Niles Johnson: Visualizations of the Hopf fibration

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  • Опубликовано: 15 ноя 2024

Комментарии • 105

  • @igorcossack8117
    @igorcossack8117 6 лет назад +27

    Thank you for making my medical licensing exam seem easy. This is the motivation I needed!

    • @PaPa-kr5yt
      @PaPa-kr5yt Год назад

      Almost same here. The animation is an art, and the lecture was a constructive (not really self-deceived lol ) procrastination and made my current work easy.

  • @sumitmatta
    @sumitmatta Год назад +1

    The best 4d space explanation video on RUclips

  • @andyeverett1957
    @andyeverett1957 12 лет назад +6

    What a privilege to view this, thanks to all that made this possible and to Professor Johnson for its clarity!

  • @seansmith5468
    @seansmith5468 6 лет назад +74

    JRE brought me some good fibrations

    • @remusilie6958
      @remusilie6958 6 лет назад

      me2

    • @HR-yd5ib
      @HR-yd5ib 5 лет назад +2

      Good old Eric weinstein and Lawrance Krauss should take a lesson from this guy on how to explain things clearly.

    • @Ramin2340
      @Ramin2340 5 лет назад +1

      It's so weird I was half paying attention when it was brought up on JRE then a yearish later it's the topic of my final year Physics dissertation wtf is going on!!

    • @nickacelvn
      @nickacelvn 4 года назад

      Ha ha you should write a song.

    • @mazumazu8873
      @mazumazu8873 4 года назад

      Geometry is difficult for me.

  • @tombarry9736
    @tombarry9736 5 лет назад +7

    This is blowing my mind right now. The geometry of how 3 dimensions can be inverted without intersecting itself.

  • @alexmitchell5339
    @alexmitchell5339 3 года назад +2

    Here from Grand Illusions! Thanks Tim!

  • @2beJT
    @2beJT 6 лет назад +20

    What a great teacher

  • @epsilonkleiner0
    @epsilonkleiner0 11 лет назад +8

    I wish I'd known about this video earlier...it makes fibrations so much easier to understand! Thanks a lot Mr. Johnson!

  • @kcflitter2879
    @kcflitter2879 8 лет назад +15

    This is the craziest coolest stuff, thanks for making these videos!!!

  • @Cognitiveleaper
    @Cognitiveleaper 6 лет назад +32

    This fellow enjoys his occupation.

  • @skyclaw
    @skyclaw 5 лет назад +1

    This has got me revisiting all the knot theory I did for my MSc.

  • @MelodyAnnHM
    @MelodyAnnHM 9 лет назад +4

    Very clear and helpful talk! Thank you for uploading!

  • @SpotterVideo
    @SpotterVideo 11 месяцев назад

    Conservation of Spatial Curvature:
    Both Matter and Energy described as "Quanta" of Spatial Curvature. (A string is revealed to be a twisted cord when viewed up close.)
    Is there an alternative interpretation of "Asymptotic Freedom"? What if Quarks are actually made up of twisted tubes which become physically entangled with two other twisted tubes to produce a proton? Instead of the Strong Force being mediated by the constant exchange of gluons, it would be mediated by the physical entanglement of these twisted tubes. When only two twisted tubules are entangled, a meson is produced which is unstable and rapidly unwinds (decays) into something else. A proton would be analogous to three twisted rubber bands becoming entangled and the "Quarks" would be the places where the tubes are tangled together. The behavior would be the same as rubber balls (representing the Quarks) connected with twisted rubber bands being separated from each other or placed closer together producing the exact same phenomenon as "Asymptotic Freedom" in protons and neutrons. The force would become greater as the balls are separated, but the force would become less if the balls were placed closer together. Therefore, the gluon is a synthetic particle (zero mass, zero charge) invented to explain the Strong Force. An artificial Christmas tree can hold the ornaments in place, but it is not a real tree.
    String Theory was not a waste of time, because Geometry is the key to Math and Physics. However, can we describe Standard Model interactions using only one extra spatial dimension? What did some of the old clockmakers use to store the energy to power the clock? Was it a string or was it a spring?
    What if we describe subatomic particles as spatial curvature, instead of trying to describe General Relativity as being mediated by particles? Fixing the Standard Model with more particles is like trying to mend a torn fishing net with small rubber balls, instead of a piece of twisted twine.
    Quantum Entangled Twisted Tubules:
    “We are all agreed that your theory is crazy. The question which divides us is whether it is crazy enough to have a chance of being correct.” Neils Bohr
    (lecture on a theory of elementary particles given by Wolfgang Pauli in New York, c. 1957-8, in Scientific American vol. 199, no. 3, 1958)
    The following is meant to be a generalized framework for an extension of Kaluza-Klein Theory. Does it agree with some aspects of the “Twistor Theory” of Roger Penrose, and the work of Eric Weinstein on “Geometric Unity”, and the work of Dr. Lisa Randall on the possibility of one extra spatial dimension? During the early history of mankind, the twisting of fibers was used to produce thread, and this thread was used to produce fabrics. The twist of the thread is locked up within these fabrics. Is matter made up of twisted 3D-4D structures which store spatial curvature that we describe as “particles"? Are the twist cycles the "quanta" of Quantum Mechanics?
    When we draw a sine wave on a blackboard, we are representing spatial curvature. Does a photon transfer spatial curvature from one location to another? Wrap a piece of wire around a pencil and it can produce a 3D coil of wire, much like a spring. When viewed from the side it can look like a two-dimensional sine wave. You could coil the wire with either a right-hand twist, or with a left-hand twist. Could Planck's Constant be proportional to the twist cycles. A photon with a higher frequency has more energy. ( E=hf, More spatial curvature as the frequency increases = more Energy ). What if Quark/Gluons are actually made up of these twisted tubes which become entangled with other tubes to produce quarks where the tubes are entangled? (In the same way twisted electrical extension cords can become entangled.) Therefore, the gluons are a part of the quarks. Quarks cannot exist without gluons, and vice-versa. Mesons are made up of two entangled tubes (Quarks/Gluons), while protons and neutrons would be made up of three entangled tubes. (Quarks/Gluons) The "Color Charge" would be related to the XYZ coordinates (orientation) of entanglement. "Asymptotic Freedom", and "flux tubes" are logically based on this concept. The Dirac “belt trick” also reveals the concept of twist in the ½ spin of subatomic particles. If each twist cycle is proportional to h, we have identified the source of Quantum Mechanics as a consequence twist cycle geometry.
    Modern physicists say the Strong Force is mediated by a constant exchange of Gluons. The diagrams produced by some modern physicists actually represent the Strong Force like a spring connecting the two quarks. Asymptotic Freedom acts like real springs. Their drawing is actually more correct than their theory and matches perfectly to what I am saying in this model. You cannot separate the Gluons from the Quarks because they are a part of the same thing. The Quarks are the places where the Gluons are entangled with each other.
    Neutrinos would be made up of a twisted torus (like a twisted donut) within this model. The twist in the torus can either be Right-Hand or Left-Hand. Some twisted donuts can be larger than others, which can produce three different types of neutrinos. If a twisted tube winds up on one end and unwinds on the other end as it moves through space, this would help explain the “spin” of normal particles, and perhaps also the “Higgs Field”. However, if the end of the twisted tube joins to the other end of the twisted tube forming a twisted torus (neutrino), would this help explain “Parity Symmetry” violation in Beta Decay? Could the conversion of twist cycles to writhe cycles through the process of supercoiling help explain “neutrino oscillations”? Spatial curvature (mass) would be conserved, but the structure could change.
    =====================
    Gravity is a result of a very small curvature imbalance within atoms. (This is why the force of gravity is so small.) Instead of attempting to explain matter as "particles", this concept attempts to explain matter more in the manner of our current understanding of the space-time curvature of gravity. If an electron has qualities of both a particle and a wave, it cannot be either one. It must be something else. Therefore, a "particle" is actually a structure which stores spatial curvature. Can an electron-positron pair (which are made up of opposite directions of twist) annihilate each other by unwinding into each other producing Gamma Ray photons?
    Does an electron travel through space like a threaded nut traveling down a threaded rod, with each twist cycle proportional to Planck’s Constant? Does it wind up on one end, while unwinding on the other end? Is this related to the Higgs field? Does this help explain the strange ½ spin of many subatomic particles? Does the 720 degree rotation of a 1/2 spin particle require at least one extra dimension?
    Alpha decay occurs when the two protons and two neutrons (which are bound together by entangled tubes), become un-entangled from the rest of the nucleons
    . Beta decay occurs when the tube of a down quark/gluon in a neutron becomes overtwisted and breaks producing a twisted torus (neutrino) and an up quark, and the ejected electron. The production of the torus may help explain the “Symmetry Violation” in Beta Decay, because one end of the broken tube section is connected to the other end of the tube produced, like a snake eating its tail. The phenomenon of Supercoiling involving twist and writhe cycles may reveal how overtwisted quarks can produce these new particles. The conversion of twists into writhes, and vice-versa, is an interesting process, which is also found in DNA molecules. Could the production of multiple writhe cycles help explain the three generations of quarks and neutrinos? If the twist cycles increase, the writhe cycles would also have a tendency to increase.
    Gamma photons are produced when a tube unwinds producing electromagnetic waves. ( Mass=1/Length )
    The “Electric Charge” of electrons or positrons would be the result of one twist cycle being displayed at the 3D-4D surface interface of the particle. The physical entanglement of twisted tubes in quarks within protons and neutrons and mesons displays an overall external surface charge of an integer number. Because the neutrinos do not have open tube ends, (They are a twisted torus.) they have no overall electric charge.
    Within this model a black hole could represent a quantum of gravity, because it is one cycle of spatial gravitational curvature. Therefore, instead of a graviton being a subatomic particle it could be considered to be a black hole. The overall gravitational attraction would be caused by a very tiny curvature imbalance within atoms.
    In this model Alpha equals the compactification ratio within the twistor cone, which is approximately 1/137.
    1= Hypertubule diameter at 4D interface
    137= Cone’s larger end diameter at 3D interface where the photons are absorbed or emitted.
    The 4D twisted Hypertubule gets longer or shorter as twisting or untwisting occurs. (720 degrees per twist cycle.)
    How many neutrinos are left over from the Big Bang? They have a small mass, but they could be very large in number. Could this help explain Dark Matter?
    Why did Paul Dirac use the twist in a belt to help explain particle spin? Is Dirac’s belt trick related to this model? Is the “Quantum” unit based on twist cycles?
    I started out imagining a subatomic Einstein-Rosen Bridge whose internal surface is twisted with either a Right-Hand twist, or a Left-Hand twist producing a twisted 3D/4D membrane. This topological Soliton model grew out of that simple idea. I was also trying to imagine a way to stuff the curvature of a 3 D sine wave into subatomic particles.
    .-----

  • @Icingde4th
    @Icingde4th 11 лет назад +9

    This was remarkably easy to understand and quite enjoyable, thanks!

    • @nickacelvn
      @nickacelvn 4 года назад +1

      Explain in back to me in a nutshell!

  • @imathaddict
    @imathaddict 11 лет назад +3

    Someone, somewhere, has passed the great opportunity to call Brougham Bridge as The Quaternion Bridge. Just saying. Neat lecture by the way. Congrats.

  • @nateschmidt5835
    @nateschmidt5835 5 лет назад +1

    Great visualization for those who can understand the math

  • @rockapedra1130
    @rockapedra1130 2 года назад

    I thought it went a little fast. Probably from trying to squeeze the whole thing into 50 minutes. I’ll prob just watch it again and pause every so often. Thanks! Loved the animation at the end!

  • @andyeverett1957
    @andyeverett1957 6 лет назад +8

    It made more sense the 2nd or 3rd time through. Clear as it can get, the rest is my fault.

  • @mtb4u
    @mtb4u 2 года назад +1

    This is fantastic. Very well explained!

  • @flashinglightstf
    @flashinglightstf 2 года назад

    I absolutely love it and I have no idea why.

  • @gewinnste
    @gewinnste 4 года назад +1

    Isn't an easier way to envision S3 to think of the 4th dimension as time and then S3 "starts" as a small version of S2 (actually a point) and this sphere grows to actual S2, then gets smaller again to a point? Well the "growing-velocity" is cosinoid, is what needs to be added.

    • @geometerfpv2804
      @geometerfpv2804 3 года назад

      The point isn't necessarily to "envision" S3. The Hopf fibration can be used to get useful topological information about S3. Having a fibration (so, not just a parametrization or graph of S3, but a way of mapping it into a base space such that the fibers are homotopic to one another) gives you information about the homotopy groups of S3 (it gives an "exact sequence"). Fibrations are an important part of homotopy theory in general.

  • @dlbattle100
    @dlbattle100 8 лет назад +3

    Hamilton's original carvings are gone, but there's a plaque.

  • @CarnifaxMachine
    @CarnifaxMachine 5 лет назад +4

    Well, it made sense the first 7 minutes or so. It's clear that to comprehend the whole thing I need to watch more Rick and Morty.

  • @jamma246
    @jamma246 12 лет назад +2

    Really nice clear talk.
    One tiny nitpicky comment is that when you are talking about fibrations, really you are just talking about fibre bundles - fibrations are a little more general.

    • @geometerfpv2804
      @geometerfpv2804 Год назад

      They are really not the same thing...a fiber bundle has homeomorphic fibers, a fibration has only homotopic fibers. That's a big difference. Oh what I would give for the path space fibration to be a fiber bundle...😂

  • @Illuminateur
    @Illuminateur 11 лет назад

    These might be the same question,
    1. Where in the 3-ball is the 2-sphere?
    2. Why are the projected circles linked?

  • @honeybunny8222
    @honeybunny8222 6 лет назад +2

    Can you please make a video on topology verify the hopf bundles are in fact bundles

  • @spam2080
    @spam2080 6 лет назад +14

    I don't speak this language. I need to learn mathematics.

  • @donrichards514
    @donrichards514 Год назад

    the difficulties in expressing irrational coordinates

  • @GeonGeorge
    @GeonGeorge 5 лет назад +1

    Where does one learn stuff like these?

  • @ivanbarbosa81
    @ivanbarbosa81 5 лет назад +1

    Great respect for teachers.

  • @NilesJohnson
    @NilesJohnson 12 лет назад +3

    @jamma246 Ah, of course you're right! I should have made that comment during the talk :)

  • @jaimeafarah7445
    @jaimeafarah7445 Год назад

    The quaternion q = W+iX+jY+kW can be expressed as a 2×2 matrix
    Q₂ₓ₂ = [ Q₁₁ , Q₁₂ ; Q₂₁ , Q₂₂ ] = [ W + iX , Y + iZ ; −Y + iZ , W − iX ]
    The determinant of Q is 1 e.g. det Q = W² + X² + Y² + Z² ≡ 1
    hence Q⁻¹ = [ W − iX , −Y − iZ ; Y − iZ , W + iX ]
    The projection η = qkq⁻¹ in matrix form is : QKQ⁻¹ ≡ H
    where K is considered a quaternion in matrix form as: K = [ 0 , i ; i , 0 ]
    since η = qkq⁻¹ = q' = ( W' , X' , Y' , Z' ) similarly
    H = QKQ⁻¹ = Q' = [ W' + iX' , Y' + iZ' ; −Y' + iZ' , W' − iX' ]
    Then, H₁₁ = (QKQ⁻¹)₁₁ = 2i(WY + XZ) = W' + iX'
    which means W' = 0 and X' = 2WY + 2XZ.
    Also, H₁₂ = (QKQ⁻¹)₁₂ = 2YZ − 2WX + i(W² − X² − Y² + Z²) = Y' + iZ'
    thus Y' = 2YZ − 2WX and Z' = W² − X² − Y² + Z²
    The determinant of H = det (QKQ⁻¹) = det Q ⋅ det K ⋅ det Q⁻¹ = (1)(1)(1) = 1
    = ( 2WY + 2XZ )² + ( 2YZ − 2WX )² + ( W² − X² − Y² + Z² )²
    which is a unit 2-sphere.

  • @SliversRebuilt
    @SliversRebuilt 2 года назад +1

    Are you fucking serious. THAT'S ALL A FIBRE IS?? WHY the HELL could NO ONE explain that before in terms that were anything less than absolutely mystifying?? Thank you so much for this lol

    • @geometerfpv2804
      @geometerfpv2804 Год назад +1

      Any textbook on topology defines a fiber as the inverse image of a point, it's not a secret 😅. If you are talking about youtube videos, most people talking about things like the Hopf fibration don't actually understand it. They are math/physics communicators, not mathematicians/physicists, and there is a big difference.

  • @nickacelvn
    @nickacelvn 4 года назад

    3.35 separated by what? Planck length?

  • @nicholaswilliams4507
    @nicholaswilliams4507 3 года назад

    Outstqanding lesson. Thanks for posting.

  • @nahkaimurrao4966
    @nahkaimurrao4966 Год назад

    I thought I would never understand what s vibration was but this is very simple

  • @humilityconsulting7499
    @humilityconsulting7499 6 лет назад +1

    28:47 so where did the S¹ come from in this fibration?

    • @Scigatt
      @Scigatt 6 лет назад

      Multiplying q by k^r(r real mod 4) from the right.

  • @chaoshengzhe
    @chaoshengzhe 12 лет назад +1

    this is damn better than the "dimension" video!!!

  • @sergeyten1683
    @sergeyten1683 4 года назад

    In most important part of lecture you mostly see only lecturer's back. He is writing something on the board but it's almost impossible to see what he is writing

  • @sorrysonofa
    @sorrysonofa 12 лет назад +1

    very good talk!

  • @darkanderson8761
    @darkanderson8761 5 лет назад

    thank you, much better than in physics lesson on medington high

  • @mihail263
    @mihail263 8 лет назад

    Thank you for your wonderful lecture. I have a question about difference between Hopf band and (S^1)x(I). Obviously(?) they are homeomorphic. But do they have different fiber structure?

    • @eddiebATL
      @eddiebATL  8 лет назад +1

      +Mihail Arabadji I'd suggest posting this question in a place like reddit.com/r/learnmath or even reddit.com/r/math if /r/learnmath doesn't work. This video is approaching 10,000 views in the course of five years, (which I am, of course, really proud of). However, those reddits can get 10,000 views in a day and from professional mathematicians.
      Great place to ask for weirder math help.

    • @sebastianschulz1583
      @sebastianschulz1583 8 лет назад +4

      +Mihail Arabadji Well, they are indeed homeomorphic and (depending on what exactly you mean by that) have the same structure as fiber bundles. However, you might be missing an important point here, namely that you really want to look at the Hopf band as embedded in $R^3$ (or $S^3$, it doesn't matter for the point to come). Hence, you would really ask for a homeomorphism $R^3
      ightarrow R^3$, that takes the Hopf band to $(S^1)x(I)$. That is impossible, as you would unlink the boundary spheres! Put differently, while the Hopf band seems trivial on its own, it becomes interesting once it is nested in $R^3$.

    • @redaabakhti768
      @redaabakhti768 5 лет назад

      @@sebastianschulz1583 however the boundary circles can be unlinked in R^4 without crossing

  • @donrichards514
    @donrichards514 Год назад

    Thank You Well said.

  • @IzzyisDizzyandFizzy
    @IzzyisDizzyandFizzy 3 года назад

    This is so helpful, thank you!

  • @alexgil4623
    @alexgil4623 4 года назад

    Muchas gracias...

  • @zachariahhanson1792
    @zachariahhanson1792 8 лет назад

    Can you trace out a shape on the 3-sphere that would not leave a circle in the stereographic projection? I'm guessing so. If this is correct then why don't we see any different shapes than circles in the (amazing) video at the end?... I'm also wondering about the line (the one circle that doesn't resemble a circle), what point on the 3-sphere is that projected from? -Surely if it's the point "at the top" (where the projecting light rays come from) then it should be out of sight infinitely far away? I'm hoping I'm making sense and that I don't sound horribly confused.
    Thanks!

    • @gastropuce9537
      @gastropuce9537 4 года назад

      That line is actually a circle of infinite radius, that's why it looks like that.

    • @geometerfpv2804
      @geometerfpv2804 3 года назад +1

      The circles aren't an effect of the projection, they are the fibers of the Hopf fibration. The Hopf map from S3->S2 has fibers (inverse images of single points) each homeomorphic to the circle.

  • @Czeckie
    @Czeckie 4 года назад

    9:00 isn't that a definition of a fibre bundle and not fibration?

  • @rhaldryn7511
    @rhaldryn7511 3 года назад

    He has a roll of toilet paper handy if some student shits themselves trying to understand this.

  • @ryder11211
    @ryder11211 3 года назад

    great teacher. thanks

  • @bencopeland4514
    @bencopeland4514 6 лет назад +12

    Got here because of the JRE. I found this explanation far more comprehensible than Eric Weinstein’s.

    • @stefan9229
      @stefan9229 6 лет назад +3

      Weinstein likes the jargon too much. Half of not understanding him is not knowing the jargon. I think he does it because he likes to sound smart which is unnecessary because he is clearly incredibly smart

    • @Adam_Boy
      @Adam_Boy 5 лет назад +1

      on the other hand this is a 45 min video in a particular format to just teach you this topic and nothing else. The JRE talk is just that - a talk, where they brush over this topic for about a measly 15 min or what? So you really can't compare apples with oranges here. Weinsteins himself also says that Joe is not a meathead that won't ever get what he is saying but they rather lack the time for him to understand the topic in that talk.

    • @georgewashington1200
      @georgewashington1200 5 лет назад

      jay maybe he meant geodude evolution

  • @kritikakhanwal614
    @kritikakhanwal614 6 лет назад

    how is the result obtained at 28.22, why is w=0

    • @NilesJohnson
      @NilesJohnson 6 лет назад +8

      Ah -- I used w to mean "the first coordinate". I should have said "this point lies in the 3-dimensional subspace where the first coordinate is 0".

    • @robinnilsson8317
      @robinnilsson8317 6 лет назад

      @@NilesJohnson Seriously? Not a single like from the lecturer answering a youtube question(s) with an direct and good answer in a channel that isn't his?
      Well Niles, just sit back - Joe Rogan has brought the views - and you shall get the recognition you deserve!

    • @NilesJohnson
      @NilesJohnson 6 лет назад +3

      @@robinnilsson8317 Ha! Not sure I would want to get what I deserve, but I'm glad people are discovering this content :)

  • @kritikakhanwal614
    @kritikakhanwal614 6 лет назад

    What is the meaning of having an inverse of q?

    • @NilesJohnson
      @NilesJohnson 6 лет назад

      q^{-1} means the multiplicative inverse, using quaternion multiplication. This is called the "reciprocal" on the wikipedia page about quaternions.

  • @Crasshopperrr
    @Crasshopperrr 11 лет назад

    22:10 Brougham Bridge in Dublin!

    • @OscarMaris
      @OscarMaris 5 лет назад

      I'm surprised they were able to stop drinking long enough to make a bridge

  • @brennaboucher999
    @brennaboucher999 6 лет назад

    Why wouldn't R x S1 be an infinite flat circle, like R x R with rounded corners? And why wouldn't S1 x S1 be an infinite sphere? Wouldn't a torus be better described by S2 x a hollow S1? Or couldn't that infinite cylinder be described by R x S2? Is there an assumption I missed?

    • @zakkatz9144
      @zakkatz9144 5 лет назад

      You have to think about the definition of the cartisian (cross) product. It takes every pair of points from S1 (a circle) and paires them to a point in R (the real line). So S1×R is the circle with strait lines going off to infinity. Im omitting some technicalities that would be covered in a rigorious math class.

    • @zakkatz9144
      @zakkatz9144 5 лет назад

      Similarily for the other cartisian products in your comment. S1×S1 is every point in S1 paired with a point of the same circle. I assume when you say infinite, you mean points of the sphere go on "forever". The teo circle, though, are "finite", so you will get a finite result, as you could verifying by googling the definition of cartisian product and testing it yourself

    • @brennaboucher999
      @brennaboucher999 5 лет назад

      ​@@zakkatz9144 It sounds like you mean that a cartesian product is specific result of two finite matrices multiplied (describing a specific points, shapes or objects), and not the product of a platonic "line" or "circle" or "sphere" per se. Am I understanding you correctly? Thanks!

    • @zakkatz9144
      @zakkatz9144 5 лет назад

      @@brennaboucher999 I'd suggest you google the definition of cartisian product. It doesnt invovle matrices, but sets. If you're unfamiliar with the notion of sets, khan academy has some great videos on the matter. If you are, then A×B takes every a in A and b in B, creates an tuple, or ordered pair, (a,b), and returns a new set with every such ordered pair. So if A= {1,2}, B={a,b}. Then A×B= {(1,a), (1,b), (2,a), (2,b)}. Note that order matters: (1,a) is not equal to (a,1). That means that A×B is not necessarily equal to B×A.

    • @zakkatz9144
      @zakkatz9144 5 лет назад

      Also note sets need nod be finite. The real line is not a collection of finite points. The "finite" part was of S1×S1. Both have finite length, so the cartisian product will be "finite" in the sense that it will also have finite length

  • @ricci1729
    @ricci1729 7 лет назад

    excellent...............................

  • @cg7622
    @cg7622 3 года назад

    I got lost at the Cartesian product stuff oop-

  • @SpontaneityJD
    @SpontaneityJD 6 лет назад +1

    Damn, I feel stupid now

  • @Hambxne
    @Hambxne 4 года назад

    oh eric, what have you done to by brain

  • @shootingwithguns3308
    @shootingwithguns3308 2 года назад

    Erases the board. *eraser marks*... Looks about right.

  • @kavishkhamesra3582
    @kavishkhamesra3582 Год назад

    #hopffibration

  • @kpw84u2
    @kpw84u2 Год назад

    H is for Hamiltonians -- which are a whole other sidebar 😂😂😂

  • @McGroppy
    @McGroppy 6 лет назад +11

    JRE brought me here...

  • @FerrariMekhari
    @FerrariMekhari 3 года назад

    A chalk board lil

  • @gert-janscharstuhl4446
    @gert-janscharstuhl4446 4 года назад

    hmmmmmm donuts

  • @dlbattle100
    @dlbattle100 8 лет назад +1

    Wake up camerman.

  • @kasparasvisockas4289
    @kasparasvisockas4289 6 лет назад

    This teacher is toochaotic, comes out as insecure of himself.