Tangent vectors

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Different definitions of tangent vectors.

A reference for this material is Chapter 3 of John M. Lee. Introduction to smooth manifolds. Second edition. Grad. Texts in Math., 218. Springer, New York, 2013. xvi+708pp. ISBN: 978-1-4419-9981-8.

Let $M$ be a smooth $n$-dimensional manifold and $p \in M$.

Tangent space The tamgenet space to $M$ at $p$ is defined as the set of linear functions $L : C^{\infty } (M) \to \mathbb {R} $ such that $L(fg) = L (f) g(p) + f(p)L(g)$ for all $ f,g \in C^{\infty }(M) $. Elements of $T_pM$ are called the derivations at $p$ or tangent vectors at $p$.

Curves define derivations Let $\gamma : I \to M$ be a smooth curve with $0 \in I$ and $\gamma (0) = p$. The map $L_{\gamma } : C^{\infty }(M) \to \mathbb {R}$ given by

\[ L_{\gamma } (f) : = (f \circ \gamma )' (0) \]

is a derivation at $p$.

Solution:: Exercise.

Equivalence of definitions The map $\gamma \mapsto L_{\gamma }$ defined in Proposition prop:curve-to-derivation is a surjective function from the set of smooth curves passing through $p$ at $0$ to $T_pM$. Moreover, $T_pM$ is an $n$-dimensional vector space.

The proof will consist of several steps. Fix $L \in T_pM$.

Derivative of constant For a constant function $c \in C^{\infty }(M)$, one has $L (c) = 0$.

Solution:: \begin{align*} L(c) & = L ( 1 \cdot c) \\ & = L (1) c + 1 \cdot L (c) \\ & = L (c) + L (c) \\ & = 2 L (c) . \end{align*}

This is only possible if $L(c) = 0$.

Locality If $f,g \in C^{\infty } (M)$ agree in a neighborhood of $p$, then $L(f) = L (g)$.

Solution:: Let $U \subset M$ be an open neighborhood of $p$ where $f$ and $g$ agree. Let $\rho \in C^{\infty } (M)$ be a bump function around $p$ supported inside of $U$. Then
\[ f-g \equiv (f -g) ( 1 - \rho ) \]
(check this!), and
\begin{align*} L( f) - L (g) & = L (f-g) \\ & = L ((f - g) ( 1 - \rho ) ) \\ & = L (f-g) ( 1 - \rho (p) ) + (f(p) - g(p)) L(1 - \rho ) \\ & = L(f-g) (0) - (0) L (1 - \rho ) \\ & = 0. \end{align*}

Restriction of derivations Let $ U \subset M$ be open containing $p$. For each $f\in C^{\infty }(U)$, define

\[ L^U (f) : = L (\tilde {f}) , \]

where $\tilde {f} \in C^{\infty } (M) $ is a function such that $f = \tilde {f} $ in a neighborhood of $p$. This is a well-defined derivation $L^U \in T_p U $.

Solution:: For $f \in C^{\infty } (U)$, the function $\tilde {f} \in C^{\infty }(M)$ can be obtained by multiplying $f$ by a bump function around $p$ supported inside of $U$. Proposition prop:locality-of-derivatives implies that $L^U (f) $ does not depend on the choice of $\tilde {f}$. The linearity and Leibniz rule for $L^U$ follow from the corresponding properties for $L$.

Restriction of derivations is injective The map $L \mapsto L^U$ is a linear isomorphism from $T_pM$ to $T_pU$.

Solution:: For any $D \in T_pU$, one can define $\tilde {D} \in T_pM$ by
\[ \tilde {D} (f) : = D (f \vert _ U ). \]
It is straightforward to show that $D \mapsto \tilde {D}$ is the inverse of $L \mapsto L^U$.

For a chart $(U , \varphi ) $ and $f \in C^{\infty }(U)$, by an abuse of notation, we often identify $f$ with $f \circ \varphi $. Because of this, we sometimes write $\partial _i f$ instead of $\partial _i (f \circ \varphi ^{-1} ) \circ \varphi $.

Proof of Theorem thm:tpm-definitions Let $(U, \varphi )$ be a chart with $\varphi (p) = 0$ and $\varphi (U) = B_1(0)$. For each $i \in \{ 1, \ldots , n \}$, notice that $\partial _i \vert _p : C^{\infty }(U) \to \mathbb {R}$ given by

\[ \partial _i \vert _p (f) : = \partial _i f (p) \]

is a derivation at $p$. Moreover, $\partial _i \vert _p = L _{\gamma _i}$, where $\gamma _i (t) = \varphi ^{-1} ( t e_i)$. We will show that

\begin{equation}\label {eq:tpm-basis} \{ \partial _1 \vert _p , \ldots , \partial _n \vert _p \} \end{equation}

is a basis of $T_pM$.

Fix $L \in T_pM$, and $L^U \in T_pU$ given by Proposition prop:restriction-of-derivatives. We claim that

\begin{equation}\label {eq:derivation-is-partials} L ^U = \sum _{i = 1} ^ n L ^U ( x_i )\, \partial _i \vert _p . \end{equation}

To see that, fix $f \in C^{\infty } (U)$. By the Fundamental Theorem of Calculus, for any $x \in B_1(0)$ we have

\begin{align*} f(x) & = f(0) + \int _0 ^1 \frac {d}{dt} (f (tx)) \, dt \\ & = f(0) + \int _0 ^1 \nabla f (tx) \cdot x \, dt \\ & = f(0) + \sum _{i = 1}^n x_i \int _0^1 \partial _i f(tx) \, dt \\ & = f(0) + \sum _{i = 1}^n x_i h_i (x) \end{align*}

for some functions $h_i \in C^{\infty } (U)$ with $h_i (0) = \partial _if (0)$. Finally,

\begin{align*} L ^U (f) & = L^U (f(0)) +\sum _{i = 1}^n L ^U \left ( x_i h_i \right ) \\ & = 0 + \sum _i ^n \left [ L ^U( x_i ) \, h_i (0) + 0 \cdot L^U (h_i) \right ] \\ &= \sum _{i = 1} ^n L^U (x_i) \, \partial _i f (p). \end{align*}

This proves (eq:derivation-is-partials). To see that (eq:tpm-basis) is linearly independent, assume

\[ \sum _{i = 1} ^n c_i \partial _i \vert _p = 0\]

for some scalars $c_1, \ldots , c_n$. Then

\begin{align*} 0 & = \sum _{i = 1} ^n c_i \partial _i \vert _p \left ( \sum _{j =1 }^n c_j x_j \right ) \\ & = \sum _{i = 1} ^n \sum _{j =1 }^n c_i c_j \partial _i \vert _p x_j \\ & = \sum _{i = 1} ^n \sum _{j =1 }^n c_i c_j \delta _{ij} \\ & = \sum _{i = 1 }^n c_i ^2 . \end{align*}

This would mean $c_1 = \ldots = c_n = 0$.

We finally note that for any scalars $a_1, \ldots , a_n$, we have

\[ \sum _{i = 1} ^n a_i \partial _i \vert _p = L_{\gamma } \]

with

\[ \gamma (t) : = \varphi ^{-1} ( ta_1, \ldots , ta_n) . \]