Question regarding the 2-Sphere example: Here, "h" obviously is a parallel projection and always covers half of the sphere. but you also could do a "beam" ("point"/"central") projection by defining a point inside the sphere (e.g. a small distance z below the north pole). This would cover most of the sphere and map it to complete R². The remaining "cap" could use some other map (e.g. by parallel projection. Now you could choose a z arbitrary low. Is this limit really a manifold on the complete sphere minus a single point ? Or does such approximation not make sense ?
Question regarding choosing Epsilon balls.: We induce a topology on the set by choosing a homeomorphism and having the open epsilon balls in R^n define the open subsets. Here we use Pythagoras aka the sum of squares of the coordinates (the squareroot does not matter). We don't need to know that this is a metric (in fact Euclidean) , so we can ignore that. But it in fact is a metric and I understand we could use any metric do define these balls. Now my goal is to understand how the "Einstein"-universe constructs the spacetime interval (i.e. how it applies the "metric of the spacetime" ) to have us see it's "reality"). Obviously we will need a locally Minkowski metric, AFAIK this is called a "pseudo Riemanian Manifiold"). Now to me it seems odd to use Euclidean metric (1,1,1,1 signature) to induce the topology and later use Minkowski metric (1, -1, -1, -1 signature) to induce the metric tensor field. So the question is "can Minkowski metric be used to construct the Epsilon balls" (e.g. by 0 < s² < Epsilon, |s²| < Epsilon, or similar) and does this in fact make any sense ? I feel "Minkowski balls" are something like light cones in the Spacetime diagram, which at least seems rather intuitively sensible (NarfWhals' idea). -Michael
8:57 Is it correct to say that U_{3,+} = {(x_1, x_2, x_3) in S^2 | x_3 >=0 } rather than the strict inequality x_3>0? If not, how does the atlas deal with points lying exactly on the equator, i.e not in the southern nor in the northern hemisphere?
Never mind. I see intuitively how {U_{i,+-}, h_{i,+-}} for i = 1, 2, 3 covers all points. With that being said, is it correct to state that a collection of the U_{3,-} you defined and the U_{3,+} I outlined above (along with sensible h's) is an atlas?
In one one the videos the Euclidean space "locally associated" to point x in the manifold X is said to have dimension n. Is in all cases the dimension of the local Euclidean spaces equal for all points x in X ? Is this dimension proven to be equal to the Lebesgue covering dimension of the Topology ? (... Maybe covered in a future video ? ) - Thanks ! - -Michael
Hey! Great video! I have a question: Is EVERY manifold Hausdorff? In my opinion, I don't think so. Consider any singleton set. It's always a 0-manifold, but is never Hausdorff! Therefore, my question is: when we talk about manifolds, is one always talking about an n-manifold for positive n? That is, non-singleton sets?
@brightsideofmaths Well, by definition, to be Hausdorff distinct elements can be "seperated" by disjoint open neighborhoods. But singleton sets have only one element inside. Am I right? Pls correct me.
That is, I am confused about the fact the one needs distinct elements and then find open and disjoint neighborhoods of those distinct points... I also found that all singleton sets are closed in a Hausdorff space. Does that contradict anything?
Great lecture! Two naive questions: 1. I am under the impression that for example (c), you used the existence of atlas to prove S^2 is indeed a 2-d manifold. Is building an atlas a common practice when proving if some topological space is a manifold or not? 2. How do you define the openness with S^2? I guess it would be any intersection of open subsets in R^3 with S^2? Please let me know if I am wrong on this one. Thanks again!
Very late response but relevant: the choice of topology is a relevant thing - they affect the choice of pseudogroup of transformations. Often subspace or another simple topology gives interesting things. However, even with these topologies, two different choice of atlas/charts may or may not give the same manifold structure overall. Apparently the 7 sphere has 15 different (smooth, standard topology?) structures based on the choice of atlas! (or more specifically, the choice of pseudogroup of transformations which a given atlas is subordinate to) Note: A pseudogroup of transformations is a set of maps Rn -> Rn satisfying some group like properties, and an atlas is subordinate to it if the transition maps are all in that pseudogroup
Exactly, it does not cover the whole manifold. So it is not an atlas. However, for integration later, it is still useful since the equator is a null set.
Great series! Keep going :) In example c) would it be possible to define the locally eucledian subsets the same way as you did but with x3 smaller OR equal to 1? And that way we would only need the U3+ and U3- to be an atlas?
Great video as always ! Just a quick question : why is the southern hemisphere of the 2-sphere an open subset of R^3 with the standard topology ? I don't see how to write it as an intersection or union of open balls...
Thank you for your reply ! So that means for every manifold, we need to define a topology on it ? Seems okay for the sphere but for more "exotic" shapes it sounds like it would be a difficult exercise !
@@brightsideofmaths there should three chart. 1 is for northern hemisphere.2 is for southern hemisphere. 3 is for equator. right?? but you have wrote total six if i combine all +,_ with 1,2,3. that is 1+,1_,2+,2_,3+,3_ .total six
@@brightsideofmaths So basically you would have two hemispheres for the top and bottom. Then another two that would wrap around vertically covering most of whats left of the equator. Then there would be two points left to cover on the equator that would be covered by an additional 2 half spheres. Hence, 2+2+2=6.
This playlist is now becoming the most epic thing ever created on youtube, congratulations!
So true. It's amazing.
These are truly top tier videos, keep up the good work!
Thanks a ton!
Thank you for your work! on 6:15 should it be {(x_1,x_2,x_3)^T\in S^2| x
Yeah, you are right!
Please pin this comment or add a correction in the video.
i havent learned about manifold but your video is really clear and interesting , thank you
Glad it was helpful!
Question regarding the 2-Sphere example:
Here, "h" obviously is a parallel projection and always covers half of the sphere.
but you also could do a "beam" ("point"/"central") projection by defining a point inside the sphere (e.g. a small distance z below the north pole). This would cover most of the sphere and map it to complete R². The remaining "cap" could use some other map (e.g. by parallel projection.
Now you could choose a z arbitrary low. Is this limit really a manifold on the complete sphere minus a single point ? Or does such approximation not make sense ?
Yes, you could choose a chart where the domain is the whole sphere S^2 without one point. You cannot do it better because S^2 is compact.
@@brightsideofmaths Is parallel projection a homeomorphism?
insanely good explanation, i was stack on this for some hours. thank you so much!
Thanks a lot and thanks for the support!
Question regarding choosing Epsilon balls.:
We induce a topology on the set by choosing a homeomorphism and having the open epsilon balls in R^n define the open subsets. Here we use Pythagoras aka the sum of squares of the coordinates (the squareroot does not matter). We don't need to know that this is a metric (in fact Euclidean) , so we can ignore that. But it in fact is a metric and I understand we could use any metric do define these balls.
Now my goal is to understand how the "Einstein"-universe constructs the spacetime interval (i.e. how it applies the "metric of the spacetime" ) to have us see it's "reality"). Obviously we will need a locally Minkowski metric, AFAIK this is called a "pseudo Riemanian Manifiold"). Now to me it seems odd to use Euclidean metric (1,1,1,1 signature) to induce the topology and later use Minkowski metric (1, -1, -1, -1 signature) to induce the metric tensor field.
So the question is "can Minkowski metric be used to construct the Epsilon balls" (e.g. by 0 < s² < Epsilon, |s²| < Epsilon, or similar) and does this in fact make any sense ? I feel "Minkowski balls" are something like light cones in the Spacetime diagram, which at least seems rather intuitively sensible (NarfWhals' idea).
-Michael
Manifolds? More like “Learning new information never gets old!” Thanks so much for sharing all these wonderful videos.
8:57 Is it correct to say that U_{3,+} = {(x_1, x_2, x_3) in S^2 | x_3 >=0 } rather than the strict inequality x_3>0? If not, how does the atlas deal with points lying exactly on the equator, i.e not in the southern nor in the northern hemisphere?
Never mind. I see intuitively how {U_{i,+-}, h_{i,+-}} for i = 1, 2, 3 covers all points. With that being said, is it correct to state that a collection of the U_{3,-} you defined and the U_{3,+} I outlined above (along with sensible h's) is an atlas?
Thanks for the question. We need open sets, so we exclude the equator.
We need open sets from which topology?
Standard topology in R^n and their subspace topologies :)@@wolfgangsullifire6158
Super. Congratulations! Thank you for your clear explanation.
I agree, its a lifeline
Please Continue !!
An Lp-norm ball is also a manifold if 1 < p < ∞, isn't it?
Sure! The maps are the same. And please note that the induced topology is the same on R^n.
@@brightsideofmaths But for 0 < p ≤ 1 and p = ∞ there are (in R³) eight and six maps, respectively?
@@PrzemyslawSliwinski the induced topology is the same, so we can use the same atlas, which means we can define an atlas with as few as 2 charts
In one one the videos the Euclidean space "locally associated" to point x in the manifold X is said to have dimension n.
Is in all cases the dimension of the local Euclidean spaces equal for all points x in X ?
Is this dimension proven to be equal to the Lebesgue covering dimension of the Topology ?
(... Maybe covered in a future video ? )
- Thanks ! -
-Michael
I appreciate the in depth examples!
Nicely done once again! 😃
Hey! Great video! I have a question: Is EVERY manifold Hausdorff? In my opinion, I don't think so. Consider any singleton set. It's always a 0-manifold, but is never Hausdorff! Therefore, my question is: when we talk about manifolds, is one always talking about an n-manifold for positive n? That is, non-singleton sets?
Why are singletons not Hausdorff?
@brightsideofmaths Well, by definition, to be Hausdorff distinct elements can be "seperated" by disjoint open neighborhoods. But singleton sets have only one element inside. Am I right? Pls correct me.
@@yoyostutoring Yes, one element, so the Hausdorff property is always satisfied :)
@brightsideofmaths Can you explain a bit in detail of this, pls? I searched some information but never found an answer to this.
That is, I am confused about the fact the one needs distinct elements and then find open and disjoint neighborhoods of those distinct points... I also found that all singleton sets are closed in a Hausdorff space. Does that contradict anything?
Great lecture! Two naive questions: 1. I am under the impression that for example (c), you used the existence of atlas to prove S^2 is indeed a 2-d manifold. Is building an atlas a common practice when proving if some topological space is a manifold or not? 2. How do you define the openness with S^2? I guess it would be any intersection of open subsets in R^3 with S^2? Please let me know if I am wrong on this one. Thanks again!
Thanks! On S^2 we have the subspace topology. Or with another viewpoint: the metric space S^2 with standard metric induces a topological space.
Very late response but relevant: the choice of topology is a relevant thing - they affect the choice of pseudogroup of transformations. Often subspace or another simple topology gives interesting things. However, even with these topologies, two different choice of atlas/charts may or may not give the same manifold structure overall. Apparently the 7 sphere has 15 different (smooth, standard topology?) structures based on the choice of atlas! (or more specifically, the choice of pseudogroup of transformations which a given atlas is subordinate to)
Note: A pseudogroup of transformations is a set of maps Rn -> Rn satisfying some group like properties, and an atlas is subordinate to it if the transition maps are all in that pseudogroup
9:58 would (U_3,+; h_3,+), (U_3,-; h_3,-) be an atlas?
I guess not, since it covers the whole S^2 except the equator where x_3 is zero.
Exactly, it does not cover the whole manifold. So it is not an atlas. However, for integration later, it is still useful since the equator is a null set.
@@brightsideofmaths Thanks!
Great series! Keep going :)
In example c) would it be possible to define the locally eucledian subsets the same way as you did but with x3 smaller OR equal to 1? And that way we would only need the U3+ and U3- to be an atlas?
No, since the U3+ and U3- would not be open (there are border points).
Amazing content, thank you!!!
1:09 chart
2:16 every subset of m is open
Great video as always ! Just a quick question : why is the southern hemisphere of the 2-sphere an open subset of R^3 with the standard topology ? I don't see how to write it as an intersection or union of open balls...
Not an open set of R^3 but an open set of S^2. You should see S^2 without the surrounding space.
Thank you for your reply ! So that means for every manifold, we need to define a topology on it ? Seems okay for the sphere but for more "exotic" shapes it sounds like it would be a difficult exercise !
@@StratosFair A manifolds needs to be a topological space. Therefore, we need a topology :)
Ungefähre Zeitangabe wann Homologientheorie kommt?
Nice video!
Why do you need all the atlasses? Wouldn't just the northern and southern hemisphere cover the set
equator would be missing :)
@@brightsideofmaths there should three chart. 1 is for northern hemisphere.2 is for southern hemisphere. 3 is for equator. right??
but you have wrote total six if i combine all +,_ with 1,2,3. that is 1+,1_,2+,2_,3+,3_ .total six
@@anowarali668 The equator would not be an open set.
@@brightsideofmaths So basically you would have two hemispheres for the top and bottom. Then another two that would wrap around vertically covering most of whats left of the equator. Then there would be two points left to cover on the equator that would be covered by an additional 2 half spheres. Hence, 2+2+2=6.
@@brofessorsbooks3352 Exactly!
Nice! :)