🧵 Manifolds
Anonymous at Thu, 7 Mar 2024 12:10:40 UTC No. 16061253
Hi, i'm studying pic related. Right now i'm getting filtered by the proof of equivalency of the tangent space. Basically i want to show that defining T_pM as a space of derivations and defining T_pM as a quotient space by an equivalence relation of curves is the same. Diff geometry brainlets will understand. Any tips about understanding better this subject?
Anonymous at Thu, 7 Mar 2024 15:04:37 UTC No. 16061404
What exactly are you not getting?
A curve is a map from the reals R to the manifold M and a function is a map from M->R, so composing the two lets you take a derivative on functions. From the other direction, pick some coordinates on the manifold, and then you can show any derivation is a directional derivative with some vector direction, and that vector maps to the equivalence class of curves with that as the tangent vector.
Maybe there are some fussy details depending on how rigorous you are trying to be, but the basic idea is easy
Anonymous at Thu, 7 Mar 2024 15:40:00 UTC No. 16061447
>>16061404
The idea behind is clear to me. It's just the technical part that bothers me, im getting there, just not yet. Thanks.
Anonymous at Thu, 7 Mar 2024 19:20:01 UTC No. 16061682
>>16061253
Uhmmm, you have to imagine it all using your hippocampus and amygdala, and then you also have to use logical reasoning with your frontal lobe...
Basically gtfo if you don't have the genes
El Arcón at Thu, 7 Mar 2024 19:29:06 UTC No. 16061691
I think the issue is that when you construct an object by either method, one corresponding to every "p in M," the objects are the same, and the total construction is the same since they assigned to the different "p in M" in the same way. Just show that each tangent space is R^n. If you want more, you can take the directional derivatives of the curves in your equivalence classes and show that they conform to the definition of a derivation, so the space of directional derivatives (which is the space constructed from the equivalence classes of curves), is automatically a space of derivations. Then make a statement like, "If A conforms to the definition of B, then A is B."
Anonymous at Fri, 8 Mar 2024 14:38:43 UTC No. 16062812
when you define a derivation you are defining a set of continuous functions that give a tangent vector. So you can look at all smooth curves on M that pass through the point P with tangent vector equal to that one you defined before. Then you define your equivalence relation on that set of curves. So you end up with gamma'(0) is a derivation at the point P=gamma(0) , and its also an equivalence class. So both ways of defining TpM give the same set of tangent vectors.