Manifolds in Maryland
Manifolds in Maryland
  • 36
  • 51 794

Відео

Scott Wolpert's advice to young geometers
Переглядів 2674 місяці тому
For the full interview see: ua-cam.com/video/r5eJC0QFVgc/v-deo.html
Beneath the Surface - interview with Scott Wolpert
Переглядів 2954 місяці тому
In the first episode of Beneath the surface, we interview Scott Wolpert. Yanir Rubinstein chats with Scott about math, magic, music, and Riemann surfaces, hearing (or not hearing) the shape of the drum and hyperbolic geometry. Scott also talks about his hobbies. Prof. Wolpert's professional website: www.math.umd.edu/~swolpert/ Prof. Wolpert's UA-cam channel: www.youtube.com/@UC5_OKn5fpweanMsQZE...
Visualizing Conic Sections Using Blender and Desmos
Переглядів 470Рік тому
In this video we take a look at the ellipse, parabola, and hyperbola through the perspective of conic sections. I utilize the free open source 3D modeling and animation program Blender and the free graphing calculator Desmos. My beginner Blender tutorials: www.youtube.com/@howtuts6591/featured Blender: www.blender.org/ Desmos: www.desmos.com/calculator
Rotations in 3D Graphics With Quaternions
Переглядів 12 тис.Рік тому
In this video we will explore the advantages of using quaternions to calculate rotations in three dimensions. For examples we utilize the open source 3D modeling and animation software Blender as it has has several options for doing rotations. It can be downloaded free here: www.blender.org/ I do beginner Blender tutorials on my other UA-cam channel here: www.youtube.com/@howtuts6591/playlists ...
Geometry With Compass and Straightedge
Переглядів 270Рік тому
In this video we will do some geometry like the ancient Greeks using just a compass and straightedge. We cover the basic rules of using these tools and go over several types of constructions. Chapters: 0:00 Introduction 0:37 Tools Required 2:03 Rules of Compass and Straightedge 3:36 Equilateral Triangle 5:25 Bisecting a Line 6:06 Drawing a Parallel Line 6:45 Inscribing a Hexagon in a Circle 8:5...
Using Voronoi Diagrams In Computer Graphics
Переглядів 294Рік тому
In this video we explore Voronoi Diagrams and their applications in computer graphics for creating textures and randomizing geometry. This video makes use of Blender, a free open source 3D modeling and animation program: www.blender.org/ I also make beginner tutorials for Blender here www.youtube.com/@howtuts6591/featured
Normal Vectors and Their Applications in Computer Graphics
Переглядів 2,7 тис.Рік тому
This video discusses normal vectors, how they work, how to calculate them, and a variety of uses in computer graphics including baking normals on a plane to fool the lighting system into thinking there is more complicated geometry on the 3D model. For the demonstrations we utilize Blender, a free open source 3D modeling and animation program: www.blender.org/ I also have another UA-cam channel ...
Euler's formula and spherical geometry
Переглядів 369Рік тому
We present Euler's formula for convex polyhedrons and give a proof using spherical geometry.
The Bergman kernel of the polydisk and the ball
Переглядів 264Рік тому
I compute the Bergman kernel of the unit polydisk and the unit Euclidean ball. For my previous video on the Bergman kernel see ua-cam.com/video/loIC28LNgNM/v-deo.html
What is the Bergman kernel?
Переглядів 510Рік тому
I introduce the Bergman kernel of a domain and study its first properties. For more on this topic see Chapter 1.4 of Krantz's "Function theory of several complex variables."
The Cartan-Hadamard theorem
Переглядів 8992 роки тому
I give a proof of the Cartan-Hadamard theorem on non-positively curved complete Riemannian manifolds. For more details see Chapter 7 of do Carmo's "Riemannian geomety". If you find any typos or mistakes, please point them out in the comments.
The Hopf-Rinow Theorem
Переглядів 9412 роки тому
We present a proof of the Hopf-Rinow theorem. For more details see do Carmo's "Riemannian geometry" Chapter 7.
Lie derivatives of differential forms
Переглядів 6 тис.2 роки тому
Introduces the lie derivative, and its action on differential forms. This is applied to symplectic geometry, with proof that the lie derivative of the symplectic form along a Hamiltonian vector field is zero. This is really an application of the wonderfully named "Cartan's magic formula" The invariance of the symplectic form almost immediately implies Liouville's theorem, an important result in...
Poincare recurrence
Переглядів 6612 роки тому
I talk about Poincare recurrence, inspired by Louiville's theorem in classical mechanics. Previous videos in this series: Lie derivatives: ua-cam.com/video/8hWOHJ4Xka8/v-deo.html Louiville's theorem: ua-cam.com/video/YABAJS3FIEA/v-deo.html
Liouville's Theorem through Symplectic Geometry
Переглядів 6782 роки тому
Liouville's Theorem through Symplectic Geometry
What is the tangent bundle?
Переглядів 3,1 тис.2 роки тому
What is the tangent bundle?
Duality in Optimal Transport
Переглядів 1 тис.2 роки тому
Duality in Optimal Transport
Optimal Transport (According to Leonid Kantorovich)
Переглядів 1,5 тис.2 роки тому
Optimal Transport (According to Leonid Kantorovich)
Optimal Transport (according to Gaspard Monge)
Переглядів 9532 роки тому
Optimal Transport (according to Gaspard Monge)
Curvature of a surface, only using calculus
Переглядів 1 тис.2 роки тому
Curvature of a surface, only using calculus
What is a manifold?
Переглядів 8572 роки тому
What is a manifold?
Domains of holomorphy and Dolbeault cohomology
Переглядів 4332 роки тому
Domains of holomorphy and Dolbeault cohomology
The Cartan-Thullen theorem
Переглядів 4133 роки тому
The Cartan-Thullen theorem
What are domains of holomorphy?
Переглядів 6123 роки тому
What are domains of holomorphy?
Analytic continuation in higher dimensions
Переглядів 1,3 тис.3 роки тому
Analytic continuation in higher dimensions
The Cobordism Exhibition
Переглядів 2,1 тис.3 роки тому
The Cobordism Exhibition
The Nakano vanishing theorem for positive line bundles
Переглядів 2223 роки тому
The Nakano vanishing theorem for positive line bundles
What is the Chern connection?
Переглядів 7753 роки тому
What is the Chern connection?
An exactness theorem in Hilbert spaces: the Hormander technique
Переглядів 2653 роки тому
An exactness theorem in Hilbert spaces: the Hormander technique

КОМЕНТАРІ

  • @nitintomar1451
    @nitintomar1451 Місяць тому

    Can you please make a lecture on Levi flat part of boundary of a domain in C^n

  • @manfredbogner9799
    @manfredbogner9799 Місяць тому

    Sehr gut

  • @curiouskoala411
    @curiouskoala411 2 місяці тому

    the graphics are amazing. how did you make them?

  • @zdspider6778
    @zdspider6778 3 місяці тому

    Interesting presentation. Well done.

  • @Markty07
    @Markty07 3 місяці тому

    a vector can be a single number if you live in 1D

  • @Jaylooker
    @Jaylooker 4 місяці тому

    I think considering reflexivity in the equivalence relation of cobordant manifolds gives some justification for why in A^1-homotopy theory the affine line A^1 replaces the interval [0,1]. A cobordism theory with this equivalence relation using the affine line describes an algebraic cobordism. The algebraic cobordisms between complex manifolds are of most interest since polynomials are holomorphic (= infinitely complex differential). Note that solutions to sets of polynomials are algebraic varieties.

    • @manifoldsinmaryland
      @manifoldsinmaryland 4 місяці тому

      thanks for these valuable remarks!

    • @Jaylooker
      @Jaylooker 4 місяці тому

      @@manifoldsinmaryland Yup 👍. I think I found a better reference for why the affine line A^1 is used in the notes of “Motives I” by Eion Mackall. There near the end he mentions: “…the cohomology we work with satisfies the property H*(X x A^1) = H*(X). One way to see this is the calculation H^1(A^1) = 0 and then use the Künneth decomposition on the product. …we consider A^1 as a suitable substitute for the interval’s role in homotopy theory.” Thom’s theorem equates cobordism to a homotopy theory. Thom’s theorem derived that cobordisms groups are homotopy groups of (universal) Thom spectra MG by π_n(MG ⋀ X) = MG_n(X) with G a topological group such as the unitary group U. The associated cohomology theory of a spectra E is of interest considering the realization functor mentioned in 4 of Mackall. Whitehead showed that any homology is E -> π_(E ∧ X) = H_n(X; C(X)) for spectrum E on CW-complex X with coefficients C(X). Brown’s representation theorem or assuming Poincaré duality between homology and cohomology (ie Weil cohomology realized by motive of rational equivalence) it is possible to derive a cohomology theory from a spectrum E such as the Thom spectrum MG.

  • @amaama4140
    @amaama4140 4 місяці тому

    Splendid !

  • @teamoverc-hellletloose
    @teamoverc-hellletloose 4 місяці тому

    at 23:30, the star of the poincare duality and the use of Hodge star operator are used in the same place, kind of confusing.

    • @manifoldsinmaryland
      @manifoldsinmaryland 4 місяці тому

      true, I wish there was something I could do about it :)

    • @teamoverc-hellletloose
      @teamoverc-hellletloose 4 місяці тому

      @@manifoldsinmaryland Anyway, great structral explaination and thank you for the video! I also did a talk on the Hodge Theory on Monday :)

    • @ficsur86
      @ficsur86 4 місяці тому

      Glad you found our summary useful.

  • @nuwahasan8532
    @nuwahasan8532 5 місяців тому

    Omg thank you so so much 😭😭💖💖💖💖there's a vfx artist who said it would be a plus learn this topic, so here am I, and you explained it perfectly 😭💖💖💖💖

  • @harald4704
    @harald4704 5 місяців тому

    thank u for this awsome video I have been trying to cauculate the normal Vectors for my 3d terrain generator on the GPU and all the methods didnt work exept your method. Great Video!

  • @depressedguy9467
    @depressedguy9467 5 місяців тому

    Amazing video ,not just directely giving definition

  • @rafitiki
    @rafitiki 5 місяців тому

    complex analysis is dope

  • @RedDesigns
    @RedDesigns 5 місяців тому

    Cool!

  • @dislasolutions
    @dislasolutions 5 місяців тому

    Nice one brotha ☀️

  • @jacekwojcieszynski8368
    @jacekwojcieszynski8368 6 місяців тому

    Cool. I just try to recall my study subject. Thanks for sharing.

  • @khoavo5758
    @khoavo5758 6 місяців тому

    Useful info, thanks!

  • @adjoint_functor
    @adjoint_functor 7 місяців тому

    7:24 why is a closed segment cobordant to zero? would the cobordism not have to meet its endpoints with edges and thus create more boundary than just the segment, making it not a cobordism?

    • @adjoint_functor
      @adjoint_functor 7 місяців тому

      nevermind, it's because a closed segment isn't a manifold. which is stupid, but true nonetheless lol

    • @manifoldsinmaryland
      @manifoldsinmaryland 7 місяців тому

      True. Thanks for visiting the cobordism exhibition!

  • @yolotaylor993
    @yolotaylor993 8 місяців тому

    so intutive

  • @philipm3173
    @philipm3173 9 місяців тому

    Phenomenal!

  • @reimannx33
    @reimannx33 11 місяців тому

    Hodge is long gone, and turning over in his grave because of the hodge-podge this old fart has created out of his ideas.

  • @alexboche1349
    @alexboche1349 11 місяців тому

    What background do I need to understand this? Learning differenential forms now. I know very basic stuff about manifolds e.g. tangent space/bundle. I don't know differential geometry.

    • @manifoldsinmaryland
      @manifoldsinmaryland 10 місяців тому

      That should be enough to get started. A good book that is suitable for your background is Wells' Differential Analysis on complex manifolds

    • @alexboche1349
      @alexboche1349 10 місяців тому

      @@manifoldsinmaryland I didn't know if you would answer so I didn't give that much info. Let me give a little more. I want to understand the generalization of Helmholtz decomposition to n dimensions. Wikipedia says that's called Hodge theory. I don't think I need too much about *complex* manifolds specifically but maybe that helps. I'm from economics / game theory not physics/math. Does your book recommendation still stand? OR is there a better book in that case? Thanks!

    • @manifoldsinmaryland
      @manifoldsinmaryland 10 місяців тому

      @@alexboche1349 It does not stand anymore. Unfortuantely, I know little to nothing about Hemholtz equations :(

  • @CameronHendry-w2f
    @CameronHendry-w2f 11 місяців тому

    Yo who else is watching this video because of our CALC 3 class?

  • @columbus8myhw
    @columbus8myhw Рік тому

    Desmos 3D got released a few days ago!

  • @jonathanbottazzi8142
    @jonathanbottazzi8142 Рік тому

    thank's very usefull and informative insight ! is there any reason good not to use quaternions then ?

    • @ishino_ki
      @ishino_ki 8 місяців тому

      too hard to learn ;p

    • @romanavr
      @romanavr 5 місяців тому

      one reason is that geometric algebra is considered even better way to rotate (but also somewhat harder to grasp), so why no learn it instead haha

  • @KabeeshS
    @KabeeshS Рік тому

    Wow great video, it's just that it was too advanced for me to understand all those mathematics for my simple brain. Can we get a video with simpler explanations with examples to visualise what's being said?

  • @VijayAVk
    @VijayAVk Рік тому

    Hey there, that an interesting explanation, I would love to see more of utilizing blender and general physics to understand complex topics visually❤❤❤❤❤❤

  • @andresl2773
    @andresl2773 Рік тому

    nice video!

  • @ryanjbuchanan
    @ryanjbuchanan Рік тому

    Crystal clear explanation, thanks doc.

  • @GrapefruitGecko
    @GrapefruitGecko Рік тому

    Beautiful aesthetic to this video! I especially loved seeing the tile floor!

  • @mystic7132
    @mystic7132 Рік тому

    Wow, this is legit amazing work. More people need to see this, it is very well made 👏👏

  • @dylanparker130
    @dylanparker130 Рік тому

    Really cool stuff. I wonder if this could help with projections of surfaces (e.g. a tube or torus) onto a 2D plane - help to distinguish between the points which are closer and points which are further from the viewer?

  • @rainbow-cl4rk
    @rainbow-cl4rk Рік тому

    Nice proof

  • @fabiancaro5311
    @fabiancaro5311 Рік тому

    ey good video, a have a quastion, which is that sofware??

  • @valentinodrachuk5692
    @valentinodrachuk5692 Рік тому

    Good video, subscribed

  • @baruchspinoza4979
    @baruchspinoza4979 Рік тому

    Wow! Excellent! Thank you.

  • @akrishna1729
    @akrishna1729 Рік тому

    These are wonderful videos. Thank you so much for them!

  • @rohingarg7236
    @rohingarg7236 Рік тому

    Extremely helpful!

  • @abdallahchaibedra5363
    @abdallahchaibedra5363 Рік тому

    Nice video very clear explanation, can you make a video on riemann metric and curvature on manifold

  • @coreconceptclasses7494
    @coreconceptclasses7494 Рік тому

    I am looking for that thanks keep it up.

  • @debraj10
    @debraj10 2 роки тому

    At 3:08 what is $X$? Is it $\{(z_1,\dots, z_n): (z_1,\dots, z_{n-1},0)\in L\cap \Omega, z_n\in \mathbb{C}\}$? We probably need to be more careful about the cutoff then.

    • @manifoldsinmaryland
      @manifoldsinmaryland 2 роки тому

      Thanks for noticing. X should be simply Omega here. Indeed, I should have said more about the construction of rho, but I am never really sure how much details is OK in math UA-cam videos.

    • @debraj10
      @debraj10 2 роки тому

      @@manifoldsinmaryland How can $X$ be $\Omega$ if the section $L\cap \Omega$ is small ? e.g. take $\Omega$ to be a ball in $\mathbb{C}^2$ and let $L=\{z_n =0\}$ be a complex line that does not pass through the center of $\Omega$ but intersects $\Omega$ in a nonempty open set. Then the formula $g(z_1,\dots, z_n)= f(z_1,\dots,z_{n-1})$ does not define $g$ on all of $\Omega$, but on a proper subset.

    • @manifoldsinmaryland
      @manifoldsinmaryland 2 роки тому

      @@debraj10 not sure what you mean. I say specifically that g is defined on a proper open subset of Omega. Then I construct the global extension using a cutoff of L cap Omega.

  • @manifoldsinmaryland
    @manifoldsinmaryland 2 роки тому

    It should be (Graph of - T)^perp around 7:40

  • @aratrikapandey7693
    @aratrikapandey7693 2 роки тому

    Where we are using exp_p is defined for all of T_pM when we are proving there exists a distance minimizing geodesics

  • @SM321_
    @SM321_ 2 роки тому

    This is wonderful

  • @cristhiangalindo4800
    @cristhiangalindo4800 2 роки тому

    Hi how are you?. I would like to know, what difference can exist between a compact Kahller manifold in X, with the HyperKahlerian manifolds, one of the cases that I consider is that for HyperKahlerian metrics X can be compact in a $\textbf{P}^{5}$-hyperplane special of the Theorem of Lefschetz-Noether, , But in an article of mine (if you want I'll pass it on to you), I was able to prove that in HyperKahlerian metrics, X is not only "compact" but also linear in $X\to{} g_{ij, k}$ where for example $X_{1}\xrightarrow{\ \sim\ }g(1) (See X1 as a Linear-Isomorphism on a generic-Kahller Einstein in g(1)) So a compact X becomes in $X+ g_{ij, k} Where, this linear-isomorphism of g(1) is a HyperKahlerian metric, which is in d= g_{ij, k}> g_{i, k} -g_{j} where it differs, only by a semi-stability, to the Kahller-Hermitic metric, In this case the compact of X is a special hyperplane, which depends on any d-degree with singularity on a curve C(g(1)) (see you also as if $X\subset{} \{\mathcal{M}_{g, d}= g_{ij, k}$) Where, well, the Noether-Lefschertz theorem of $\textbf{P}^{5}$ is generalized, for any $d\geq{3}$, with D-degree in the singularity of a curve, for $d>3+ d<3,$ in this case is true for example that S5, S8, S(7,8) and Sg (6,9), can produce forms of that hyperplane of NL, and are identical to a HyperKahlerian metric on X, In this case we can write to X(g(5, (7,8))$ where X_{1} as the structure compact of an isomorphism in the Metric Hermitic-Variety Kahller for example in d<0= k[I, J]- (I, J) (see the square bracket of a quaternion, for Lie-algebras), The idea is in $X_{1}$ is true an theorem of Brill-Noether? if only considered an idea for fixed-curve invariant-Donaldson and Thomas?, then, is a DT-invariant for an irreducible B-model also a hyper-Kahllerian metric, too? and I got this question too, and if it is true because a DT-invariant is equal to 0, in A-reducible models?. This is part of a general model of how these very high-dimensional manifolds can generally be differentiated.

  • @laurenpinschannels
    @laurenpinschannels 2 роки тому

    ooh found this via the recommender

  • @Czeckie
    @Czeckie 2 роки тому

    this is really cool, thanks. Any chance you would cover Chern-Moser theorem on normal forms of hypersurfaces?

    • @manifoldsinmaryland
      @manifoldsinmaryland 2 роки тому

      thanks for the suggestion!

    • @kilogods
      @kilogods Рік тому

      @@manifoldsinmarylandI agree with Czeckie.a vid of Chern Moser normal forms would be fantastic!

  • @arkapointer
    @arkapointer 2 роки тому

    Hi Firstly thanks for such amazing videos Can you please make a video on Kahler identities? There's this operator called lambda which is adjoint of the operator L ( L is the operator that wedges with the Kahler form). Can you make videos on that please?

  • @biksoonsia7874
    @biksoonsia7874 2 роки тому

    Very Nice Summary. Please make more video to explain maths

  • @SilboMovies
    @SilboMovies 2 роки тому

    What an incredible video! I was reading about cobordisms in the context of TQFT and found this gem. I wish I could visit the Cobordism Exhibition in real life...

  • @cristhiangalindo4800
    @cristhiangalindo4800 2 роки тому

    Good can you make a talk, about the Hodge conjecture, I am working the CH for example for a dimension p-addic "convergences-thus falls" where the CH is true only under a 2-cycle Hodge, because I found that p- adic ($P= N_{|-1|}$ as an isomorfism of the algebraic in $N\subset{} K$) where the $Map_{K}=1$ an also this contains to $Map_{K}: \sum_{j}^{\infty= - i}$ where the P-adic case on the dimensión $q(2n)> p$ it is well a complement with the subvariety (n-1) where the contains "symmetric" is localy finite in $\delta\mathfrak{H} \to (\delta, - \delta)$ (which takes for a finite thing, its value in the regular Homotopia), Where if the Homotopy coincides with Mod p (1), then a discrete set of normal walls is produce.

    • @manifoldsinmaryland
      @manifoldsinmaryland 2 роки тому

      Thanks for the comment. Unfortunately I know very little about algebraic geometry over the p-adics.

    • @cristhiangalindo4800
      @cristhiangalindo4800 2 роки тому

      @@manifoldsinmaryland Yes, of course, but knowing things about what a 2-cycle is, for example from Hodge, your P-adic space can be an application of how this is continuously pasted into a complex N_ {| -1 |} submanifold.

    • @cristhiangalindo4800
      @cristhiangalindo4800 2 роки тому

      @@manifoldsinmaryland For example, you can consider the space P-adic (which in general is $Q^{*}(H)$-adic or also as $Q^{*+n}$-adic of cycle ) On some standard volume (group SU3 in the hyperplane convex) , which corresponds to a Hodge-structure, I for example have been studying the P-adic for Hodge- 2-dimensional structures, like $ P: Hdg_ {2} here you see that also P \ xrightarrow {\ \ cong \} Hdg_ {2}, where well any P-adic volume of a Hodge-structure is identical to an isomorphism -linear, Thanks to this dear colleague, it has been possible to generalize the P-adic volume (see it as a Module-potent oin SU_{Z3} =Mod-p^{2} (x;i)) where all the Q(H,n)-cycles of structure linear are compatible with the linear part of some P-adic volume, from here for example the Hodge-conjecture arises but in this case for example the P-adic is wrapped in P = | -1 |\ in {} N (see N as the submanifolds of N_{g^{p,q}}$) where the P-adic corresponds to a Hodge structure of class q (2n)> p (case of 2 Hodge cycles), where well the volume of all Hodge structures is generated in a semi-stable abelian manifold, see how P (adc) = N_ {| -1 |} \ to \ log_ {q} (where \ log_ {q} = \ log_ {, (2n) - xi}) which is where all logarithmic abelian manifolds admit some volume "in q" of the structures of 2 -Hodge cycle, for example, you can write to q: = P * = P (adic). Currently, I am working on a paiper, (which if you want I can send you), with the CH (conjecture-Hodge), for P-adic which are associated functions of P_ {l <0} (where l is some cohomology group in Hdg_ {2n}), my approach is that for the spaces $ P_ {l <0} - $ adic the preserved volumes (see it as an indicator of symmetry) are true for a subgroup of functions l g3 = 1, or if the P_ {l <0} \ to g3 = Mod 1 where well all the projections of a subgroup in 3 are relevant, in the volume P_ {l <0}, this case for example leads us to think of the previous idea as the "extension "of a semi-stable manifold $ \ log_ {q} $ must be identical to $ q (2n)> p $ geometrically representing the P-adic in this case, only as a volume of symmetries locally. This as the volume of symmetries locally is this maximum smooth subgroup g3 in a local finite Map_ {K} = (n-1). That is why I tell you that the P-adic space of a dense volume of symmetries is of intense understanding within the CH, and Hodge's own theory. So my general version of the, CH is the following consider whether $g3 \{ Mod |p, +1|\forall{} N_{|-1|}\} or g3= Mod|p, +1|\times{} \mathcal{N}_{|-1|} where see you that, the $\mathcal{N}_{|-1|}$ of an structure of submanifolds in $P_{*}$ (see you only as $\mathcal{N}\subset{} P_{*}$) is the standard case of, admitting P-adic in some N submanifold, Even since the P-addiction is of a soft or projective class, aEven since the P-addiction is of a soft or projective class, is traced a $P(adic)= P_{('(2n)+1)} but if $P_{*}\in{} N_{|-1|} then you written to $P(adic)= P_{(, (2n)-1)}$ or also written as $P(adic)= P_{l<0}$ where l are all dense and maximal correspondence functions, identical with modulo in a submanifold n-1, Proving that the CH is only true if g3 = Mod | p, +1 |. but not in 3 nor its present dimensions. I await your answers, if you want me to send you my paiper. greetings friend

    • @98danielray
      @98danielray 10 місяців тому

      100% a crank