(mathematical) applications of the tensor product.

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  • Опубликовано: 20 дек 2024

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  • @danebryant2535
    @danebryant2535 2 года назад +98

    Yes , I absolutely want to see “universally enveloping algebra “ and how it connect to Lie algebra. I don’t even know what universally enveloping algebra is. But I really want to learn .Thank for great lessons .Professor Penn you’re Awesome!

    • @i_amscarface_the_legend9744
      @i_amscarface_the_legend9744 2 года назад +1

      Also, what is Lie algebra Michael please do a video about the subject..

    • @redapplefour6223
      @redapplefour6223 2 года назад +1

      he has a video on that one now (and to the other guy, several on lie algebras but im not sure which the defining one is, if any?)

  • @habermasnyc
    @habermasnyc 2 года назад +57

    I would love to see more videos along these lines.

  • @goodplacetostop2973
    @goodplacetostop2973 2 года назад +14

    16:18

  • @Noiralef
    @Noiralef 2 года назад +30

    Small but important nitpick at 6:50: It is not correct that any element of (V tensor V) can be written as (ax + by) tensor (cx + dy). For example, (x tensor y + y tensor x) can not be written in this form. It is true though that x^2, xy, yx and y^2 form a basis, which is used in the following.

    • @bentoomey15
      @bentoomey15 2 года назад +16

      The correct statement would be that any element of (V tensor V) is a *linear combination* of elements that can be written as (ax+by) tensor (cx+dy), right? And that's why we still get the same basis.

    • @the_hidden_library
      @the_hidden_library 2 года назад

      Absolutely.

    • @johannesmoerland5438
      @johannesmoerland5438 2 года назад +4

      This exactly gives raise to entanglement on a bipartite quantum system, where each system is modelled by a different Hilbert space, and the entangled states are exactly the vectors from the tensor product space that do not factorize.

  • @mMaximus56789
    @mMaximus56789 2 года назад +21

    This shed quite a lot of light into the construction of the Clifford algebra! Thank you!

  • @akubaxx
    @akubaxx 2 года назад +28

    that’s extremely helpful and so well explained, thank you! love advanced topics being discussed on the channel

  • @JM-us3fr
    @JM-us3fr 2 года назад +12

    It would be pretty awesome to see this go down the representation theory route. I'm currently writing my thesis on Galois representations and Artin L-functions, and it heavily utilizes the Induced Representation and Frobenius Reciprocity.

  • @evanev7
    @evanev7 2 года назад +15

    More algebra please!
    Λ(x_1,...,x_n) has nCr basis elements of degree r since we can't repeat basis elements, so Λ(x_1,....,x_n) has dimension Σ nCr = 2^n.

  • @andreben6224
    @andreben6224 2 года назад +11

    WOAH this series of videos on algebras is so cool. YES definitely a big YES to the Lie Algebras video.
    So for the dimension of the exterior algebra I just try and count the anticommuting monomials of each degree. It turns out that for a finite dimension vector space, spanned by x_1, ... , x_n, there are only {k choose n} monomials in the x_i of degree k, since we can re-arrange the terms, up to changing the sign, so for instance
    zyx = -zxy = xzy = -xyz (for V=span{x,y,z})
    and there are no powers of these monomials, so for instance
    xyx = -x²y = 0
    Thus the exterior algebra for a vector space V of dimension k is spanned by
    sum_{k=0}^n {k choose n} = 2^n
    vectors. I omit a lot of details though they can "seem obvious" to be completed: the polynomials of non-repeating variables span the exterior algebra but are they free? Eh I'll leave it at that :P

    • @drmathochist06
      @drmathochist06 2 года назад

      You can simplify the argument. You've done great by collecting them together by degree to get n-choose-k, but what if you just throw all the monomials into a pile?
      Well, like you said, we can rearrange them. No generator can show up more than once, or else you can bring two copies together and get 0. If we pick an order on the generators we can just put the ones that show up in the monomial into increasing order. The only thing that matters to a monomial is whether a generator shows up or not, meaning that monomials correspond to ("are in bijection with") subsets of the set of generators!
      But that's easy to count: for each of the n generators, you have the choice to include or exclude it. The fundamental theorem of arithmetic tells us that we get 2^n subsets, and thus 2^n monomials.
      So now turn back to the way you calculated it, as the sum of n-choose-k for all k. You've now given a proof that the sum actually equals 2^n, which pops out for free by counting in two different ways!

  • @charlottedarroch
    @charlottedarroch 2 года назад +4

    I'm loving these videos! It's great to have a clear and concise, abstract construction of these algebras from vector spaces and tensor products of such spaces. I've come across these constructions before (with most focus on the exterior algebra), but I haven't seen such a clear description of the universal enveloping algebra, so I'm very keen to see that!

  • @tune_fisch0269
    @tune_fisch0269 2 года назад +3

    I would love to see some Videos on Representation Theory!

  • @TheMauror22
    @TheMauror22 2 года назад +5

    Please keep up with these tensor videos! I would like to see what's the connection between this abstract construction of the tensor product and the tensors used throughout physics

  • @gilmonat
    @gilmonat 2 года назад +5

    Hi Michael, love those kind of videos!
    As a physics student I've heard about Lie algebra but never studied it. You should definitely make a video about it.
    Also, I would love some explanations for people who don't have a very deep base in algebra.
    (Such as physicists and engineers as uppose to mathematicians)
    Keep up the good work!

  • @ukaszpawlak6953
    @ukaszpawlak6953 2 года назад +4

    The dimension is 2^n because the vectorspace is spanned by any maximal collection of words which use every letter no more than once, because words using the same letters being the same up to a minus sign. So there exactly as many such words as subsets of the set of variables.
    Please make Lie Algebra video :)

  • @bentoomey15
    @bentoomey15 2 года назад +3

    To see the dimension of an n-variable anticommuting polynomial ring is 2^n, we will count the number of terms in a generic element of the ring. Each term is of the form constant*(some product of our n variables, to the first power), just as in the two-variable case.
    How many monomials are possible of the form constant * (k variables), for a given k? Well, n choose k, since we're choosing k distinct variables from among our n. The total number of monomials in a generic element is therefore the sum as k ranges from 0 to n of the n-th binomial coefficients, which is well-known to be 2^n.

  • @stevenwilson5556
    @stevenwilson5556 2 года назад +3

    If you want to see a video about… Yes. Suffice to say, make a video about everything you know in math. This will live on as your legacy long after you are dead and can motivate generations of people to learn and appreciate math. Note, I have watched many videos on math and science that were at least 10+ years old, and I have no idea if some of the video creators are even still alive.

  • @pacificll8762
    @pacificll8762 2 года назад +1

    We want the video ! Thank you so much !

  • @jaredsimpson5672
    @jaredsimpson5672 2 года назад +1

    Super interesting. I went down a Geometric Algebra rabbit hole a couple years ago. And I feel like a got a decent intuition of the algebra and geometry, but none of it stuck. But seeing the construction of Sym versus ^ and how the choice of quotient determines semantics is really hitting home for me. I feel like I get how you could "endow" something now. That word really means something to me now!!

  • @brattok
    @brattok Год назад

    Dimension of external algebra of a span with n vectors can be calculated from the following equality: (0, n) + (1, n) + ... + (n, n) = 2^n, where (n, m) is a binomial coefficient. This follows from the fact that cardinality of a set with n objects is 2^n and also a sum of the cardinalities of all the subsets. And the cardinality of a subset with k objects equal to (k, n)

  • @lilyhayden5732
    @lilyhayden5732 2 года назад +1

    Very interested in seeing more about this! Been studying algebra in a very haphazard way where I have slight tangential familiarity with almost all of these ideas, so it's helpful to have them connected like this.

  • @dehnsurgeon
    @dehnsurgeon 2 года назад

    to find a basis, you can take the product over any subset of {x,y,...} so 2^n (if you repeat a term it will give zero)

  • @stervi2
    @stervi2 Год назад

    Interestingly the tensor algebra is used in quantum mechanics and quantum field theory to describe systems of a variable number of particles. In physics it is typically called a Fock space and the construction is slightly different when dealing with indistinguishable particles as each term in the direct sum is symmetrised if the particles are bosons and anti-symmetrised if they are fermions.

  • @drmathochist06
    @drmathochist06 2 года назад +31

    The core application was put best to me by Roger Howe: "The tensor product is a machine for replacing bilinear functions by linear functions."

    • @samueldeandrade8535
      @samueldeandrade8535 Год назад

      I never understood why people like statements such as the one you mentioned. I mean, it basically says nothing.

    • @drmathochist06
      @drmathochist06 Год назад

      ​@@samueldeandrade8535 It says that there is a natural bijection between two different sets.
      On the one side you have bilinear functions taking two arguments from vector spaces U and V and returning a result in vector space W.
      On the other side you have linear functions taking one argument from the vector space U⊗V and returning a result in vector space W.

    • @samueldeandrade8535
      @samueldeandrade8535 Год назад

      @@drmathochist06 I disagree. I disagree the statement says all that. I know what the tensor product is. Or, to be more precise, what is taught about it. But I never felt satisfied. One reason, maybe THE reason, is because we don't talk about the origin of tensor products. Coming back to my criticism about the statement, one word I dislike is "machine". It is just a substitute for "function". Or, which would be more appropriate in this case, "functor". Anyway. Just ... sleepy. Good night.

    • @drmathochist06
      @drmathochist06 Год назад

      @@samueldeandrade8535 Calm down, dude. It's a slogan, designed to remind you about the fact of the bijection. The bijection *is* the defining property of the tensor product; it's why we care about the tensor product in the first place.

    • @samueldeandrade8535
      @samueldeandrade8535 Год назад

      @@drmathochist06 hahaha. Calm down? Did I was rude or something?

  • @gnomeba12
    @gnomeba12 2 года назад

    Yes we do want to see that video!

  • @jamiewalker329
    @jamiewalker329 2 года назад

    We also use the tensor product in physics - in fields like electromagnetism, where we wire it up to a battery to create light.

  • @TopCuber
    @TopCuber 2 года назад +1

    I'm loving the higher algebra videos, I would definitely watch anything like that

  • @zhuolovesmath7483
    @zhuolovesmath7483 2 года назад

    LOVE YOU Michael! I really wanna see more videos on higher algebra like this.

  • @maurocruz1824
    @maurocruz1824 2 года назад

    Sure, I really need more of this!

  • @MrOvipare
    @MrOvipare 2 года назад

    The explanation of Sym(V) is crystal clear despite my relative ignorance of abstract algebra. Awesome video!
    Looking forward to see more about Lie algebras!

  • @charlottedarroch
    @charlottedarroch 2 года назад +2

    For the exterior algebra in characteristic 2, do you instead quotient by the ideal generated by v⊗v for all v in V, since that ideal contains v⊗w+w⊗v for all v,w in V, whereas the ideal generated by v⊗w+w⊗v for all v,w in V doesn't contain v⊗v for all v in V?

  • @qschroed
    @qschroed 2 года назад

    More videos like this please! They give a niece sneak peek into the world of more complex mathematics

  • @mfavier
    @mfavier 2 года назад +1

    I think it is important also to say that T(V) has an obvious product so the ideals for the constructions of the algebras are really ideals and not just sub-modules

  • @jzngk
    @jzngk 2 года назад

    Great video! Keep up the good work

  • @ismaelcastillo188
    @ismaelcastillo188 2 года назад

    I am learning about tensor products and linear algebra in general, but on a more geometrical perspective (for theoretical physics). It's nice to see some other applications

  • @C3POXTC
    @C3POXTC 2 года назад +1

    You are pumping out videos faster than I can understand them

  • @Holital
    @Holital 2 года назад

    Yes (to the end question)!!! I love all the algebra videos

  • @tuaigetsdj3699
    @tuaigetsdj3699 2 года назад

    You explain so well

  • @NutziHD
    @NutziHD 2 года назад

    Awesome! I would really appreciate more tensor videos (topic suggestions: more about the exterior algebra, clifford algebra, relation to the tensors used in physics). Keep it going! :)

  • @renpnal229
    @renpnal229 2 года назад

    Yes more videos like this please!

  • @m1323fj
    @m1323fj 2 года назад

    Excellent video! I would love to see the followup.

  • @tzimmermann
    @tzimmermann 2 года назад

    Hmm. I like where this is going! You're planing to talk about the construction of Verma modules using universal enveloping algebras, aren't you?
    Love your videos, definitely waiting for more!

  • @chalkchalkson5639
    @chalkchalkson5639 2 года назад +1

    I'd be very interested in seeing more Lie algebra and leading into representation theory!

  • @luciuskhor554
    @luciuskhor554 2 года назад

    Yes more video about this

  • @qschroed
    @qschroed 2 года назад

    At 12:05, shouldn't v,w be from T(V)? otherwise it only commutes for the first tensor powers of V

  • @eschudy
    @eschudy 2 года назад +1

    Yes please do the Lie Algrebra and Enveloping Algebra videos!

  • @philipschloesser
    @philipschloesser 2 года назад

    Well, the Clifford algebra is also a relevant algebra you can construct out of T(V)

  • @jplikesmaths
    @jplikesmaths Год назад

    5:13 can someone help with constructing an isomorphism?

  • @Budha3773
    @Budha3773 2 года назад +1

    Please make a video about the universal enveloping algebra

  • @nathanisbored
    @nathanisbored 2 года назад +1

    more abstract algebra!

  • @erikdavis8901
    @erikdavis8901 2 года назад

    Here to mention that I do want to see a follow up video that you mentioned in the final 20 sec

  • @jpay
    @jpay 2 года назад

    "Okay, nice." 🙂3:09

  • @brooksbryant2478
    @brooksbryant2478 2 года назад +1

    I’d be interested in the Lie algebra video!

  • @CrazyShores
    @CrazyShores 9 месяцев назад +1

    Great 🎉

  • @nablahnjr.6728
    @nablahnjr.6728 2 года назад

    spank{x} is a very interesting notation

  • @lucachiesura5191
    @lucachiesura5191 2 года назад

    Very important for the theory of the relativity

  • @Manuel-mw3id
    @Manuel-mw3id 2 года назад

    What are applications of the kronecker product

  • @__gavin__
    @__gavin__ 2 года назад +2

    Here is the video mentioned in the intro:
    ruclips.net/video/K7f2pCQ3p3U/видео.html

  • @djsmeguk
    @djsmeguk 2 года назад

    Is there a relationship between the exterior algebra and differential forms? It seems they're very similar.

  • @benhsu42
    @benhsu42 2 года назад

    Does anyone have a link to the "previous video" Professor Penn alluded to? I looked for the word "tensor" in recent videos but couldn't find it
    thank you

    • @benhsu42
      @benhsu42 2 года назад

      Was it this one? ruclips.net/video/K7f2pCQ3p3U/видео.html

  • @chrstfer2452
    @chrstfer2452 2 года назад

    Professor Penn: i know youre not a physics guy, but do you think it'd be possible to connect this abstract construction of the tensor algebra and these related algebras from this and your previous video to the way tensors are used in physics or computer science/AI? I know there are some videos that try to do this but i havent found any that click in my mind, whereas i havent felt so many math clicks in my mind as when i watch your videos since i was in undergrad.

  • @kevincardenas6629
    @kevincardenas6629 2 года назад

    Please do physics applications! :’p

  • @mathunt1130
    @mathunt1130 2 года назад +1

    Don't you mean x wedge x=0 rather than x\otimes x=0?

    • @MichaelPennMath
      @MichaelPennMath  2 года назад +2

      This depends on if you change notation while passing to the quotient... you are right that \wedge is generally in the quotient but not totally necessary

  • @xizar0rg
    @xizar0rg 2 года назад

    I'm glad the video has a meaningful name, but would still like to see the videos enumerated for easy reference.

  • @ruinenlust_
    @ruinenlust_ 2 года назад

    I spent an _embarassingly_ long time thinking why V tensor 0 would be equal to its ground field but then I finally realised that it was a different notation...

  • @mario30893
    @mario30893 2 года назад

    We want to see said video

  • @rafaelgcpp
    @rafaelgcpp 2 года назад

    Dr Penn, could you make a video on tensor decompositions and tensor trains? It seems to simplify some kinds of projective operations.
    It might not be your field, so feel free to decline!

  • @asdf8asdf8asdf8asdf
    @asdf8asdf8asdf8asdf 2 года назад

    “Dr Mike, when ya gonna put yer drppin’ swag up for sale?”
    Would you consider affiliate marketing or at least providing pointers to where you get some of your apparel?
    😎

  • @raa9558
    @raa9558 2 года назад

    Self duality is interesting

  • @f5673-t1h
    @f5673-t1h 2 года назад +1

    Thought this was an X-Men video from the thumbnail

  • @lucabellitti6893
    @lucabellitti6893 2 года назад

    try to solve the Diophantine equation 2x ^ 2 -1 = y ^ 2

    • @noahtaul
      @noahtaul 2 года назад +1

      Hello, the answer is y=+-((1+sqrt(2))(3+2
      sqrt(2))^k+(1-sqrt(2))(3-2sqrt(2))^k)/2, x=((1+sqrt(2))(3+2
      sqrt(2))^k-(1-sqrt(2))(3-2sqrt(2))^k)/(2sqrt(2)) for some integer k and some choice of sign for y.

  • @sanch023
    @sanch023 2 года назад

    This video should have been in the other channel, shouldn't it? V good video btw

  • @RedDaLord
    @RedDaLord 2 года назад +1

    Yes, more on lie algebra

  • @mcqueen424
    @mcqueen424 2 года назад

    Thought it was X-Men at first

  • @mrl9418
    @mrl9418 2 года назад +1

    Non-mathematical applications of maths offend me

  • @kilianklaiber6367
    @kilianklaiber6367 Год назад

    Why is the tensor product not defined as the product between tensors but between "tensor spaces"? I find that to be purposely obfuscating something simple in order to make it appear smart. No love for this kind of math from me.