Exercise 11.21 (Properties of the Differential)

Question

extbf{Exercise 11.21.} Prove Proposition 11.20.

	extbf{Proposition 11.20 (Properties of the Differential).} Let $M$ be a smooth manifold with or without boundary, and let $f, g \in C^{\infty}(M)$.
\begin{enumerate}[label=(\alph*)]
\item If $a$ and $b$ are constants, then $d(a f+b g)=a d f+b d g$.
\item $d(f g)=f d g+g d f$.
\item $d(f / g)=(g d f-f d g) / g^2$ on the set where $g 
eq 0$.
\item If $J \subseteq \mathbb{R}$ is an interval containing the image of $f$, and $h: J ightarrow \mathbb{R}$ is a smooth function, then $d(h \circ f)=\left(h^{\prime} \circ fight) d f$.
\item If $f$ is constant, then $d f=0$.
\end{enumerate}

Elu Thingol · Accepted Answer

ewcommand{\diff}{\mathrm{d}}

\begin{proof}
Let $p \in M$ be arbitrary, i.e., we look at one covector $\diff_{p} f $ at a time from the covector field $df$.
\begin{enumerate}[label=(\alph*)]
\item 	extit{If $a$ and $b$ are constants, then $\diff_{p}(a f+b g)=a \diff_{p} f+b \diff_{p} g$.}
ewline
For all inputs $v \in T_p M$, we have
\begin{equation*}
    \diff_{p}(a f+b g) (v) = v(a f+b g) = a v(f) + bv(g) =a \diff_{p} f+b \diff_{p} g
\end{equation*}
where the equality in the middle is the direct consequence of the linearity of elements $v$ of $T_pM$.

\item 	extit{$\diff_{p}(f g)=f \diff_{p} g+g \diff_{p} f$.}
ewline
We verify the identity for all inputs $v \in T_p M$:
\begin{equation*}
    \diff_{p}(fg)(v) = v(fg) = v(f) \cdot g(p) +  f(p) \cdot v(g) = f(p) \diff_{p} g(v) +g(p) \diff_{p}f(v)
\end{equation*}
where the equality in the middle is the direct consequence of the Leibniz rule for elements $v_p$ of $T_pM$.

\item 	extit{$\diff_{p}(f / g)=(g \diff_{p} f-f \diff_{p} g) / g^2$ on the set where $g 
eq 0$.}
ewline
Let $v \in T_p M$ be an arbitrary input. By \href{./exercise_3-5}{Lemma 3.4}, $0 = v\left(g \cdot \frac{1}{g}ight) = \frac{1}{g(p)}v(g) + \left(\frac{1}{g}ight)$; hence, we deduce that $v\left(\frac{1}{g}ight) = -\frac{1}{g(p)^2}v(g)$, and thus

\begin{align*}
    \diff_{p}(f / g) (v) = v(f/g) 
    &= f(p) v \left(\frac{1}{g}ight) + \frac{1}{g(p)} v(f)\
    &=  f(p) \cdot \left(-\frac{1}{g(p)^2}v(g)ight) + \frac{1}{g(p)} v(f)\
    &= \dfrac{g(p)v(f) - f(p)v(g)}{g(p)^2}\
    &=\dfrac{g(p) \diff_{p} f(v)-f(p) \diff_{p} g(v)}{g^2(p)}.
\end{align*}

\item 	extit{If $J \subseteq \mathbb{R}$ is an interval containing the image of $f$, and $h: J ightarrow \mathbb{R}$ is a smooth function, then $\diff_{p}(h \circ f)=\left(h^{\prime} \circ fight) \diff_{p} f$.}
ewline
Again, for all $v \in T_pM$, we have
\begin{align*}
    \diff_{p}(h \circ f)(v)
    = v(h \circ f)
    = (h' \circ f) \cdot v(f)
    =\left(h^{\prime} \circ fight) \diff_{p} f(v).
\end{align*}
where in the equation in the middle we have used the chain rule for derivations $v \in T_pM$ (which is proven in local coordinates).

\item 	extit{If $f$ is constant, then $\diff_{p} f=0$.}
ewline
If $c \in \mathbb{R}$ is the constant value of $f$, then for all inputs $v \in T_p M$:
\begin{equation*}
    \diff_{p}(c)(v) = v(c) = 0
\end{equation*}
where the last equality follows by the fact that derivations of constant functions are zero (\href{./exercise_3-5}{Lemma 3.4}).
\end{enumerate}
\end{proof}

Exercise 11.21 (Properties of the Differential)

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Exercise 11.21 (Properties of the Differential)

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