Exercise 2.1 ($C^{\infty}(M)$ is commutative and associative algebra)

Question

Let $M$ be a smooth manifold with or without boundary. Show that pointwise multiplication turns $C^{\infty}(M)$ into a commutative ring and a commutative and associative algebra over $\mathbb{R}$. (See Appendix B, p. 624, for the definition of an algebra.)

Elu Thingol · Accepted Answer

\begin{proof}
$C^{\infty}(M)$ is a subspace of the vector space $\mathbb{R}^M$ of all functions from $M$ to $\mathbb{R}$. Therefore, we only demonstrate that $C^{\infty}(M)$ is closed under addition, scalar multiplication and multiplication. Consider arbitrary  $f, g \in C^{\infty}(M)$ and $c \in \mathbb{R}$.

We demonstrate that $(f+g):M 	o \mathbb{R}$ is smooth as well. Let $p \in M$ be arbitrary. Since $f$ is smooth, there is a chart $(U,\varphi)$ containing $p$ such that $f \circ \varphi^{-1}: \varphi(U) 	o \mathbb{R}$ is smooth. By \href{./exercise_2-3}{Exercise 2.3}\footnote{It is impossible to prove this without using any of the results related to the Exercise 2.3.}, $g$ must be smooth for $(U,\varphi)$ as well. Thus, the function
\begin{align*}
    (f+g) \circ \varphi^{-1} &= f \circ \varphi^{-1} + g \circ \varphi^{-1} : \varphi(U) 	o \mathbb{R}
\end{align*}
must be smooth since it is the sum of two smooth functions in the calculus sense. In other words, $f+g$ is smooth.

The proofs for $fg: M 	o \mathbb{R}$ and $cf: M 	o \mathbb{R}$ follow similarly.

Commutativity and associativity follow from the commutativity and associativity on the range $\mathbb{R}$.
\end{proof}

Exercise 2.1 ($C^{\infty}(M)$ is commutative and associative algebra)

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Exercise 2.1 ($C^{\infty}(M)$ is commutative and associative algebra)

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