any errors/typos/confusions you find! The academic level of the text is somewhere between beginner/intermediate. I spend a LOT of time on the mathematics of elliptic integrals and functions, which is pure applied math. I have another book about the standard model that is also ready to go, if you are interested.

[@levels1937]

+Donez Horton-Bailey Hey wow small world!Sorry to spy on your profile but I transfer to UC Santa Cruz this fall!

[@fawzyhegab]

I will be interested to get a copy of both texts and I promise I will keep them for myself. Here is my email: [e-mail]

[@georgefrank7405]

This is so helpful. Crystal clear, thank you so very much.

[@XylyXylyX]

You are welcome. I'm glad someone is watching these videos.

[@TheBigBangggggg]

I don't think it's crystal clear when the notation (of maps) is changed after some videos without mentioning it! Very confusing.

[@lbomcarvalho]

Thanks for your nice playlist of tensors. What is a good reference book on this material ?

[@1994sarsar]

Thanks for this valuable videos . I appreciate your work

[@iliassuvanov8539]

Brilliant explanation!!!

[@LibertyAzad]

The trick you refer to is fine for me NOW that I've watched your video filling in the gap. Once again, thanks!

[@magicandmagik]

completely agree with you on the last part there, thank you a lot.

[@paras3681]

I don't know if its too dumb to ask but I could not get how did you come up with the answer at 12:25. How did you solve that kronecker delta into that answer?

[@crehenge2386]

I just wish teachers wheren't allergic to actual examples... That would reduce the handwaving alot...

[@alexanderbecker4360]

To me too all these Videos are really helpful Thanks so much!!

[@XylyXylyX]

I’m glad, thanks for watching.

[@atnn4698]

Hi, at 11:00...you write < b_eta.e_eta , e_miu > . shouldn't it be < e_miu , b_eta.e_eta > ? and going on further in this order. Thank you very much by the way, your videos help grasp this notions. I got into this because I could not find derivation for some equation involving the strain tensor. more precisely, i wanted to understand an equation relating measured strain in some direction in a 3D Cartesian coordinate system to the strain tensor. then i saw in a book: "scalar product of the strain tensor with the unit vector (in some direction) is STRAIN_phi,psi = STRAIN)_i,j * UNITvector_i * UNITvector_j " well, i learned a lot :) but still haven't grasped this equation, more disturbing i can not understand why my way of doing it (projecting all the strain tensor elements onto the coordinate system axes and then each one onto the desired direction unit vector is not getting me the same result. anyway, very off topic the message, except the first two rows, sorry for that

[@NavrasNeo]

I was wondering the same. He did sth similar last video too. Have you resolved whether this is the right way or not? Though in the end the result of the calculation would be the same regardless, as it seems.

[@kamatikos]

Is the rank classification of tensors a legacy convention that originated in a specific use of tensors? It seems to me like there are better ways to classify tensors without making them a subset of multilinear products.

[@XylyXylyX]

Hmm....I don't know the history, but I am aware that the word "rank" is used a few different ways. Sometimes it is just the total number spaces in the tensor product, and sometimes it is reported as a pairing of the total number of vector spaces and covector spaces. What surprised me was that this convention led to the strange convention that a "tensor" can not have some arbitrary sequence of vector and covector spaces even though such space are not a problem to use at all.

[@kingplunger6033]

Very cool title text

[@mcslasher]

Great.Thank you Very much

[@jordantistetube]

At 11:50, why does the B eta e eta covector come first in the subsequent bracket notation? Isn't the vector mapping the covector to a real number in this case? Are the two equivalent?

[@XylyXylyX]

jordantiste It is just convention to have the covector first. < , vector> is the mapping that eats a covector and < covector, > is the mapping that eats a vector.

[@jordantistetube]

Okay, I see. In the end I guess the notation doesn't make a difference. Thanks for the quick answer!

[@davidthue]

I'm struggling to accept the convention that "< , vector> is the mapping that eats a covector and is the mapping that eats a vector" (as you mentioned in a reply to an earlier comment) because it seems to allow for ambiguity. Consider a vector v and a covector v*. If we use v to map v* to a real number, the convention would seem to say that we should write it as , following "< , vector>". ...but if we use v* to map v to the reals instead, then the convention seems to lead us to write it in the same way: , following "", even though we're talking about an entirely different mapping. Using the stated convention, it seems like might refer to either of these two mappings, which seems like an undesirable ambiguity. Is this ambiguity resolved somehow, or am I misunderstanding something along the way? Thanks very much for this series of videos!

[@XylyXylyX]

We are taking advantage of the fact the by construction both mappings are the same. If w* is a covector and v is a vector, w* mapping v will give the same result as v mapping w* and that is what allows this notation to be unambiguous. I see why it isn't super obvious however. w* is, by definition a map from V to R and v is by definition just a member of V. But v can be considered in a strangely obvious way as a map from V* (the covector space) by simply saying that v will map members of V* to R by allowing those covectors to act on v! That is, by definition v(w*) = w*(v)! The right hand side make perfect sense because w* is defined to be a map V->R. We *define* the nature of the mapping executed by v to be given by this super simple expression.

[@davidthue]

Thanks very much for the quick and extremely clear reply. It's really helpful to understand that the two mappings are the same because of how the mapping executed by v is defined, namely, that "v will map members of V* to R by allowing those covectors to act on v". I'm looking forward to continuing with this series!

[@qorilla]

@@XylyXylyX It's like the visitor pattern in object-oriented programming.

[@jasonbroadway8027]

This video is a masterpiece, but it is unfortunate that Einstein wanted "be cute" and drop the indices.

[@XylyXylyX]

Thank you for your kind comment. I am sure that the component-only system is far easier to work with but it tends to conceal the underlying mathematical principles. I am glad you found this useful.

[@jordantistetube]

If (and take X here to represent the circle X) V* X V X V is not considered a tensor because of the ordering of the vector and covector spaces, does that mean that a tensor product like we saw in previous lectures, like V X V* X V X V* is not a tensor either because of the ordering?

[@XylyXylyX]

jordantiste That's right. But be aware that this issues is entirely semantic. All these objects are entirely legitimate multilinear maps and all use the appropriate transformation rule of tensors. It is probably entirely fine to call all of these objects "tensors." For some reason we like to have a consistent structure to associate with the word. I'm not sure it matters much, but I mentioned it because it appears in some of the more math-centric books on the subject.

[@jordantistetube]

Okay, thanks for the fast response! Took me seven lectures to find one point that was not entirely made clear to me so thanks for being so comprehensive in your videos :)

[@alexekkis3548]

+XylyXylyX Very helpful videos, but one question: The tensor product is defined by the multiplication defined in the field R, which is commutative. Therefore, the tensor product spaces V, V, V* and V, V*, V and V*, V, V are all equivalent, aren't they? They are not identical of course, but they are equivalent in the sense that only the dimensions are flipped around. They are ordered and thus not identical, but all of them generate fundamentally the same space regardless of the order. The tensor products they define are all equivalent, because they are defined using the multiplication in R and thus are commutative. Hence, for the rank of any given TPS it only matters how many vectors and covectors are in the tensor product, but it doesn't matter what order they are in. Or am I getting this totally wrong here? I'd be grateful for your feedback!

[@XylyXylyX]

Alex Ekkis You are indeed getting it totally wrong! :) There are two sorts of multiplication floating around: scalar multiplication and the tensor product. You are confusing the two! Any tensor product space is ultimately a vector space and as such it has a scalar multiplication which allows us to multiply any member of the space by a scalar, usually a scalar drawn from R. That is indeed commutative. However, the tensor product is something that acts *between different vector spaces* and it is not commutative at all. A good practice is to always fully understand the domain and range of every operation in sight.

[@mridulsinghal5661]

I think the tensor product spaces V,V,V* and V,V*,V will give the same real number in R when operated on corresponding cartesian product domain.

[@themathtutorphd]

You are really the best!!! I've followed the tensor and exterior derivatives contents coz I'm taking a differential topology class. You are the greatest. I wish you could do something on algebraic topology. Keep doing the good work. Cheers!!!

[@ThomasImpelluso]

So I get it that vectors come first and dual vectors second. I also get that there is only happenstance to the difference bewteen vectors and covectors (SOMETHING had to have come first, so we just list the vectors first). So far, so good. Now, I know the force can be considered a one form as it produces the scalar "work" with operating on the vector displacment. And, we can "see" its one-form-ness explicitly stated as the gradient of a potential function. And, THIS, now make me think that "dual-ness" is not as happenstance as we'd like to believe. And THAT makes me wonder WHY it comes about that when we see a tensor, and if there exist contravariatn components, that those always appear first? WHY do we never see or deal with tensors where the contravariant components come last? In other words, at 5:55 I am not happy with your statement ("I think the idea of a rank needs to have specificity to the ordering such that the vector comes first.")

[@XylyXylyX]

Thomas Impelluso Of all the topics of lessons, this lesson should be the least interesting because it deals more with nomenclature than with substance. Every tensor space is characterized by the ordering of the tensor product and that ordering can be any ordering at all. The "vectors first" rule is entirely arbitrary and is there simply so we can use the "( p, q )" nomenclature for rank. THat is what I meant at 5:55. Regarding the physics, you bring up a nice point I will address soon in a new lesson. All of these mathematical objects are more or less suited to model things in the real world. It is our job to figure out which mathematical object work best for which physical concepts we want to model. Force works well as a 1-form in the coordinate basis, as you point out. The magnetic field works well as a 2-form in three dimensional space or as a combined field tensor in 4-dimensional spacetime. Figuring out what models work best is what science is....according to some.

[@lbomcarvalho]

What is a reference book on this subject with the same level?

[@XylyXylyX]

Try “Introduction to Vectors and Tensors” by Bowen and Wang (Dover Press)

[@lbomcarvalho]

@@XylyXylyX thank you very much

[@XylyXylyX]

@@lbomcarvalho It is not an easy book, but it is the one I used for all of the notation and lesson planning. Good Luck!

[@lbomcarvalho]

@@XylyXylyX I already ordered at amazon web site and for sure I Will like it.

[@lbomcarvalho]

@@XylyXylyX after following your nice lectures on tensors, and after I saw you pointing out the correct interpretation of what is a tensor, I looked at some linear algebra book, and so far Gelfand it is wonderfull. What a nice book as well as your set of lectures.

[@nafriavijay]

Thanks you made it easy boss

[@Channel-zb1fi]

Hey, what does your username mean?

[@climatechangedoesntbargain9140]

12:37 Why these specific indices?

[@TiredGradStudent]

Hey, I don't know if you resolved this issue, but I think it has to do with the Kronecker deltas and the fact that we choose to focus on the coefficients that survive the K-delta. Since we know the only non-zero coefficients that survive from the arbitrary basis vectors from the Cartesian Product are the same as our mapping basis vectors, we show these mapping basis vector coefficients only. If you replace the Greek letters in the mapping tensor with 0, 1, and 3, then the surviving coefficients will have sub/superscripts of 0,1, and 3. Terribly sorry if this was poor wording, and/or you already found your answer!

[@johnlie8586]

You explained rank of tensor. I have been told real numbers have zero rank. Could you explain more the rank of the real numbers?Please. Secondly how we can write real numbers rank? Is (0,0) right? Many thanks

[@kamilkonieczny3613]

Imho if you treat R as scalars then it is 0 rank or (0,0), but then it is not a tensor, because a tensor must be from vector space . If you treat R as vector space with base vector e0 = 1, then it is (1,0) rank tensor. Hmm... i am not so sure about this.

[@sheafofthings]

Shouldnt it be "TPS" in title and description? :)

[@joelhaggis5054]

Zero dislikes

[@MathwithMing]

What books would you recommend for this subject?

[@XylyXylyX]

“Introduction to Vectors and Tensors” by Bowen and Yang (Dover). My favorite.

[@MathwithMing]

XylyXylyX Thank you very much for your reply! Love your videos!

[@eamon_concannon]

Thanks very much.

[@_DD_15]

So is it just a convention that V* ⊗ V doesn't give a Tensor because of the way the Rank was introduced? Does it mean that the Rank is undefined so the operation makes no sense?

[@XylyXylyX]

DD I think that is correct. THe idea of “rank” is just a tool to help us quickly identify what tensor product space we are talking about. We chose to use “(p,q)” as the rank so obviously if we have any multilinear map structure that has a more complex structure then “rank” is meaningless.