Very good explanation of a very difficult topic. One item: unless I missed it there is never a single screen that says: A fiber bundle is: and then listing the characteristics of what makes for a fiber bundle. Towards the end of the lecture you list the properties but other than hearing you say fiber bundle in describing the elements that are needed it's not on the screen. I think it would be nice to have a summary screen that in fact says something like: Fiber Bundles consist of: then listing/naming/numbering the elements in a column with a brief explanation of what the element means. It would be a nice thing to go back to later to kick in the concepts.
By far the best explanation of fiber bundles anywhere including You Tube, math and physics textbooks or lecture notes. I like how you draw where the maps go from one space to another, especially the inverse images. I am doing self study on differentiable manifolds and struggled with fiber bundles. Your video helped me tremendously. Can you do videos on Principle G Bundles?
I came here after trying to read about 10 books and lecture notes with a hope of understanding those objects, in vain. I really wonder why the heck all math & physics books tries to write in such an abstract way that strips off all of the intuition and the deepens of the original thinkers that invented those definitions, it is very stupid on all those professors and teachers that I feel sometimes that they have no clue themselves what they talking about. It is like if they will be shamed if they explain the origin of the idea and how to visualize it, really illogical. Thank you very much for a such a clear explanation, and I really appreciate your efforts to cover every question that may popup in my mind during the lecture, like when you explaining the difference between entire space and trivialization, I am very thankful, great work.
Good point! The funny part is that the definition of a fiber bundle does not have that many parts. By the end of this video a quick list would have been easy to make. A manifold B called the "base space", a manifold E called the "fiber bundle", a standard fiber F, and a projection map pi: E-->B and that B is covered by n-hoods U_i such that pi^(-1)(U_i) is homeomorphic to U x F. Also, the projection onto U executed after the homeomorphism must yield the same U_i as the n-hood.
Excellent lectures! I have a silly question: a point P at the Base is related to a fiber F. BUT at 54:24 a point P that belongs to π^(-1) {Uij} is mapped to 2 different points in the same fiber F?. A point on the cylinder would not have the same point on the fiber? Thank you for the lectures!!
I suppose that's the reason why he mentioned the structure group has only one element {1} for the cylinder but has 2 elements {1,-1} for the Mobius strip. (I haven't checked the next lecture though). The implication of having only one element {1} for the structure group is the transition function of F is simply an identity, namely the two points are essentially the exact same point.
At 31:48 There is a P in the base; Point P is mapped only to the blue. A region around P -- ui -- is mapped to inside the red dotted line in E. So, far, so good. But inside that ui, there are other points: P1, P2, P3 (an infinity of them) EACH of those points will have a BLUE region of its own in E. Will ALL of those blue regions be entirely contained in the red? If so, it is getting MUCH clearer.
Thomas Impelluso Yes! What you said is correct. And furthermore, each of the regions \pi^{-1}(p_i) will be disjoint. Each such region is "the fiber over p_i" and each such region is homeomorphic to F.
It is quite interesting, that two bundle charts with different E -> F maps, map intersection region \pi^-1(u_ij) on E to the same intersection region on B, u_ij. But the difference can be seen only \Psi and \Phi map the point on E to the generic fiber F.
I have not examined that book. Did you try Nash and Sen "Topology and Geometry for Physicists"? It is a decent introduction and smooth read. Missing some things, but a nice place to start.
@@XylyXylyX Nash and Sen's book is frequently referred in Nakahara book. I only have a quick look, which is friendly to start. Nakahara book is more used among physicists, hopefully I do not need so much formal details since I do not work on high energy physics, but quantum information.
Thank you so much! This really helps me in my self-studies on Loring Tu's introduction to manifolds. I just wonder - what would "standard fiber" F look like in general? Is this some natural thing, or an extra condition I should impose on Tangent bundle? In Tu's book, I think F is presented as \mathbb{R}^r which I think can be thought of as a chart on E.. So I was a bit confused when you said the bundle map is totally different from manifold structure. (Maybe I misunderstood)
Each fiber must be "the same" in a mathematical sense. The only way two fibers can be the same is if they are homeomorphic. So all fiber on the manifold must be homeomorphic with each other, and therefore they are all homeomorphic to on particular topological space we can call "F".
Try this: every point in the base space has a pre-image. That pre-image is the “fiber” associated with that point in the base space. Every point in the base space has a fiber. Now, every fiber must be “the same”. What we formally mean by “the same” is that every fiber must be diffeomorphic to F. And F is the “standard fiber”.
At 3:37, is an angular coordinate? Then S_1 is a set of numbers from 0 to 2*/pi? Then can I define the biggest manifold as 2d cylinder as the radius is fixed?
I chose [0, 2\pi) as the coordinate interval strongly suggesting an angular coordinate, but that was an arbitrary choice. The radius is irrelevant, which I think you are pointing out.
Tursinbay Oteev In order to be a fiber bundle you must define *some* base manifold and that manifold will have a coordinate system by the definition of a “manifold”. So if I understand your quesiton correctly, the answer is “no.”
Sorry, I am giving many questions. But this is a tough topic. At 26:30, by "from there" do you mean {f^-1(u_i) x F}? If yes, is it because F is diffeomorphic to itself, and f and f^-1 are continuous? At 30:45, does "gets mapped" mean projected?
At the second time stamp: no. But i note that I did say it incorrectly! I should have said: \phi is a mapping from the inverse of a single point ( “p” in this case) to the generic fiber F. So: \phi: \pi^{-1} (p) -> F. This is the mapping that tells us that each fiber over each point in the base space is diffeomorphic to a certain generic fiber F. Instead i think I said that \phi was a map from the inverse projection of U_i, and that is not correct.
Hi, in 32:41, you have two cartesian product, ui X F and uj X F, how do you know they are the same F? Or they are different F associated with each i and j, i.i. Fi, Fj?
F is the "standard fiber" which means all fibers are diffeomorphic to F. So the answer to your question is that we know the *by definition* of what a fiber bundle is. If every fiber is not diffeomorphic to a standard fiber F then you don't actually have a fiber bundle!
I see. In there earlier part in the example of example of the cylinder. You specifically distinguish the Fiber (the real line) at point p and at point q, so that's only for illustrative purpose? Eventually there is only 1 fiber which is the real line?
Not it is not only for illustrative purposes. Each fiber is distinct! However they are all *copies* of F and F is called the "standard fiber". For the cylinder the standard fiber is the real line. the inverse of the projection map acting on a point of the base space must always yield a set that is homeomorphic to the standard fiber.
Thanks for the replies. Ok now i understand. The fiber F is a collection of distinctive fibers and these fibers are diffeomorphic to the 'standard fiber', so they are diffeomorphic to each other too. 1 more question: How do we know these fibers are forming a smooth manifold? The map from F->F by the 'fiber transaction'( psi, phi in 59:33) is smooth but how do they cooperate with the idea of open sets?
Ka ho Wong If I remember correctly, in one of the upcoming lectures I discuss the topology of manifolds. In short, every fiber bundle can be given a certain topology which is created in order to make the projection map \pi continuous. From that all else will follow.
Hi Xylyxylyxy, In 25:08, you said the inverse map pi of the base space S1 is those strips, Are we considering the inverse map of a point on S1, or an interval on S1?
It is fantastic series ! , I would like to apply differential geometry and topological processing to medical imaging domain (I am physycian) would you recommend a source with algorithms for automatic learning of the properties like path connectedness curvature of a mesh (or any other data structure derived from point cloud) ...
First of all thank you for the great explanation ! Could you help me a bit understand the main difference between the section and the inverse mapping ? They both map from the base space to the total space and the projection of both takes us back to the base point. I understand that a section is a continuous function and can be defined locally in a neighborhood or (in rare cases) globally, but I still would benefit from an intuition on what the section can do that the inverse map cannot. Thanks in advance ! :)
Hmm….the inverse mapping from a point in the base space is an entire copy of the fiber. The section map from the base space is a *single point* in a fiber. I think that answer addresses what you are asking?
There seems to be a problem for me at around 31:00 . I think that you have defined (π,ψ) globally on the whole total space when you say that ψ is the same diffeomorphism for every preimage of a basepoint. So maybe the definition should be something like ψ_i : π^{-1} -----> u_i x F. I am guessing it's a notation issue.... I am not sure, though, of my claims, so any answer is going to be appreciated. Either way thank you for your very helpful series of videos.
Well, I wondered about this also at first time he wrote that, however, latter on, he says that it is just a (clumsy) notation: since we have "i" in (pi_i,xi), they usually do not put "i" for "xi" part, so it is not globally defined as may it seem. So do not worry, it is just a usual abuse of notion, that I really hate toooooo much.
You are correct. PI is the only global mapping over E. I am using \phi and \psi to name the maps associate with U_i Dan U_j. THe capital \Psi_i notation is what you are looking for,
This is the best and clearest introduction to Fiber Bundles I've ever found!!! Thank you so much. Also a curiosity, if you wish to answer it: which is the software you use to produce these educational videos? More to the point, thanks again for your wonderful channels, as many others, I'm really grateful to you!!!
In your response to my last two posts, you shined a light: I now see that the diffeomorphism between the fibers and the real line is a CRITICAL observation. I see when you are drawing the triangle, that you have the generic fiber in mind. Your video here, at about 7 minutes does explain this but I overlooked its significance. Now I see. Thank you. So may I ask a question to nail this shut? The use of the word TRIVIAL at 16:49. I am compelled to ask: what is NON-trivial? It seems to me that you gloss over the significance of this word. Why did you not call it... oh, I don't know... "pumpkin?" I am serious: words have purpose. IF you are going to use a word like TRIVIAL, you must distinguish it from NON-trivial (and this has not yet been done for categorical purposes, so could you?) Are you suggesting that the "triviality" here is that the fiber consists of only real numbers? For example, the base could be the coordinate patch, and the fibers could be the covectors. And, in that case, it is not so EASY to identify the covector with a line, the way you can identify the point P, with a Real numbered fiber. Is THAT what you mean by TRIVIAL(=EASY) OR.... Is the nature of the word TRIVIAL to be found at 26:38. (Which is also something clearer to me due to your previous response). Namely, that the TRIVIALITY consists of the EASE of "UNWRAPPING" E into a product space: Ui x F? Namely (and ultimately), the ease of extracating "coordinate information" and "tangent space information", from a "bundle of information."
Thomas Impelluso The word "trivial" is intended to mean "has the structure of a Cartesian product." A manifold is "trivial" it is can be reduced to a *single* bundle map and this is sometimes called "globally trivial". All manifolds, by definition, are "locally trivial" and can be covered with several bundle charts. If the structure group which connects the bundle charts is the Identity group {1}, then the manifold is actually globally trivial.
At 33 minutes, I see only ONE triangle, but TWO small rectangles. Is this because the maps to all possible fibers are diffeomorphic to Rn? Also, I am guessing right now that the Base is the coordinates. And the fiber (once endowed with a specific chart) becomes the tangent vector for a field? Is this correct?
Each rectangle corresponds to a different bundle chart: from a different U_i. The triangle represents the "typical fiber" and pi^inverse(p) is a region of E that is always homeomorphic to the "typical fiber" F. The two arrows from E to F should both originate in the blue shaded region (that is, phi should be a bit longer and reach the shaded region like psi does). Each may have a different homeomorphism because psi and phi are associated with different bundle charts.
As a math hobbyist I kinda wish there were a few specific examples. This stuff is sort of "what we would like to have". I know the essence of math is abstraction but I like to see cogs turning cogs and stuff working. Also, when was this manifold theory developed? I now Riemann started it in the mid 19th century and Whitney did some stuff in the 20th.
Check out the next lecture, it is dedicated to a classic example of the tangent space to a circle. I plan two more: I just started the Möbius strip example, and I will also do the Hopf fibration if I have time.
I can now finally poinpoint my confusion (the cause of all my questions) At 28:00 On the left, the map to the triangle is from the BLUE region. That means it is from ONE fiber (The fiber originating at P on the base, B) Logically, then, the triangle represents all the points on that one fiber. That implies the map (which you will later call PHI) must be a map to a specfic point on that one fiber. So I assume that means that PHI maps to a point on the fiber, and the triangle represents all the points in ONE fiber: the fiber of the blue dotted region. Yet... On the right, the trivial map takes you from the RED dotted-square. And since that RED square originates from a region surrounding a point in the Base , then it must include other points, and, thus, include many fibers. So the RED region is many fibers. That implies that the map to the trivial product (which you will call (pi, phi) ) is NOT to a specific point on a fiber but to a family of fibers for the different points in the RED region So it seems that the F in the triangle is DIFFERENT from the F in the trivial product. The triangle's F represents ONE fiber and a point on that fiber. The trivial product's F represents MANY fibers. I am confused. Further, at 36:11, you draw (a,b) in green next to the trivial product (Uj x F) at the top. But this cannot be. Because that map comes from the green dotted which necessarily includes many ponts, so there is no way you can identify a (b) on one fiber. I think it comes down to: when are you using F to designated ONE fiber, or a point on ONE fiber, or, perhaps both a sepcific fiber and a point on it. At the conclusion of this video (60 minutes or so), it seems that the triangle represents BOTH a fiber and a point on the fiber.
Thomas Impelluso You are definitely approaching this the correct way: identifying all the maps and struggling to understand the domain and range of everything. Let's start with psi and phi. The map pi^(-1)(p) is the "fiber" over the point p. It is a set of points because the mapping pi is one-to-many. The "fiber" of p lives in E and is the blue region. Now you must understand that in order for this structure to be a true fiber bundle, every fiber over every point must be homeomorphic to a "standard fiber" which we call "F". That is what the triangle represents: the "standard fiber." The map \psi is the homeomorphism from the fiber over p to the standard fiber F in the chart U_i. Similarly for the map \phi for the chart U_j. Since \psi is a homeomorphism it is one-to-one and onto and invertible. Regarding 36:11... any set which can be expressed as A x B is, by definition of the Cartesian product, as set of ordered pairs (a,b). In this case "a" is an element of U_j and "b" is an element of F, and (a,b) is an element of U_j x F. The "bundle" chart is the homeomorphism that takes regions of E and "flattens them out" so to speak to look like U_j x F.
You wrote: "Now you must understand that in order for this structure to be a true fiber bundle, every fiber over every point must be homeomorphic to a "standard fiber" which we call "F"." OK, so that means that the map "psi" or "phi" is a map to a POINT on a generic fiber (which is diffeomorphic to the standard Rn fiber)
Around the 11 minute mark, you said you've defined the product topology. However I'm fairly certain you have defined the box topology, not the product topology. There are many cases where these are the same, however there are also many cases where they are not. Please correct me if I am wrong about that. I think it might also be important to note that the projection mapping \pi need not be invertible, so talking about \pi inverse as a map rather than just considering specific pre-images could be misleading. With the example of the cylinder, the projection mapping is many-to-one, and hence is certainly not invertible.
rinwhr The way I have defined Phi gives Phi_Inverse: U x F --> pi_inverse(U), so yes, I think you are correct. Some reference use other directions for the basic mappings, so it can get confusing.
Also i have a question. When we say a function is diffeomorphic doesn't it have to be from Rd ---> Rd to be differentiable? how the bundle chart is differentiable?
Diffeomorphism applies to two differentiable manifolds. The language, as I use it, goes like this: Two differentiable manifolds are "diffeomorphic" if we can find a function from one to the other that is a "diffeomorphism". A function between two manifolds is a "diffeomorphism" if it is 1-1, onto, bicontinuous, and "differentiable." A function between two differentiable manifolds is "differentiable" if the chart representative of the function (\phi o f o \gamma^{-1}) is differentiable in the sense of ordinary calculus. (Here \gamma is the chart map of one manifold and \phi is the chart map of the other manifold. The fiber bundle must be a differentiable manifold as a whole. That is, the total space is a differentiable manifold. Does that answer your question?
+XylyXylyX I used to think (phi o f o gamma^-1) is differentiable because it's so in the calculus way but now i understand if the latter is true then f alone is differentiable and so it is homeomorphism (if also 1 to 1 and onto). following this logic we can announce (pi,phi) to be diffeomorphism and the 2 manifolds of the total space E and the cartesian product Ui x F are diffeomorphic. is that correct?
+John Ron .....but only for a *region* of the total space corresponding to a *region* of the base space! That is very important. E is usually not globally trivial!
I get very confused at 37:06. Your language becomes a bit vague and the cursor is swooping around. You say "I can do this projection process with pi by eliminating the fiber part" But you have not defined "eliminating" and the cursor is moving around. The you use the cursor to point to BOTH projectios pi and pi inverse THen you say "by adding the fibers back" while your cursor is swooping over pi-inverse. It is a little fast, jerky and vague... but it is important. Could you elaborate?
Thomas Impelluso THe projection map is easy to execute if you are working in U_j x F because the projection map pi simply involves picking out the point in U_i and dropping the part in F. That is, the map pi o (pi,phi) has a domain of U_i x F and if (p, f) is a point in U_i x F then the map pi o (pi , phi) acting on the point (p,f) simply returns p.
I'm finally catching up with u.It's all clear but i'm a bit confused with the names (the trivialisation thing and its derivatives). Can't wait for the next lecture but in the meantime i may find some related exercises. Do u recommend any decent source, series, videos...whatever?
There are so many! Some are more mathematical than others. I will be doing some examples and will work on an example lecture today. Let me think about it. I want my lectures to simply be a way to help you through ANY other book or online video lecture.
You explained this very well. But it still caused me some confusion that you named an element of the base set B with lowercase a and not b. Fortunately you correct this later. Another thing I noticed is that it kind of gets lost why this is called a fibre bundle. I can see the idea of a fibre, but where is the bundle? It also threw off my comprehension when you draw a fibre as a triangle at some point. Then you draw an actual fibre bundle (lines on a cylinder with the S1 circle bundling them) and immediately wipe it out. There went my visual clue. Later it becomes less clear what a fibre is since you never draw any. Instead a fibre is supposed to be something in that triangle labelled F. Never mind. I want to learn this stuff and this was helpful that you tried very well to explain this, while I’m still somewhat confused.
@@GamingBlake2002 They know him because first year students ask about him! I find incredible that he can get around throwing random words around and so many on the web get fascinated: "wow he must be a genius, so many words I don't understand"! no dummy! he is just BSing you!
@@StephenCrowley-dx1ej no there are many coordinate systems we could build on any fiber bundle because ultimately they are manifolds which allow for arbitrary coordinate.
I just found this channel. I'm overjoyed at the thought of the understandings my future self is going to obtain. Thanks for your hard work.
Very good explanation of a very difficult topic. One item: unless I missed it there is never a single screen that says: A fiber bundle is: and then listing the characteristics of what makes for a fiber bundle. Towards the end of the lecture you list the properties but other than hearing you say fiber bundle in describing the elements that are needed it's not on the screen. I think it would be nice to have a summary screen that in fact says something like: Fiber Bundles consist of: then listing/naming/numbering the elements in a column with a brief explanation of what the element means. It would be a nice thing to go back to later to kick in the concepts.
By far the best explanation of fiber bundles anywhere including You Tube, math and physics textbooks or lecture notes. I like how you draw where the maps go from one space to another, especially the inverse images. I am doing self study on differentiable manifolds and struggled with fiber bundles. Your video helped me tremendously. Can you do videos on Principle G Bundles?
Wonderful explanation dude, I feel like i just found a goldmine :D ty
Thank you for your kind comment.
I came here after trying to read about 10 books and lecture notes with a hope of understanding those objects, in vain. I really wonder why the heck all math & physics books tries to write in such an abstract way that strips off all of the intuition and the deepens of the original thinkers that invented those definitions, it is very stupid on all those professors and teachers that I feel sometimes that they have no clue themselves what they talking about. It is like if they will be shamed if they explain the origin of the idea and how to visualize it, really illogical.
Thank you very much for a such a clear explanation, and I really appreciate your efforts to cover every question that may popup in my mind during the lecture, like when you explaining the difference between entire space and trivialization, I am very thankful, great work.
Thank you for your kind comments. Someday I will redo all the lectures based on feedback and make them better.
I love it. I am learning myself. I don’t know but it appears to resemble something expressed in GU. Which I know 0 about.
Those are so simple concepts but mathematicians try to obfuscate them. Thank you for keeping it simple
I was thinking of y-shaped cylinder in my head on the later part of the video. Thanks for a good and kind explanation.
Good point! The funny part is that the definition of a fiber bundle does not have that many parts. By the end of this video a quick list would have been easy to make.
A manifold B called the "base space", a manifold E called the "fiber bundle", a standard fiber F, and a projection map pi: E-->B and that B is covered by n-hoods U_i such that pi^(-1)(U_i) is homeomorphic to U x F. Also, the projection onto U executed after the homeomorphism must yield the same U_i as the n-hood.
Maybe this is trivial, but homeomorphic to Ui x F with which topology on U x F? The product topology?
Excellent lectures! I have a silly question: a point P at the Base is related to a fiber F. BUT at 54:24 a point P that belongs to π^(-1) {Uij} is mapped to 2 different points in the same fiber F?. A point on the cylinder would not have the same point on the fiber? Thank you for the lectures!!
I suppose that's the reason why he mentioned the structure group has only one element {1} for the cylinder but has 2 elements {1,-1} for the Mobius strip. (I haven't checked the next lecture though). The implication of having only one element {1} for the structure group is the transition function of F is simply an identity, namely the two points are essentially the exact same point.
At 31:48
There is a P in the base;
Point P is mapped only to the blue.
A region around P -- ui -- is mapped to inside the red dotted line in E.
So, far, so good.
But inside that ui, there are other points: P1, P2, P3 (an infinity of them)
EACH of those points will have a BLUE region of its own in E.
Will ALL of those blue regions be entirely contained in the red?
If so, it is getting MUCH clearer.
Thomas Impelluso Yes! What you said is correct. And furthermore, each of the regions \pi^{-1}(p_i) will be disjoint. Each such region is "the fiber over p_i" and each such region is homeomorphic to F.
It is quite interesting, that two bundle charts with different E -> F maps, map intersection region \pi^-1(u_ij) on E to the same intersection region on B, u_ij. But the difference can be seen only \Psi and \Phi map the point on E to the generic fiber F.
This is really nice explanation. I have hard time reading Nakahara book as a physicist.
I have not examined that book. Did you try Nash and Sen "Topology and Geometry for Physicists"? It is a decent introduction and smooth read. Missing some things, but a nice place to start.
@@XylyXylyX Nash and Sen's book is frequently referred in Nakahara book. I only have a quick look, which is friendly to start. Nakahara book is more used among physicists, hopefully I do not need so much formal details since I do not work on high energy physics, but quantum information.
Thank you so much! This really helps me in my self-studies on Loring Tu's introduction to manifolds. I just wonder - what would "standard fiber" F look like in general? Is this some natural thing, or an extra condition I should impose on Tangent bundle? In Tu's book, I think F is presented as \mathbb{R}^r which I think can be thought of as a chart on E.. So I was a bit confused when you said the bundle map is totally different from manifold structure. (Maybe I misunderstood)
Each fiber must be "the same" in a mathematical sense. The only way two fibers can be the same is if they are homeomorphic. So all fiber on the manifold must be homeomorphic with each other, and therefore they are all homeomorphic to on particular topological space we can call "F".
At 21:02, what is the standard fiber? How should it be understood? Is it a projection of the preimage on F, fiber manifold?
Try this: every point in the base space has a pre-image. That pre-image is the “fiber” associated with that point in the base space. Every point in the base space has a fiber.
Now, every fiber must be “the same”. What we formally mean by “the same” is that every fiber must be diffeomorphic to F. And F is the “standard fiber”.
At 3:37, is an angular coordinate? Then S_1 is a set of numbers from 0 to 2*/pi? Then can I define the biggest manifold as 2d cylinder as the radius is fixed?
I chose [0, 2\pi) as the coordinate interval strongly suggesting an angular coordinate, but that was an arbitrary choice. The radius is irrelevant, which I think you are pointing out.
Can it be thought without involving the coordinate as a point in S_1 manifold?
Tursinbay Oteev In order to be a fiber bundle you must define *some* base manifold and that manifold will have a coordinate system by the definition of a “manifold”. So if I understand your quesiton correctly, the answer is “no.”
Sorry, I am giving many questions. But this is a tough topic. At 26:30, by "from there" do you mean {f^-1(u_i) x F}? If yes, is it because F is diffeomorphic to itself, and f and f^-1 are continuous? At 30:45, does "gets mapped" mean projected?
At the first time stamp: “from there” meant “from the inverse projection of U_i”. So that would be \pi^{-1} ( U_i).
At the second time stamp: no. But i note that I did say it incorrectly! I should have said: \phi is a mapping from the inverse of a single point ( “p” in this case) to the generic fiber F. So: \phi: \pi^{-1} (p) -> F. This is the mapping that tells us that each fiber over each point in the base space is diffeomorphic to a certain generic fiber F.
Instead i think I said that \phi was a map from the inverse projection of U_i, and that is not correct.
At 18:32, why the inverse projection of a point in the total space it is projected on base space?
The inverse projection of a point in base space will be a copy of the fiber space.
@@XylyXylyX , it sounded as any point in total space not just in base space
Hi, in 32:41, you have two cartesian product, ui X F and uj X F, how do you know they are the same F? Or they are different F associated with each i and j, i.i. Fi, Fj?
F is the "standard fiber" which means all fibers are diffeomorphic to F. So the answer to your question is that we know the *by definition* of what a fiber bundle is. If every fiber is not diffeomorphic to a standard fiber F then you don't actually have a fiber bundle!
I see. In there earlier part in the example of example of the cylinder. You specifically distinguish the Fiber (the real line) at point p and at point q, so that's only for illustrative purpose? Eventually there is only 1 fiber which is the real line?
Not it is not only for illustrative purposes. Each fiber is distinct! However they are all *copies* of F and F is called the "standard fiber". For the cylinder the standard fiber is the real line. the inverse of the projection map acting on a point of the base space must always yield a set that is homeomorphic to the standard fiber.
Thanks for the replies. Ok now i understand. The fiber F is a collection of distinctive fibers and these fibers are diffeomorphic to the 'standard fiber', so they are diffeomorphic to each other too.
1 more question: How do we know these fibers are forming a smooth manifold? The map from F->F by the 'fiber transaction'( psi, phi in 59:33) is smooth but how do they cooperate with the idea of open sets?
Ka ho Wong If I remember correctly, in one of the upcoming lectures I discuss the topology of manifolds. In short, every fiber bundle can be given a certain topology which is created in order to make the projection map \pi continuous. From that all else will follow.
Could we also choose our base space to be R, and have our fibers be copies of S^1?
theleastcreative Yes. That would be fine. In that case the base space is unbounded, but that is not a problem. It needs to be a manifold.
Hi Xylyxylyxy,
In 25:08, you said the inverse map pi of the base space S1 is those strips,
Are we considering the inverse map of a point on S1, or an interval on S1?
An interval on S_1 has an inverse of a "strip". I drew lines and said "strips", sorry :). The inverse image of a point in S_1 would be a single line.
It is fantastic series ! , I would like to apply differential geometry and topological processing to medical imaging domain (I am physycian) would you recommend a source with algorithms for automatic learning of the properties like path connectedness curvature of a mesh (or any other data structure derived from point cloud) ...
First of all thank you for the great explanation ! Could you help me a bit understand the main difference between the section and the inverse mapping ? They both map from the base space to the total space and the projection of both takes us back to the base point. I understand that a section is a continuous function and can be defined locally in a neighborhood or (in rare cases) globally, but I still would benefit from an intuition on what the section can do that the inverse map cannot. Thanks in advance ! :)
Hmm….the inverse mapping from a point in the base space is an entire copy of the fiber. The section map from the base space is a *single point* in a fiber. I think that answer addresses what you are asking?
There seems to be a problem for me at around 31:00 . I think that you have defined (π,ψ) globally on the whole total space when you say that ψ is the same diffeomorphism for every preimage of a basepoint. So maybe the definition should be something like ψ_i : π^{-1} -----> u_i x F. I am guessing it's a notation issue.... I am not sure, though, of my claims, so any answer is going to be appreciated. Either way thank you for your very helpful series of videos.
Well, I wondered about this also at first time he wrote that, however, latter on, he says that it is just a (clumsy) notation: since we have "i" in (pi_i,xi), they usually do not put "i" for "xi" part, so it is not globally defined as may it seem. So do not worry, it is just a usual abuse of notion, that I really hate toooooo much.
You are correct. PI is the only global mapping over E. I am using \phi and \psi to name the maps associate with U_i Dan U_j. THe capital \Psi_i notation is what you are looking for,
Thank you both very much!!!
This is the best and clearest introduction to Fiber Bundles I've ever found!!! Thank you so much.
Also a curiosity, if you wish to answer it: which is the software you use to produce these educational videos?
More to the point, thanks again for your wonderful channels, as many others, I'm really grateful to you!!!
So well explained!!!! 🎉
In your response to my last two posts, you shined a light: I now see that the diffeomorphism between the fibers and the real line is a CRITICAL observation. I see when you are drawing the triangle, that you have the generic fiber in mind. Your video here, at about 7 minutes does explain this but I overlooked its significance. Now I see. Thank you.
So may I ask a question to nail this shut? The use of the word TRIVIAL at 16:49. I am compelled to ask: what is NON-trivial? It seems to me that you gloss over the significance of this word. Why did you not call it... oh, I don't know... "pumpkin?"
I am serious: words have purpose. IF you are going to use a word like TRIVIAL, you must distinguish it from NON-trivial (and this has not yet been done for categorical purposes, so could you?)
Are you suggesting that the "triviality" here is that the fiber consists of only real numbers?
For example, the base could be the coordinate patch, and the fibers could be the covectors. And, in that case, it is not so EASY to identify the covector with a line, the way you can identify the point P, with a Real numbered fiber. Is THAT what you mean by TRIVIAL(=EASY)
OR....
Is the nature of the word TRIVIAL to be found at 26:38. (Which is also something clearer to me due to your previous response). Namely, that the TRIVIALITY consists of the EASE of "UNWRAPPING" E into a product space: Ui x F? Namely (and ultimately), the ease of extracating "coordinate information" and "tangent space information", from a "bundle of information."
Thomas Impelluso The word "trivial" is intended to mean "has the structure of a Cartesian product." A manifold is "trivial" it is can be reduced to a *single* bundle map and this is sometimes called "globally trivial". All manifolds, by definition, are "locally trivial" and can be covered with several bundle charts. If the structure group which connects the bundle charts is the Identity group {1}, then the manifold is actually globally trivial.
Please can you suggest me a few books on this topic?
At 33 minutes, I see only ONE triangle, but TWO small rectangles. Is this because the maps to all possible fibers are diffeomorphic to Rn?
Also, I am guessing right now that the Base is the coordinates. And the fiber (once endowed with a specific chart) becomes the tangent vector for a field?
Is this correct?
Each rectangle corresponds to a different bundle chart: from a different U_i. The triangle represents the "typical fiber" and pi^inverse(p) is a region of E that is always homeomorphic to the "typical fiber" F. The two arrows from E to F should both originate in the blue shaded region (that is, phi should be a bit longer and reach the shaded region like psi does). Each may have a different homeomorphism because psi and phi are associated with different bundle charts.
As a math hobbyist I kinda wish there were a few specific examples. This stuff is sort of "what we would like to have". I know the essence of math is abstraction but I like to see cogs turning cogs and stuff working. Also, when was this manifold theory developed? I now Riemann started it in the mid 19th century and Whitney did some stuff in the 20th.
Check out the next lecture, it is dedicated to a classic example of the tangent space to a circle. I plan two more: I just started the Möbius strip example, and I will also do the Hopf fibration if I have time.
inverse of pi is one to many, does that create any problem at all? It is almost as if there is no definite number on that fibre on the real space
Maps can be one to many. They are not functions, but the map yields a set for each elements of it’s domain. It isn’t a bug, it is a feature! :)
I can now finally poinpoint my confusion (the cause of all my questions)
At 28:00
On the left, the map to the triangle is from the BLUE region.
That means it is from ONE fiber (The fiber originating at P on the base, B)
Logically, then, the triangle represents all the points on that one fiber.
That implies the map (which you will later call PHI) must be a map to a specfic point on that one fiber.
So I assume that means that PHI maps to a point on the fiber, and the triangle represents all the points in ONE fiber:
the fiber of the blue dotted region.
Yet...
On the right, the trivial map takes you from the RED dotted-square.
And since that RED square originates from a region surrounding a point in the Base , then it must include other points, and, thus, include many fibers.
So the RED region is many fibers.
That implies that the map to the trivial product (which you will call (pi, phi) ) is NOT to a specific point on a fiber but to a family of fibers for the different points in the RED region
So it seems that the F in the triangle is DIFFERENT from the F in the trivial product.
The triangle's F represents ONE fiber and a point on that fiber.
The trivial product's F represents MANY fibers.
I am confused.
Further, at 36:11, you draw (a,b) in green next to the trivial product (Uj x F) at the top. But this cannot be. Because that map comes from the green dotted which necessarily includes many ponts, so there is no way you can identify a (b) on one fiber.
I think it comes down to: when are you using F to designated ONE fiber, or a point on ONE fiber, or, perhaps both a sepcific fiber and a point on it.
At the conclusion of this video (60 minutes or so), it seems that the triangle represents BOTH a fiber and a point on the fiber.
Thomas Impelluso You are definitely approaching this the correct way: identifying all the maps and struggling to understand the domain and range of everything. Let's start with psi and phi.
The map pi^(-1)(p) is the "fiber" over the point p. It is a set of points because the mapping pi is one-to-many. The "fiber" of p lives in E and is the blue region. Now you must understand that in order for this structure to be a true fiber bundle, every fiber over every point must be homeomorphic to a "standard fiber" which we call "F". That is what the triangle represents: the "standard fiber." The map \psi is the homeomorphism from the fiber over p to the standard fiber F in the chart U_i. Similarly for the map \phi for the chart U_j. Since \psi is a homeomorphism it is one-to-one and onto and invertible.
Regarding 36:11... any set which can be expressed as A x B is, by definition of the Cartesian product, as set of ordered pairs (a,b). In this case "a" is an element of U_j and "b" is an element of F, and (a,b) is an element of U_j x F. The "bundle" chart is the homeomorphism that takes regions of E and "flattens them out" so to speak to look like U_j x F.
You wrote: "Now you must understand that in order for this structure to be a true fiber bundle, every fiber over every point must be homeomorphic to a "standard fiber" which we call "F"."
OK, so that means that the map "psi" or "phi" is a map to a POINT on a generic fiber (which is diffeomorphic to the standard Rn fiber)
Thomas Impelluso The way to think about it is "\phi and \psi are one-to-one maps (because they are, at least, homeomorphisms."
Around the 11 minute mark, you said you've defined the product topology. However I'm fairly certain you have defined the box topology, not the product topology. There are many cases where these are the same, however there are also many cases where they are not. Please correct me if I am wrong about that. I think it might also be important to note that the projection mapping \pi need not be invertible, so talking about \pi inverse as a map rather than just considering specific pre-images could be misleading. With the example of the cylinder, the projection mapping is many-to-one, and hence is certainly not invertible.
Isn't phi^{-1} the bundle chart in the context of diff geometry? Charts are maps from $M -> R$. Phi is from Phi: Ux F -> M.
rinwhr The way I have defined Phi gives Phi_Inverse: U x F --> pi_inverse(U), so yes, I think you are correct. Some reference use other directions for the basic mappings, so it can get confusing.
I believe in your notation, most books define that as a "parametrization"
Perhaps it is more straightforward if ( pi , psi) can be interpreted as a category theoretical notation: morphism psi ⭕️ pi (psi after pi).
My understanding is that *everything* can more easily be interpreted using category theory. I really should brush up on that. :)
Also i have a question. When we say a function is diffeomorphic doesn't it have to be from Rd ---> Rd to be differentiable? how the bundle chart is differentiable?
Diffeomorphism applies to two differentiable manifolds. The language, as I use it, goes like this:
Two differentiable manifolds are "diffeomorphic" if we can find a function from one to the other that is a "diffeomorphism". A function between two manifolds is a "diffeomorphism" if it is 1-1, onto, bicontinuous, and "differentiable." A function between two differentiable manifolds is "differentiable" if the chart representative of the function (\phi o f o \gamma^{-1}) is differentiable in the sense of ordinary calculus. (Here \gamma is the chart map of one manifold and \phi is the chart map of the other manifold.
The fiber bundle must be a differentiable manifold as a whole. That is, the total space is a differentiable manifold. Does that answer your question?
+XylyXylyX I used to think (phi o f o gamma^-1) is differentiable because it's so in the calculus way but now i understand if the latter is true then f alone is differentiable and so it is homeomorphism (if also 1 to 1 and onto).
following this logic we can announce (pi,phi) to be diffeomorphism and the 2 manifolds of the total space E and the cartesian product Ui x F are diffeomorphic.
is that correct?
+John Ron Yes, that is correct. And that diffeomorphism is called a "local trivialization".
+John Ron .....but only for a *region* of the total space corresponding to a *region* of the base space! That is very important. E is usually not globally trivial!
+XylyXylyX okay! noted! thank you VERY MUCH....!
Can you please tell the name of the book which you are mostly following? Thanks. Btw it was a very nice explanation.
I use many different ones. I like Darling’s book “Differential Forms and Connections”.
@@XylyXylyX Okay. Thanks
can you specifically refer me a book from where I can read this fiber bundle as you have explained
@@Jana-pi3ub Try, "Geometrical Methods of Mathematical Physics" by Bernard Schutz.
Thank you
What is the first list to begin with: tensors? manifold? or they are independent?
and what we need in order to begin with as pre...
Skander TALEB HACINE Either one. They are independent. It is helpful to have completed up to 11 of the Manifold series first, if you can choose
I get very confused at 37:06. Your language becomes a bit vague and the cursor is swooping around. You say "I can do this projection process with pi by eliminating the fiber part"
But you have not defined "eliminating" and the cursor is moving around.
The you use the cursor to point to BOTH projectios pi and pi inverse
THen you say "by adding the fibers back" while your cursor is swooping over pi-inverse.
It is a little fast, jerky and vague... but it is important. Could you elaborate?
Thomas Impelluso THe projection map is easy to execute if you are working in U_j x F because the projection map pi simply involves picking out the point in U_i and dropping the part in F. That is, the map pi o (pi,phi) has a domain of U_i x F and if (p, f) is a point in U_i x F then the map pi o (pi , phi) acting on the point (p,f) simply returns p.
Would it be helpful to explain much of this with more appeal to physics? E.g. , thinking in terms of flow fields on manifolds?
rewtnode Naw....I want to keep this abstract. I think leaning on physical intuition causes more problems in the long run.
I'm finally catching up with u.It's all clear but i'm a bit confused with the names (the trivialisation thing and its derivatives).
Can't wait for the next lecture but in the meantime i may find some related exercises. Do u recommend any decent source, series, videos...whatever?
There are so many! Some are more mathematical than others. I will be doing some examples and will work on an example lecture today. Let me think about it. I want my lectures to simply be a way to help you through ANY other book or online video lecture.
You explained this very well. But it still caused me some confusion that you named an element of the base set B with lowercase a and not b. Fortunately you correct this later. Another thing I noticed is that it kind of gets lost why this is called a fibre bundle. I can see the idea of a fibre, but where is the bundle? It also threw off my comprehension when you draw a fibre as a triangle at some point. Then you draw an actual fibre bundle (lines on a cylinder with the S1 circle bundling them) and immediately wipe it out. There went my visual clue. Later it becomes less clear what a fibre is since you never draw any. Instead a fibre is supposed to be something in that triangle labelled F. Never mind. I want to learn this stuff and this was helpful that you tried very well to explain this, while I’m still somewhat confused.
I LOVE YOU!!!!!!!!!
I love you too, pal. Thanks for watching.
god damn that's well explained
These videos are probably gonna catch a lot more traction now that Eric Weinstein is releasing geometric unity
Get into any department of physics in any University and ask about Eric Weinstein, the guy is a joke!
@@tobia5416 He doesn't seem very popular outside of the internet. I doubt the average physics professor even knows who he is
@@GamingBlake2002 They know him because first year students ask about him! I find incredible that he can get around throwing random words around and so many on the web get fascinated: "wow he must be a genius, so many words I don't understand"! no dummy! he is just BSing you!
@@tobia5416 How do you know that first year students around the country are asking about him? Where are you getting this information?
@@GamingBlake2002 that's my job, I teach and do research in physics
: D
Are fiber bundles naturally related to Bessel functions since the cylindrical coordinate system appears?
@@StephenCrowley-dx1ej no there are many coordinate systems we could build on any fiber bundle because ultimately they are manifolds which allow for arbitrary coordinate.