Search Results

Found 1 results for "cc948c3a3e8480d0b89cb10b8cb1d0bf" across all boards searching md5.

Anonymous /sci/16683009#16708400
6/27/2025, 4:16:24 AM
Pretty sure this is a simple question
For the top pic, tangent vectors of [math] \mathbb{R} [/math] are supposed to be objects like [math] a \tfrac{\partial}{\partial t} \in T_p\mathbb{R} [/math], and tangent vectors of [math] M [/math] are supposed to be like [math] \sum b^i \tfrac{\partial}{\partial x^i} \in T_pM [/math]. But "canonical" time derivative, the author says, is defined to be the pushforward differential applied to the tangent vector "1", so [math] \dot\gamma(t) := D\gamma(1) [/math]. This is confusing because to me it'd make more sense for it to be [math] \dot\gamma(t) := D\gamma( \tfrac{\partial}{\partial t} ) = \sum \tfrac{d\gamma^i}{dt} \tfrac{\partial}{\partial x^i} [/math], cause at least now it's working on an actual tangent vector. Now, the "obvious" reason might be that he's just referring the tangent vector 1 means the coordinate 1 for basis [math] \tfrac{\partial}{\partial t} [/math], as there's an objious bijection there, but the guy never mentions a damn thing about this, nor has he ever done it before!

Similarly, for the lower pic, normally is should be that [math] Df(X)(g) = X(g \circ f) [/math], where [math] g : \mathbb{R} \rightarrow \mathbb{R} [/math]. To get close to what he wrote, it would work when [math] g(x) = x [/math], so something like [math] Df(X)(1) = X(1 \circ f) [/math], but except for the problem mentioned above, the author never talks about this intention or wtv they actually mean.

Outside of these two situations, he's been fairly straightforward about the notation. Is he really using "shorthand" for the 1-vector in the top and the 1-function (identity) in the bottom?

thx