Lecture 6: Fields (International Winter School on Gravity and Light 2015)

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Recommends mike · 2016-04 · Mathematics Stack Exchange · on “Vector fields as section of tangent bundle”

tl;dr: The injectivity of the section boils down to TM being the union of the tangent space at every point. If you now map from or to $TM$, this domain/range contains all of $M$! You are talking about a vector field being a section $TM \to M$,.. If I'm not mistaken, this way round goes the surjective projection $\pi: TM \to M$, because taking a an element from $TM$ (vector with basepoint) it maps them to a point in the manifold. The projection is surjective, because you took $T_pM$ for $\forall p \in M$! Now the section $s: M \to TM$ can at most be injective. This is obvious, since TM contains all the vectors at each point, but with your field you just want to show the "user" one distinct ve

Recommends san · 2019-06 · Mathematics Stack Exchange · on “Smooth vector fields form a $C^\infty (M)$-module”

The lecturer answers exactly that question of a student five minutes later [this video] "Because if you want to write just any vector field as a linear combination of this, then at the point where one of these vanishes, then at that point you will not be able to reproduce a component in that direction." This argument proves that if any vector field necessarily vanishes at some point, then there is no basis for the module of smooth vector fields with only $d$ elements, where $d$ is the dimension of $M$ . In fact, if one of the vector fields vanishes at $P$ , then the other $d-1$ supposed members of the basis would span only a $d-1$ dimensional vector space inside $T_p M$ . So any vector field $X$

Recommends Níck · 2020-04 · Mathematics Stack Exchange · on “Is There a Basis for a Smooth Vector Fields on the 2-Sphere?”

I've been watching Frederic Schuller's course on "The Mathematics and Physics of Gravity and Light" and at a moment during Lecture 6 [this video] he claims there is no basis for $\Gamma(TS^2)$ (the space of smooth sections $\sigma \colon S^2 \to TS^2$ ) when considered as a $C^\infty(S^2)$ -module and gives an example through the claim that since every vector field must vanish somewhere, you can't get a basis. During his example, he uses a vector field which vanishes at points $(\pm 1, 0, 0)$ and $(0, 0, \pm 1)$ and says that since on, say, $(0, 0, \pm 1)$ you have only one nonvanishing vector field, and thus a linear combination can't point towards a different direction. My first question is: ad

Recommends Clemens Bartholdy · 2021-05 · Mathematics Stack Exchange · on “Relating the traditional definition of a vector field in terms of function with the differential geometry definition involving fibers and bundles”

My first exposure to the concept of vector fields was in highschool physics courses, which had a simple intuitive idea of being a function which aassociates a point in a given region in space (domain where function is defined) with an arrow pointing in some direction. For example, I can give the famous electric field of point charge centered at origin as vector field: $$ \vec{E}(r) = \hat{r} \frac{kq}{r^2} $$ This is fine. Recently, I came with a more mathematically sophisiticated definition when going through some lectures on the mathematical side of general relativity. The following definition is given: A smooth vector field $\chi$ is a smooth map that is a section of the map $TM \xrightar

Recommends Clemens Bartholdy · 2021-05 · Mathematics Stack Exchange · on “Intuitively explanation on spherical basis vectors?”

What I do not get my head around is how a different radius or angle θ can change the 2nd and 3rd basis vector and thus the vector A as a whole, if I set θ=0: Good. You self discovered the hairy ball theorem [LINK] . (*) "If f is a continuous function that assigns a vector in R3 to every point p on a sphere such that f(p) is always tangent to the sphere at p, then there is at least one pole, a point where the field vanishes (a p such that f(p) = 0)." Follow-up question: Is a vector with the same spherical components seen differently by different points in space? In the way we view intrinsic geometry, given a point on the surface, we draw a tangent plane on this surface and use the tangent vec

Cites HerrWarum · 2019-01 · Mathematics Stack Exchange · on “Smooth vector fields form a $C^\infty (M)$-module”

I am watching these [this video] lectures and am confused at this point [this video] about the counterexample to the module of smooth vector fields having a basis. The speaker invokes the Hairy Ball Theorem to say that every vector field on $S^2$ must vanish, but it is not clear to me why that prevents any vector field from being a part of a basis.

Cites Clemens Bartholdy · 2022-04 · Mathematics Stack Exchange · on “Could we construct an infinite chain of tangent bundle on a manifold? Would there be any use of this?”

In this lecture by Frederic Schuller [this video] , it is discussed how we can turn the tangent bundle could be turned into a manifold through charts of the manifold. So, this made me wonder, couldn't we repeat this process again on the tangent bundle infinitely and keep layering levels of tangent bundle? Is there any practical use of this viewpoint? One thing to note is that it's pretty clear that the dimension of the tangent bundle rises by $2^j r$ where $r$ is the dimension of initial manifold, and j is the number of times we took the tangent bundle. So, for instance tangent bundle of manifold is $2r$ , tangent bundle of the tangent bundle of manifold is of dimension $2^2r$ and so on