I plan on posting educational material here. As of now, there are videos on analysis, linear algebra, category theory, probability, differential equations, differential topology, and a LaTeX tutorial, for example. There are also some research talks. I hope to expand the list of topics in the near future.
Edit: Oh I have found the description and the notes answer my first question. Also, I think I figured out the answer for my second question. Because the critical points are the ones which have differential of rank less than m, the regular points are precisely the ones where the differential at these points does have rank m. I think in a sense the set of critical points is at most countable countable in general, so it has measure zero as you alluded before, or is that not true in general? Shouldn't the regular points of f @9:53, be R \ {(x,y) \in R^2 : x = y} since all those points map to 0? And @11:05, what does it mean for the rank of the function to be full? Doe that mean we require the differential of the function to have rank m? I do not see where we considered that in the definitions we made before the theorem or if we required that in the statement of the theorem at the end, could you please clarify?
Depending on what definition of a stochastic map you are familiar with, that is the subject of this video. Here, I took a stochastic map from X to Y to mean that it is a function that associates to each element x in X a probability measure, what I write as f_x, on Y. The value of this probability measure on the event {y} is written as f_{yx}. Depending on the notation used for stochastic matrices, the elements f_{yx} form a stochastic matrix, or a set of conditional probabilities y given x, which is an alternative definition of a stochastic map. The notion of "correspondence" I am referring to here is captured mathematically by the notion of a fully faithful functor. A fully faithful functor is one kind of generalization of the idea of a bijection between sets to the setting of morphisms in categories. In this example, positive unital maps between commutative finite-dimensional C*-algebras correspond to stochastic maps between finite sets under this fully faithful functor.
I can only provide a vague connection, since I am not too familiar with free probability theory. Categorical probability theory, as it currently stands, is primarily concerned with formulating classical probabilistic and statistical concepts in the language of category theory, as well as providing a more "synthetic" approach towards the results of classical statistics in a way that removes some of the technicalities normally observed when approaching the subject from a measure-theoretic point of view. Categorical probability theory is not concerned with non-commutative probability, and hence, it does not address free probability. This is because free probability theory is a subject inside non-commutative probability theory that is sort of midway between classical and fully non-commutative (the freeness condition is a restriction, as far as I am aware). That being said, there has been some work on extending categorical probability to include non-commutative probability, which should encompass both free probability theory and categorical probability together under the same framework (eg. my work on quantum Markov categories). However, this is a fairly new subject and still requires a lot of further development.
I take two explicit vector fields v_1=(y, -x, 0) and v_2=(-y,x,0) and calculate the index of each vector field at isolated singularities. These ones are exactly the first one in you lecture. I'm just thinking about another explicit tangent vector field to the sphere to illustrate that the Poincaré-Hopf theorem does not depend on (tangent) vector fields.
Hi, why the second vector field (x^2-y^2, 2xy) is tangent to the sphere? Thank you! (I'm seeing that this vector field is tangent to the sphere only at the points (0, 0, 1), (0, 0, -1) and the ones of the circle (x = 0, y^2+z^2 =1).
Thank you for your question. The vector field (x^2-y^2, 2xy) that I described is actually defined on R^2, rather than on the sphere. I meant to draw such a vector field on R^2 to illustrate how a vector field with index 2 at a single point would look like locally on the sphere, but I did not intend to write an explicit formula for such a vector field on the sphere. Writing down an explicit expression for such a vector field on the sphere sounds like a good exercise!
@@ArthurParzygnat by my calculations, a vector field v tangent to the sphere must be of the form (u(x,y,z), v(x,y,z), 0). If we want to consider the vector field (x^2-y^2, 2xy) on the sphere, so for it being tangent to the sphere, I don't know how, because the vector field (x^2-y^2, 2xy, 0) is tangent to the sphere only at two points (0, 0, 1), (0, 0, -1) and at the circle (x = 0, y^2+z^2 =1). I don't know if I had some mistakes with my computation.
Thank you for your question. I believe the answer is contained in the description to this video: "This is an intermediate step in trying to find a polynomial that interpolates several points, which itself is used for computing f(A) for a function f defined on the eigenvalues of the matrix A." Does this answer your question?
a good lecture but lacking some points that should be there because these are not obvious first for around 3:40 tv1 + (1-t)v2 = tv1 + v2 - tv2 = t(v1-v2) + v2. v1 - v2 is the line vector connecting v1 and v2. so v2 plus any combination of that will give the whole line of vectors passing through v1 and v2. and second for around 6:36 say we want affine span of those 3 vectors v1,v2,v3 then it will be,let's call lambda l for simplicity, l1v1 + l2v2 +l3v3 such that summation l =1. now to explain why the lines that connect v1,v2 or v2,v3 or v1,v3 are in the span,you can take for eg. l1=0 and l2 and l3,then sum of l2 and l3 is 1 for v2 and v3 and that will give the line which is in the affine span of our original v1,v2,v3. then also why the affine span of points on these lines,which will again be lines,is in our original affine span? because say you take the affine span of these points say x1,x2 and give them coefficients t and 1-t. then say x1= l1v1 + l2v2 +0v3 and x2 = 0v1 + l2v2 +l3v3 multiply x1 by t and x2 by 1-t and add them and you get (t l1 ) v1 + (t l2 + (1-t)l2) v2 + ((1-t) l3 ) v3 .now it can be verified that sum of these coefficients of v1 v2 and v3 is 1. just add them and use the fact that l1 +l2 =1 and l2+l3=1. this shows an affine combination of v1 v2 v3 exists that contains the affine combination of points on those lines.
Thanks for the comment. It's great to see your thinking process here. In fact, both perspectives are totally valid! It seems to me that your explanation seems more in line with your way of thinking about things. I think it's great to compare the two viewpoints to clarify in case other people have similar thoughts. As for the affine span of v1 and v2, the formula I wrote gives you the interpretation of "connecting the two points v1 and v2 by a straight line." In your version of the formula, which is t(v1-v2) + v2, the interpretation is "starting at v2, go forwards and backwards along the v1-v2 direction." I would not say that one is more obvious than the other, but would say that they are two ways of looking at the same thing. As for the comment about 3 vectors, my purpose here was to give an intuition, rather than a proof, so I am happy to see you thinking about this and supplying a proof. This is exactly what one should be doing when reading papers/books, watching lectures, or just learning!
In this video, I describe affine subspaces of R^n (or more generally vector spaces). An affine space can be defined more abstractly without viewing it as living inside of R^n (or a vector space). In this case, an affine space consists of points, rather than vectors. This is a subtle distinction! For example, imagine if you could draw an infinitely long straight line on an infinitely large paper. Then this line is an affine space. It consists only of points, not of vectors. But, if you draw a dot somewhere on the page, then your line is an affine subspace of R^2, where the dot is the zero vector. Now, every point on your line acquires a directionality with respect to the reference point you drew (the dot). Does this help clarify things for you? Or perhaps I misunderstood your question?
@@ArthurParzygnat Nice!! To recreate all R3 with affine combinarmos, do i need 2 planes like X on each other ? Otherwise can i say, i need 6 points not colinear ?
I realized that needs only four. Because the intersection of the 2 planes is a line and have 2 point in common to make this line. So need only 2 more not colinear to create 2 different planes
@@arthurlbn If you take affine combinations, then yes, you will get all of R^3. One way to see this is to pick a random point in the empty region away from these planes. Can you find a straight line through this point such that this line intersects two other points on these planes?
You are correct! If you check the descriptions of some of my videos, you will find comments on things I forgot to mention or if I made mistakes. In particular, this video mentions exactly these two identities.
Hey Arthur, Thank you for these videos that are absolutely stunning and very helpful. I was interested in the proof of this formulation of Bayes' theorem so I had a look at the sources you wrote in the description, but they all seem much less accessible than your videos (which shows one more time the great job you did), except for Culbertson and Sturtz's, but who refer to another complicated paper to prove their theorem 3.1, used to prove theorem 3.2 that appears to be analogous to this one. Maybe you could give me hand finding a prove as comprehensible as your approach? Thank you again for your work and I hope we can talk soon :)
Hi. Thank you for your interest! From what I recall, Culbertson and Sturtz prove the theorem based on the existence of disintegrations. On slide 20 (page 33 of the pdf) of arthurparzygnat.com/work/Category_Theory_2019_Edinburgh.pdf there is a sketch of a proof (which I believe is the essence of their argument---take this with a grain of salt because it's been a while since I thought about it!). I really need to emphasize that this depends on the existence of a disintegration, and I do not know how to prove such existence categorically. Based on Fritz' work, such an existence can be viewed as an axiom that one might want a Markov category to have (eg. the existence of conditionals). So eventually, one needs to prove something using analysis, but the point is to isolate those few key ideas categorically, and then prove subsequent results using only those categorical axioms. And likewise! Keep in touch, feel free to send an email just to say hello!
I honestly can't easily answer your question because it largely depends on your learning style. I often go through many books without being satisfied with one that covers a particular topic. I would look for a book that (a) covers (at least most of) the material and (b) is also compatible with your learning style. In this course, I used Spivak's "Calculus on Manifolds" and Kolmogorov and Fomin's "Introductory Real Analysis." However, I would instead recommend Munkres' "Analysis on Manifolds" and Bryant's "Metric spaces: iteration and application" because I ended up liking that combination more. I also added a bit of my own view on things and wrote rough lecture notes, which are available at arthurparzygnat.com/wp-content/uploads/2019/12/3151Spring2017Notes.pdf.
@@nammashraqi4029 Yes, it diverges from Spivak's in both content and presentation style. I used a total of four books: Spivak, Munkres (Analysis on Manifolds), Kolmogorov-Fomin, and Bryant (Metric Spaces: Iteration and Application). I also added my own personal style in terms of category theory. I also wrote lecture notes, which are available on my personal website under Courses Taught - Analysis II.
In wiki “vector field”, I saw these words: “it is equal to +1 around a source, and more generally equal to −1^k around a saddle that has k contracting dimensions. In paper “Umbilic points on Gaussian random surfaces”, I saw: the index of a monkey saddle (S shape) equals -0.5. However, by definition, I count the index of a monkey saddle as -2. What did I missed?
I found that, if I go through a small circle centered on the umbilic point of a monkey saddle gradient. The index count is -2. However, the vector field discussed in paper "Umbilic..." is not about gradient directions, but curvature directions. That is very different. To my understanding, gradient is about R2->R function, but curvature is about R2 manifold. I found a way to conclude -1/2 on curvature directions, but I am not sure yet, the paper is very abstract.
Another great video, and I think you ended at a great point with the discussion of states. The one thing that I felt was missing was around 7:30 or so, I would say that you can use the condition <v,Tv> ≥0 and a polarization-type trick to show that this implies that T is self-adjoint. (There's also an approach via eigenvectors and induction on the dimension iirc.)
Nice video! It's always difficult to motivate the basic definition and examples of C*-algebras but I think you did a really great job. A few small suggestions to make it easier for non-C*-algebraists: 0) In the case of functions on a finite set, and even in the more general case of continuous functions vanishing at infinity on a locally compact Hausdorff space, the supremum of any positive function is attained, so you can use the more familiar term "maximum." While your usage is certainly correct, I've found that non-experts can be confused by use of a technical term like supremum when a marginally less-technical term like "maximum" is around. 1) In your catalog of examples, I would mention that the algebra \mathbb{C}^X becomes, in the case when X is the model finite set {1,2,...n}, just the standard complex vector space \mathbb{C}^n, which is a bit more familiar. Under this identification, your basis functions e_x correspond with the standard basis vectors e_1,...,e_n, which are well-known. I would also do an example like computing the norm of (1+i, 3) or something. 2) In the same page, it might be worth mentioning that the norm of an n-by-n matrix (even a non-diagonalizable one) is simply the largest singular value, something that people may have more familiarity with. You could mention that there isn't a neat formula for the norm of an n-by-n matrix, as opposed to the simplicity of your C^X example. 3) You could also mention that C is just the algebra of 1-by-1 matrices, or the set of functions on a singleton set, and that, in the last example, setting n_j = 1 for all j gives us the second example back, so that the fourth example encompasses all the others. 4) A few times you mention that the norm, either on C^X or on M_n(C) can be defined in a few different ways. This is true, but it might be worth mentioning that only one C*-norm can ever be defined on an algebra. For example, the very familiar L^2 norm on C^n, the latter being treated as an algebra, is not a C*-norm, whereas the sup/max norm is (as you point out). 5) It could be worth mentioning the converse to your homomorphism theorems, namely that any conjugate of a diagonal map actually does define a unital *-homomorphism. You could also mention the fact no unital *-homomorphisms exist going either way between M_2(C) and M_3(C). I would also mention that a decent amount of work goes into proving these results, and that they aren't meant to be obvious to beginners. Great video! There is a dearth of quality, approachable C*-algebra material on RU-vid but this video is very well-done.
Yes! You didn't miss the point! But when you have an *infinite* number of rectangles, you can't just say the intersection is one of the R_{m}'s, because for each m, you can always go further out in the sequence if they're progressively getting smaller. Infinity can be very tricky sometimes...
@@ArthurParzygnat Thank you, I see the problem now! If we simply treat the intersection of nested retangles as the area of the most inner one, the infinite lead to a conclusion of empty area! Just like 1/2^n=0 when n->infinite. But this lecture shows a different conclusion.
@@ArthurParzygnat Is this proving course saying "an inifinte small close rectangle is a point"? Then, does a nested sequence of non-empty open rectangles has empty intersection?
@@zz1774 That's reasonable intuition, but imagine taking intervals of the form [1-1/n,2+1/n]. What is the intersection of these intervals as n varies over positive integers? Does the intersection have empty area?
@@zz1774 As for your question about open rectangles, consider my example again but this time use (1-1/n,2+1/n). What is the intersection of these open intervals?
Hi and thanks for the question. If I understood your question correctly, it is actually not sufficient. Take X={a,b,c}, Y={1,2}, and U=X. Take \pi_{X} to be the identity map and take \pi_{Y} to be the map sending a to 1, b to 2, and c to 2. Then both \pi_{X} and \pi_{Y} are surjective, but (U,\pi_{X},\pi_{Y}) is not the product of X and Y. To see that (U,\pi_{X},\pi_{Y}) doesn't satisfy the universal property of the product, consider V={*} a set with a single element so that any map from V to X and Y picks an element of X and Y, so we can call these maps x and y, respectively. For (U,\pi_{X},\pi_{Y}) to be a product, there must be a unique map from V to U such that the two "projection" maps agree. Sometimes, such a map does not exist. For example, if x=a and y=2, no such map exists---do you see why? One can also construct examples where many such maps exist, so that uniqueness fails. Is this what you were asking about?
@@ArthurParzygnat I see, neat example! I find out that this test is not easy, since what we use to test "U" must be a product itself! Thank you very much for your teaching!
Closed rectangles of the form [a, +\infty ) are not compact as they are not bounded. However, what is said here is still true, since you are only considering rectangles which have real number; rather than extended real delimiters? Therefore, I find it odd that you discuss boundedness and then don't use it as a condition within the Heine-Borel Theorem for R^n.
Thanks for the question. Indeed, I believe I defined "closed rectangle" in such a way so that it is bounded (see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-v4zkqqYtH2o.html). It's conceivable (and reasonable) that there exist people who use the same terminology to allow for unbounded rectangles and then they impose an additional compactness/boundedness assumption when necessary. Does this clarify your question?
Mr.Parzygnat, I’d like to hug you for this video; but I do not know where you are rn :) You have saved my research. I have been trying to do a practical proof for one of my theoretical stuff and I have been trying to understand that diffeomorphism psi(which is exp map in my case) and its relation to both associated flow curves and vector fields. Finally, I can go back to my pytorch code and get it over with for good hopefully :) Even though I cannot hug you physically; I have been planning to add you to my “Acknowledgement section” dedicated to only my youtubers who have been helping me a lot for the last six years of my math studies. It has been really tough for an engineer; yet it has turned out to be easy with the aid of people like you :) Thank you once again! One more moment of enlightment for me, stemming from one of my excellent youtuber Professors :) Hope to meet you on one day for a face-to-face gratitude expression
Can you please elaborate on your claim? I believe it's correct as written. Note that the factor of 1/2 comes from P^{-1} and not the multiplication P*f(D).
