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Suppose $A$ is a commutative Banach algebra. By Gelfand duality there is a compactum $X$ such that $A = C(X)$ is the ring of continuous functions. The space $X$ can be recovered as the space of characters on $A$. That is to say multiplicative linear functionals $A \to \mathbb R$ under the weak$^*$ topology. Observe this topology on the space of characters does not depend on the topology of $A$.

Now let $f \in C(X)$ be any non-invertible element. In other words $f$ has a zero. We can localise the ring $C(X)$ at $f$ to get the ring of 'formal fractions' $C(X)_f = \displaystyle \{\frac{g}{f^n} \colon g \in C(X), n \in \mathbb N\}$.

There is a natural embedding $C(X) \to C(X)_f$ but I am unaware if $C(X)_f$ carries a compatible Banach algebra structure. By this I mean a norm under which it is complete and the embedding is an isometry. Nevertheless we can consider the space of characters on $C(X)_f$ and give that the weak$^*$ topology.

  1. Under what conditions is the character space of $C(X)_f$ some compactum $Y$?

  2. When will we have $C_f(X) = C(Y)$?

  3. Does $Y$ have a topological characterisation in terms of the space $X$ and function $f$?

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  • $\begingroup$ In algebraic geometry, inverting $f$ in a ring of functions corresponds to removing the solution set to $f=0$ from a space. I imagine a similar thing happens here. $\endgroup$ – Hurkyl Feb 22 '17 at 16:31
  • $\begingroup$ Geometrically speaking, it seems like this might correspond to a weighted norm where the weight vanishes on the zeroes of $f,$ since if $g$ is positive near a zero of $f,$ in order for $g/f$ to have a finite norm, you would need to make sure that the singularities aren't overly penalized. In fact, the decay near the zeroes of $f$ would probably need to be asymptotically equivalent to the decay of $|f|.$ $\endgroup$ – RideTheWavelet Feb 22 '17 at 17:08
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There are some problems here. It is not true that commutative Banach algebras are isomorphic to $C(X)$ for some $X$. In the case where $A$ is semi-simple, we have an embedding with dense range. There are examples of infinite-dimensional commutative Banach algebras with a unique character!

Localisations do not have Banach-algebra structure because they are fields. By the Mazur-Gelfand theorem, the field of complex numbers is the only commutative Banach algebra that is also a field.

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