A Concrete Introduction to Tensor Products

[@MuPrimeMath]

The definition of tensor product for modules over a commutative ring is the same as for vector spaces. For noncommutative rings, we replace the scalar multiplication relations by the relation (ar) ⊗ b = a ⊗ (rb). There's one very important thing that I didn't cover in this video: the universal property of tensor products! Basically, the universal property says that bilinear maps on V × W are the same as linear maps on V ⊗ W. This comes from the fact that bilinear maps also work like multiplication. For example, bilinear maps have the property f(a+b,c) = f(a,c) + f(b,c), just like the distributive law for multiplication. In fact, the simplest example of a bilinear map ℝ × ℝ → ℝ is f(x,y) = xy.

[@CrazyShores]

Hi what is a good book to learn this?

[@user-qy8ib4ef1g]

@@CrazyShores for commutative algebra which studies commutative rings, their modules, their ideals you have the book by atyia and the book by eisenbud both are great and they complement each other well. note that you would probably need some abstract algebra knowledge besides linear algebra

[@zhuolovesmath7483]

Oh my goodness your ability to deliver ideas in a clear way is beyond description. I'm 1000% sure that you'll become a great professor or researcher. Keep going!

[@oyugioyugi6018]

I cannot thank you enough for this amazing lecture. Your ability to provide clarity is insane. Thank you, professor, for not only this video but for your truly remarkable channel as well. I look forward to future uploads

[@boriskogan666]

that's just too funny. came here to say exactly the opposite. i thought i knew the definition of vector spaces, but he lost me from moment one. defining Z (vector space) in terms of VxW (also vector spaces) is circular. his basis definition (of Z) was also tortured. after that my eyes glazed over and i never made it to the definition of the tensor product or the examples (which i really was hoping would give me some deeper understanding). all in all came away with no clearer understanding than i had before.

[@anirudhvaradarajan5635]

the only lecture that my brain gladly accepted for the explanation of tensor products, thanks a lot !

[@Minstorm34]

far and away the best explanation i've seen of this construction. i've gone back to this several times. many thanks

[@jackmeyergarvey]

You are an insanely good instructor. Thank you

[@adz5841]

Showing the analogical construction of direct sum of two vector spaces is really a nice pedagogical device.

[@peterengel8601]

This is much more digestible than what my textbook says. Thanks

[@danialhussain1336]

You made my life easier man. I really appreciate your teaching methodology.

[@chrstfer2452]

Bout halfway through, only note (other than you might want to look at cleaning up your board work, lining things up a bit more, but that'll come with practice making videos and stuff, no worries) is that you left out the summation indices and that was a bit confusing/seemingly ambiguous around 19:30. Awesome video, keep it up, and thanks for the work you do.

[@manstuckinabox3679]

three seconds into the lecture and I knew it was gonna be fire, auto sub.

[@jonashallgren4446]

You outclassed gpt 4 on this one, banger video

[@kristianfella-glanville]

Omg thank you so much for actual examples and calculations. Most textbooks have barely any

[@Karim-nq1be]

So basically tensor product is like generalizing multiplication to vector spaces?

[@luisbelgois2102]

Exactly

[@hwang1439]

Thank you very much for this insightful, carefully prepared video, I learned a lot.

[@mechadraws]

this is video is like finding pure gold on a river,thanks

[@deceo2119]

Brilliant video. Helped me so much with representation theory. Thank you :)

[@AnastasiiaPetrova-v7c]

omg, this is really perfect explanation, thank you very much!!!

[@geochum]

Awesome job, my man.

[@friedrichfreigeist3292]

I think a nice follow up to this would be the Clebsch Gordan decomposition

[@andrewdirr824]

Yes, but fortunately, someone has done much of the work for us and we have nifty tables. 😀

[@samsullivan2434]

At 22:41, I was under the impression that the previous two arguments showing v⨂0 = 0 relied on v being a vector in the vector space V and 0 being the zero vector in the vector space W. Why are we allowed to write v⨂1 and v⨂(-1)? What is the interpretation of 1 and -1 as elements of the vector space W?

[@MuPrimeMath]

You are correct, in general we cannot say that 1 is an element of W. The correct thing would be to write v⨂0 = v⨂(-w + w) = v⨂(-w) + v⨂w = -(v⨂w) + v⨂w = 0 where w is any element of W.

[@eddiejennings5262]

thank you, great explanation

[@mykewl5406]

Top tier explanation

[@massimilianoluppi9816]

Thank you for this excellent lesson. I would like to ask about the set Z that you mentioned at the very beginning of the lesson. Every element is a linear combiation of the basis elements. Are they just finte sums? Or can they be even infinite sums? I wonder if infinite sums in Z could be a problem for the construction of the tensor product coming next.

[@MuPrimeMath]

Like I mention at 2:45, we're only considering finite sums. This is because, by definition, a basis must span the vector space by only finite linear combinations of the basis vectors.

[@stephentashiro5177]

It would help if you took care of a few formal details. For example in order to construct a quotient object from two mathematical objects, we should say what type of objects they are. For example, the Z defined in the introduction is defined as a set of ordered pairs of vectors and apparently we implicity assume Z is a vector space. I suppose it is a "free" vector space on those ordered pairs. Do the relations we mod out by generate a subspace of Z? Are we taking the quotient of a vector space by a subspace? According to ChatGPT, there is a difference between the tensor product of two vectors versus an element of the tensor product of two vector spaces. Is that correct? For people who play your video at a high volume (due to background noise from fans etc) it would help to reduce the sibilance by applying some sort of "de-esser" software to the audio. The exposition of tensor manipulations is very good.

[@ויקטורגורביץ]

I don't understand why are you saying that (v,w) and (2v,w) are lineary independent vectors? Can you please give an example of what you meaning?Thanks a lot.

[@danilbutygin238]

(v,w) and (2v,w) is not a vectors it's just a ordered pairs with no extra structure

[@viglioni]

Very enlightening, thank you!

[@sergioquijanorey7426]

Thanks for this great video!

[@naveedthalappil7976]

Amazing, keep going.

[@programmer1379]

How can we be sure that the pairing of elements from V and W forms a well-defined basis for Z ?

[@MuPrimeMath]

Proving linear independence of the elements is typically done using the universal property of the tensor product, which I will cover in a future video.

[@StratosFair]

This lecture is pure gold and I can't thank you enough for it. Haven't quite finished the whole video yet but a couple questions have popped up in my mind : 1) seems like V and W need to be vector spaces over the same field, right ? Or is it possible to define a scalar product for vector spaces over different fields ? 2) your construction works very nicely for finite dimensional vector spaces, but what about infinite dimensional ones ? Namely, for Hilbert spaces or Banach spaces with countable Schauder basis, can we do a similar construction and is the Schauder basis of the tensor product again given by the pairwise tensor products of the original basis elements ?

[@MuPrimeMath]

1) Yes, both need to be vector spaces over the same field. 2) None of the steps rely on any dimensionality assumption, so everything applies in the infinite-dimensional case (countable or uncountable). I have a proof of the basis result using the universal property in my tensor products playlist: kzbin.info/aero/PLug5ZIRrShJHCtzgzZyRqSdzr8wYlN2qk

[@StratosFair]

@@MuPrimeMath amazing, thanks for your answer ! i will check out that other video soon :)

[@GlockenspielHero]

A couple clarifications that would have helped me a lot before watching: 1) The first 10 minutes of the video are **not** defining the tensor product ⨂, they are defining the direct sum ⨁. The construction is analogous, (so maybe it is a helpful jumping-off point to someone already familiar with the direct sum), but just keep in mind those 10 minutes are not defining a tensor product (yet!). 2) There are really two distinct operators, both named "tensor product" and both written ⨂, which are defined and used in the video. The first introduced is V⨂W, an operation combining two **vector spaces** into one larger space. The second introduced is v⨂w, an operator combining two individual **vectors** . Both are used throughout the video, and his capitalization is consistent enough to distinguish V⨂W from v⨂w. As you'll see eventually, the elements of the space V⨂W can be written as linear combinations of elements of the form v⨂w. At least, this is my current understanding. Let me know if there are any errors!

[@mathsandsciencechannel]

GREAT VIDEO

[@eugenemars]

Hello, thank you for this great work. However, I ask myself a question. You say Z is a space vector. Could you be more accurate about its elements ? are they (v,w) ? could you define the addition law (u,v) + (r,s)= ? the external law c(u,v) = ? Thank you

[@tim-701cca]

How can we define the vector addition and scalar multiplication of tensor product space VxW, where x is the symbol of the tensor product? How to compute (v1 x w1) + (v2 x w2)?

[@drlangattx3dotnet]

right away, I am confused. What do you mean that one basis vector comes from two different vector spaces? Are you saying that this set of basis vectors is a basis for both V and W?

[@MuPrimeMath]

The set is NOT a basis for V or W; these new basis vectors are not even elements of V or W. We are simply using the elements of V and W to index this set of basis vectors for a new vector space.

[@boriskogan666]

considering how many positive comments he received about this lecture, i am forced to conclude that it's us and not him who lack the prerequisites to digest this. i am exactly in the same boat as you are.

[@Hank-ry9bz]

Since the dimension of Z is dim(V)*dim(W), shouldn't there be a basis vector in Z only for each pair of basis vectors in V and W? That is to say, a basis vec in Z of the form (e_{v_i,w_j}) where v_i is one of V's basis vector and w_j is one of W's basis vectors (since otherwise dim(Z) will be much too large)? Or is that because it is prior to the construction of the quotient space?

[@MuPrimeMath]

The dimension of Z is not dim(V)*dim(W) because Z is not the tensor product. What you say is true of the tensor product, but Z is only an intermediate step in the construction, and often will be infinite-dimensional.

[@prakharpandey8968]

Sir please explain the base system and conversion btw different bases

[@raulbeienheimer]

I didn't learn about tensors in LA but in Tensor Calculus.

[@AksharaPrasadRED]

Thank you for the detailed explanation. That was very helpful! One small suggestion is to not put the cap on the pen everytime you stop writing. Since you hold your hand close to your mic, the sound of closing the cap and opening it can be annoying.

[@ali_ahmed31899]

Excuse me please Do you have a playlist in tensor plz ? Or can you make a playlist to explain tensor and tensor Division...etc please

[@maciej12345678]

What look elemnts of Qutient Space c*span{......} that is 0 hm ?

[@nocomment000]

Doesn't defining Z as a vector space with basis vectors {e, w} at the beginning confer the operations of addition and multiplication by a scalar on the basis vectors by definition, by virtue of {e, w} existing in the vector space Z? I.e. you can't have a vector space without those operations defined on the elements of the vector space. Also, isn't the quotient just the set of equivalence classes, i.e. the quotient doesn't automatically infer it equals zero (maybe this is just true under the Topology definition and it implies zero for vector spaces)?

[@MuPrimeMath]

Your first statement is correct. For the second question, the quotient is indeed the set of equivalence classes. We say that an element equals zero in the quotient if its equivalence class is equal to the equivalence class of zero.

[@methatis3013]

If Im not mistaken, you didn't actually show that the tensor product of V and W is a vector space itself?

[@MuPrimeMath]

The quotient of a vector space by a subspace is always a vector space. The big space Z is defined as a vector space consisting of a certain basis, and we quotient by the subspace generated by a certain set of elements, so the quotient is necessarily a vector space.

[@methatis3013]

@MuPrimeMath right, this was skimmed over a bit, but it needs to be shown that a quotient of a vector space by a subspace is, again, a vector space. Although I do understand that would go beyond the scope of the video. I guess it's a good homework exercise

[@coolbepis9301]

I think it would be best to start with the universal property as a definition of tensor product, then show why this construction satisfies that definition and why it's unique.

[@edgarosvaldo9217]

27:50 How can we proove that?

[@MuPrimeMath]

The proof comes from the universal property of the tensor product, which I mentioned in the pinned comment. For each pair e,f of basis vectors from the original vector spaces V,W there is a bilinear map p : V × W → F satisfying p(e,f)=1 and p(a,b)=0 if a,b ≠ e,f. This corresponds to a linear map V ⊗ W → F which maps e ⊗ f to 1 and a ⊗ b to zero if a,b ≠ e,f. We can use the existence of that map on V ⊗ W to prove linear independence, since applying the map to a linear combination of tensors of the form e ⊗ f simply outputs the coefficient of that particular tensor.

[@gustafkugelberg3906]

This is a great video, but I wonder, why do we first introduce Z as this massive vector space where the basis is every single pair of vectors in V and W, only to quotient away almost all of it? Couldn’t we just start with defining the space whose basis is all the pairs of BASIS vectors in V and W? Isn’t that what we end up with anyway?

[@MuPrimeMath]

One of the key properties of the tensor product is the universal property. I explain the universal property and its applications in this video: kzbin.info/www/bejne/rIvdi5uagaiSj7M The universal property has to do with bilinear maps, and it turns out that the bilinearity properties are easy to check with the construction I use in this video. You could define the tensor product as you describe, using just a set of basis vectors from each space. However, this doesn't generalize well to modules, since a module doesn't always have a basis. The construction I use in the video works for modules as well.

[@gustafkugelberg3906]

@@MuPrimeMath OK thanks!

[@jimnewton4534]

Question about your accent: do you pronounce peer tensor and pure tensor the same way? At first I thought you were saying "peer tensor"

[@brycejobe1181]

well done

[@kapoioBCS]

Weird timing, I was just reviewing some tensor products over rings for my course on representation theory. What a nightmare that the are the antisymmetric/symmetric tensors over rigs O>O

[@coreymonsta7505]

Direct sum is more than just the sum of vector spaces

[@omaral-odat2019]

Isn't (1,0) _tensor (1,0) = 0?

[@MuPrimeMath]

In the context of tensor products, (1,0) tensor (1,0) is a basis vector, so it does not equal zero.

[@juandavidrodriguezcastillo9190]

Gracias

[@bigredone1030]

sublime

[@sdsa007]

thanks!

[@ladnir]

Go beaves!

[@tomkerruish2982]

Masterfully done! Now use it to build a Poisson algebra from a Lie algebra.😁 (Seriously, I bet you'll be able to do this within a year, especially after taking QM and learning aaallll about commutator brackets.)

[@ShivamJha00]

bhai kehna kya chahte ho😭

[@whereami475]

Bro I remember I had this problem on my practice test to get into a school in 7th grade

[@lucasgroves137]

Lmao.

[@ko-prometheus]

☢☢☢☢☢

[@atonumukherjee4398]

too many proofs of v tensor 0 is 0 😂

[@nathanisbored]

so um if you write R × R thats the same as R², but for R ⊗ R you have to do the stuff in this video right?

[@MuPrimeMath]

Right! As a vector space, the Cartesian product R × R or R² is the same as the direct sum R ⊕ R. The tensor product R ⊗ R is a different set and a different vector space. R ⊗ R is sometimes written as R^(⊗2), and similarly R ⊕ R is written R^(⊕2).