31:401:05:00 I'm not convinced that the examples he's giving are manifolds. For the first one it looks like it's not locally homeomorphic to R^2 at point q, and for the second the dimension of any open set that contains the boundary seemingly isn't consistent - it's homeomorphic to R^(dim M + 2) to the left and R^(dim M + 1) to the right.
summarize some mistakes: 9:20 It should be S1 and figure 8 is not homeo. to circle 32:43 E is not a manifold 50:58 (u,f) continuous 57:20 Mobius strip and cylinder are not homeo. and isomorphic as bundle 1:00:36 restricted bundle is bundle isomorphic to some "subbundle" of (E′, π′, M′) 1:05:00 E is not a manifold
Wow, what a remarkable and fascinating explanation about the chart representation of a continuous curve! I haven't seen before like this. With aid of this lecture, I can completely understand the purpose of manifold concept and its usefulness.
This is what I was thinking. Except that q is dim 0 and the rest are dim 1. Aren't manifolds supposed to have constant dimension, which is why the notion of dimension is well-defined on a manifold by definition?
Also the example given at 1:04 is not a bundle of topological manifolds, because at some points it is of real dimension 3 and in others it is of real dimension 2. This is clarified in lecture 10, at 1:32:22.
@@curtjaimungal The rest are indeed dim 2, locally around p the manifold looks like R x S1 , so the surface of a cylinder which is dim 2. And for q its more accurate to say there is no meaningful dimension, since there is no open neighborhood which maps homeomorphically to R^d for any d. Either way, you are both right, the example fails to be a manifold. Edit: I drew a picture for fun imgur.com/a/fRGBHA9 its like a strip of paper pinned to the tip of a cone.
I am interested in algebraic geometry and want to go into schemes. So I wanted to learn manifolds and vector bundles etc properly. Well I stumbled upon this series, and now all of a sudden I want to devote my full energy to learning manifolds. Thanks youtube for this recommendation. Professor, you have become one of my favorites. Great job.
Correct, I think that was just a slip of the tongue. He probably meant to say that although, when regarded as fiber bundles, the Mobius strip and cylinder have homeomorphic base spaces and typical fibers, they are still not isomorphic as fiber bundles.
@@cartmansuperstar Usually, to prove that two topological spaces are not homeomorphic, you look for a property that remains invariant under isomorphism (i.e. all homeomorphic spaces have it equal) and notice that for a specific pair of spaces it is actually different. The boundary of the cylinder consists of two disjoint circles, top and bottom; the boundary of a Möbius Strip is actually homeomorphic to a single circle. If they were homeomorphic, their boundaries would be homeomorphic.
@@brani114 he said as fibre bundles they were not homeomorphic which is correct they’re not-but he should have stated that’s true for the total space as well-is that what you’re saying?
@@eastwestcoastkid Yes, he was trying to give an example of non isomorphic fiber bundles that have homeomorphic total spaces and homeomorphic base spaces. But the möbius strip/cylinder example is just not a good example, since the total spaces are not homeomorphic.
@@eastwestcoastkid A better example would have been: The circle once as fiber bundle over itself by the identity mapping and once as fiber bundle over itself by the mapping z^2 (the circle covering itself twice).
I do not see how a total space may be a manifold if its fibers are not homeomorphic. Particularly, is the space on 32:13 a manifold? It is not locally homeomorphyc to R^2 around point q.
1:00:36 There is a mistake in the definition of locally isomorphic bundles. Rather, the restricted bundle should be isomorphic to the subbundle (u(preim_π(U)), π′|_u(preim_π(U)), f(U)) of (E′, π′, M′) where (u,f) is a bundle isomorphism. The fact that (u,f) is a bundle isomorphism means that u and f are assumed to be homeomorphisms with the relevant commutative property. In this way, f(U) is a neighbourhood of f(p) just as U is a neighbourhood of p and moreover, these neighbourhoods are isomorphic. It is also clear that (u(preim_π(U)), π′|_u(preim_π(U)), f(U)) is indeed a subbundle of (E′, π′, M′) by definition of u and f. Hence, this definition makes sense. Intuitively, one can see that this definition makes sense since if (E, π, M) is the cylinder and (E′, π′, M′) is the Möbius strip then they are clearly locally isomorphic bundles. On the other hand, unfortunately, Schuller’s given definition does not make sense as this would imply that the cylinder and the Möbius strip are not locally isomorphic as there is no subbundle of the cylinder that is isomorphic the Möbius strip. Actually, now that I’ve written this, maybe a cleaner definition should be (E, π, M) and (E′, π′, M′) are locally isomorphic bundles if for every point in E there is an open neighbourhood of p such that the restricted bundle is bundle isomorphic to some subbundle of (E′, π′, M′). Ah, yes, the words “some subbundle of (E′, π′, M′)” would have saved the day here. Enjoy my thought processes TH-camrs!
General of your mom that may be true for the circle and the square he drew, but for last object, which looked like a loop in space, he didn’t make any indications about it being somehow filled. The loop itself is definitely S1. If that thing is supposed to be S2 it ought to be drawn different so there is an interior visible.
@Bennabi Mehdi If it is supposed to be a disk then it is a different set usually called D^2. It has the same definition as the S sets but change the x^2 + y^2 =1 to x^2 + y^2
The cylinder is understood as vertical intervals attached around the circle. The projection would then be : take the "above" points and map them down to the circle, and take the "below points and map them to the circle. This is the same for all points. However, for the moebius strip we have that the "above" points end up becoming the "below" points during the twist. Think of the paper model. So now this change in placement needs to be taken care of in the projection mapping in a continuous way..again thinking of the continuous twisting of the paper model.
It is indeed a great fact that he is teaching topology. The fact... U r not bright, aren’t u? U think watching a sitcom can make u a physicist? Dumb idiot
I forwarded straight to the part on bundles, and was very impressed with your lecture. As Amara Katabarwa said below, this is the best discussion I have seen on the subject. I didn't have time to watch all of it, but I have liked and subscribed, and will definitely be back. I'm sure you'll have much more that will be of help to me. Thanks!
Question: The definition of a topological manifold as presented here does not include manifolds with boundary. So, the Mobius strip should not be a manifold as points on the boundary do not have an open neighbourhood homeomorphic to R^n, right? Or we can take the boundary to be open, which would give the interval (-1, 1) at 29:38.
Can someone explain how E at 32:18 would constitute a manifold? To the left of q, any point on those circles would be locally homeomorphic to R except for the point that joins the circle to the real line, which would the homeomorphic to R^2. On q, it looks like R and then to the right of q, again we have points that locally look like R^2. The dimension does not appear to be well-defined...
I completely agree! Was wondering that myself as I was reviewing the notes. Since the total space needs to be a manifold, I think this example would technically not be a bundle. I wonder if the fibers need to be homeomorphic even if it is not a fiber bundle in order for the total space to be a manifold. Have you had any more thoughts on this?
29:59 This is not a manifold, but a manifold with boundary. For points on the upper edge of the square won't have euclidean neighbourhoods (after identification)
At 1:44:24, I think the reason of the original r is continuous if xr is continuous is that x is a homomorphism. In general, x being only a continuous map is not enough (unless the topology on M is as weak as the weak-topology induced by x).
He probably meant S^1 is homeo to the square('s bounday) is homeo to the (boundary curve) of the wavy disc embedded in R^3, esp since later he says it is not homeo to R because it is compact, like he's assuming the same dimension, otherwise S^2 would be even more obviously nonhomeo to R on basis of dimension.
Shouldn't the total space be also a manifold? Why the example in ~31:50 the total space E does not seem to be a manifold because it somehow has intersections?
it isn't intersecting he's drawing an illustration of the fibres of the points (turns out not to be a manifold, though, because at q your local open set is not homomorphic to R^d but to R^d-1)
I think you're close. We want the pullback of the section to be a section itself, meaning it takes on values in E', but u^-1 can give us a set. For example, if m1' and m2' are both mapped by f to the same value in M (that is, f(m1') = f(m2') = m for some m in M) then u^-1(sigma(f(m1'))) = u^-1(sigma(f(m2'))) =u^-1(sigma(m)) must be a set containing (m1', sigma(m)) and (m2', sigma(m)) since both are elements of E' and u((m1', sigma(m))) = u((m2', sigma(m))) = sigma(m). What I think we want to do instead is let sigma' be defined by sigma'(m') = (m', sigma(f(m'))), which ensures that sigma' is a section even if f is not injective and that the graph commutes.
Andre Caldas it's not a manifold if you take his drawing as being a subset of R^2 with the induced topology, but that's not what he was trying to illustrate.
39:36 how is section diff from preimage of pi? is it a generalization? example ofna map from base top.nmanifold to a total top manifold that is a section but not a preimage?
42:00 section vs a function 53:00 isomorphic as bundles. defn. qs: what eiuld be a degn of equivalence. isimorphism btw two fynctions. in particular teo prob dists ove the ame fomain. 55:00 same E, same base manifd. but if the twisting is encoded doff via thevorojection then the two bundles are not isomorphic as bundles 58:30 defn of local isomorphic as a bundle qs: grn. models (kapping density function) is a surjective map?? 1:08:00 wave fuctíon. local tivialjty of a bundle 1:09:00 pull back bundle of a bundle. import. notion fefn and how to construct it can construct a morphism based on the constructed pullback bundle 1:18:00 swich from bundles to "viewing manifods from atlases"
40:20 when he defines the section of a bundle his explanation is clear but leaves me still puzzled why this construction is called a section? I suppose it makes sense when I consider some kind of line connecting all the points in E that are the images of Sigma applied to the base space M in his sketch.
Sections are also sometimes called cross-sections. They generalize the notion of the graph of a function. So for example in R^2, with the base space being the x-axis and the fibers being the y-axis, we define the standard fiber bundle with projection pi(x, y)->x. With this definition, a section is any function of the form sigma(x)=(x,f(x)), which is exactly how we're used to plotting the graph of a function. Proving that sigma is a section is easy since pi(sigma(x))=pi(x,f(x))=x which is the identity function. It gets more interesting when the fiber bundle is twisted or distorted in some way, allowing us to "graph" sections along the base space similar to how we graph functions in R^2.
He mentioned in the last lecture that the cartesian product top. space of two paracompact top.spaces may not be paracompact. But in this lecture, he defined the product manifold as a cart. product of two top. mfld.s. In this sense product space may not be paracompact and so not a mfld.. :/ on the other hand, all manifolds are paracompact by his definition... I feel there is something wrong there:/
=( you're right! I guess that's why most authors consider manifolds to be second countable, or else the product manifold is well defined only if one of the factors is compact =/.
It is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Smirnov metrization theorem) -> Product of two metrizable spaces is a metrizable space -> Every metrizable space is paracompact -> Product is paracompact.
Has anybody understood (42:15 - 45:55) why the wave function is not a function, but rather a section over a C-bundle? Isn't it the trivial product CxR^3? Why you need the concept of bundle here? Probably the answer is that a wave function is defined modulo a phase (a phase is not an observable), hence there is this phase arbitrariness of the "wave functions" that make them not actually behave like real functions, but I'm not really sure about that.
@8:40 He said that since a circle can be continuously deformed into a square, it is topologically homeomorphic. But, shouldn't that be homotopic as the latter is a weaker condition than former?
You are right he should have said that there is a bijective map continuous in both directions, a homeomorphism, between them. I think lots of people say "continuously deformed" when they mean homeomorphism. You should be careful however since for a homotopy the ambient space and how the two subsets are embedded is important. For example the square {(x, y) in R^2 | max(|x|, |y|) = 1} and the circle {(x, y) in R^2 | x^2 + y^2 = 1} are not homotopic inside R^2 minus the point (0.8, 0.8). In this sense saying "homotopic is a weaker condition than homeomorphic" as homotopy relies on an ambient space and how the sets are embedded while homeomorphic is just between the abstract topological spaces.
@@tonymok7752 Yeah for sure, I guess I didn't make that clear and just gave the images of the curves. Maybe there is a sense in which sets rather than maps can be deformed, but I'm not aware of it. If I recall correctly there is possibly a metric on the collection of compact subsets of a metric space, but I don't know for sure.
It's not a line in the sense of a continuum that can be given an ordering that is compatible with the field operations. This is obvious. But in this context, line is referring to the dimension of the fibers in a line bundle, where by definition a line is a topological space with dimension 1. And the vector space of complex numbers (with appropriate topology) is 1-dimensional over the field of complex numbers. Hence, a complex line, so to speak.
1:44:44 "Imagine a Bird flying through this Room" : * "Room" is isomorphic to "Theoretical Physics" embedded in "TH-cam". * "Bird" is isomorphic to "Dr F. Schuller".
No, there is no need that U and V must be continuous. A bundle morphism is simply a different word for a bundle map. One could of course consider special cases where u and v are assumed to be continuous, or diffeomorphisms, or are smooth etc.
This is a great lecture series. Not too hard to follow and very systematic. Still, it's so easy to forget the details - there are simply too many definitions in this field, and if I return a week later to this topic, in my mind much of it has turned to mush. What I'm missing are compelling drawings to illustrate many of the proofs. The symbols and fancy notation is necessary but somehow isn't good enough. Too rarely does he try to draw, but when he does, it's quite helpful. Is there a good straight forward book or lecture notes that would roughly follow the same material in this order?
21:30 great lecture, but jeez, he can’t draw a Moebius strip. I begin to get an intuition why many mathematicians love extremely elaborate formal constructs. The funny thing is that at this point he gives a very good motivation for bundles. Just try to draw a Moebius strip as a product manifold - it doesn’t work.
Also I think he didn't defined correctly what locally homeomorphic means for the manifold (in the definition for the manifold). Because locally homeomorphic is not equivalent to the existence of a local homeomorphism. Two spaces are homeomorphic iff there exists a homeomorphism between them, but it's NOT true that two spaces are locally homeomorphic iff there exists a local homeomorphism between them.
30:20 - No, I don't see. I understood all the things you just said, but it looks entirely arbitrary and pointless to me. Why is it of value? How can I put it into an application context that makes it clear what benefits its bringing to me? This is really the whole reason I watched this course up to now - to get an understanding of this whole "fiber bundle" business that I'm constantly running into in my studies. But I'm not getting any enlightenment here.
31:40 1:05:00 I'm not convinced that the examples he's giving are manifolds. For the first one it looks like it's not locally homeomorphic to R^2 at point q, and for the second the dimension of any open set that contains the boundary seemingly isn't consistent - it's homeomorphic to R^(dim M + 2) to the left and R^(dim M + 1) to the right.
summarize some mistakes:
9:20 It should be S1 and figure 8 is not homeo. to circle
32:43 E is not a manifold
50:58 (u,f) continuous
57:20 Mobius strip and cylinder are not homeo. and isomorphic as bundle
1:00:36 restricted bundle is bundle isomorphic to some "subbundle" of (E′, π′, M′)
1:05:00 E is not a manifold
Btw, at 9:20, that's not a figure 8. More like the perimeter of a Pringles chip. Thanks for the summary though!
I've always wanted to learn about topology. Your video playlist and the Covid confinement gave me a great opportunity :) Thanks again!
Watching these lectures in my first week of grad school. Many thanks, Dr. Schuller!
Wow, what a remarkable and fascinating explanation about the chart representation of a continuous curve! I haven't seen before like this. With aid of this lecture, I can completely understand the purpose of manifold concept and its usefulness.
27:51 bundle of top. mfd
35:19 fibre bundle
49:18 subbundle
51:35 bundle morphism
57:36 bundle isomorphism
1:00:31 locally isomorphic as bundle
1:04:20 trivial, locally trivial
1:17:58 pullback bundle
these are wonderful lectures, they have the best discussion of fibre bundles I have seen
In the example at 33:40 i think that E is not a manifold because at the point q its dimension is 1 while everywhere else it's 2
This is what I was thinking. Except that q is dim 0 and the rest are dim 1. Aren't manifolds supposed to have constant dimension, which is why the notion of dimension is well-defined on a manifold by definition?
Came here to say this. You are exactly right.
Also the example given at 1:04 is not a bundle of topological manifolds, because at some points it is of real dimension 3 and in others it is of real dimension 2. This is clarified in lecture 10, at 1:32:22.
@@curtjaimungal The rest are indeed dim 2, locally around p the manifold looks like R x S1 , so the surface of a cylinder which is dim 2. And for q its more accurate to say there is no meaningful dimension, since there is no open neighborhood which maps homeomorphically to R^d for any d. Either way, you are both right, the example fails to be a manifold. Edit: I drew a picture for fun imgur.com/a/fRGBHA9 its like a strip of paper pinned to the tip of a cone.
I am interested in algebraic geometry and want to go into schemes. So I wanted to learn manifolds and vector bundles etc properly. Well I stumbled upon this series, and now all of a sudden I want to devote my full energy to learning manifolds. Thanks youtube for this recommendation. Professor, you have become one of my favorites. Great job.
I believe the cylinder and the Möebius strip are not homeomorphic. (57:20)
Correct, I think that was just a slip of the tongue. He probably meant to say that although, when regarded as fiber bundles, the Mobius strip and cylinder have homeomorphic base spaces and typical fibers, they are still not isomorphic as fiber bundles.
came here to say that.
how can one reason that ?
@@cartmansuperstar Usually, to prove that two topological spaces are not homeomorphic, you look for a property that remains invariant under isomorphism (i.e. all homeomorphic spaces have it equal) and notice that for a specific pair of spaces it is actually different.
The boundary of the cylinder consists of two disjoint circles, top and bottom; the boundary of a Möbius Strip is actually homeomorphic to a single circle.
If they were homeomorphic, their boundaries would be homeomorphic.
These lectures are great. His mistakes are (mostly) obvious and forgivable.
What mistake has he made?
@@eastwestcoastkid At 57:15 he said for example that the cylinder was homeomorphic to the möbius strip, which is not true.
@@brani114 he said as fibre bundles they were not homeomorphic which is correct they’re not-but he should have stated that’s true for the total space as well-is that what you’re saying?
@@eastwestcoastkid Yes, he was trying to give an example of non isomorphic fiber bundles that have homeomorphic total spaces and homeomorphic base spaces. But the möbius strip/cylinder example is just not a good example, since the total spaces are not homeomorphic.
@@eastwestcoastkid A better example would have been: The circle once as fiber bundle over itself by the identity mapping and once as fiber bundle over itself by the mapping z^2 (the circle covering itself twice).
I do not see how a total space may be a manifold if its fibers are not homeomorphic. Particularly, is the space on 32:13 a manifold? It is not locally homeomorphyc to R^2 around point q.
1:00:36 There is a mistake in the definition of locally isomorphic bundles. Rather, the restricted bundle should be isomorphic to the subbundle (u(preim_π(U)), π′|_u(preim_π(U)), f(U)) of (E′, π′, M′) where (u,f) is a bundle isomorphism.
The fact that (u,f) is a bundle isomorphism means that u and f are assumed to be homeomorphisms with the relevant commutative property.
In this way, f(U) is a neighbourhood of f(p) just as U is a neighbourhood of p and moreover, these neighbourhoods are isomorphic.
It is also clear that (u(preim_π(U)), π′|_u(preim_π(U)), f(U)) is indeed a subbundle of (E′, π′, M′) by definition of u and f.
Hence, this definition makes sense.
Intuitively, one can see that this definition makes sense since if (E, π, M) is the cylinder and (E′, π′, M′) is the Möbius strip then they are clearly locally isomorphic bundles.
On the other hand, unfortunately, Schuller’s given definition does not make sense as this would imply that the cylinder and the Möbius strip are not locally isomorphic as there is no subbundle of the cylinder that is isomorphic the Möbius strip.
Actually, now that I’ve written this, maybe a cleaner definition should be (E, π, M) and (E′, π′, M′) are locally isomorphic bundles if for every point in E there is an open neighbourhood of p such that the restricted bundle is bundle isomorphic to some subbundle of (E′, π′, M′).
Ah, yes, the words “some subbundle of (E′, π′, M′)” would have saved the day here.
Enjoy my thought processes TH-camrs!
Isn't the S^2 at 8:26 supposed to be S^1?
Yes, it should be.
no, you are wrong... the interior is not empty
General of your mom that may be true for the circle and the square he drew, but for last object, which looked like a loop in space, he didn’t make any indications about it being somehow filled. The loop itself is definitely S1. If that thing is supposed to be S2 it ought to be drawn different so there is an interior visible.
@Bennabi Mehdi If it is supposed to be a disk then it is a different set usually called D^2. It has the same definition as the S sets but change the x^2 + y^2 =1 to x^2 + y^2
@@mappingtheshit he clearly says at 10:24 that locally it is part of the real line, so he means s1
This is my favorite kind of Biology.
By the way, when are we going to the gym tomorrow? LOL.
Lmao whenever
Wait you said tomorrow but its the end of the universe let postpone to yesterday (makes sense if time is cyclic)
Makes sense if noting makes sense
I love these so much, I have looking for these stuff explaining connection between math and physics which make me so clear for so long. Thanks !
57:50 - Why is the projection function different for the cylinder and the Moebius strip? I thought it just pointed at the base space.
The cylinder is understood as vertical intervals attached around the circle. The projection would then be : take the "above" points and map them down to the circle, and take the "below points and map them to the circle. This is the same for all points. However, for the moebius strip we have that the "above" points end up becoming the "below" points during the twist. Think of the paper model. So now this change in placement needs to be taken care of in the projection mapping in a continuous way..again thinking of the continuous twisting of the paper model.
Great lectures on topology, i like the fact that he exaplains, the topological spaces. One of my favorite subjects,
It is indeed a great fact that he is teaching topology. The fact... U r not bright, aren’t u? U think watching a sitcom can make u a physicist? Dumb idiot
I forwarded straight to the part on bundles, and was very impressed with your lecture.
As Amara Katabarwa said below, this is the best discussion I have seen on the subject. I didn't have time to watch all of it, but I have liked and subscribed, and will definitely be back. I'm sure you'll have much more that will be of help to me. Thanks!
Question: The definition of a topological manifold as presented here does not include manifolds with boundary. So, the Mobius strip should not be a manifold as points on the boundary do not have an open neighbourhood homeomorphic to R^n, right? Or we can take the boundary to be open, which would give the interval (-1, 1) at 29:38.
Can someone explain how E at 32:18 would constitute a manifold? To the left of q, any point on those circles would be locally homeomorphic to R except for the point that joins the circle to the real line, which would the homeomorphic to R^2. On q, it looks like R and then to the right of q, again we have points that locally look like R^2. The dimension does not appear to be well-defined...
I completely agree! Was wondering that myself as I was reviewing the notes. Since the total space needs to be a manifold, I think this example would technically not be a bundle. I wonder if the fibers need to be homeomorphic even if it is not a fiber bundle in order for the total space to be a manifold. Have you had any more thoughts on this?
29:59 This is not a manifold, but a manifold with boundary. For points on the upper edge of the square won't have euclidean neighbourhoods (after identification)
At 1:44:24, I think the reason of the original r is continuous if xr is continuous is that x is a homomorphism. In general, x being only a continuous map is not enough (unless the topology on M is as weak as the weak-topology induced by x).
Where can I find the problem sheets for this lecture?
Great lecture! Is it supposed to be S1(not S2) at 8:30?
no, you are wrong... the interior is not empty
Shahid Naqvi Yes
Yes. S^1 is the "circle".
@@mappingtheshit Well that doesn't make sense, if the interior were not empty it would be D2 not S2, surely he meant S1.
He probably meant S^1 is homeo to the square('s bounday) is homeo to the (boundary curve) of the wavy disc embedded in R^3, esp since later he says it is not homeo to R because it is compact, like he's assuming the same dimension, otherwise S^2 would be even more obviously nonhomeo to R on basis of dimension.
Can someone please post the problem sheet, please...!
I'd like to see the problem sheets, too.
@@paulmcc8155 drive.google.com/file/d/1nchF1fRGSY3R3rP1QmjUg7fe28tAS428/view?usp=sharing lecture notes
This is an awesome lecture to revisit topological bundles. Thanks!
Shouldn't the total space be also a manifold? Why the example in ~31:50 the total space E does not seem to be a manifold because it somehow has intersections?
it isn't intersecting he's drawing an illustration of the fibres of the points (turns out not to be a manifold, though, because at q your local open set is not homomorphic to R^d but to R^d-1)
at 1:16:50 … Is it like this, pullback of section = u^-1(sigma(f(m'))) . Can anyone please suggest some correction if it's wrong. Thank you.
I think you're close. We want the pullback of the section to be a section itself, meaning it takes on values in E', but u^-1 can give us a set. For example, if m1' and m2' are both mapped by f to the same value in M (that is, f(m1') = f(m2') = m for some m in M) then u^-1(sigma(f(m1'))) = u^-1(sigma(f(m2'))) =u^-1(sigma(m)) must be a set containing (m1', sigma(m)) and (m2', sigma(m)) since both are elements of E' and u((m1', sigma(m))) = u((m2', sigma(m))) = sigma(m). What I think we want to do instead is let sigma' be defined by sigma'(m') = (m', sigma(f(m'))), which ensures that sigma' is a section even if f is not injective and that the graph commutes.
21:55, 25:32, 29:27, 30:23, 33:30, 38:59, 41:53, 43:08, 48:56, 54:15, 1:13:12, 1:14:51, 1:19:59, 1:22:18
In 30:13, must the Mobius strip be the result of this construction? Why can't a cylinder be a result of this construction?
I'm confused. Shouldn't it be like product bundles (not mfds) < fibre bundles < bundles?
Yes, but he has in the back of his mind physics where pretty much all topological spaces of interest are manifolds.
I guess the space E in example at 31:40 is not a topological manifold for the same reason the "cone" isn't.
Also, at 1:05:00.
Andre Caldas it's not a manifold if you take his drawing as being a subset of R^2 with the induced topology, but that's not what he was trying to illustrate.
In 49:21, how do you know that the pre image is in fact a manifold?
He did not explain it, but in fact the fibers on a fibered manifold (which seems to be his construction) are indeed submanifolds of the total space.
His definition of fibre bundle is not enough to prove that. It should be restricted to locally trivial fiber bundle.
39:36 how is section diff from preimage of pi? is it a generalization? example ofna map from base top.nmanifold to a total top manifold that is a section but not a preimage?
41:26 conditioning or marginalizing a joint prob. distribution is this special case
42:00 section vs a function
53:00 isomorphic as bundles. defn. qs: what eiuld be a degn of equivalence. isimorphism btw two fynctions. in particular teo prob dists ove the ame fomain.
55:00
same E, same base manifd. but if the twisting is encoded doff
via thevorojection then the two bundles are not isomorphic as bundles
58:30 defn of local isomorphic as a bundle
qs: grn. models (kapping density function) is a surjective map??
1:08:00 wave fuctíon. local tivialjty of a bundle
1:09:00 pull back bundle of a bundle. import. notion
fefn and how to construct it
can construct a morphism based on the constructed pullback bundle
1:18:00 swich from bundles to "viewing manifods from atlases"
1:29:00
c0 compTibility of two charts (at the sMe point p on top manifold)
1:48:30 for new gan proposal
Had a small doubt(might be trivial)
Is the union of the fibres(as sets) for all the points in the base space equal to the total space?
yes, by the fact that the projection map is defined on all of the total space
40:20 when he defines the section of a bundle his explanation is clear but leaves me still puzzled why this construction is called a section? I suppose it makes sense when I consider some kind of line connecting all the points in E that are the images of Sigma applied to the base space M in his sketch.
Sections are also sometimes called cross-sections. They generalize the notion of the graph of a function. So for example in R^2, with the base space being the x-axis and the fibers being the y-axis, we define the standard fiber bundle with projection pi(x, y)->x. With this definition, a section is any function of the form sigma(x)=(x,f(x)), which is exactly how we're used to plotting the graph of a function. Proving that sigma is a section is easy since pi(sigma(x))=pi(x,f(x))=x which is the identity function. It gets more interesting when the fiber bundle is twisted or distorted in some way, allowing us to "graph" sections along the base space similar to how we graph functions in R^2.
@@alexsiryj Ah! Now I can appreciate it better. Thank you
That's a naming convention from category theory. Just search for "section category theory".
Also there is a good explanation for this name in the book "conceptual mathematics" in the context of Set.
I am glad he clarified the Möbius strip is not a product manifold.
He mentioned in the last lecture that the cartesian product top. space of two paracompact top.spaces may not be paracompact. But in this lecture, he defined the product manifold as a cart. product of two top. mfld.s. In this sense product space may not be paracompact and so not a mfld.. :/ on the other hand, all manifolds are paracompact by his definition... I feel there is something wrong there:/
=( you're right! I guess that's why most authors consider manifolds to be second countable, or else the product manifold is well defined only if one of the factors is compact =/.
It is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Smirnov metrization theorem) -> Product of two metrizable spaces is a metrizable space -> Every metrizable space is paracompact -> Product is paracompact.
In 1:21:15 technically the components functions x^i it should be the composition map proj_i o x? because its input is an element of M.
Yes, he defines x^i as the composition.
Has anybody understood (42:15 - 45:55) why the wave function is not a function, but rather a section over a C-bundle? Isn't it the trivial product CxR^3? Why you need the concept of bundle here? Probably the answer is that a wave function is defined modulo a phase (a phase is not an observable), hence there is this phase arbitrariness of the "wave functions" that make them not actually behave like real functions, but I'm not really sure about that.
See this later lecture th-cam.com/video/C93KzJ7-Es4/w-d-xo.html
@8:40 He said that since a circle can be continuously deformed into a square, it is topologically homeomorphic. But, shouldn't that be homotopic as the latter is a weaker condition than former?
You are right he should have said that there is a bijective map continuous in both directions, a homeomorphism, between them. I think lots of people say "continuously deformed" when they mean homeomorphism.
You should be careful however since for a homotopy the ambient space and how the two subsets are embedded is important. For example the square {(x, y) in R^2 | max(|x|, |y|) = 1} and the circle {(x, y) in R^2 | x^2 + y^2 = 1} are not homotopic inside R^2 minus the point (0.8, 0.8). In this sense saying "homotopic is a weaker condition than homeomorphic" as homotopy relies on an ambient space and how the sets are embedded while homeomorphic is just between the abstract topological spaces.
@@00TheVman when talking about homotopy, the circle and the square should be interpreted as curves in R2, isn't it?
@@tonymok7752 Yeah for sure, I guess I didn't make that clear and just gave the images of the curves. Maybe there is a sense in which sets rather than maps can be deformed, but I'm not aware of it. If I recall correctly there is possibly a metric on the collection of compact subsets of a metric space, but I don't know for sure.
What is complex line? How can complex numbers be arranged in a line?
I too have the same question.
It's not a line in the sense of a continuum that can be given an ordering that is compatible with the field operations. This is obvious. But in this context, line is referring to the dimension of the fibers in a line bundle, where by definition a line is a topological space with dimension 1. And the vector space of complex numbers (with appropriate topology) is 1-dimensional over the field of complex numbers. Hence, a complex line, so to speak.
1:44:44 "Imagine a Bird flying through this Room" :
* "Room" is isomorphic to "Theoretical Physics" embedded in "TH-cam".
* "Bird" is isomorphic to "Dr F. Schuller".
31:13 The bundles are the generalization of the idea of taking a product.
Any good reference which closely follows this ?
The requirement is missing that the image of the coordinate map must be open in R^d, in the definition of manifold.
+ANSIcode x is homeomorphism :)
Homeomorphism over its image! In principle, the image needs not to be an open set.
In the definition he gave _any_ subspace of R^d would count as a topological manifold (take each U as that subspace, and x as the identity).
Thank you so much! This is so clear and helpful!
What is a composition of a function and a curve gives? Example??
Hi professor, why do you have one def for topological manifold in this video different from the one you introduced in another video?
Wish my teachers were this great
The end points of 375 will not agree with three, so it gives a twist that affects the mobius strip.
Great Lecture. Many thanks Sr.
The maps @50:55 need to be continuous
Thanks. I was just wondering why he talks about homeomorphism in bundle isomorphism. If u,v are continuous, then it makes sense.
No, there is no need that U and V must be continuous. A bundle morphism is simply a different word for a bundle map. One could of course consider special cases where u and v are assumed to be continuous, or diffeomorphisms, or are smooth etc.
best exposition ever
This is a great lecture series. Not too hard to follow and very systematic. Still, it's so easy to forget the details - there are simply too many definitions in this field, and if I return a week later to this topic, in my mind much of it has turned to mush. What I'm missing are compelling drawings to illustrate many of the proofs. The symbols and fancy notation is necessary but somehow isn't good enough. Too rarely does he try to draw, but when he does, it's quite helpful. Is there a good straight forward book or lecture notes that would roughly follow the same material in this order?
Lets be honest forget about illustrations for these sort of abstract concepts. However Frankels book is good
Sir. Roger Penrose's book : "The road to Reality"
I found his drawing is extremely good and expressive.
why the section is called sigma and not just pi^-1 ? maybe a naive question
I dont get how we know that the for every p the preimage of {p} under π forms a manifold ? (in the case of the fiber bundle ) :S
21:30 great lecture, but jeez, he can’t draw a Moebius strip. I begin to get an intuition why many mathematicians love extremely elaborate formal constructs. The funny thing is that at this point he gives a very good motivation for bundles. Just try to draw a Moebius strip as a product manifold - it doesn’t work.
thanks
Also I think he didn't defined correctly what locally homeomorphic means for the manifold (in the definition for the manifold). Because locally homeomorphic is not equivalent to the existence of a local homeomorphism.
Two spaces are homeomorphic iff there exists a homeomorphism between
them, but it's NOT true that two spaces are locally homeomorphic iff
there exists a local homeomorphism between them.
Nice sir tq
Chat happening how
8:23
The definition of submanifold given by the lecturer is not good
30:20 - No, I don't see. I understood all the things you just said, but it looks entirely arbitrary and pointless to me. Why is it of value? How can I put it into an application context that makes it clear what benefits its bringing to me? This is really the whole reason I watched this course up to now - to get an understanding of this whole "fiber bundle" business that I'm constantly running into in my studies. But I'm not getting any enlightenment here.
Frederic Schuller should change his name to Lehrer!
who the hell would dislike this????
1:26:46 U are certainly compatible with someone if u never meet that person lol
HOCAM YA HOCAM
1:17:23 It took me far longer than I’m going to admit to realise σ′(m′):=(m′, σ(f(m′))).
"physics example" oh yeah it's a physics course