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$\newcommand\abs[1]{\lvert{#1}\rvert}$Let $X$ be a compact hausdorff space, and put $C(X)$ for the $\mathbb{R}$-algebra of continuous maps from $X$ to $\mathbb{R}$.

For each point $x$, there is a multiplicative semi-norm $\abs-_x$ on $C(X)$, where $\abs f_x = \abs{f(x)} \in \mathbb{R}_{\geq 0}$. That is,

1. $\abs{f}_x \geq 0$.

2. $\abs{cf}_x = \abs c \cdot \abs f_x$.

3. $\abs{fg}_x = \abs{f}_x\cdot\abs{g}_x$.

4. $\abs{f + g}_x \leq \abs f_x + \abs g _x$.

Let $Y$ be the set of multiplicative semi-norms on $C(X)$ mod the equivalence relation where $\abs- \sim \abs-'$ when each is bounded by the other up to a constant. We have just constructed a function $\nu : X \rightarrow Y$. My questions are:

1. How might we put a topology on $Y$ using only the information of $C(X)$?

2. Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism?

So in this view, multiplicative seminorms are like points or generalized points. Interestingly, the submultiplicative norm $\abs- : C(X) \rightarrow \mathbb{R}$ sending $f$ to $\operatorname{sup}_x \abs f_x$ seems like it could be the supremum of multiplicative norms in $Y$. That way, multiplicative norms are the "local" versions of submultiplicative norms, which would be "global".

$\newcommand\abs[1]{\lvert{#1}\rvert}$Let $X$ be a compact hausdorff space, and put $C(X)$ for the $\mathbb{R}$-algebra of continuous maps from $X$ to $\mathbb{R}$.

For each point $x$, there is a multiplicative semi-norm $\abs-_x$ on $C(X)$, where $\abs f_x = \abs{f(x)} \in \mathbb{R}_{\geq 0}$. That is,

1. $\abs{f}_x \geq 0$.

2. $\abs{cf}_x = \abs c \cdot \abs f_x$.

3. $\abs{fg}_x = \abs{f}_x\cdot\abs{g}_x$.

4. $\abs{f + g}_x \leq \abs f_x + \abs g _x$.

Let $Y$ be the set of multiplicative semi-norms on $C(X)$ mod the equivalence relation where $\abs- \sim \abs-'$ when each is bounded by the other up to a constant. We have just constructed a function $\nu : X \rightarrow Y$. My questions are:

1. How might we put a topology on $Y$ using only the information of $C(X)$?

2. Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism?

So in this view, multiplicative seminorms are like points or generalized points. Interestingly, the submultiplicative norm $\abs- : C(X) \rightarrow \mathbb{R}$ sending $f$ to $\operatorname{sup}_x \abs f_x$ seems like it could be the supremum of multiplicative norms in $Y$. That way, multiplicative norms are the "local" versions of submultiplicative norms, which would be "global".

$\newcommand\abs[1]{\lvert{#1}\rvert}$Let $X$ be a compact hausdorff space, and put $C(X)$ for the $\mathbb{R}$-algebra of continuous maps from $X$ to $\mathbb{R}$.

For each point $x$, there is a multiplicative semi-norm $\abs-_x$ on $C(X)$, where $\abs f_x = \abs{f(x)} \in \mathbb{R}_{\geq 0}$. That is,

1. $\abs{f}_x \geq 0$.

2. $\abs{cf}_x = \abs c \cdot \abs f_x$.

3. $\abs{fg}_x = \abs{f}_x\cdot\abs{g}_x$.

4. $\abs{f + g}_x \leq \abs f_x + \abs g _x$.

Let $Y$ be the set of multiplicative semi-norms on $C(X)$ mod the equivalence relation where $\abs- \sim \abs-'$ when each is bounded by the other up to a constant. We have just constructed a function $\nu : X \rightarrow Y$. My questions are:

1. How might we put a topology on $Y$ using only the information of $C(X)$?

2. Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism?

So in this view, norms are like points or generalized points. Interestingly, the submultiplicative norm $\abs- : C(X) \rightarrow \mathbb{R}$ sending $f$ to $\operatorname{sup}_x \abs f_x$ seems like it could be the supremum of multiplicative norms in $Y$. That way, multiplicative norms are the "local" versions of submultiplicative norms, which would be "global".

$\newcommand\abs[1]{\lvert{#1}\rvert}$Let $X$ be a compact hausdorff space, and put $C(X)$ for the $\mathbb{R}$-algebra of continuous maps from $X$ to $\mathbb{R}$.

For each point $x$, there is a multiplicative semi-norm $\abs-_x$ on $C(X)$, where $\abs f_x = \abs{f(x)} \in \mathbb{R}_{\geq 0}$. That is,

1. $\abs{f}_x \geq 0$.

2. $\abs{cf}_x = \abs c \cdot \abs f_x$.

3. $\abs{fg}_x = \abs{f}_x\cdot\abs{g}_x$.

4. $\abs{f + g}_x \leq \abs f_x + \abs g _x$.

Let $Y$ be the set of multiplicative semi-norms on $C(X)$ mod the equivalence relation where $\abs- \sim \abs-'$ when each is bounded by the other up to a constant. We have just constructed a function $\nu : X \rightarrow Y$. My questions are:

1. How might we put a topology on $Y$ using only the information of $C(X)$?

2. Is $\nu$ above a bijection? In light of the last question, is it a homeomorphism?

So in this view, multiplicative seminorms are like points or generalized points. Interestingly, the submultiplicative norm $\abs- : C(X) \rightarrow \mathbb{R}$ sending $f$ to $\operatorname{sup}_x \abs f_x$ seems like it could be the supremum of multiplicative norms in $Y$. That way, multiplicative norms are the "local" versions of submultiplicative norms, which would be "global".