“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).

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“Representability”

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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