Differential structures: the pivotal concept of tangent vector spaces - Lec 09 - Frederic Schuller

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Cites Antoni Parellada · 2017-09 · Mathematics Stack Exchange · on “Tangent vector in differential geometry: combining direction and derivative”

Frederic Schuller [this video] explains the directional derivative operator at a point $p$ along a curve $\gamma$ on a manifold $M$ as the linear map $X_{\gamma\ p}: C^\infty (M) \to \mathbb R$ defined as the mapping of a smooth function $f\mapsto (f\circ\gamma)'(0),$ with $0$ being the parameter value at which the curve goes through the point $p.$ He goes on to state that in differential geometry, $X_{\gamma\ p}$ is usually called the tangent vector to the curve $\gamma$ at point $p.$ Immediately afterwards he points out [this video] that in differential geometry There is no longer an independent notion of direction and an independent notion of derivative; there is only the combination of the two. Wh

Cites Taylor Rendon · 2021-03 · Mathematics Stack Exchange · on “Notation for taking the derivative of $x \circ \delta$, where $(\mathcal{U},x)$ is a chart”

I am currently studying differentiable manifolds and have been using this definition for a chart of a manifold $M$ : Let $(M,\mathcal{O})$ be a manifold od dimension $d$ . The the pair $(\mathcal{U},x)$ where $\mathcal{U}\in \mathcal{O}$ and $x:\mathcal{U} \to x(\mathcal{U}) \subseteq \mathbb{R}^d$ , $p \mapsto x(p):=(x^{1}(p),...,x^{d}(p))$ is called a chart of the manifold. The component functions of $x$ , $x^{i}: \mathcal{U} \to \mathbb{R}$ for $i=1,...d$ , are then called the coordinates of the point $p\in M$ with respect to the chart $(\mathcal{U},x)$ . With this definition in mind, I have been working on reading through some lecture notes (from here [this video] at $34:23$ ) and noticed the

Cites Wyatt Kuehster · 2022-07 · Mathematics Stack Exchange · on “How can a vector field eat a smooth function?”

You can find this in Schuller's lecture on Differential Structure [this video] during the aside on derivations and algebras. Recall that a tangent vector $X_{\gamma,p}\in T_pM$ is nothing but a derivation $D:C^\infty(M)\rightarrow \mathbb{R}$ defined by $X_{\gamma,p}\mapsto X_{\gamma,p}f = (f\circ \gamma)'(0)$ . When you apply $X\in \Gamma(TM)$ to a smooth function on $M$ , you obtain a real number for each $p\in M$ , i.e. a smooth function on $M$ . So a vector field on $M$ is nothing but a derivation $D:C^\infty(M)\rightarrow C^\infty(M)$ defined exactly as in the comments: $X\mapsto Xf$ where $(Xf)(p) = X(p)f=X_{\gamma,p}f$ for all $p\in M$ .

Cites Kleber · 2022-10 · Mathematics Stack Exchange · on “Are tangent spaces at different points disjoint according to this definition?”

In this [LINK] lecture by Prof. F. P. Schuller (I've included the correct time stamp) it is claimed that two tangent spaces at different points are disjoint, i.e. $T_pM\cap T_qM =\emptyset$ for $p,q\in M$ and $p\neq q$ . In this [this video] lecture, the tangent space at a point $p\in M$ was defined as $$ T_pM:=\{X_{\gamma,p}|\gamma \,\mathrm{is\, a\, smooth\, curve\, through\,p}\} \quad,$$ i.e. the collection of the directional derivatives [this video] along all (smooth) curves through $p$ : $$X_{\gamma,p}:\mathcal C^\infty(M) \longrightarrow \mathbb R$$ with $$ X_{\gamma,p} : f\mapsto \left(f\circ\gamma\right)^\prime(0) \quad ,$$ where $\gamma(0)=p$ . I don't understand the aforementioned claim. If