tangent space

Definition

$T_{p}M=\{\phi \in {\mathcal{O}}_{M,p}^{\ast }|\phi (\alpha \beta )=\phi (\alpha )\beta (p)+\alpha (p)\phi (\beta )\forall \alpha ,\beta \in {\mathcal{O}}_{M,p}\}$ Notes:

• $T_{p}M$ is the collection of all the derivations
• $\phi \in {\mathcal{O}}_{M,p}^{\ast }⟹\phi :C^{\infty }(U\subseteq M)\to \mathbb{R}$
• In the notation above ${\mathcal{O}}_M$ denotes the sheaf of smooth functions on M, and ${\mathcal{O}}_{M,p}$ is the stock where ${\mathcal{O}}_{M,p}=({⨆}_{p\in U\subset M}{\mathcal{O}}_M(U))/\sim$ Where $(U,f)\sim (V,g)$ if there exists and open neighborhood $W\subset U\cap V$ such that $p\in W$ and $f{|}_w=g{|}_w$. This takes care of compatibility of functions across open sets.

Tangent space of ${\mathbb{R}}^n$

We can show that tangent space at a point $x\in {\mathbb{R}}^n$ is isomorphic to another copy of ${\mathbb{R}}^n$.

Theorem

Consider the map

\Phi: \mathbb{R}_a^n &\to T_a\mathbb{R}^n\\ v &\mapsto D_v|_a . \end{split}

Where $D_v{|}_a$ is the directional derivative in the direction $v$. $\Phi$ is an isomorphism of vector spaces.

Proof

Linearity: We know that $\Phi$ is linear because

\Phi(\vec v + \vec w)(f) &= D_{\vec v+\vec w}|_a(f) \\ & = \nabla f\cdot (\vec v+ \vec w)\\ & = \nabla f \cdot \vec v + \nabla f \cdot \vec w\\ & = \Phi(\vec v)(f) + \Phi(\vec w)(f) \end{align*}$$ With a similar argument for scalar multiplication. **Injective:** Need to show that $\ker \Phi = \{0\}$. First, note that $\Phi(\vec 0) = \nabla \cdot 0 = 0$. Next, let $\vec v \in \ker \Phi$. Then $$0 = D_v|_a(f) = \nabla f \cdot v \quad \forall f: \mathbb{R^n} \to \mathbb{R}.$$ Let $f$ be the coordinate function that takes the $i$th coordinate to, and all others to 0. By the above, $v_i = 0$ for all $i$, so $v = \vec 0$. **Surjective:** Let $w \in T_x\mathbb{R}^n$. Thus, $w: \mathcal{O}_{\mathbb{R}^n} \to \mathbb{R}$ that is linear and $w(fg) = w(f)g(x) + f(x)w(g)$. For any function $f \in \mathcal{O}_{\mathbb{R}^n}$, we can take the Taylor series expansion with 2nd order remainder $$ \begin{split} f(\vec x) &= f(\vec a) + J(\vec a)(\vec x-\vec a) + R(\vec x) + \\ &= f(a) + \sum_i \frac{\partial f}{\partial x_i}(x_i - a_i) + \sum_{i,j} (x_i - a_i)(x_j - a_j) \int_0^1(1-t)\frac{\partial^2f}{\partial x_i \partial x_j} (a + t(x-a))\end{split}$$ Because $w$ is linear we can look at each term. Looking at the last term because $w$ is a derivation, and $f(a)$ is a constant $w\big(f(a)\big) = 0$. Also, for the remainder term, $$ \begin{align}& w\bigg(\sum_{i,j} (x_i - a_i)(x_j - a_j) \int_0^1(1-t)\frac{\partial^2f}{\partial x_i \partial x_j} (a + t(x-a)) \bigg) \\ &= \sum_{i,j} w\bigg((x_i - a_i)(x_j - a_j) \int_0^1(1-t)\frac{\partial^2f}{\partial x_i \partial x_j} (a + t(x-a)) \bigg)\\ &= \sum_{i,j}w\big((x_i - a_i)\big)(a_j - a_j) \int_0^1(1-t)\frac{\partial^2f}{\partial x_i \partial x_j} (a + t(x-a)) \bigg)\\ & \qquad + \sum(a_i - a_i)w\bigg( (x_j - a_j) \int_0^1(1-t)\frac{\partial^2f}{\partial x_i \partial x_j} (a + t(x-a)) \bigg)\\ &= 0 \end{align}$$ So we only need to look at the middle term. Therefore, $$\begin{align} w(f) &= w\bigg(\sum_i \frac{\partial f}{\partial x_i}(x_i - a_i))\bigg)\\ &= \sum_i \frac{\partial f}{\partial x_i}\big(w(x_i - a_i)\big)\\ &= \sum_i \frac{\partial f}{\partial x_i}\big(w(x_i) - w(a_i)\big)\\ &= \sum_i \frac{\partial f}{\partial x_i}\big(w(x_i)\big)\\ \end{align}$$ Let $v_i = w(x_i)$ then $\Phi(v) = w$ so $\Phi$ is surjective. Hence, it is an isomorphism of vector spaces. # Tangent spaces of manifolds For a [[202403052130|topological manifold]] $M$, the $\dim T_pM = \dim M$. # Change in coordinates Given smooth charts $(U, \varphi)$ and $(V, \psi)$ on $M$, we can change between partial derivatives in the two systems. Let $(x^1, \dots, x^n)$ be the coordinate functions for $\varphi$, and $(\tilde x^1, \dots, \tilde x^n)$ be coordinate functions for $\psi$.

\begin{align*} \frac{\partial}{\partial x^i} &= d(\varphi^{-1}){\varphi(p)}\left( \frac{\partial}{\partial x^i} \bigg|{\varphi(p)}\right) \ &= d(\psi^{-1}){\psi(p)} \ \circ \ d(\psi \circ \varphi^{-1}){\varphi(p)} \left( \frac{\partial}{\partial x^i}\bigg|{\psi(p)}\right)\ &= d(\psi^{-1}){\psi(p)}\left( \frac{\partial \tilde x^j}{\partial x^i}(\varphi(p)) \frac{\partial}{\partial \tilde x^j}\bigg|_{\psi(p)}\right)\ &= \frac{\partial \tilde x^j}{\partial x^i}(\hat p) \frac{\partial}{\partial \tilde x^j}\bigg|_p \end{align*}

--- ### References [[@lee2013]] provides a good proof of why this works. Though technically he doesn't use the [[202403081144|sheaf]] of smooth functions taking the [[202403081145|stock]] as I learned from Peter.