Thanks for the clear videos! I do have one small question. When you are drawing the vectors on the charts, you are treating the d/d(theta) basis vectors as if they are equivalent to the usual polar theta_hat basis vector. Does this require assuming an extra convention? (Since really these vectors are more abstract living in a vector space of partial derivatives)
What actually is an example of this arbitrary function that the vector acts on, why does there have to be a function composed with the parameterised curve, what is this function f that is defined on the manifold?
According to my understanding, the function f is quite common in the sense of real problems like physics where some functions are defined on the manifold. Once the function f is fixed, then together with the chart (coordinate system you defined), the derivative of the function can be easily computed in the form of directional derivative on the chart with the help of the "components part “ which actually forms the tangent space/vector field of the manifold.
Only in the very special case where the manifold is already R^n, since then the (otherwise abstract) points of the manifold already have an intrinsic coordinate description (taking the chart to be the identity). Otherwise, we require a chart (coordinates), since one should really view vectors (at least in this description) as arising after we equip a manifold with smooth differentiable charts. One then realises the `differential structure' on the manifold, that the derivatives of the chart functions (the coordinate basis derivatives) carry a natural vector space structure (the tangent space). So, in order to talk concretely about vectors it is best to realise them as arising \emph{after} the manifold is equipped with more structure (namely a differentiable chart), which gives us a concrete way to talk about vectors in terms of components and basis vectors. Going even further this one can then realise vectors in a slightly more abstract way as derivations (satisfying the Leibniz property) of smooth functions on M and even even further as arising from infinitesimal diffeomorphisms (in a very similar way to how I discussed the velocity vector of a curve previously). However, all of these can only be defined once M is equipped with a suitable smooth differentiable collection of charts! I hope to discuss these more abstract realisations when I move on to talk about Lie derivatives etc.
So that mean’s that the manifold and their respective charts can be of different dimensions,I cannot see why that make sense, it’s like how points on a 2 dimensional surface such as the torus can be mapped onto a 1 dimensional line. And one more question I have seen people use two overlapping charts of an Atlas to prove the differentiability of the manifold , my question is can one chart not completely suffice for that, since in my opinion the differentiability of the manifold must be independent of how many charts you use to describe it, and then also don’t you need coordinates on the manifold to describe the notion of differentiability on the manifold. And lastly when you will be continuing on these videos. Thanks
@@muddassirtariq6530 @MUDDASSIR TARIQ No it definitely doesn't make sense! All charts must be the same dimension as the manifold. Since dimension is a topological invariant, and in defining a chart we require that a local neighbourhood of some point in the manifold be homeomorphic to an R^d of the same dimension, otherwise the two could not be homeomorphic!(differing on a topological invariant) Regarding your second question, the statement of differentiability in this context is really the statement that the transition functions themselves are differentiable, so it only really makes sense when there are more than one chart. Hopefully very soon! I'm just finishing up my masters thesis (using a LOT of really nice differential geometry/relativity) and after than I plan to make videos continuing this series to eventually reach the stuff I've been working on! Roughly september onwards I'll be free to make more vids but I'm getting new ideas for content all the time!
@@muddassirtariq6530 Also I would add we can always construct arbitrarily as many charts as we like, in fact one constructs the 'maximal' atlas of all possible charts and then shows that this can be refined to a suitably differentiable set of charts thus defining the differentiability of the manifold. Weaker notions would be just a continuous set of charts or C^k differentiable, but the differentiable always refers to the transition functions!
@@WHYBmaths So if you say that the dimensions of the chart and dimensions of the manifold are the same that means that we can always consider a coordinate system on the manifold itself, because in the first reply you gave me you said that to be able to do this the manifold must have a special case where the it has to have a dimension R^n(same as chart) and now you are saying it has always the same dimension as the chart. Secondly why does the chart transition function refers to differentiability in the manifold. I am basically trying to learn Einstien's GR in it's full glory, so it would be helpful if you could give an advice on how much mathematical abstractness do I need and how much can I avoid to do that.
This helped me clear up some misunderstandings about coordinate expressions of basis vector fields. Thanks so much!
Thank you! Glad it helped
Hooray, an example!
I too find my own lack of examples frustrating, but I'll try to do more in future videos!
I need help to find the general expression for X belongs to the vector field manifold which satisfies : [d/dx + d/dy , X ]=X
Thanks for the clear videos! I do have one small question. When you are drawing the vectors on the charts, you are treating the d/d(theta) basis vectors as if they are equivalent to the usual polar theta_hat basis vector. Does this require assuming an extra convention? (Since really these vectors are more abstract living in a vector space of partial derivatives)
No they absolutely are the same thing! The theta hat basis vectors are derived by considering this vector field d/d(theta)
What actually is an example of this arbitrary function that the vector acts on, why does there have to be a function composed with the parameterised curve, what is this function f that is defined on the manifold?
According to my understanding, the function f is quite common in the sense of real problems like physics where some functions are defined on the manifold. Once the function f is fixed, then together with the chart (coordinate system you defined), the derivative of the function can be easily computed in the form of directional derivative on the chart with the help of the "components part “ which actually forms the tangent space/vector field of the manifold.
Can we use the manifold directly as our natural coordinate system, and do all the calculations directly on that.
Only in the very special case where the manifold is already R^n, since then the (otherwise abstract) points of the manifold already have an intrinsic coordinate description (taking the chart to be the identity). Otherwise, we require a chart (coordinates), since one should really view vectors (at least in this description) as arising after we equip a manifold with smooth differentiable charts. One then realises the `differential structure' on the manifold, that the derivatives of the chart functions (the coordinate basis derivatives) carry a natural vector space structure (the tangent space). So, in order to talk concretely about vectors it is best to realise them as arising \emph{after} the manifold is equipped with more structure (namely a differentiable chart), which gives us a concrete way to talk about vectors in terms of components and basis vectors.
Going even further this one can then realise vectors in a slightly more abstract way as derivations (satisfying the Leibniz property) of smooth functions on M and even even further as arising from infinitesimal diffeomorphisms (in a very similar way to how I discussed the velocity vector of a curve previously). However, all of these can only be defined once M is equipped with a suitable smooth differentiable collection of charts! I hope to discuss these more abstract realisations when I move on to talk about Lie derivatives etc.
So that mean’s that the manifold and their respective charts can be of different dimensions,I cannot see why that make sense, it’s like how points on a 2 dimensional surface such as the torus can be mapped onto a 1 dimensional line. And one more question I have seen people use two overlapping charts of an Atlas to prove the differentiability of the manifold , my question is can one chart not completely suffice for that, since in my opinion the differentiability of the manifold must be independent of how many charts you use to describe it, and then also don’t you need coordinates on the manifold to describe the notion of differentiability on the manifold. And lastly when you will be continuing on these videos.
Thanks
@@muddassirtariq6530 @MUDDASSIR TARIQ No it definitely doesn't make sense! All charts must be the same dimension as the manifold. Since dimension is a topological invariant, and in defining a chart we require that a local neighbourhood of some point in the manifold be homeomorphic to an R^d of the same dimension, otherwise the two could not be homeomorphic!(differing on a topological invariant)
Regarding your second question, the statement of differentiability in this context is really the statement that the transition functions themselves are differentiable, so it only really makes sense when there are more than one chart.
Hopefully very soon! I'm just finishing up my masters thesis (using a LOT of really nice differential geometry/relativity) and after than I plan to make videos continuing this series to eventually reach the stuff I've been working on! Roughly september onwards I'll be free to make more vids but I'm getting new ideas for content all the time!
@@muddassirtariq6530 Also I would add we can always construct arbitrarily as many charts as we like, in fact one constructs the 'maximal' atlas of all possible charts and then shows that this can be refined to a suitably differentiable set of charts thus defining the differentiability of the manifold. Weaker notions would be just a continuous set of charts or C^k differentiable, but the differentiable always refers to the transition functions!
@@WHYBmaths So if you say that the dimensions of the chart and dimensions of the manifold are the same that means that we can always consider a coordinate system on the manifold itself, because in the first reply you gave me you said that to be able to do this the manifold must have a special case where the it has to have a dimension R^n(same as chart) and now you are saying it has always the same dimension as the chart. Secondly why does the chart transition function refers to differentiability in the manifold. I am basically trying to learn Einstien's GR in it's full glory, so it would be helpful if you could give an advice on how much mathematical abstractness do I need and how much can I avoid to do that.
Did the other dog die?
at the intro, does your dog eat the chalk?
It looks like it! But it was a piece of kibble don't worry ;)