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A reference for this material is Chapter 8 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.
Vector field Let \(M\) be a smooth manifold and \(\pi : TM \to M\) its tangent bundle. A vector
field is a smooth map \(V : M \to TM\) with \(\pi \circ V = \text {Id}_M\). We denote by \(\mathfrak {X}(M)\) the set of vector fields on \(M\).
Global derivation Let \(M\) be a smooth manifold. A global derivation is a linear map \( L : C^{\infty } (M) \to C^{\infty }(M) \)
such that
\[ L(fg) = L(f) g + f L(g) \]
for all \(f,g \in C^{\infty }(M)\).
Vector fields induce global derivations Let \(V \in \mathfrak {X} (M)\) and \(f \in C^{\infty } (M)\). Then the function \(\tilde {V}f : M \to \mathbb {R}\) given by
\[ \tilde {V}f (x) : = V(x) (f) \]
is
smooth. Moreover, the map \(f \mapsto \tilde {V}f\) is a global derivation.
Solution:
Consider \((U, \varphi )\) a chart of \(M\) and \((\pi ^{-1}(U), \tilde {\varphi }) \) the corresponding chart of \(TM\). In these coordinates, \(V\) is given
by
Since \(V\) is a smooth map, the components \(V^1, \ldots , V^n\) depend smoothly on \((x_1, \ldots , x_n)\). This means that
in \(U\), the vector field \(V\) is given by
Global derivations are vector fields For any global derivation \(L : C^{\infty }(M) \to C^{\infty }(M)\), there is a vector field \(V \in \mathfrak {X}(M)\)
with \(\tilde {V} = L\).
Solution:
For any \(p \in M\), the map \(L_p: C^{\infty }(M) \to \mathbb {R}\) given by \(L_p (f) : = L(f)(p)\) is a derivation at \(p\). Lnearity of \(L_p\) is immediate, and for the
Leibniz rule:
If we define \(V : M \to TM\) by \(V(p) : = L_p\), then by construction \(\tilde {V} = L\). All we need to show is \(V\) is smooth.
Fix \(p \in M\) and let \((U, \varphi )\) be a chart on \(M\) around \(p\). For each \(i \in \{ 1 , \ldots , n \} \), choose \(h_i \in C^{\infty } (M)\) with
\[ h_i (x_1, \ldots , x_n ) = x_i\]
near \(p\). In \(U\), the function \(V : M \to TM\)
has the form
\[ V ( x ) = \sum _{ i = 1 }^n V^i (x) \partial _i \vert _x . \]
Consequently, for each \(j \in \{ 1 , \ldots , n \} \) and \(x\) near \(p\) one has
\begin{align*} V^j (x) & = \sum _{ i = 1 }^n V^i (x) \partial _i h_j (x) \\ & = V (x) (h_j ) \\ & = L_ x (h_j) \\ & = L (h_j) (x). \end{align*}
Since \(L(h_j)\) is smooth, so is \(V^j\) around \(p\). Since \(p\) was arbitrary, \(V\) is smooth.
Let \(f : M \to N\) be smooth, \(X \in \mathfrak {X}(M)\), and \(Y \in \mathfrak {X}(N)\). We say \(X\) and \(Y\) are \(f\)-related if
\[ f_{\ast } X(p) = Y(f(p)) \]
for all \(p \in M\).
\(X\) and \(Y\) are \(f\)-related if and only if for all \(h \in C^{\infty }(N)\) one has
\[ (\tilde {Y}h) \circ f = \tilde {X}( h \circ f). \]
Solution:
For each \(p \in M\),
\begin{align*} [ f_{\ast } X (p) ] (h) & = X(p) (h \circ f) = \tilde {X} ( h \circ f ) (p) , \\ Y(f(p)) ( h ) & = (\tilde {Y}h) ( f(p)) . \end{align*}
Hence \(f_{\ast } X(p) = Y(f(p))\) if and only if \(\tilde {X} (h \circ f) (p) = (\tilde {Y}h) ( f(p)) \) for all \(h \in C^{\infty }(N)\). Quantifying over \(p\) we get the result.
From now on, for \(V \in \mathfrak {X}(M)\), we often denote \(\tilde {V} \) simply by \(V\).
Smooth manifolds