Tensors/tensor products demystified

Help me keep making videos: paypal.me/mlbakermath
I'm baaaaack! ;-)
Achieving a full understanding of the formalism of tensors is one thing that COMPLETELY transformed my view of linear algebra; thereafter every concept in differential geometry, representation theory, etc. seemed absolutely natural. I want to share this elegant, unified viewpoint with you (*cough* and people are always badgering me to repost rather embarrassing videos I recorded on the topic as a young and naive undergrad, which I REFUSE to do, so there).
Now I just need to make some videos on geometric algebra/calculus... then things will get really fun.

[@alexanderstephens822]

Great stuff. My introduction to the tensor product was the formal construction found on Wikipedia, and while I understand it, it totally misses the POINT of tensors, which is the universal property that reduces multilinear algebra to linear algebra.

[@alexanderstephens822]

And I should mention, the formal construction on Wikipedia is still quite interesting, and it did ultimately lead to me learning a lot about free objects, which in turn led me to universal properties. So it all worked out in the end.

[@Sky-pg6xy]

You might be interested in the categorical construction. There’s a really interesting construction of the tensor product wherein it’s defined as a “universal pair” (a multi-linear map and a mediating vector space), such that every multi linear map on an n-tuple of vectors can be written as a multi-linear map from the Cartesian product of vector spaces to that mediating vector space, and then a linear map from the mediating vector space to the co-domain of the original map. All of that gobbledygook just to say, it’s a construction that shows that any multi linear map is multi linear in the exact same way, and the unique features of the map are determined by a simple linear transformation.

[@m322_yt]

20 minutes in and I feel like the tensor product has been motivated very adequately and in an easy to understand manner! Thank you.

[@DumblyDorr]

How do you recognize a structuralist? In explaining mathematical objects, they give you a commutative diagram for the universal property and call everything else "symbol-pushing crap" 😂 But hey... I completely agree 😁

[@rfabbri]

Useful as an overview among many passes of overviews before getting a solid grasp of the subject. mlbaker: Thanks for not being shy to state your views, even at the expense of not being completely clear or risking making mistakes.People: don't expect to understand everything that is being said, expect come back to watch this again as your knowledge evolves.

[@litsky]

I think we can all agree that any viewer who understands absolutely everything in this video can be assured they possess a very adequate command of this material.

[@goldjoinery]

+mlbaker At around 33 minutes you go over type (_m,n_) tensors. From en.wikipedia.org/wiki/Tensor_(intrinsic_definition) a type (_m,n_) tensor is an element of *V ⊗ ... ⊗ V ⊗ V' ⊗ ... ⊗ V'*, with _m_ copies of *V* and _n_ copies of *V'*. Do you have them swapped around here, or is this just a matter of convention? (*V'* is the dual space of *V*.)

[@litsky]

They're just using the opposite convention. It's a completely arbitrary choice anyway.

[@goldjoinery]

mlbaker Cheers. PS, I hope you make more videos if you can, they're excellent.

[@viktor1098]

Watched this back in September and didn't get much. After a course of abstract algebra, I returned and could easily follow. Very helpful stuff!

[@litsky]

That's how it goes!

[@no3339]

I’m a mechanical engineering undergrad that has taken intro to Lin alg and currently complex analysis. I must admit that some of these explanations are above the scope of my knowledge and bit confusing. I think it would’ve helped if I had learned Lin alg and geometric algebra in high school. I’ll definitely have to come back to this video in a while to fully understand everything. Thanks for the explanation!

[@differentialtopology]

Practice some more linear algebra, work on learning some abstract algebra too and then u can learn even more linear algebra and eventually multilinear algebra:3

[@deltalima6703]

@no3339 this video is worth a second look. Come watch it again now, see if it makes more sense.

[@InventiveHarvest]

welcome back! I like the new direction this is going!

[@zolokur6702]

Best explanation on tensor product ever.

[@litsky]

Thanks!

[@MikhailBarabanovA]

Really clean one!

[@ianprado1488]

This is a good video. I liked your conversational approach in explaining the frustrations surrounding learning tensors

[@chimetimepaprika]

Hell yeah, brother man. This opened up some new ground in my journey.

[@Par3val]

"Everything is just about things that are eating other things. Thats what tensor algebra is" just great

[@jameschen2308]

Quote of the year.

[@cloneofgramsci]

This was a really good explanation and by far the best video on the subject that I found. Really really well explained and helpful.

[@RD-fv2bf]

really???

[@callanmcgill]

The perfect video to share with our friends in physics.

[@oyugioyugi6018]

RIP to this masterpiece of a youtube channel. Hoping for a comeback next year.

[@lopezb]

Brilliant lecture. This guy is an excellent teacher.

[@notgoodatmathmmm6185]

20:47 started to study dualspaces and tensorproducts todayy and what you said there was soooo cool

[@seonjoonpark7841]

It's really helpful to catch intuition behind tensor product. Thanks for upload.

[@pankajb64]

As someone from CS field I found it a bit advanced yet much better than other explanations out there. Cheers! @mlbaker

[@nbrader]

This made a surprising amount of sense to me. Thanks!

[@tighemcasey7589]

Best explanation I've seen

[@TheEzz09]

This video came just in time! I was just introduced to tensor products... will have to return to this video once I get a better grasp... but still helpful and interesting!

[@zelosmalum2645]

+Esmer Tremb I found it easier to get as a quotient module

[@zman2168]

totally!!!

[@niccolozanotti]

Best video on the subject. Thanks!!

[@lopezb]

For another view of the statement at 28:00 that Hom(V,W) is isomorphic to V*tensor W, well, an element of the first is a matrix; but that gives a bilinear map on V* times W, by putting a row vector on the left and a column vector on the right.Then, by the universal property he covered, we are done. OK it uses bases but still it's maybe helpful!

[@alexheaton2]

This is a really great video. I watched it a month ago, and was clueless, but I could tell I needed some more background information, so I slowly read and worked through on paper some of Vinberg's "Course in Algebra" and then came back today, and now this is putting everything into place for me. Thanks!

[@ianprado1488]

I'm a x math student and I have been very much enjoying listening to the lectures you give. Are thinking about uploading more videos of you addressing topics?

[@ptahtheeldaoud]

Very insightful!

[@rudypieplenbosch6752]

I was following a channel on tensors, he kept going on and on, without really getting to a point ever, so i stopped it. Seems there is still a chance to understand tensors, at the moment I'm studying spinors, which are a bit less confusing. But you did clarify quite some things in this lecture already, of course i need zo watch it again. I like your passionate way of explaining, its very motivating.

[@litsky]

Spinors are far more complicated objects!

[@rudypieplenbosch6752]

@@litsky Really ?, then it must be because i found a good explanation on them, to me tensors are more confusing, but you gave already some good insights. Too bad it seems you stopped your explanation videos, you are very good at it.

[@rewtnode]

Watched the whole thing, could follow for the most part, and feel like I learned a bit. Yes it is a helpful video< thank you. A little more about those wedge product spaces would be great though. ..

[@deltalima6703]

Its a shame I did not see this earlier. KZbin does people dirty serving up far more sketchy descriptions of this content instead of this. This is quite clear. I like how he does not use abbreviations. Its worth it to a general audience to just write "isomorphism", imho. Great presentation, just shares knowledge, not trying to look smart. 👍

[@stevebrown2557]

I first watched this a while back when my exposure to the tensor product was a couple of paragraphs from Tu's Intro to Manifolds. After the first watching(s), I read Gowers' write-up and a few other things and got a sense that tensor products are what let you reduce multilinear things to linear things. Just this past couple days, I read through section 10.4 of Dummit and Foote (on tensor products of modules) a couple times, then wanted to watch this video again. And so I did....and this time, nothing you said seemed foreign anymore. The lesson to be learned (which I'm saying to those that say things like "I didn't understand this video at all"): (1) Don't stop studying, and (2) if you try to take huge leaps forward, don't expect progress to be any quicker than when you take little steps instead. Also, I'm curious where to find some mathy discussion of quantum information. What you said toward the end of the video resonated with me, as I've been reading a bit about information theory over the past few months, wondering if the definition of information was going to translate to something like "w/e it takes to specify a state...which in QM terms probably means the coefficients of a representation of some state function, in some eigenbasis." (...or something like that. I'm new to this.) Your mention of a complex hilbert space made me wonder if I'm on the right track.

[@litsky]

Have a look at Nielsen-Chuang.

[@apotheosys1]

Thanks, helped me a lot!

[@jemehrtestobesser262]

Thank you very much for this!

[@ChrisDjangoConcerts]

I would like to ask if you can do a geometric algebra video with lots of concrete examples and pictures, such that people like me can understand it.

[@matron9936]

Nice intuition thanks!

[@forgetfulfunctor2986]

This is legit the best explanation around

[@abbamed269]

magnifique!

[@davidgillies620]

Wish I'd seen this 35 years ago when the linear algebra portions of my physics degree were making my brain explode.

[@maciejmarek551]

What do you mean when you say that defining tensor product as a set of bilinear maps works for "evil reasons"? BTW, thanks a lot for this video. It put one of the final pieces to my understanding of tensor products. The classical definition/construction (with quotient spaces) bugged me for a very long time...

[@lucaolmastroni6270]

Some time ago I saw a video of yours on Hilbert Spaces which I can't reach anymore. Is it still accessible? Could you please give me the link to it? Thanks. Luca

[@miguelaphan58]

good intent baker...but clear ideas in your mind and set pedagogic objectives neatly

[@zy9662]

Very insightful explanations. Maybe later you can do a complementary video showing how some of the isomorphisms you showed would be (concretely) defined at the element level. For instance what would be the matrix corresponding to an element in Hom(V, W) that corresponds to an element of V*(tensor)W, which is a vector. Or how it looks a generic element in V(tensor)W in terms of the elements of V and W

[@litsky]

I think it's clear enough in the video. A matrix A=(A^i_j) corresponds to the element \sum_{i,j} A^i_j e^j \otimes e_i of the tensor product. That is, the entries of the matrix are literally the coefficients of the element's expansion in the natural basis. There's not much else to say.

[@zy9662]

@@litsky I see, thanks. Maybe your exposition assumes that the audience already saw the operation v \otimes w represented as the outer product of the vectors v and w. Which is fair enough. Personally, my "first" introduction to tensor products was as the universal object you showed (in my current Homological Algebra class) but defined for Modules and we never saw the outer product realization. Before that, I just saw them in a didactical physics video, about how they are generalizations of matrices, and honestly I didn't imagine they were the same thing of the Homological Algebra class. Today, and thanks to your video, I went again to the wikipedia entry and I saw the outer product formula, now is easier for me to understand the identifications you showed. PS. I wasn't expecting such a fast reply, thanks for that. Specially since this video is from 4 years ago. I'm curious why you haven't uploaded more videos, given that you have a decent number of subscriptions and the topics you covered are kind of lacking in youtube.

[@litsky]

​@@zy9662 Regarding why I haven't uploaded more videos, the answer is simply that perfect is the enemy of done. Every time I tried to record lectures recently I just agonized and agonized over the exposition and eventually tossed it aside.

[@charlescrawford1103]

Great video, man! Thanks! I was under the impression that the dual space vectors in V* are contravariant.

[@litsky]

+Charles Crawford One should be careful here. Although the functor that takes V to V* is indeed a contravariant functor, in the sense of tensor "transformation laws" that physicists often speak of, it is the elements of V that are contravariant (those of V* are covariant).

[@charlescrawford1103]

+mlbaker Thanks for the replies. I'm trying to teach myself differential geometry. I've used some of these concepts as tools before, but I'm trying to get a more complete understanding of what these objects are and where they come from. The book (Jeevanjee's Intro to Tenors and Group Theory) I'm using follows along this more mathematical formulation, e.g. looking at tensors as eating vectors. I appreciate your articulation on this topic! Thanks again!

[@litsky]

+Charles Crawford Yeah, the whole "active" perspective on tensors (viewing them as operating on vectors, etc.) is motivated by physics/differential geometry. From an abstract standpoint, though, tensors are just a completely formal construction, as can be seen in the formal definition of the tensor product using quotient spaces.

[@timholzschuh7545]

Could you also construct (V tensor W) as the free vector space over the cartesian product of a basis of V and a basis of W? Although it depends on choosing a basis for each of them.

[@goldjoinery]

Yeah. More generally, if one views V as the free vector space on S and W free on T, then one may define F(S) ⊗ F(T) = F(S × T).

[@fernandojackson7207]

Great presentation.I'm confused. Maybe because I haven't done Abstract Algebra in a while. Please be patient. that for a bilinear map you use F(x1,x2,..,axi+xi',.,xn ), at around 18:01 . Wouldn't a bilinear map be defined on a pair of vector spaces, and thus be defined on pairs(x1,x2), rather than on n-ples (x1,x2,..,xn)? Maybe in 32:41, you can define the map on the basis elements, the pure tensors f(x)w ; f in V*, w in W that are a basis for V*(x)W and extend by linearity to V*(x)W, i.e., to the "non-pure" elements?

[@uzivatel123]

thank you

[@Tarzan_of_the_Ocean]

great video. especially the part where you explained how to think of tensors as things that eat a bunch of vectors and/or covectors until they spit out a scalar, that really made it click for me. but I have one question: why are they defined that way, if many intersting objects (like linear maps, cross product, stress tensor, etc) are not equal but only isomorphic to a tensor defined that way (map to the underlying field)? could they also be defined as multilinear maps to another vector space? I am an engineering student (who is also interested in math and physics) so my „theoretical math“ knowledge is quite limited.

[@Tarzan_of_the_Ocean]

does sb know the reason behind this?

[@rhodesmusicofficial]

Dear mlbaker, is it possible to give me [[the access]] to watch Galois Talk #3?? The first two are really freaking good and the other ones seem great too but it feels bad that idk anything from the third vid :(((

[@ChrisDjangoConcerts]

I'm a theoretical physicist who is struggling with this stuff. In the beginning when you explained the dual within linear algebra context i didn't completely followed, making the rest of the talk hard to follow. I still got a lot out of it. Maybe some more explicit examples (of duals) would make it less abstract for me

[@dennisestenson7820]

Look into geometric algebra. It makes this stuff obviously beautiful.

[@ChrisDjangoConcerts]

@@dennisestenson7820 Yes, I have looked into it. Just read the book by John Vince. I you have any other sources that you can recommend please do.

[@farhadtowfiq6767]

This is very useful! Would you care to make a written summary to be used as a reference?

[@DarioOliveri]

I need to compute a tensor product. Can I just say that given two tensors of size respectively 4x3x7x5 and 4x3x5x6 the output is a tensor of size 4x3x7x6 where we have basically 4x3 lined up products of matrices of size 7x5 and 5x6????? Is my understanding correct??

[@Palisade5810]

7x6

[@aziz0x00]

👏👏👏👏👏👏👏

[@octaviocarpinetti4326]

I have a question. (Here V(x)V is V tensor V, I don't know how to make the symbol) You defined tensors such that they linearize bilinear maps. So, in the case of a bilinear form B:VxV->K. we have B corresponds to a linear transform L:V(x)V->K. But after, you said B is identified with an element of V*(x)V*. Why did we do linearization in the first place if we won't be using linear transformations? And also, how does that linear transformation L corresponds to the element of V*(x)V*? Is [V(x)V]*=V*(x)V*?

[@rajarshichatterjee3281]

Are you doing your research/studies from MIT? Anyways.. Great lecture

[@litsky]

I wish. Just visited once.

[@charlescrawford1103]

So, is pairing essentially saying that V and V* are 1 to 1 which then implies that V and V* are onto?

[@litsky]

+Charles Crawford "1-to-1" and "onto" are both properties of a function, not of a vector space. The pairing between V and its dual yields a map V -> V** given by sending each vector to "the thing that evaluates its argument at v", the latter being a functional on V*. It is pretty easy to see this map V -> V** is injective, at which point you can just remember that dim V = dim V* = dim V** and thereby deduce surjectivity with no effort.

[@M0rph1sm]

With a finite basis for V, then maybe V iso Hom(k, V) implies Char(k) > 0 ?

[@litsky]

Hom(k, V) and V are always isomorphic as vector spaces over k; the characteristic of k is of no relevance. Consider the map that takes an element f of Hom(k, V) and sends it to f(1). One sees immediately that this map is an isomorphism.

[@ChrisDjangoConcerts]

4:37 why is V = Hom_k (k,V) ? Can you give an example of that? Also, can you give an example of phi(v) = k without a bilinear form being present?

[@litsky]

Q1: well, k itself is 1-dimensional (as a vector space over k). so as soon as you know where a linear map k->V sends 1, you know what map it is. to really spell it out: the isomorphism T:Hom_k(k,V)->V is given by sending a linear map f:k->V to the element f(1) of V. check that T is linear. now, T is an isomorphism because there is also a linear map S:V->Hom_k(k,V) given by sending a vector v to the unique linear map k->V taking 1 to v, and S and T are clearly mutually inverse. Q2: i mean okay, if V is just an abstract vector space, you're going to have to make a choice - some choice - to define a functional. but really this is all silly because if V is finite-dimensional we know it's just k^n in disguise (viewed, say, as a space of column vectors), and the functionals on that thing are precisely the row vectors... (remember that a 1*n matrix times an n*1 matrix gives you a 1*1, namely a scalar)

[@ChrisDjangoConcerts]

@@litsky That is a great and long answer, which my physics mind needs to study a bit, before I can understand it. :)

[@ChrisDjangoConcerts]

@@litsky question 2: How is that different from an inner product. (v_1,v_2) \cdot (w_1,w_2) = v_1 w_1 + v_2 w_2 and at the same time through invoking the transpose (v_1,v_2) (w_1,w_2)^T = v_1 w_1 + v_2 w_2

[@litsky]

​@@ChrisDjangoConcerts well, sure, it's not really. but the point is that R^n - being a space of literal and concrete n-tuples of real numbers - carries a standard inner product, namely the one you pointed out = v^T w (where v,w are considered as column vectors, i.e. n*1 matrices). you can bend over backwards to write down a linear functional phi seemingly without invoking an inner product, but at the end of the day, it's futile: if i fix ANY inner product , then no matter what phi you picked, i will be able to find a vector v such that your phi is nothing more than . (by this i mean phi(w) = .) this phenomenon persists even in infinite dimensions, in the sense formalized by the riesz representation theorem. anyway, just after the time you mentioned, i define the dual basis, which gives you a bunch of functionals none of whose definitions make any reference to an inner product. (but again, you still have to fix a basis on V...)

[@litsky]

some more meta-mathematical intuition for why we seem plagued by these choices: any linear functional on V has a kernel - the set of all vectors it sends to 0 - which (unless the functional is identically 0) has to be a hyperplane in V. so, if you have a vector space V and you have no idea what it is (it's just a totally arbitrary abstract vector space), and you refuse to even introduce a coordinate system to V (that is, choose a basis), then it makes sense that you can't hope to specify a functional, because you obviously can't even hope to specify a hyperplane. that is: you have no sense of what anything is, and no sense of direction at all, so why would any particular hyperplane stand out from the others?

[@williamhepfer8956]

Doesn't "over-mystified" mean the subject is made more mystifying? Is that your purpose?

[@breeeesh]

water water water, loo loo loo!

[@azeds]

Man do you have something to follow you .

[@jameschen2308]

No, thank YOU.

[@rewtnode]

I wonder how often I have seen this all explained. And every time I got confused again about some notation. Like in this case, in the first minutes you write down this symbol Hom_k . I kind of know what it must be, but only because I know already what a dual vector space is. Oh it's homomorphism. Why then the subscript k, when k is already an argument. What is k any ways ? Sorry I'm just here to ask the stupid questions

[@rewtnode]

Ok , never mind . You said it. K is the underlying field. E.g. Real numbers

[@TheBigBangggggg]

Now try this in ordinary language ;-)

[@davidwilkie9551]

..something in nothing in something in nothing is an echo chamber of standing waves. States of position and rates of change are secondary characteristics, so the symbolic representation/nomenclature should "recognize" the origins of the information structure. The systems we have are derived from cultural evolution, perception organized to do a job, applying techniques to technology. Any attempt to refine the process is to be supported.

[@litsky]

Don't post spam on my videos.

[@davidwilkie9551]

You are mistaken, and how is supporting your intention spam?

[@litsky]

In that case, it's quite unclear what your comment was trying to convey.

[@davidwilkie9551]

In the context of the video, I was sharing your proposition that the subject was over-mysterfied. It's normal to reduce a body of work, with all the conventions, references and habitual practices to a form that new students can deal with. I thought you were doing that sort of reduction to a more workable notation. For myself, much older, I have continued to look for the bottom of the stack but it is mostly useless to anyone else, that's why I like what you are doing here, advancing in smaller stages.

[@maximuslord8058]

I'm sorry bro but you should start with basic idea of tensor's, specially for beginners... However Thanks for the video

[@godfreypigott]

I guess I'll have to watch this video on Tenthors and Vector Spathes.

[@ANSIcode]

Great video, appreciate it! This might be somewhat exotic, but I'm looking for a proof that the isomorphism V->V* is not natural. Clearly this is something that can be proven in the category of vectorspaces, I've seen it claimed multiple times, but the proof doesn't seem that easy. I'm wondering if the idea of proving that some isomorphism is NOT natural is in the spirit of the whole concept of naturality. Why does noone ever bother to prove that isomorphisms are/aren't natural? Is this something that one should have a failproof intuition for? Also: Ever think of making marks on the board to indicate the screen capture?

[@litsky]

+ANSIcode There are two things to appreciate here. Firstly, there is no privileged isomorphism between V and V*, so your use of the definite article "the" there is misplaced. Secondly, when people say things like "V and V** are naturally isomorphic", they are really referring to a relationship existing between two *endofunctors* of the category of finite-dimensional vector spaces over a fixed field, namely the identity functor and the double-dual functor. The crucial point is that it makes no sense to speak of a single isomorphism between two fixed vector spaces being "natural" or more generally, between two algebraic structures -- groups, rings, whatever. "Natural isomorphism", or more generally "natural transformation", describes a correspondence that exists between two "constructions", that is, between two *functors*. The natural isomorphism consists of a big giant package of linear isomorphisms V->V**, one for EVERY vector space V, such that a coherence/compatibility condition is satisfied with regards to linear maps V->W (a certain square commutes). Now, for the V* case, the functor in question is contravariant, and so it doesn't even make sense to formulate this naturality condition! However you may have read that a nondegenerate bilinear form on V yields a "natural isomorphism between V and V*". To make sense of what "natural" means in this context requires the more exotic concept of a "dinatural transformation". See Mac Lane's book "Categories for the Working Mathematician" for more details. I hope that helps to clear things up. I don't know what you mean by your final sentence.

[@ANSIcode]

+mlbaker Thank you for your quick comment. I tried reading the book of Saunders Mac Lane once, but being a physics student found it extremely technical and didn't get very far. I do remember though, that the concept of natural transformations can be defined for covariant->contravariant functors as well. The diagramm to draw is in any case straightforward. After just looking it up, it does seem like most people don't call this a natural transformation (and treat it as a special case of the "dinatural transformation" you mentioned, a concept I've not heared of before). Although what I meant in my comment was looking for a proof that V and V* are not naturally isomorphic (in the above somewhat unusual terminology) and didn't mean to refer to a concrete isomorphism, the wiki-article on natural transformation actually talks about finding a single automorphism of an object, that doesn't "commute" with the isomorphism, the general case of objects not being naturally isomorphic being proven by providing one such automorphism for every given isomorphism... I'm quite sure, in any case, the non-naturality of the isomorphy V~V* can be proven in this sense, just haven't been able to do it yet. I guess having a functor (like the dual functor) that's naturally equivalent to equality would be pretty pointless anyway but still... Maybe you could still comment on the question, if wanting to prove naturality whenever it is claimed (everyone always claims and never proves it) is the right way of thinking? I'm not sure how much "it doesn't seem to involve artificial choices" qualifies as a foolproof way to tell. The last sentence means that you could put some marks on the board so you don't have to check if what you're writing is still visible in the video.

[@litsky]

+ANSIcode I'm pretty sure the "naturality square" arising from the zero mapping V->V on any nonzero vector space will fail to commute.

[@ANSIcode]

Yes, that seems to be the case for every possible choice of infranatural transformation. Guess I was doing something really weird when I tried to prove this (a while ago), when it's so easy. Thanks for the help!

[@madmath1971]

Actually it can happen that there is a "natural" isomorphism between V - being a finite dimensional vector space (over R or C) - and V* : indeed in this case V is a Hilbert space too and riesz representation theorem holds. And once you have fixed a basis for V, you can write down the representation in a constructive way.

[@JustNow42]

I never understood why people start in the new univers they try to explain and expect those in the old univers to understand the lingo. If you start in the old univers people can develop the new univers themselves,

[@litsky]

Not sure what you mean. If you want a more primitive/computational (rather than conceptual) treatment of tensors there are plenty of other videos that do just that. My intention with this video was to fill that niche. I really dislike the habit of certain people who seem almost to "take offense" upon being exposed to a certain language or viewpoint. This is something that still plagues category theory today, and as we all recall the term "Gruppenpest", plagued representation theory in the past. I feel it's something that we as a community would really do well to drop, and learn instead to appreciate the merit and benefits of diverse perspectives...

[@clickaccept]

haha

[@BBB32648]

"Just some category theoretic garbage." Really?

[@zelosmalum2645]

+BRANDON THOMAS VAN OVER Yepp!

[@modolief]

Chalkboards ... Makes me feel like I'm back in the year 1300. There has to be a better way.

[@litsky]

Have fun getting laughed out of every mathematics video on KZbin.

[@modolief]

content is free = you get what you pay for

[@notgoodatmathmmm6185]

i love the blackboard

[@2coryman]

PRACTICALLY SPEAKING WHAT GOOD DOES IT ALL DO?

[@paulp1204]

Have a four year bachelor degree in electronics engineering which is effectively minor in mathematics - didn't understand a thing. This video is from a clearly, very clever guy, who has not realised the yawning divide between himself and the students (including me at closer to double his age) he is attempting to educate.

[@gdsfish3214]

This video is not aimed at ppl like you. Engineers don't do abstract maths. This is only relevant for people getting a degree in mathematics and maybe some physics students. Other than that no one should really watch this video since you need already need some background in abstract algebra/linear algebra to understand what he is saying.

[@paulp1204]

​@@gdsfish3214 My point is the title of the video is totally off-base. It does not demystify tensors, let alone their products. The video title attracted me because I'd like to learn what they are, as I have not background with them. The presenter clearly does, but does not bring the newcomer on the journey. You did not, nor anybody else, have a background in tensors before someone explained it to you, either.

[@gdsfish3214]

@@paulp1204 yeah unfortunately tensors "mean" different things for people that do not study pure mathematics. In order to understand what a tensor really is you need a small introduction to abstract linear algebra. I've heard plenty of engineers just say "a tensor is just a matrix", which is fine, I mean this stuff does not have too much importance for their field so it's not worth learning. This is a case of a mathematician talking about tensors. but with the way he talks he already assumes a big chunk of knowledge so this is not even suited as an introduction. This video is more for students who learned that stuff some time ago and thought "yeah I never really knew what that meant exactly".

[@litsky]

As @GDSFish points out, yeah, this is a video on what a tensor "really is", in the deeply conceptual, coordinate-free sense they would be conceived of by a pure mathematician. (We are not content with "a tensor is a giant ungodly gadget with loads of indices that transforms according to this disgusting formula I have not motivated to you at all".) That is, it's unapologetically an abstract algebra lecture. The applications of tensors in physics and engineering are myriad, and they are not addressed at all here, not even for the sake of giving some intuition. In fact the only applications I really mention are to further areas of pure mathematics or theoretical physics, like differential geometry and general relativity.

[@MarcusAndersonsBlog]

I'm SOOOOOOOOOooooooooooooooooo............................ glad I didn't do a maths degree.

[@godfreytshehla2291]

Hahaha because Maths is chowing...I didn't even understand a single thing in this video.

[@aaabbb-kj7lk]

pls unprivate your multilinear algebra videos

[@litsky]

No. Those never existed.

[@bonbonpony]

"Demystified" is a lie. Thumbed down.

[@litsky]

¯\_(ツ)_/¯

[@bonbonpony]

OK, so let me be a bit more specific: When seeing a word "demystified", one expects all the mysteries around a subject to fade out into oblivion, not to see the thing A "explained" by introducing things B, C, D, E and F without any explanations or even definitions whatsoever of what do these additional concepts mean. This video is like trying to explain how to add 2+2 by using string theory. I would rather call it "over-mystified" than "demystified".

[@litsky]

Can you give an example of these additional concepts you're referring to?

[@bonbonpony]

Pretty much everything in this video. But if you want some examples, then ehhh... ok, here you have some: _"So if you're learning the theory of smooth manifolds for the first time..."_ I would probably want to know what is a _smooth manifold_ in the first place (recall, I'm learning it _for the first time_, as the assumption in the quote says). But this notion is not explained before referring to it in the video. _"you may be kind of overwhelmed by the things people are talking about."_ Yup, that's pretty much how I feel when watching this video: overwhelmed by all these notions which has not been introduced before use. _"In _*_linear algebra_*_ you have _*_linear operators_*_, you have _*_linear maps_*_ between _*_vector spaces_*_, you have _*_vectors_*_, you have _*_linear functionals_*_, ok, so the elements of the _*_dual space_*_"_ 7 undefined terms in one sentence! Thank Celestia that I already know what they mean, but if I didn't, your video wouldn't help me in any way. Sure, we can assume that the viewer knows what these are, but can the viewer be sure if he knows the exact same things you're talking about? What if they're just coincidentally named the same way? (I've seen this problem so many times in my life as a source of all the confusion...) That's why terms should be defined before use, to ascertain that we're on the same page at the very least. Look, for example, how Euclid's "Elements" or all those great books from 19th century are structured: definitions and axioms first, propositions next (along with proofs), and they are always introduced BEFORE using them for the first time anywhere else. Then it gets even worse, because you're just opening a Pandora's Box full of intricately-named notions, like "dual spaces", "double-dual spaces", "exterior algebras", "exterior powers", "differential forms", "finite-dimensional vector space", "natural isomorphism" etc. I came here to understand the tensor product, to have it "demystified" for me, but instead I got a whole lot of mathematical gobbledegook and now I'm starting to wonder: do I really need to understand all that stuff first to understand tensor product? Isn't it just "take every element from one bucket and combine it with every element from the other bucket in every possible way"? Then at 01:28 you write some symbols without explaining what do they mean. And they can mean many different things in different contexts (especially the stars). But you didn't specify what should they mean and how should they be interpreted in this particular case. Then I'm 5 minutes in and still nothing "demystified" (nor anything about tensor product), on the contrary actually: I'm starting to loose my understanding of linear algebra :P which is not good. But it's perhaps because of the fact that I prefer the "Look how easy it is!" approach than the "Look how a smartass I am" approach... Nevertheless, nothing is being "demystified" in this video, but the opposite is true.

[@litsky]

I honestly can't tell if you're trolling or not. With regards to defining everything before use, terms such as "vector space", "linear map", "dual space", etc. are *so*. *utterly*. *ubiquitous*. throughout all of mathematics that there is absolutely no need to define them again, unless I was actually intending to teach *elementary* linear algebra, which in this video I certainly was *not*. In fact, such needless pedantry can often be gratingly boring for the audience. Indeed, as you become more familiar with the mathematical literature, you will very rarely see these terms defined in all but the most elementary books (unless you read books published before the advent of the Internet and so on, in which case they often _will_ be defined, but for the entirely different reason that they want their exposition to be completely self-contained, so that at least in theory one could read the book without knowing the first thing about linear algebra). If you are so unreasonably bent on having all of basic linear algebra expounded to you, all the way back to the bloody ZFC axioms, then you will find the overwhelming majority of textbooks (indeed, perhaps all but Bourbaki's works!) simply unacceptable. In case it's not clear, I'm strongly advising you to develop some thicker skin so that you never find yourself in this situation. Euclid's days are over, my friend: mathematics is far larger than it was then, and we have _things_ to _do_. Finally, let me just explicitly point out how absurd it is to expect the same conventions to be followed in video lectures as in textbooks. If it was done your way, this video would be 10-15 times longer. I'd have trouble staying awake long enough to finish speaking, and I really doubt anyone has the time to watch a video that long (in fact, *as it is*, it's probably too long for many). In summary, my answer to your large paragraph is a resounding "YES, the viewer CAN be sure". At 01:28, I was speaking while writing down that equation (thus effectively removing _any_ possible ambiguity), in case you somehow missed that. Or were you watching the video with your sound muted? Should I have warned the viewer about that too, after I was finished reciting the Principia Mathematica, just to "make sure we were on the same page"? The rest of your post is basically just you complaining about words and phrases that appeared in brief (one or two-sentence!) asides, none of which were even _part_ of the exposition. Smooth manifolds were mentioned (again, one single _sentence_) because differential geometry is when the overwhelming majority of math students have to really grapple with tensors for the first time. If you don't know what "finite-dimensional vector space" or "dual space" means, then YES, you SHOULD go learn those things before trying to understand tensor products.

[@RD-fv2bf]

this explantation is a disaster

[@ianprado1488]

*disaster

[@RD-fv2bf]

indeed :P i have corrected it

[@vtrandal]

@@RD-fv2bf why bother? You’re both a couple of losers. It’s easy to drop in and criticize. The hard thing to do is offer Solutions. Where are your solutions? I thought so. Losers! And another thing! This reminds me of a story.

[@MrSidTaylor]

How to "demystify" a subject by making it ten times more complicated!

[@eglintonflats]

".. this so natural that everybody should understand it..." Yeah, that wearing body mike is mandatory if you want anybody to follow your videos, this is 21st century.