I don't have much background in functional analysis, so wanted to check whether my thinking about the weak topology and Gelfand topology is correct when it comes to C* algebras/Banach spaces. My understanding is that:

  • The weak topology is defined on a Banach space $X$, and is the weakest topology such that every element of the dual space $f\in X^*$ is continuous.
  • On the other hand the Gelfand topology is defined on commutative C* algebras (which are Banach spaces)- say the algebra A- and is such that the subspace of the double dual A** corresponding to the bounded linear maps $\hat{x}:S(A)\rightarrow \mathbb{C}$ where $S(A)$ is the space of all continuous homomorphisms of A, $\phi : A \rightarrow C$, $\phi(ab)=\phi(a)\phi(b)$, with the maps defined using the canonical isomorphism $\hat{x}(\phi)=\phi(x)$, every such map $\hat{x}$ in this subspace of A** is continuous.

Thus, it seems to me that the Gelfand topology is contained in the weak topology, and as such is weaker. Because it only requires a subspace of the double dual A** to be continuous, when acting on a subspace of A*. Namely the subspace of linear homomorphisms of A, and not merely the linear functionals on A.


1 Answer 1


These two things do not really mix much. The weak topology is defined on a topological vector space: you need a vector space to be able to define linear functionals. The weak$^*$ topology makes sense only on duals.

The Gelfand topology is used to give a topology to the set of characters of an abelian C$^*$-algebra $A$ (it can be done for a Banach algebra, too). What you do is you consider the characters as a subset $\Sigma$ (not a subspace!) of the dual $A^*$ of $A$, and you endow $\Sigma$ with the relative weak$^*$-topology (which is simply pointwise convergence).

The weak$^*$ topology is nice because it makes closed balls compact, which is often very useful. In particular it makes $\Sigma$ above compact when it's closed (which is precisely when $A$ is unital), and locally compact in general.


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