Really fantastic videos! I've watched so many different lectures and youtube videos from different people and yours are excellent, easily in contention with the best I've seen.
Aww I loved Lula getting impatient around 20 minutes in. Also, I think almost all of your comments are this, but you do an amazing job demystifying things that authors of all of the smooth manifolds books seem to be unable to.
I've been reading sean carrolls book on spacetime and general relativity which i think is excellent but you have really explained some of the 'big machinary' he refers to in his book. Thanks!!! Also I think your perspective on what a tangent vector really is helped me.
@WHYB Maths At 5:11, I'm having trouble distinguishing f and phi(lambda) because they both seem to be elements of the set of all smooth curves on M. What are the input and output of each, and what is their relationship to each other? Thanks so much for taking the time to make this series.
The function f is meant to represent an arbitrary (smooth) function acting on the points in the manifold, that maps points in the manifold into some R space. So in this scenario, phi(lambda) defines a curve (subset of points) throughout the manifold, which is then taken as the domain for the function f(phi(lambda)) -> R. In a lot of cases, when defining the tangent space it is customary to just use f = id the identity, so that the tangent vectors are obtained directly from the phi(lambda) themselves, and thus are quite intuitively the 'tangent' vectors to the smooth curves on the manifold. But the essence of what I'm trying to convey here is that we should stop thinking about vectors as arrows, but rather as differential operators that act on functions defined over a manifold, which is why the arbitrary f is used. Hope that helps, let me know if you are still confused!
Yes you are correct, my mistake! Everything else I end up saying is correct, but just with replacing f(x^\mu) with f(\phi(\lambda)) - essentially they are both talking about the same object (the image of the curve under the function f) but one uses the image of the curve in the chart to be the domain rather than the actual curve itself. Thanks for spotting it!
18:50, vector operating on a function ... why is this stated this way? I assumed other way round? Isn't An operator a map from one vector space or module to another. (1) So what is meant exactly with "the Vector" just before the summary at 19:15 , isn't it rather something that creates those "vectors" we are interested in ? (I think I got confused with what the partial derivative/basis of tangent space means exactly, as normally a basis would not look like this, that is, it should look like a special vector in that sense instead of something that is trying to act on something else) Different question (2): What about all other possible curves on the manifold, did we now restrict ourselves to just one or is it representative of them all through P ?
Hi, thanks for the video! At t=16m you write f o x^-1 = f(x^mu), but it can't be the same f in the right hand side, can it? Do you just mean that it is another arbitrary function like f (which we might call, say, g, and g != f)?
Technically yes it would be better to call it some other arbitrary function, since really the function is F(x^\mu) = f o x^-1 (x^\mu). So whilst it is a bit sloppy to just write f(x^\mu), since the point in the chart and manifold must both map to the same point under the function f, it is usually easier in practice just to write it this way since using the other F would simply amount to having more chart functions in the chain rule.
Hi. Thanks. At 4:55 you begin to define a vector by having the vector "act" on a function. Isn't the vector already defined as the velocity of the curve phi? I thought the tangent vector WAS the velocity vector at each point...that is...the derivative of phi. I must be missing something. Also, can you explain exactly what it means for a vector to "act" on a function?
Thanks a lot for the video. I'm still confused about the motivation for introducing the "f". If I understood correctly, you are conveying that the parameter derivative of a function acting on a curve of the manifold equals to a directional derivative of another(derived) function on the chart. But from the last video, I learned that we need the velocity vectors(without applying "f") of the curves on the original manifold to construct the tangent space. I'm wondering how are the two different objects connected? (sorry I don't come with a strong math background 😥).
If I understand ur question: function is required because, the curve gives a point or points on the manifold from a parameter in R. The function works on points on the manifold and outputs a number In R. To take a derivative we need a curve to specify adjacent points and the function will provide values as we move through the curve. So it is called directional derivative. The curve is map from R to Manifold, and function maps points on manifold to R.
I've been struggling with the tangent space definition a lot. Majorly bc it implies 4 different definitions, in 2 of these considering curves to be members of the tangent space, an the other 2 - linear operators (derivatives) And I'm still facing issues with understanding what is a member of the tangent space: an operator or a curve... Why is it so hard to perceive
Thanks, great video. I'm a little concerned by what looks like a subtle shift in meaning in the notation. First the function 'f' is a function on the manifold to R. Then, when you start rearranging the algebra, it gets reinterpreted as a function from R^2 (the image space of the chart) to R. That's fine to do conceptually, but the same 'f' symbol is used for both and that gets me a little worried. I've seen the same subtle shift elsewhere. Why not use a different symbol for what are actually different functions? Or am I misunderstanding something? Thanks if you can explain.
Geometry topology and physics is great from Nakahara, mostly a pure maths textbook focusing on all aspects of differential geometry and topology. Spacetime and geometry by Sean Carrol is another good textbook but is more focused on physics and relativity but has a decent level of differential geometry also. I'm also starting a series on relativity which will eventually fully cover GR!
Hi, thanks for the video! I have a question at the 15:42: you wrote d(fox^-1)/d(x^mou) * d(x^mou)/d(lamda). But if you eliminate the d(x^mou) you do not end up with the previous eq (the term (x^mou(lamda)) is missing). Is that true or am I missing something?
It's slightly misleading how I wrote things, I should have left the argument in the second derivative term because it's really d(f o x^-1)/d(x^mu) * d ( x^mu(lambda))/d(lambda) so we cannot simply use the chain rule and eliminate the d (x^mu), notice it is still present in both numerator and denominator of the following expression. I probably could have used a slightly better name for the parameterised curve which I'm calling x^mu (the same as the coordinates themselves) say gamma^mu, then the final equation would be d (gamma^mu)/d (lambda) * d f/dx^mu, hope that helps let me know if you're still confused!
@@WHYBmaths That makes a bit more sense but then the x^mu and gamma^mu are different parameterized curves in the eq: d (gamma^mu)/d (lambda) * d f/dx^mu? And if yes what is their difference? Thank you!
@@stergiosverros5000 So I mean to replace the curve (previously denoted x^mu(lambda)) entirely with gamma^mu(lambda), you will still have x^mu which are referring to the chart map functions themselves. It was a bad choice to denote the curve as x^mu(lambda) since I was inconsistent with writing the argument so it can be confused with the coordinates x^mu. To make sense out of what I had written; if x^mu is the numerator in a derivative it refers to the curve, and if in the denominator it refers to the coordinates. Hope that helps!
Excellent explanation. I like the colors and fast forwarding. However if f(x_inv) = f(x) then x_inv = x, assuming f is injective. This would imply that x is the identity map. If x is the identity map then dx/d(lamda) = 0.
See my reply to a question posted below by PDiracDelta: At t=16m you write f o x^-1 = f(x^mu), but it can't be the same f in the right hand side, can it? Do you just mean that it is another arbitrary function like f (which we might call, say, g, and g != f)? Response: Technically yes it would be better to call it some other arbitrary function, since really the function is F(x^\mu(lambda)) = f o x^-1 (x^\mu(lambda)) = f(phi(lambda)). So whilst it is a bit sloppy to just write f(x^\mu), since the representation of the point in the chart x^mu(lambda) and the actual point in the manifold phi(lambda) must both map to the same point under the function f, it is usually easier in practice just to write it this way since using the other F would simply amount to having more chart and inverse chart functions in the chain rule which would ultimately cancel. (At least, I think that is how it works out if I can remember this correctly, as it was awhile ago now when I filmed this! Please let me know if you are still confused however!)
How do we know for a fact that a single chart can contain the entire curve? Doesn't the fact that we are working with a manifold only tell us that locally there exists some chart (aka homomorphism to R^d)? How can we assume that the entire curve will be in the domain of that chart?
In general a single chart won't contain the entire curve as you said, and yes in full generality only locally does the manifold map to R^d, this is essentially the essence of Riemannian geometry in that these tangent spaces only exist locally and are independent at every point in the manifold. In the case where a manifold has multiple charts all we can expect is that the charts agree and hence the tangent spaces to any curve passing through the intersecting chart must also agree. Usually for simplicity when beginning this construction it is simpler to think of a single chart containing the curve in question, the generalisation then simply deals with each portion of the curve separately in the respective charts (no more complex conceptually but just a computational and notational headache!)
Hi! I have a question. I didn't understand why the partial derivative with respect to x^(mou) becomes the basis, how can we see that? Can someone please answer me thanks. Also, love your lessons.
Still a student! Studying a masters in mathematical physics, and I'm currently working on a project investigating boundary conditions in gauge theories, focusing on developing the Einstein-Cartan formalism of GR in this context! Hoping to make many videos about what I'm doing when I have the chance!
So I'm guessing this is the algebraically-defined tangent space: which is the vector space of derivations of the ring of germs or T_p^{alg}(M), where M is some manifold and p\in M?
Thanks! Could you clarify why "the vector should be a linear map, i.e. it has to map some kind of object to produce a tangible result", why doesnt it make sense to talk about a vector just as en element of a vector space, why does it need to be applied to a function to make sense? Also, the basis is just \delta_mu right? not V^mu \delta_mu right? thanks!
Hey, I am working on something called tangent vectors over a space( I did the translation from my frensh book) but anyway it involves curves and stuffs.. I just wanted to know if you have any ideas about it.. Thank you!
I respect this style but I do not like it and I do not think it is the best way to learn for most people. I think it is pedagogically bad to go through a maze of algebra and then seek to interpret it. I think it is better to start with the concept and its interpretation and then do the algebra: The tangent space is a set of contravariant vectors associated with a point on a manifold. The contravariant vectors themselves are what is called 'derivations'. These 'derivations' are actually functions from the set of continuous functions in the manifold to the reals. The interpretation of the derivations is that they are like little machines that measure how much any input function on the manifold changes for the given increment and direction associated with the derivation (aka tangent vector). These derivations are naturally and intuitively a linear operator. The tangent space is the set of all those tangent vectors together at your point. They naturally and intuitively are a vector space. Then you can go back and do the complicated algebra....
Hi I just have a small question in the beginning of the previous video I understood that smooth parameterizations are on C ᪲ but from. H here I understood the ones on C ᪲ are the derivatives, are both of them part of C ᪲ or did I get something wrong?
Your videos on these topics are by far the best I have seen on the internet, your are a master teacher, thank you so much
Thank you! I'm so glad they are helpful
This is by far the best video i've found about this
Thank you!
Really fantastic videos! I've watched so many different lectures and youtube videos from different people and yours are excellent, easily in contention with the best I've seen.
Wow, thank you so much! Really happy my videos are helping people!
Same herr. I totally agree 😊
this is so so helpful, such a good way of teaching, also the cute dog is a site of comfort!
Aww I loved Lula getting impatient around 20 minutes in.
Also, I think almost all of your comments are this, but you do an amazing job demystifying things that authors of all of the smooth manifolds books seem to be unable to.
What a great teacher you are. Thanks a lot for the videos ❤😊
I've been reading sean carrolls book on spacetime and general relativity which i think is excellent but you have really explained some of the 'big machinary' he refers to in his book. Thanks!!!
Also I think your perspective on what a tangent vector really is helped me.
Excellent! Now I get why they are called vectors! Thanks!!
very very helpful thank you!!!!!
@WHYB Maths At 5:11, I'm having trouble distinguishing f and phi(lambda) because they both seem to be elements of the set of all smooth curves on M. What are the input and output of each, and what is their relationship to each other? Thanks so much for taking the time to make this series.
The function f is meant to represent an arbitrary (smooth) function acting on the points in the manifold, that maps points in the manifold into some R space. So in this scenario, phi(lambda) defines a curve (subset of points) throughout the manifold, which is then taken as the domain for the function f(phi(lambda)) -> R.
In a lot of cases, when defining the tangent space it is customary to just use f = id the identity, so that the tangent vectors are obtained directly from the phi(lambda) themselves, and thus are quite intuitively the 'tangent' vectors to the smooth curves on the manifold. But the essence of what I'm trying to convey here is that we should stop thinking about vectors as arrows, but rather as differential operators that act on functions defined over a manifold, which is why the arbitrary f is used.
Hope that helps, let me know if you are still confused!
At 16:01, should it not be f(\phi(\lambda)) above the curly brackets? This confused me a little...
Yes you are correct, my mistake! Everything else I end up saying is correct, but just with replacing f(x^\mu) with f(\phi(\lambda)) - essentially they are both talking about the same object (the image of the curve under the function f) but one uses the image of the curve in the chart to be the domain rather than the actual curve itself. Thanks for spotting it!
@@WHYBmaths Thanks for your prompt response! I've been really enjoying this playlist, and cannot wait to check out the others :)
@@WHYBmaths Also, Lula and Maeby are the best :D
18:50, vector operating on a function ... why is this stated this way? I assumed other way round? Isn't An operator a map from one vector space or module to another. (1) So what is meant exactly with "the Vector" just before the summary at 19:15 , isn't it rather something that creates those "vectors" we are interested in ? (I think I got confused with what the partial derivative/basis of tangent space means exactly, as normally a basis would not look like this, that is, it should look like a special vector in that sense instead of something that is trying to act on something else) Different question (2): What about all other possible curves on the manifold, did we now restrict ourselves to just one or is it representative of them all through P ?
Hi, thanks for the video! At t=16m you write f o x^-1 = f(x^mu), but it can't be the same f in the right hand side, can it? Do you just mean that it is another arbitrary function like f (which we might call, say, g, and g != f)?
Technically yes it would be better to call it some other arbitrary function, since really the function is F(x^\mu) = f o x^-1 (x^\mu). So whilst it is a bit sloppy to just write f(x^\mu), since the point in the chart and manifold must both map to the same point under the function f, it is usually easier in practice just to write it this way since using the other F would simply amount to having more chart functions in the chain rule.
awesome
Thanks for this video bro
Hi. Thanks. At 4:55 you begin to define a vector by having the vector "act" on a function. Isn't the vector already defined as the velocity of the curve phi? I thought the tangent vector WAS the velocity vector at each point...that is...the derivative of phi. I must be missing something.
Also, can you explain exactly what it means for a vector to "act" on a function?
Thanks a lot for the video. I'm still confused about the motivation for introducing the "f". If I understood correctly, you are conveying that the parameter derivative of a function acting on a curve of the manifold equals to a directional derivative of another(derived) function on the chart. But from the last video, I learned that we need the velocity vectors(without applying "f") of the curves on the original manifold to construct the tangent space. I'm wondering how are the two different objects connected? (sorry I don't come with a strong math background 😥).
If I understand ur question: function is required because, the curve gives a point or points on the manifold from a parameter in R. The function works on points on the manifold and outputs a number In R. To take a derivative we need a curve to specify adjacent points and the function will provide values as we move through the curve. So it is called directional derivative. The curve is map from R to Manifold, and function maps points on manifold to R.
I've been struggling with the tangent space definition a lot. Majorly bc it implies 4 different definitions, in 2 of these considering curves to be members of the tangent space, an the other 2 - linear operators (derivatives)
And I'm still facing issues with understanding what is a member of the tangent space: an operator or a curve...
Why is it so hard to perceive
Thanks, great video. I'm a little concerned by what looks like a subtle shift in meaning in the notation. First the function 'f' is a function on the manifold to R. Then, when you start rearranging the algebra, it gets reinterpreted as a function from R^2 (the image space of the chart) to R. That's fine to do conceptually, but the same 'f' symbol is used for both and that gets me a little worried. I've seen the same subtle shift elsewhere. Why not use a different symbol for what are actually different functions? Or am I misunderstanding something? Thanks if you can explain.
what are good books to learn differential manifold theory for GR??
Geometry topology and physics is great from Nakahara, mostly a pure maths textbook focusing on all aspects of differential geometry and topology. Spacetime and geometry by Sean Carrol is another good textbook but is more focused on physics and relativity but has a decent level of differential geometry also. I'm also starting a series on relativity which will eventually fully cover GR!
Hi, thanks for the video! I have a question at the 15:42: you wrote d(fox^-1)/d(x^mou) * d(x^mou)/d(lamda). But if you eliminate the d(x^mou) you do not end up with the previous eq (the term (x^mou(lamda)) is missing). Is that true or am I missing something?
It's slightly misleading how I wrote things, I should have left the argument in the second derivative term because it's really d(f o x^-1)/d(x^mu) * d ( x^mu(lambda))/d(lambda) so we cannot simply use the chain rule and eliminate the d (x^mu), notice it is still present in both numerator and denominator of the following expression. I probably could have used a slightly better name for the parameterised curve which I'm calling x^mu (the same as the coordinates themselves) say gamma^mu, then the final equation would be d (gamma^mu)/d (lambda) * d f/dx^mu, hope that helps let me know if you're still confused!
@@WHYBmaths That makes a bit more sense but then the x^mu and gamma^mu are different parameterized curves in the eq: d (gamma^mu)/d (lambda) * d f/dx^mu? And if yes what is their difference?
Thank you!
@@stergiosverros5000 So I mean to replace the curve (previously denoted x^mu(lambda)) entirely with gamma^mu(lambda), you will still have x^mu which are referring to the chart map functions themselves. It was a bad choice to denote the curve as x^mu(lambda) since I was inconsistent with writing the argument so it can be confused with the coordinates x^mu. To make sense out of what I had written; if x^mu is the numerator in a derivative it refers to the curve, and if in the denominator it refers to the coordinates. Hope that helps!
@@WHYBmaths Ohh OK. That makes more sense! Thank you very much :)
Excellent explanation. I like the colors and fast forwarding. However if f(x_inv) = f(x) then x_inv = x, assuming f is injective. This would imply that x is the identity map. If x is the identity map then dx/d(lamda) = 0.
I dont understand how f of x inverse is f(x mu) at 16:00
can someone expain why that is true?
See my reply to a question posted below by PDiracDelta: At t=16m you write f o x^-1 = f(x^mu), but it can't be the same f in the right hand side, can it? Do you just mean that it is another arbitrary function like f (which we might call, say, g, and g != f)?
Response: Technically yes it would be better to call it some other arbitrary function, since really the function is F(x^\mu(lambda)) = f o x^-1 (x^\mu(lambda)) = f(phi(lambda)). So whilst it is a bit sloppy to just write f(x^\mu), since the representation of the point in the chart x^mu(lambda) and the actual point in the manifold phi(lambda) must both map to the same point under the function f, it is usually easier in practice just to write it this way since using the other F would simply amount to having more chart and inverse chart functions in the chain rule which would ultimately cancel. (At least, I think that is how it works out if I can remember this correctly, as it was awhile ago now when I filmed this! Please let me know if you are still confused however!)
@@WHYBmaths thanks for the prompt response. I appreciate it.
How do we know for a fact that a single chart can contain the entire curve? Doesn't the fact that we are working with a manifold only tell us that locally there exists some chart (aka homomorphism to R^d)? How can we assume that the entire curve will be in the domain of that chart?
In general a single chart won't contain the entire curve as you said, and yes in full generality only locally does the manifold map to R^d, this is essentially the essence of Riemannian geometry in that these tangent spaces only exist locally and are independent at every point in the manifold. In the case where a manifold has multiple charts all we can expect is that the charts agree and hence the tangent spaces to any curve passing through the intersecting chart must also agree. Usually for simplicity when beginning this construction it is simpler to think of a single chart containing the curve in question, the generalisation then simply deals with each portion of the curve separately in the respective charts (no more complex conceptually but just a computational and notational headache!)
Hi! I have a question. I didn't understand why the partial derivative with respect to x^(mou) becomes the basis, how can we see that? Can someone please answer me thanks.
Also, love your lessons.
th-cam.com/video/UPGoXBfm6Js/w-d-xo.html (takes roughly 40 mins)
Are you currently a student or are you a teacher?
Still a student! Studying a masters in mathematical physics, and I'm currently working on a project investigating boundary conditions in gauge theories, focusing on developing the Einstein-Cartan formalism of GR in this context! Hoping to make many videos about what I'm doing when I have the chance!
@@WHYBmaths Sounds very interesting. keep up the great videos!
So I'm guessing this is the algebraically-defined tangent space: which is the vector space of derivations of the ring of germs or T_p^{alg}(M), where M is some manifold and p\in M?
Yes exactly, although I didn't want to throw everyone in quite such deep water at this stage!
Thanks! Could you clarify why "the vector should be a linear map, i.e. it has to map some kind of object to produce a tangible result", why doesnt it make sense to talk about a vector just as en element of a vector space, why does it need to be applied to a function to make sense? Also, the basis is just \delta_mu right? not V^mu \delta_mu right? thanks!
Okay, I think you clarify in 22:25
Hey, I am working on something called tangent vectors over a space( I did the translation from my frensh book) but anyway it involves curves and stuffs.. I just wanted to know if you have any ideas about it.. Thank you!
Tangent vectors over a space does certainly sound like the tangent space so I'd say they are talking about the same thing!
@@WHYBmaths okay thanks a lot !!
lockdown math
I respect this style but I do not like it and I do not think it is the best way to learn for most people. I think it is pedagogically bad to go through a maze of algebra and then seek to interpret it. I think it is better to start with the concept and its interpretation and then do the algebra:
The tangent space is a set of contravariant vectors associated with a point on a manifold. The contravariant vectors themselves are what is called 'derivations'. These 'derivations' are actually functions from the set of continuous functions in the manifold to the reals. The interpretation of the derivations is that they are like little machines that measure how much any input function on the manifold changes for the given increment and direction associated with the derivation (aka tangent vector). These derivations are naturally and intuitively a linear operator. The tangent space is the set of all those tangent vectors together at your point. They naturally and intuitively are a vector space.
Then you can go back and do the complicated algebra....
The poor sound quality substantially reduces the content quility.
No it doesn't
Hi I just have a small question in the beginning of the previous video I understood that smooth parameterizations are on C ᪲ but from. H
here I understood the ones on C ᪲ are the derivatives, are both of them part of C ᪲ or did I get something wrong?