0:00 Start (Review of differentiable Manifolds). 3:56 Curves As a Function on R. 6:28 Curve Function on R as defined by the variable, Lamda traverses a number of charts in the Topological Space. 8:24 Curves in normal calculus (e.g. R2 with axes X and Y). 10:53 How to assign coordinates in R2 of the curve in the Topological Space using the composition of two functions R to T to R2. 12:19 The same method can be applied to a different chart in the Topological Space. 13:11 One can take the derivative of the curve in R2 as a function of Lambda. 15:27 Coordinates. 17:12 Coordinate Functions. 18:50 Tradition treatment of coordinate function in physics textbooks on Relativity. 22:02 More details on these coordinate mappings. 23:50 what are the old coordinates in the traditional physics textbook approach. 27:09 The second coordinate is derived by much the same. 27:33 Using the other way the old coordinates can also be derived from the new coordinates. 27:56 One need keep these in mind that everything is chart dependent. 28:51 One can then employ normal calculus here and transitions functions depend on the underlying Manifold. 30:00 Another way of understanding coordinate systems by back projecting coordinates to the manifold. 32:11 Back projecting coordinates from another chart in the atlas to the same manifold. 35:21 More on smoothness in regards to maps and charts and functions between manifolds. 38:57 Diffeomorphic Spaces: Homomorphic plus differentiable. 43:07 Summary: Curves, Coordinate Systems, and Diffeomorphisms.
The distinction between coordinates, manifold, and transition functions and the proper domains and co-domains is made with beautiful clarity! Thank you for lifting the haze left by most textbooks.
Damn ! That was an eye opener .. the way we connect the idea of different charts on a manifold in our heads with the diffrerent coordinate systems, you don't find such an explanation in ordinary text books on the subject. I appreciate your work, respect !
This comments apply to all the lectures out in manifolds and tensor sections. Best of all the topology and tensor lectures out here in or elsewhere in univs . Thumbs up. It cleared a lot of confusion I had about these topics. If you write a book that would probably be very good for the community too :)
At 38:44 for the discussion aren't we only concerned with the mapping of a single chart and not the intersection of two different charts in X? I suspect if we did use the intersection charts of A,B multiple paths from gamma(V) to phi(V) could be created.
Mappings are always from a single chart and the discussion you cite uses the charts (U,\gamma) and {V,\phi). The intersections are there to illustrate that X and Y are manifolds and that we can always change charts.
I'm not understanding how polar coordinates work here. In the chart at 34:32, the polar coordinates r and theta are drawn in a rectilinear way. Why would we do this? I understand how to visualize the open ball topology with rectilinear coordinates but not polar coordinates.
It is an interesting and important question you ask! Ultimately EVERY manifold must go back to a Cartesian R^n. The polar coordinates you are thinking of are actually an expression of the fact that R^n itself is a manifold and the chart mappings can label each point with (R,theta,phi), but the other side of the final chart mapping is always Cartesian and that is where the actual calculus is happening. Remember….ultimately we only know how to do calculus in a *Cartesian* setting. Even when we do integrals in polar coordinates, if you take them apart fully you see that we are actually integrating coordinate by coordinate as though we were in a Cartesian system. An open ball in the system I have drawn is literally a ball, just like in ANY Cartesian system. .
At 39:30: How can "f" be differentiable if it is a map bewteen topological spaces? I imagine PHI composed f composed gamma-inverse can be differentiable... but not f alone.
Thomas Impelluso Yes, that is a "abuse of language" to say "f is differentiable". However, it is common language to say "f is differentiable" and all the machinery of the atlas' charts is hidden beneath the language as an assumption. We must always, in the back of our minds, remember that a function literally between topological spaces is not in-and-of-itself "differentiable". Only that function composed with the various chart maps is differentiable.
Hmm at 15:05 can I wrap up this way. As you mentioned in the last video, a set X is a model of real world(could be an apple, banana, a pile of manure or whatever) and it is an abstract set that contains information that we are interested in. But we want to analyze those abstract sets with mathematical tools. To define differentiation we need an argument variable and continuous function in metric space. so f that maps R to X is a set-up process for using parameter lamda as our argument variable, adoption of function gamma and phi is to give out metric values w.r.t specific coordinates
Thanks, I have one more question. you said we cannot differentiate a function on topological space X and that is why we define a function that maps X onto R. However, at 40:52, if diffeomorphism, we should be able to differentiate the whole chunk of composite functions. How can we differentiate f where it is a function that maps X onto Y? is function f itself is a composite of some functions that allows differentiation?
Jay Lee I'm not sure I exactly understand your question but the *definition* of the differentiability of a function like f is that functions like \phi o f o \gamma^{-1} are differentiable in the sense of normal calculus. That is the ONLY way "calculus" of functions between sets can be understood. No manifold, no calculus.
Thanks so much for this series. It is absolutely top notch teaching. A question: I am struggling to understand the discussion around 43:00 which seems to imply that homeomorphisms (and diffeo) between two manifolds can map onto Euclidean spaces of different dimension. Is there an example of this?
There is not example of this because it is not true! Homeomorphisms and diffeomorphisms must map manifolds of the same dimensionality. The drawing at that time stamp was regarding a *general* mapping f. If f has certain properties the two manifolds will be diffeomorphic. Those requirements ( one to one and onto in particular) will force the two manifolds to have the same dimension. Thanks for watching!
I guess one of the reasons these lectures work out so well is that you tireless draw those diagrams of maps. I guess it would be a big challenge to try present the same material in a book though. This is almost like learning an instrument, where the best way is not to read a book but to get constant feed back from an instructor by examples and demonstrations.
Thanks a lot! These lessons are extremely valuable. I have a question: In this video, it seems that diffeomorphism is a property of two manifolds examined together. If I remember correctly, in a previous video, it was mentioned that if the transition function is differentiable then we have diffeomorphism. Transition functions consider different charts of just one manifold, right?. So is diffeomorphism is a property of each manifold on its own? or a property which is decided upon examining two manifolds together? Sorry if the question is utterly messed up
At 38:00 you say that the function (gamma ∘ f ∘ phi⁻¹) maps ℝᵈ to ℝᵇ where d is the dimension of X and b is the dimension of Y. But is that really generally true? Surely d and b don't have to have the same dimension as the manifold group since we're not actually mapping to or from the manifold, just spaces that our atlases let us access? For example, X and Y could be of dimension 3 but the map 'gamma' takes X -> ℝ² and the map 'phi' takes Y -> ℝ. That way (gamma ∘ f ∘ phi⁻¹) would map ℝ to ℝ², which would still be a valid map but have nothing to do with the dimension of X and Y. Am I wrong here?
The dimension of X as a manifold is literally *defined* to be d! And the map gamma from X to R^d *must* be a homeomorphism by the definition of a manifold. This forces the conditions I indicated. Now, you could take a subset of X and map that subset to a lower dimensional Euclidean space, but then you might as well just redraw everything replacing X with the relevant subset.
Thank you for your reply! In that case, perhaps I am getting confused about the definition of 'dimension'. Take the stereographic projection as an example (mapping a region in R^3 to a region in R^2). This is a homeomorphism - does that mean that the sphere (missing the single point) in R^3 is actually of dimension 2? I guess so since we can create a sphere in R^3 with just two parameters for the angular position. And it would follow that maps from regions in R^d to regions in R^b could be diffeomorphisms if those *regions* had the same dimension?
Glitches0and0stuff You are correct about the sphere: when we consider its *surface* which is what we do for stereographic projection. And yes we can make a diffeomorphism from R^a to R^a! That is an easy one :)
"Open or closed" @7:18. Can a topological manifold have any closed subsets? I though by definition of a topology all subsets of the manifold must be open.
Note the language! “All the elements of a topology are open sets”. And, “A topological manifold is a manifold with a topology”. So...being a “topological manifold” is not the same a “being a topology”. Any element of the manifold’s topology is also a subset of the manifold, but there are many subsets of the manifold which are NOT elements of the topology. That is the same as saying “Not all subsets of the manifold are open sets.” One definition of a closed set is that its complement is open. So take any open set of the manifold, and it’s complement will be closed!
@@XylyXylyX Thank you I get it! You then go on to say that closed set goes across chart regions which by construction, since the chart regions have bijections defined between them and R^d, these chart regions must be part of the topology of the manifold and they must be open?
All these things which we have discussed in the last two videos work only in those regions which are intersections of two charts? So, are there regions in the manifolds which can not be described as two different charts' intersection, or the hole manifold can be described as intersection of different charts? And if the answer is no, these transitions are valid only if we speak about intersections or these properties does not necessarily depend on intersections? I know that these questions were at least partially answered, but I would like to be sure that I understand correctly.
You have run into a common misconception about this material: you are failing to think abstractly enough! You are imagining a particular set with a particular collection of charts. You must think in terms of all *possible* charts. The nature of a manifold is that there will always be overlapping charts *in principle* and if so, all these charts must follow the rules set forth in the definition of a manifold. Since a local region of a manifold can be homeomorphically mapped to R^n, there are always many ways to do this (just shif the origin of the mapping, for example) so there are always intersecting charts, in principle.
Hey! Great video! Thanks a lot. Can you give an example of the gamma function (here , the function that maps a point from a manifold to a euclidean n-space which gives the point its coordinates)?
Shadab Shaikh Sure. Consider the set of all triples of points such that X^2 + Y^2 + Z^2 =1. That set is the manifold. For one chart Gamma might simply take (X, Y, Z) and assign the coordinate in R^3 of (X, Y, Z) or (R, \theta, \phi), or (R, \theta, Z).
Thanks. But do those set of triples form a manifold? Because, AFAIK, a sphere is 2-dimensional manifold and hence the gamma function for such a sphere will take in a 2-dimensional point
Shadab Shaikh Yes, those triples form a manifold. The chart mapping will produce a point in R^2 for each of the triples. The triples are severely constrained by the equation of the sphere and although there are three numbers for each point in the manifold, the equation makes is to there are only really two degrees of freedom.
I wish I knew...sigh. I am making these lectures so a person can read any book on the subject and not be starting from scratch. I would pick a book that is somewhat technical. There is no point in only learning a *little* about such a deep subject. One book I like is called "Differential Geometric Structures" by Walter Poor (Dover). It isn't an easy text, though. Let me think about it. Maybe some other subscribers will have a suggestion?
It must be a enlightening exposition but a bit confusing witout suitable examples. For examples the curves drawn in topological spaces which are merelly sets with some topological assumptions on the super structure. If we assume parametric function fixes an element in the set how the near ness of the elements can be imagined . Moreover if the smoothness or differentiability is the character we assigne in the image space in RxRx..from R, how it reveals the reality of the real world ( the primal set and the topology therewith ) ?
Abhijit Chaudhuri The nearness of elements can be understood topologically or metrically. Topologically, nearness is established by how points can be separated by open sets. Metrically, nearness is established by literally how far apart they are according to a metric. Our discussion here is mostly topological and very abstract. Regarding your second question: we are modeling the real world with differentiable manifold. It is up to us to decide if this model if useful and accurate. It may not be. That is the art of mathematical modeling. We only understand calculus in R^n. If we think calculus means something for some other set then need to invest calculus for that set. As far as I know, the only way to invent calculus on any other set becides R^n is to show that the set is locally homeomorphic to R^n, find an atlas of local homeomorphism and use that atlas to pull the calculus of R^n onto the set. There have been attempts to modify the theory of manifolds by putting a different topology on R^n. In particular the "light cone topology" has been used on R^n for spacetime physics.
Ok thanks. But I have another doubt. When we were trying to check f's differentiability we took γ^-1ofoφ but after taking γ^-1of this point might fall in the mapping region of 2 or more charts in Y. How do we choose what our φ should be?
The curve that is shown at 7.01is in X but the charts U and V are shown with respect to f(lambda). As there are no coordinates at this stage how does one know which chart the curve passes through so that the correct mapping for that portion of the curve can be used?
L Nigogosian Hmm...I think I understand your question. My answer is that if we have (X, T_x, A) then we DO have coordinates. At the time you cited them, I did not have them depicted yet, but they exist because the atlas A is associated with the manifold. When I say "X is a manifold" I mean that it has a coordinate system available over each of its charts. F(\lambda), as a practical matter is usually x^\mu (f(\lambda)=x^\mu(\lambda). That is, the curve is given as the coordinates of the curve as a function of \lambda. This implicitly uses the chart mappings, but they are not shown. That causes some confusion when learning this material!
L Nigogosian Also, it is possible to have f(\lambda) explicitly return points in X without reference to the coordinates. Each of the charts are subsets of X and the atlas covers X so it should not be a problem to identify which charts each point of f(\lamba) lives in. That is if U is a subset of X we presume we understand what is in and what is not in U. Once we know what chart region a point lies inside, we have access to a chart mapping to determine its coordinates.
I am finding it difficult to get my head round this problem. Perhaps an example would put things in perspective. Let us say that we plot f(lambda) in R^2. We get a curve because R^2 has a coordinate system. Later on in the video you present the method where we get coordinates associated with each chart mapped back from R^2. If we plotted the same coordinates used for R^2 on the charts U and V would we get a representation of the curve in X? How is the intersection section depicted as there would be two representations one for U and one for V? I am a retired engineer trying to understand the basics of General Relativity. I find your videos very useful and well thought out.
L Nigogosian Manifolds are difficult for engineers because they are so abstract and I'm not helping by keeping them that way. Consider the sphere: all point such that x^2 + y^2 +z^2 = 1. Consider those points to be the set X. Now create a collection of overlapping sections of those points that cover X. Those are the chart regions. Now for each region arbitrarily create a mapping to R^2. It it trivial to determine which regions any particular triplet (x, y, z) is in. Using each such region's mapping you can get a local coordinate for the triple (x,y,z) in R^2. Now f(\lambda) gives a triple (x,y,z) in X and the coordinates of that point in R^2 would be \gamma(f(\lambda)).
If a function maps to the real numbers, does the derivative *of that function* also map to real numbers? The answer is apparently yes, but my question is "Why?" I don't see anything readily intuitive regarding the idea of 'applying some operator (in this case the derivative) to a function' and being guaranteed that the resulting, new function should also map to the real numbers.
Perhaps I am misunderstanding the definition of a function. I did not think that a function necessarily had to output to the real numbers. Couldn't it output to some other space?
Samuel Cramer a function may map any space to any space, but if you have a function from a manifold to the real numbers it derivative will also be a function from the manifold to the real numbers.
Great, but my question is 'why?' Is that just how the derivative mapping (because the derivative is technically 'a mapping', right? ...sort of like how you defined velocity in the next lecture) is defined?...i.e. the derivative takes, as input, a function that maps real numbers to real numbers, and then outputs a *new* function that maps real numbers to real numbers?
The function does not only need to map TO the real numbers, but also FROM the real numbers. The short answer is that the derivative is defined as such, the long answer gives some motivation about the ' why' : If you have seen multivariate calculus, i.e. differentiable functions f: R^n to R^m, then the derivative (at a point x) is defined as a matrix, usually the jacobi matrix, (or if you will a linear map) such that it is the best linear approximation to f around the point x. The idea of the derivative is that it is a very good linear approximation of some function which maps from R^n to R^m. Now, if you want to approximate a function which takes its values in R^m then you would at the very least expect the derivative to take on values in the same space, namely R^m. Mathematicians found that the idea of a derivative is so nice that instead of redefining it to map to other spaces, we use the charts on a manifold in order to still talk about functions from R^n to R^m, while actually taking the derivative on the manifold.
So around 15 minutes, I see two maps to Rn, a manifold and an originating R (time). Later, (35 minutes) we will see two manifolds and ALSO another R line. On the one hand, I see distinctions between all of these things, with the manifold being the "papa" But on the other hands, I see just nothing but spaces. And maps between spaces. And, in some sence, ALL the spaces are manifolds and the "papa" loses privilege. But then again, the "papa" is the Real World, so maybe there is a privilege. Could you elaborate? In other words, at 39 minutes, I see two manifolds and FOUR spaces of Rn. But in reality, isn't everything a manifold?
Thank you for replying .But please reply to this also. I have written it as best as I can. Thanks again for replying let me consider an example of a differentiable manifold. Now take Euclidian R which is 1 dimensional line and multiply all the rational numbered point by -1 .SO now the point that was at 0.9 now it will be at -0.9. and vice versa. so now let us assume that this resultant topological space exhibits homeomorphism in this way. now we already have a topological space that is a differentiable manifold and its atlas consists of only one chart and that is all of the manifold and it is mapped in the above given way to all of Euclidian R.
Now we will discard this atlas and come up with another atlas. This atlas will consist of open sets that can be labeled by an integer n. now nth open set will have all the irrational numbers between n and n+1.5 . and it will have all the rational numbers between -n-1.5 and -n . now this will cover all of R. and also each each set can easily be mapped to an open set in Euclidian R. for example nth open set will be homomorphically mapped to (n,n+1.5) open set in Euclidian R . so using this atlas we will say that this is a differentiable manifold. now please form a differentiable curve on this and try to make it a differentiable with respect to all the charts that it passes through. please reply and clear this dought.
I order to be a differentiable manifold the charts must be homeomorphic to R^n and R^n must have the standard topology of open balls. This severely restricts the allowed topology on the manifold itself. The situation you describe could never happen because the charts would not be homeomorphic to R^n. That is, the open sets you defined for the manifold are not homeomorphic to R^n as required.
@@XylyXylyXagain thanks for replying.but still my dought is not cleared. but there is one to one and onto mapping between every open set from the above topological R to ecludian R. if u concider above transformation function that multiplys all the rational numbers by -1 and if u choose an open set in ecludian R then there will be an image open set in the above R. for example (a,b) is an open set in R the preimage of this is just remove all the rational numbers in that range and multiply them by -1 . U will end up getting an open set in the above topological R. Repeat this process for every open set in ecludian R u will always endup getting an open set in topological R. So all the open sets in ecludian R will be easily mapped to all the possible open sets in topological R.I don't see how is it not homoeomorphic . would u please give a counter example of this to prove that the above topological R is not homomorphic. But the real question is to prove that each open set of each chart that I hv concidered is homomorphic to an open set of ecludian R. So let us concider some nth chart .so it will have all the irrational numbers between n and n+1.5. and it will have all the rational numbers between -n-1.5 and -n. Now remember n is an integer so it can be negetive. Now what I claim is that nth chart's open set will exhibit homomorphism with (n,n+1.5) open set in ecludian R under the function that multiplys all rational numbers by -1and leaving all irrational numbers untouched.now would u please give a counter example that proves that this is not a homomorphism. Again thanks for replying .I am unable to find the answer for this anywhere in the net. I hope u will answer.
0:00 Start (Review of differentiable Manifolds).
3:56 Curves As a Function on R.
6:28 Curve Function on R as defined by the variable, Lamda traverses a number of charts in the Topological Space.
8:24 Curves in normal calculus (e.g. R2 with axes X and Y).
10:53 How to assign coordinates in R2 of the curve in the Topological Space using the composition of two functions R to T to R2.
12:19 The same method can be applied to a different chart in the Topological Space.
13:11 One can take the derivative of the curve in R2 as a function of Lambda.
15:27 Coordinates.
17:12 Coordinate Functions.
18:50 Tradition treatment of coordinate function in physics textbooks on Relativity.
22:02 More details on these coordinate mappings.
23:50 what are the old coordinates in the traditional physics textbook approach.
27:09 The second coordinate is derived by much the same.
27:33 Using the other way the old coordinates can also be derived from the new coordinates.
27:56 One need keep these in mind that everything is chart dependent.
28:51 One can then employ normal calculus here and transitions functions depend on the underlying Manifold.
30:00 Another way of understanding coordinate systems by back projecting coordinates to the manifold.
32:11 Back projecting coordinates from another chart in the atlas to the same manifold.
35:21 More on smoothness in regards to maps and charts and functions between manifolds.
38:57 Diffeomorphic Spaces: Homomorphic plus differentiable.
43:07 Summary: Curves, Coordinate Systems, and Diffeomorphisms.
Thanks!
Thanks for the list!
Also, did you mean homEomorphic and differentiable (for timestamp 38)
wow.. you really like physics
The distinction between coordinates, manifold, and transition functions and the proper domains and co-domains is made with beautiful clarity! Thank you for lifting the haze left by most textbooks.
Shahid Naqvi I'm glad it helped, but don't ignore the textbooks! Go back to them. We learn by repetition and re-visitation! It is time to revisit! :)
Damn ! That was an eye opener .. the way we connect the idea of different charts on a manifold in our heads with the diffrerent coordinate systems, you don't find such an explanation in ordinary text books on the subject. I appreciate your work, respect !
This comments apply to all the lectures out in manifolds and tensor sections. Best of all the topology and tensor lectures out here in or elsewhere in univs . Thumbs up. It cleared a lot of confusion I had about these topics. If you write a book that would probably be very good for the community too :)
Great Great lecture series. Thank you for your hard work and service to the academic community. Please make more series on more topics
Wow! This was very clear! You are an awesome teacher!
Thank you for your kind comment.
At 38:44 for the discussion aren't we only concerned with the mapping of a single chart and not the intersection of two different charts in X? I suspect if we did use the intersection charts of A,B multiple paths from gamma(V) to phi(V) could be created.
Mappings are always from a single chart and the discussion you cite uses the charts (U,\gamma) and {V,\phi). The intersections are there to illustrate that X and Y are manifolds and that we can always change charts.
Thanks a lot.
You are literally saving lives(mathematically speaking :)
At about timestamp 35:40, is the argument presented related to 'push-forward' and 'pull-back' between manifolds?
No. That is a different concept.
@38:10 can we also use the mapping Psi•f•inv(G) (Where suppose 'Psi' is a map from the second Manifold into the second Euclidean space)?
I'm not understanding how polar coordinates work here. In the chart at 34:32, the polar coordinates r and theta are drawn in a rectilinear way. Why would we do this? I understand how to visualize the open ball topology with rectilinear coordinates but not polar coordinates.
It is an interesting and important question you ask! Ultimately EVERY manifold must go back to a Cartesian R^n. The polar coordinates you are thinking of are actually an expression of the fact that R^n itself is a manifold and the chart mappings can label each point with (R,theta,phi), but the other side of the final chart mapping is always Cartesian and that is where the actual calculus is happening. Remember….ultimately we only know how to do calculus in a *Cartesian* setting. Even when we do integrals in polar coordinates, if you take them apart fully you see that we are actually integrating coordinate by coordinate as though we were in a Cartesian system.
An open ball in the system I have drawn is literally a ball, just like in ANY Cartesian system. .
At 39:30:
How can "f" be differentiable if it is a map bewteen topological spaces?
I imagine PHI composed f composed gamma-inverse can be differentiable... but not f alone.
Thomas Impelluso Yes, that is a "abuse of language" to say "f is differentiable". However, it is common language to say "f is differentiable" and all the machinery of the atlas' charts is hidden beneath the language as an assumption. We must always, in the back of our minds, remember that a function literally between topological spaces is not in-and-of-itself "differentiable". Only that function composed with the various chart maps is differentiable.
Hmm at 15:05 can I wrap up this way.
As you mentioned in the last video, a set X is a model of real world(could be an apple, banana, a pile of manure or whatever) and it is an abstract set that contains information that we are interested in. But we want to analyze those abstract sets with mathematical tools.
To define differentiation we need an argument variable and continuous function in metric space. so f that maps R to X is a set-up process for using parameter lamda as our argument variable, adoption of function gamma and phi is to give out metric values w.r.t specific coordinates
Jay Lee That sounds right. THe key point being that changing the coordinates should not change anything important.
Thanks, I have one more question. you said we cannot differentiate a function on topological space X and that is why we define a function that maps X onto R. However, at 40:52, if diffeomorphism, we should be able to differentiate the whole chunk of composite functions. How can we differentiate f where it is a function that maps X onto Y?
is function f itself is a composite of some functions that allows differentiation?
Jay Lee I'm not sure I exactly understand your question but the *definition* of the differentiability of a function like f is that functions like \phi o f o \gamma^{-1} are differentiable in the sense of normal calculus. That is the ONLY way "calculus" of functions between sets can be understood. No manifold, no calculus.
Thanks now I get it :)
Thanks so much for this series. It is absolutely top notch teaching.
A question: I am struggling to understand the discussion around 43:00 which seems to imply that homeomorphisms (and diffeo) between two manifolds can map onto Euclidean spaces of different dimension. Is there an example of this?
There is not example of this because it is not true! Homeomorphisms and diffeomorphisms must map manifolds of the same dimensionality. The drawing at that time stamp was regarding a *general* mapping f. If f has certain properties the two manifolds will be diffeomorphic. Those requirements ( one to one and onto in particular) will force the two manifolds to have the same dimension. Thanks for watching!
I guess one of the reasons these lectures work out so well is that you tireless draw those diagrams of maps. I guess it would be a big challenge to try present the same material in a book though. This is almost like learning an instrument, where the best way is not to read a book but to get constant feed back from an instructor by examples and demonstrations.
Physics I think you are right. I have abandoned all inhibition to repeat myself and no-body really complains. :) I like it that way.
Thanks a lot! These lessons are extremely valuable. I have a question: In this video, it seems that diffeomorphism is a property of two manifolds examined together. If I remember correctly, in a previous video, it was mentioned that if the transition function is differentiable then we have diffeomorphism. Transition functions consider different charts of just one manifold, right?. So is diffeomorphism is a property of each manifold on its own? or a property which is decided upon examining two manifolds together?
Sorry if the question is utterly messed up
At 38:00 you say that the function (gamma ∘ f ∘ phi⁻¹) maps ℝᵈ to ℝᵇ where d is the dimension of X and b is the dimension of Y.
But is that really generally true? Surely d and b don't have to have the same dimension as the manifold group since we're not actually mapping to or from the manifold, just spaces that our atlases let us access?
For example, X and Y could be of dimension 3 but the map 'gamma' takes X -> ℝ² and the map 'phi' takes Y -> ℝ. That way (gamma ∘ f ∘ phi⁻¹) would map ℝ to ℝ², which would still be a valid map but have nothing to do with the dimension of X and Y. Am I wrong here?
Or rather (phi ∘ f ∘ gamma⁻¹) is the way you had it in the video, but you get my point
The dimension of X as a manifold is literally *defined* to be d! And the map gamma from X to R^d *must* be a homeomorphism by the definition of a manifold. This forces the conditions I indicated.
Now, you could take a subset of X and map that subset to a lower dimensional Euclidean space, but then you might as well just redraw everything replacing X with the relevant subset.
Thank you for your reply! In that case, perhaps I am getting confused about the definition of 'dimension'. Take the stereographic projection as an example (mapping a region in R^3 to a region in R^2). This is a homeomorphism - does that mean that the sphere (missing the single point) in R^3 is actually of dimension 2? I guess so since we can create a sphere in R^3 with just two parameters for the angular position.
And it would follow that maps from regions in R^d to regions in R^b could be diffeomorphisms if those *regions* had the same dimension?
Glitches0and0stuff You are correct about the sphere: when we consider its *surface* which is what we do for stereographic projection.
And yes we can make a diffeomorphism from R^a to R^a! That is an easy one :)
"Open or closed" @7:18.
Can a topological manifold have any closed subsets? I though by definition of a topology all subsets of the manifold must be open.
Note the language! “All the elements of a topology are open sets”. And, “A topological manifold is a manifold with a topology”. So...being a “topological manifold” is not the same a “being a topology”. Any element of the manifold’s topology is also a subset of the manifold, but there are many subsets of the manifold which are NOT elements of the topology. That is the same as saying “Not all subsets of the manifold are open sets.” One definition of a closed set is that its complement is open. So take any open set of the manifold, and it’s complement will be closed!
@@XylyXylyX Thank you I get it! You then go on to say that closed set goes across chart regions which by construction, since the chart regions have bijections defined between them and R^d, these chart regions must be part of the topology of the manifold and they must be open?
All these things which we have discussed in the last two videos work only in those regions which are intersections of two charts? So, are there regions in the manifolds which can not be described as two different charts' intersection, or the hole manifold can be described as intersection of different charts? And if the answer is no, these transitions are valid only if we speak about intersections or these properties does not necessarily depend on intersections?
I know that these questions were at least partially answered, but I would like to be sure that I understand correctly.
You have run into a common misconception about this material: you are failing to think abstractly enough! You are imagining a particular set with a particular collection of charts. You must think in terms of all *possible* charts. The nature of a manifold is that there will always be overlapping charts *in principle* and if so, all these charts must follow the rules set forth in the definition of a manifold.
Since a local region of a manifold can be homeomorphically mapped to R^n, there are always many ways to do this (just shif the origin of the mapping, for example) so there are always intersecting charts, in principle.
@@XylyXylyX Thank you for your answer.
What software do you use to make this?
It is called "Vittle" I got it from Mac App store. Luuuuuv it.
Hey! Great video! Thanks a lot.
Can you give an example of the gamma function (here , the function that maps a point from a manifold to a euclidean n-space which gives the point its coordinates)?
Shadab Shaikh Sure. Consider the set of all triples of points such that X^2 + Y^2 + Z^2 =1. That set is the manifold. For one chart Gamma might simply take (X, Y, Z) and assign the coordinate in R^3 of (X, Y, Z) or (R, \theta, \phi), or (R, \theta, Z).
Thanks. But do those set of triples form a manifold? Because, AFAIK, a sphere is 2-dimensional manifold and hence the gamma function for such a sphere will take in a 2-dimensional point
Also, do you have a tutorial on partitions of unity used for patching together charts?
Shadab Shaikh Yes, those triples form a manifold. The chart mapping will produce a point in R^2 for each of the triples. The triples are severely constrained by the equation of the sphere and although there are three numbers for each point in the manifold, the equation makes is to there are only really two degrees of freedom.
Shadab Shaikh No, I didn't cover that important proof. Probably should.
hello
what is best book to understanding manifold and differentiable geometry for general relativity
I wish I knew...sigh. I am making these lectures so a person can read any book on the subject and not be starting from scratch. I would pick a book that is somewhat technical. There is no point in only learning a *little* about such a deep subject. One book I like is called "Differential Geometric Structures" by Walter Poor (Dover). It isn't an easy text, though. Let me think about it. Maybe some other subscribers will have a suggestion?
+XylyXylyX thank you sir,
It must be a enlightening exposition but a bit confusing witout suitable examples. For examples the curves drawn in topological spaces which are merelly sets with some topological assumptions on the super structure. If we assume parametric function fixes an element in the set how the near ness of the elements can be imagined . Moreover if the smoothness or differentiability is the character we assigne in the image space in RxRx..from R, how it reveals the reality of the real world ( the primal set and the topology therewith ) ?
Abhijit Chaudhuri The nearness of elements can be understood topologically or metrically. Topologically, nearness is established by how points can be separated by open sets. Metrically, nearness is established by literally how far apart they are according to a metric. Our discussion here is mostly topological and very abstract.
Regarding your second question: we are modeling the real world with differentiable manifold. It is up to us to decide if this model if useful and accurate. It may not be. That is the art of mathematical modeling. We only understand calculus in R^n. If we think calculus means something for some other set then need to invest calculus for that set. As far as I know, the only way to invent calculus on any other set becides R^n is to show that the set is locally homeomorphic to R^n, find an atlas of local homeomorphism and use that atlas to pull the calculus of R^n onto the set. There have been attempts to modify the theory of manifolds by putting a different topology on R^n. In particular the "light cone topology" has been used on R^n for spacetime physics.
Why is it so that for a manifold to be diffeomorfic to another both of them have to be differentiable manifolds?
That is just what “diffeomorphic” means: a homeomorphism between diffeomorphic manifolds.
Ok thanks. But I have another doubt. When we were trying to check f's differentiability we took γ^-1ofoφ but after taking γ^-1of this point might fall in the mapping region of 2 or more charts in Y. How do we choose what our φ should be?
Madhumita Dutta please send me a time stamp for the video and I’ll have a look
37:55
The curve that is shown at 7.01is in X but the charts U and V are shown with respect to f(lambda). As there are no coordinates at this stage how does one know which chart the curve passes through so that the correct mapping for that portion of the curve can be used?
L Nigogosian Hmm...I think I understand your question. My answer is that if we have (X, T_x, A) then we DO have coordinates. At the time you cited them, I did not have them depicted yet, but they exist because the atlas A is associated with the manifold. When I say "X is a manifold" I mean that it has a coordinate system available over each of its charts. F(\lambda), as a practical matter is usually x^\mu (f(\lambda)=x^\mu(\lambda). That is, the curve is given as the coordinates of the curve as a function of \lambda. This implicitly uses the chart mappings, but they are not shown. That causes some confusion when learning this material!
L Nigogosian Also, it is possible to have f(\lambda) explicitly return points in X without reference to the coordinates. Each of the charts are subsets of X and the atlas covers X so it should not be a problem to identify which charts each point of f(\lamba) lives in. That is if U is a subset of X we presume we understand what is in and what is not in U. Once we know what chart region a point lies inside, we have access to a chart mapping to determine its coordinates.
I am finding it difficult to get my head round this problem. Perhaps an example would put things in perspective. Let us say that we plot f(lambda) in R^2. We get a curve because R^2 has a coordinate system. Later on in the video you present the method where we get coordinates associated with each chart mapped back from R^2. If we plotted the same coordinates used for R^2 on the charts U and V would we get a representation of the curve in X? How is the intersection section depicted as there would be two representations one for U and one for V?
I am a retired engineer trying to understand the basics of General Relativity. I find your videos very useful and well thought out.
Just a correction to my entry. The second paragraph should read "let us say we plot f(lambda) o (gamma) in R^2.
L Nigogosian Manifolds are difficult for engineers because they are so abstract and I'm not helping by keeping them that way. Consider the sphere: all point such that x^2 + y^2 +z^2 = 1. Consider those points to be the set X. Now create a collection of overlapping sections of those points that cover X. Those are the chart regions. Now for each region arbitrarily create a mapping to R^2. It it trivial to determine which regions any particular triplet (x, y, z) is in. Using each such region's mapping you can get a local coordinate for the triple (x,y,z) in R^2.
Now f(\lambda) gives a triple (x,y,z) in X and the coordinates of that point in R^2 would be \gamma(f(\lambda)).
If a function maps to the real numbers, does the derivative *of that function* also map to real numbers? The answer is apparently yes, but my question is "Why?"
I don't see anything readily intuitive regarding the idea of 'applying some operator (in this case the derivative) to a function' and being guaranteed that the resulting, new function should also map to the real numbers.
The derivative of a function, if it is differentiable, will always be another function. I’m not sure of any other way of addressing your question.
Perhaps I am misunderstanding the definition of a function. I did not think that a function necessarily had to output to the real numbers. Couldn't it output to some other space?
Samuel Cramer a function may map any space to any space, but if you have a function from a manifold to the real numbers it derivative will also be a function from the manifold to the real numbers.
Great, but my question is 'why?' Is that just how the derivative mapping (because the derivative is technically 'a mapping', right? ...sort of like how you defined velocity in the next lecture) is defined?...i.e. the derivative takes, as input, a function that maps real numbers to real numbers, and then outputs a *new* function that maps real numbers to real numbers?
The function does not only need to map TO the real numbers, but also FROM the real numbers.
The short answer is that the derivative is defined as such, the long answer gives some motivation about the ' why' :
If you have seen multivariate calculus, i.e. differentiable functions f: R^n to R^m, then the derivative (at a point x) is defined as a matrix, usually the jacobi matrix, (or if you will a linear map) such that it is the best linear approximation to f around the point x.
The idea of the derivative is that it is a very good linear approximation of some function which maps from R^n to R^m. Now, if you want to approximate a function which takes its values in R^m then you would at the very least expect the derivative to take on values in the same space, namely R^m.
Mathematicians found that the idea of a derivative is so nice that instead of redefining it to map to other spaces, we use the charts on a manifold in order to still talk about functions from R^n to R^m, while actually taking the derivative on the manifold.
So around 15 minutes, I see two maps to Rn, a manifold and an originating R (time). Later, (35 minutes) we will see two manifolds and ALSO another R line.
On the one hand, I see distinctions between all of these things, with the manifold being the "papa"
But on the other hands, I see just nothing but spaces. And maps between spaces. And, in some sence, ALL the spaces are manifolds and the "papa" loses privilege.
But then again, the "papa" is the Real World, so maybe there is a privilege.
Could you elaborate?
In other words, at 39 minutes, I see two manifolds and FOUR spaces of Rn. But in reality, isn't everything a manifold?
Thomas Impelluso The manifold is the object that has the atlas. Everything else is somehow associated with the atlas.
Thank you
Thanks a lot for these!
Thank you for replying .But please reply to this also. I have written it as best as I can. Thanks again for replying
let me consider an example of a differentiable manifold. Now take Euclidian R which is 1 dimensional line and multiply all the rational numbered point by -1 .SO now the point that was at 0.9 now it will be at -0.9. and vice versa. so now let us assume that this resultant topological space exhibits homeomorphism in this way. now we already have a topological space that is a differentiable manifold and its atlas consists of only one chart and that is all of the manifold and it is mapped in the above given way to all of Euclidian R.
Now we will discard this atlas and come up with another atlas. This atlas will consist of open sets that can be labeled by an integer n. now nth open set will have all the irrational numbers between n and n+1.5 . and it will have all the rational numbers between -n-1.5 and -n . now this will cover all of R. and also each each set can easily be mapped to an open set in Euclidian R. for example nth open set will be homomorphically mapped to (n,n+1.5) open set in Euclidian R . so using this atlas we will say that this is a differentiable manifold. now please form a differentiable curve on this and try to make it a differentiable with respect to all the charts that it passes through. please reply and clear this dought.
I order to be a differentiable manifold the charts must be homeomorphic to R^n and R^n must have the standard topology of open balls. This severely restricts the allowed topology on the manifold itself. The situation you describe could never happen because the charts would not be homeomorphic to R^n. That is, the open sets you defined for the manifold are not homeomorphic to R^n as required.
@@XylyXylyXagain thanks for replying.but still my dought is not cleared.
but there is one to one and onto mapping between every open set from the above topological R to ecludian R. if u concider above transformation function that multiplys all the rational numbers by -1 and if u choose an open set in ecludian R then there will be an image open set in the above R. for example (a,b) is an open set in R the preimage of this is just remove all the rational numbers in that range and multiply them by -1 . U will end up getting an open set in the above topological R. Repeat this process for every open set in ecludian R u will always endup getting an open set in topological R. So all the open sets in ecludian R will be easily mapped to all the possible open sets in topological R.I don't see how is it not homoeomorphic . would u please give a counter example of this to prove that the above topological R is not homomorphic.
But the real question is to prove that each open set of each chart that I hv concidered is homomorphic to an open set of ecludian R. So let us concider some nth chart .so it will have all the irrational numbers between n and n+1.5. and it will have all the rational numbers between -n-1.5 and -n. Now remember n is an integer so it can be negetive.
Now what I claim is that nth chart's open set will exhibit homomorphism with (n,n+1.5) open set in ecludian R under the function that multiplys all rational numbers by -1and leaving all irrational numbers untouched.now would u please give a counter example that proves that this is not a homomorphism.
Again thanks for replying .I am unable to find the answer for this anywhere in the net. I hope u will answer.
I am still expecting a reply.
I think the problem of homeomorphism lies in this case with countability equivalence of the two charts