At 8:30, equivalent points differ from one to the next another with a step equal to 1. Why is not 2\pi a step here? At 25:54, how are you defining the finest topology? At 37:40, is this a definition of an open set on A/_\tilde? If so it is a union of all points belonging to each equivalent class/subsets of u, on IR. Do those points belong to the topology of IR? At 55:59, p^-1(u) are discrete points on IR, and why you are considering them as open intervals on IR?
0:38:00 The union of all points [a] in U is simply U, of course. When describing/defining the topology of the quotient set, you probably meant the union of all p-1([a]) such that [a] is in U, where this union is open in A. (Another clue lies in the fact that, there can be no union of points [x] that belongs to A, as these are points in the quotient space.)
You are mentioning p-1([a]) , but could you define your p?is this the projection but then I dint understand what function should be represented as a 1. Is it the unit matrix? Still you wouldn’t write then 1([-]).. and since when a matrix can define a set of numbers....
It has been cool to hear them. I tried many books and lectures, what i can say is that its really good and you have worked hard on it. I am going to see all of the stuff and I wish if you can give lecture on homotopy theory, representation theory, algebraic topology in a full manner.
Is your listing of sets in the quotient space really correct? I have the feeling that the open ball topology on R would imply in the quotient space the sets {[x], a
@@XylyXylyX you sir are amazing! I'm just a random stranger on the internet to you, but you still answer quicker than most of my profs at university. And thank you a lot for your awesome lectures. They save my day while studying general relativity! It's at 14:27. I think I need to refine my question: If you have open-ball topology on R, is it even possible to map single points? Doesn't the map only take intervalls on R, which are then mapped to intervalls on Q?
Viktor Zelezny Maps can be quite arbitrary if you don’t care about things like continuity, for example. Maps connect *elements* of one set to another regardless of the topology. Thank you for your kind comment. :)
So when one says [0,1) is not the quotient space, it's to say that it is not homeomorphic to |R / ~ in the way S_1 is, and this is also assuming we've agreed on a particular topology of [0,1), the subset topology inherited from the standard open-ball topology of |R, for instance. My thought is that there may be a way of defining open sets in [0,1) so that the 'closeness' of [0] and [1-epsilon] can be captured, and thereby make all three spaces homeomorphic: |R / ~, S_1, and [0,1). So might one give [0,1) a different topology which would make it homeomorphic to the quotient space |R / ~?
Robert Bowman It is difficult to speculate on all the possibilities of all the possible topologies. However I don't think anyone could build the structure you are thinking of. Whatever plan one might have to re-topologize [0,1) as you suggest will probably just be equivalent to S_1 somehow.
I don't think so. Let me think this through: We know p is continuous because we are using the quotient topology on the quotient space. Therefore, the inverse of p is an open map and a closed map because we know that the continuity of p demands that the inverse of p be both an open and a closed map by definition. However, it may still be that there is an open set in A that is the pre image of a non-open set in the quotient space. If so, then p *is not* a quotient map. That is, for a continuous map, there may still be open sets in the domain that are not the pre images of any open sets in the range, but all of open sets of the range *do* have open pre-images. For a quotient map, *EVERY* open set in the domain must correspond to an open pre-image. In fact, because of this, a quotient map is getting closer to being bicontinuous and therefore closer to being a homeomorphism. What is missing is that neither p nor p inverse is expected to be injective so neither map is bijective.
If the quotient topology is essentially a union of all equivalence classes, how does it differ form the original space? Wouldn't you just end up with all the numbers of R in the quotient topology in this case? When you say an equivalence class is an element of a candidate set for the quotient topology, does that mean ALL the elements of the equivalence class have to be contained in u? If so wouldn't that make any u candidate for the quotient topology infinite?
The circle, that is (?) a model of a quotient space can be defromed by stretching and twisting, can't it? Can it be lifted into 4D, knotted and dropped back unto an euclidean plane. Would it still be a valid phase space trajectory? Would algebraic topology help here?
Slawomir P Wojcik The process you describe would not be a homeomorphic ammping of the circle. Your mapping introduces new intersections and such a mapping will never be a homeomoprhism.
When I take up a piece of string forming a circle from a sheet of paper, that it lies upon and knot it up without cutting the loop open, I do not introduce any intersection in the "real thing" but, merely in the projection. I suppose in 4D I can do more, that is I can knot the loop up in such a way, that it won't be homeomorphic to an unknot in 3D but back in 4D I still can untie it. To be more precise "untying" is an ambient isotopy ( en.wikipedia.org/wiki/Ambient_isotopy ) where [0.1] gives a fibration and lifts us into one dimension up, but (en.wikipedia.org/wiki/Knot_theory#Higher_dimensions) I don't have to be confined to the ambient space and make my fibration [0,1]x[0,1] where I can "do the trick" of homotopy.
The lessons are nice to get a first idea, but tbh I will hardly be able to tackle and solve problems and exercises afterwards. It lacks a little bit of deepness. Otherwise, for getting the ideas, they are really fine.
Agreed. There is nothing in these lessons that replaces a full course with a good textbook and exercises. I try to point that out from time to time and I try to encourage everyone to self study with a serious textbook and use the lessons as a supplement.
Yile Ying Oh, I see. I think I made some annotations that have been eliminated by TH-cam upgrades. Now I forgot what the errors were, but they were small errors.I wont fix it very soon. I have to keep going with the newer lectures. I will update the comment and explain my current thinking on this.
whoa! So in the last video, you mentioned you would discuss the mobius strip. But now this: quotient spaces and equivalence classes. In another topic to me, you mentioned you jump around (in the tensor one, you jumped to Lie Derivative and then back to forms). But here, I suspect there is a reason you are doing this as it will help with Mobius strip. But I see NO justification. Is this going to be only an excursion into topology? What is the use of this? Do I need to learn it? Can you put this into context of, say, Tangent bundles? Right now, it just seems like it came out of nowhere.
Thomas Impelluso This is required to understand the Möbius Strip. Quotient spaces is a fundamental and elementary topic in this type of work. I had to do it before Möbius strips because the Möbius strip is homeomorphic to a quotient space. It is an important exercise in abstraction.
Sorry but could you define your set of index. As long you haven’t done that your choice of equivalent elements doesn’t make sense to me. Therefore it’s totally possible that your defined equivalence relation is reflexive. (If 0 is contains in the index set) furthermore I don’t understand why it would be rational or effective to create an uncountable set of equivalence classes...
You're talking about p & q as if they are invertible mapping which they definitely aren't. You can put some restrictions on q to make it invertible but p necesssarily can't be because there are infinitely many points in each equivalence class so like it's impossible for it to be injective The homeomorphism can't be q\circ p^-1 it has to be some other function f:R/~ to S^1 such that p\circ f=q. We're guaranteed this by a famous topology thm math.stackexchange.com/questions/495924/how-does-the-quotient-mathbbr-mathbbz-become-the-circle-s1 but f is definitely neither p^-1\circ q or q^-1\circ p because neither are injective let alone homeomorphisms
acousticsoundwave93 Overall this is a bit of a sloppy lecture. I made several annotations to correct other errors. It is too long and I messed up the end a bit. For example, the arrows on the two functions you are talking about are backwards! I am going to set the lecture to "private" for now and redo it this weekend. Thank you for comment!
You are truly gifted at presenting this stuff, I'm astounded
These lectures are really great, I hope you will keep doing them.
Burak Çopur Thanks. Note the comment below..this particular lecture I have been meaning to redo.
At 8:30, equivalent points differ from one to the next another with a step equal to 1. Why is not 2\pi a step here? At 25:54, how are you defining the finest topology? At 37:40, is this a definition of an open set on A/_\tilde? If so it is a union of all points belonging to each equivalent class/subsets of u, on IR. Do those points belong to the topology of IR? At 55:59, p^-1(u) are discrete points on IR, and why you are considering them as open intervals on IR?
0:38:00 The union of all points [a] in U is simply U, of course. When describing/defining the topology of the quotient set, you probably meant the union of all p-1([a]) such that [a] is in U, where this union is open in A. (Another clue lies in the fact that, there can be no union of points [x] that belongs to A, as these are points in the quotient space.)
You are mentioning p-1([a]) , but could you define your p?is this the projection but then I dint understand what function should be represented as a 1. Is it the unit matrix? Still you wouldn’t write then 1([-]).. and since when a matrix can define a set of numbers....
0:32:54 Are the "surjective" and "continuous" essentially the same in this context?
It has been cool to hear them. I tried many books and lectures, what i can say is that its really good and you have worked hard on it. I am going to see all of the stuff and I wish if you can give lecture on homotopy theory, representation theory, algebraic topology in a full manner.
At 1:06:00 you have the directions of p o q^-1 and q o p^-1 the wrong way around
Yes, I noticed that. It is one of the annotations I need to make. Thank you.
I made the annotation. Good catch, thanks.
I appreciate the general description, although general relativity isn’t everyone’s goal some contact is generous of your time.
Is your listing of sets in the quotient space really correct? I have the feeling that the open ball topology on R would imply in the quotient space the sets {[x], a
Please cite the time in the video relevant to your question and I will have a look. Thanks!
@@XylyXylyX you sir are amazing! I'm just a random stranger on the internet to you, but you still answer quicker than most of my profs at university. And thank you a lot for your awesome lectures. They save my day while studying general relativity!
It's at 14:27. I think I need to refine my question: If you have open-ball topology on R, is it even possible to map single points? Doesn't the map only take intervalls on R, which are then mapped to intervalls on Q?
Viktor Zelezny Maps can be quite arbitrary if you don’t care about things like continuity, for example. Maps connect *elements* of one set to another regardless of the topology.
Thank you for your kind comment. :)
So when one says [0,1) is not the quotient space, it's to say that it is not homeomorphic to |R / ~ in the way S_1 is, and this is also assuming we've agreed on a particular topology of [0,1), the subset topology inherited from the standard open-ball topology of |R, for instance. My thought is that there may be a way of defining open sets in [0,1) so that the 'closeness' of [0] and [1-epsilon] can be captured, and thereby make all three spaces homeomorphic: |R / ~, S_1, and [0,1). So might one give [0,1) a different topology which would make it homeomorphic to the quotient space |R / ~?
Robert Bowman It is difficult to speculate on all the possibilities of all the possible topologies. However I don't think anyone could build the structure you are thinking of. Whatever plan one might have to re-topologize [0,1) as you suggest will probably just be equivalent to S_1 somehow.
At 33:40 you could alternatively just say; P sends closed sets to closed sets.
I don't think so. Let me think this through: We know p is continuous because we are using the quotient topology on the quotient space. Therefore, the inverse of p is an open map and a closed map because we know that the continuity of p demands that the inverse of p be both an open and a closed map by definition. However, it may still be that there is an open set in A that is the pre image of a non-open set in the quotient space. If so, then p *is not* a quotient map.
That is, for a continuous map, there may still be open sets in the domain that are not the pre images of any open sets in the range, but all of open sets of the range *do* have open pre-images. For a quotient map, *EVERY* open set in the domain must correspond to an open pre-image.
In fact, because of this, a quotient map is getting closer to being bicontinuous and therefore closer to being a homeomorphism. What is missing is that neither p nor p inverse is expected to be injective so neither map is bijective.
+XylyXylyX ah yes your right, I think I should have said P^-1 sends closed sets to closed sets (the contrapositive)
If the quotient topology is essentially a union of all equivalence classes, how does it differ form the original space? Wouldn't you just end up with all the numbers of R in the quotient topology in this case? When you say an equivalence class is an element of a candidate set for the quotient topology, does that mean ALL the elements of the equivalence class have to be contained in u? If so wouldn't that make any u candidate for the quotient topology infinite?
The circle, that is (?) a model of a quotient space can be defromed by stretching and twisting, can't it? Can it be lifted into 4D, knotted and dropped back unto an euclidean plane. Would it still be a valid phase space trajectory? Would algebraic topology help here?
Slawomir P Wojcik The process you describe would not be a homeomorphic ammping of the circle. Your mapping introduces new intersections and such a mapping will never be a homeomoprhism.
When I take up a piece of string forming a circle from a sheet of paper, that it lies upon and knot it up without cutting the loop open, I do not introduce any intersection in the "real thing" but, merely in the projection. I suppose in 4D I can do more, that is I can knot the loop up in such a way, that it won't be homeomorphic to an unknot in 3D but back in 4D I still can untie it. To be more precise "untying" is an ambient isotopy ( en.wikipedia.org/wiki/Ambient_isotopy ) where [0.1] gives a fibration and lifts us into one dimension up, but (en.wikipedia.org/wiki/Knot_theory#Higher_dimensions) I don't have to be confined to the ambient space and make my fibration [0,1]x[0,1] where I can "do the trick" of homotopy.
Thankyou
The lessons are nice to get a first idea, but tbh I will hardly be able to tackle and solve problems and exercises afterwards. It lacks a little bit of deepness. Otherwise, for getting the ideas, they are really fine.
Agreed. There is nothing in these lessons that replaces a full course with a good textbook and exercises. I try to point that out from time to time and I try to encourage everyone to self study with a serious textbook and use the lessons as a supplement.
when will you redo this video?
I did not have any plans to redo it. Does it need to be redone?
in the description of the video you said you are gonna redo it )))
Yile Ying Oh, I see. I think I made some annotations that have been eliminated by TH-cam upgrades. Now I forgot what the errors were, but they were small errors.I wont fix it very soon. I have to keep going with the newer lectures. I will update the comment and explain my current thinking on this.
whoa!
So in the last video, you mentioned you would discuss the mobius strip. But now this: quotient spaces and equivalence classes.
In another topic to me, you mentioned you jump around (in the tensor one, you jumped to Lie Derivative and then back to forms).
But here, I suspect there is a reason you are doing this as it will help with Mobius strip. But I see NO justification. Is this going to be only an excursion into topology? What is the use of this? Do I need to learn it?
Can you put this into context of, say, Tangent bundles? Right now, it just seems like it came out of nowhere.
Thomas Impelluso This is required to understand the Möbius Strip. Quotient spaces is a fundamental and elementary topic in this type of work. I had to do it before Möbius strips because the Möbius strip is homeomorphic to a quotient space. It is an important exercise in abstraction.
IT is also critical to understand interesting spacetimes in general relativity.
Sorry but could you define your set of index. As long you haven’t done that your choice of equivalent elements doesn’t make sense to me. Therefore it’s totally possible that your defined equivalence relation is reflexive. (If 0 is contains in the index set) furthermore I don’t understand why it would be rational or effective to create an uncountable set of equivalence classes...
You're talking about p & q as if they are invertible mapping which they definitely aren't. You can put some restrictions on q to make it invertible but p necesssarily can't be because there are infinitely many points in each equivalence class so like it's impossible for it to be injective
The homeomorphism can't be q\circ p^-1 it has to be some other function f:R/~ to S^1 such that p\circ f=q. We're guaranteed this by a famous topology thm math.stackexchange.com/questions/495924/how-does-the-quotient-mathbbr-mathbbz-become-the-circle-s1
but f is definitely neither p^-1\circ q or q^-1\circ p because neither are injective let alone homeomorphisms
acousticsoundwave93 Overall this is a bit of a sloppy lecture. I made several annotations to correct other errors. It is too long and I messed up the end a bit. For example, the arrows on the two functions you are talking about are backwards! I am going to set the lecture to "private" for now and redo it this weekend. Thank you for comment!
that's fair! it's really easy to be sloppy with quotient spaces and my topology course definitely was
That moment when the Integers are denoted using “I” instead of “Z” 😮