# “Representability”

Let $M$ be a compact smooth manifold. Show that any $\Bbb{R}$-algebra morphism $C^\infty(M)\to\Bbb{R}$ is of the form $\mathrm{ev}_p$ for some $p\in M$, where $\mathrm{ev}_p(f)=f(p)$.

## 2 thoughts on ““Representability””

1. nachodarago says:

Let $M,N$ be compact smooth manifolds. By Gelfand duality, an $\mathbb{R}$-algebras map $\mathcal{C}^\infty(N)\to\mathcal{C}^\infty(M)$ is induced by the pullback of a map $\varphi:M\to N$. In our case, $\mathcal{C}^\infty(\{*\}) = \mathbb{R}$ and thus an $\mathbb{R}$-algebra morphism $\mathcal{C}^\infty(M)\to\mathbb{R}$ is the pullback of $\{*\}\to M$, which simply is evaluation at a point.

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2. eborghini says:

Necroposting. Let $F:{\mathcal C}^{\infty}(M) \to \mathbb{R}$ an $\mathbb{R}$-algebra morphism. Let’s establish in first place the following fact: there exist $p \in M$ and an open neighbourhood $U \ni p$ such that whenever $f \in {\mathcal C}^{\infty}(M)$, ${\rm supp}(f) \subseteq U$ is $F(f) = f(p)$. Indeed, if it wasn’t the case, by compactness there should exist a finite cover of $M$ by proper open sets $U_1, \dots, U_n$ together with smooth functions $f_1, \dots, f_n$ such that ${\rm supp}(f_i) \subseteq U_i$ and $F(f_i) \neq 0$. By hypothesis, there is also a smooth function $g \in {\mathcal C}^{\infty}(M)$ supported on $M \setminus \cap_{i=1}^{n} U_i$ with $F(g) \neq 0$. Then, $h := \left( \prod_{i=1}^{n} f_i \right) g$ is identically $0$ but $F(h) \neq 0$, a contradiction.
Now, the conclusion follows easily. Let $f \in {\mathcal C}^{\infty}(M)$ and take $\varphi$ a smooth function supported on $U$ with $\varphi(p) = 1$. By the choice of $p$ and the open set $U$, $F(\varphi f) = ( \varphi f) (p) = f(p)$. Since $F(\varphi) = \varphi(p) = 1$, this implies $F(f) = f(p)$.

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