# Questions tagged [gelfand-duality]

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20
questions

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votes

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### Non-commutative duality II: The two algebras of a groupoid

This question is in a sense the continuation of my previous one, Non-commutative duality I: Which C*-algebras are (isomorphic to a) convolution algebra?. Reading the great answers and the equally good ...

**14**

votes

**3**answers

889 views

### Non-commutative duality I: Which C*-algebras are (isomorphic to a) convolution algebra?

Many interesting C*-algebras can be realized as convolution algebras over a groupoid, an idea introduced in 1980 by Jean Renault (this entry in nLab provides plenty of context to the general approach ...

**0**

votes

**1**answer

52 views

### Involution in a commutative unital real C* algebra

It follows immediately from Gelfand duality that the involution in a commutative unital real C* algebra is the identity. Is there a direct proof from the axioms of C* algebras?

**10**

votes

**1**answer

170 views

### Fundamental group under Gelfand duality

Gelfand duality states that the functor of continuous functions $C(-)$ from compact Hausdorff topological to commutative $C^*$-algebras is an equivalence of categories. In other words, all topological ...

**5**

votes

**1**answer

285 views

### Generalized Gelfand Triples

Normally, when we talk about Gelfand triple we have three Hilbert spaces
$$\newcommand{\X}{\mathcal{X}}
\X_+ \subset \X_0 \subset \X_-
$$
such that the subsets are dense, the embedding mappings are ...

**4**

votes

**0**answers

67 views

### Characterizing the separability of the Gelfand space of a semisimple commutative Banach algebra

Problem. Is the separability of the Gelfand space of a semi-simple commutative Banach algebra $A$ equivalent to the existence of a countable family $\{\varphi_n\}_{n\in\omega}$ of multiplicative ...

**5**

votes

**1**answer

155 views

### Are there any non-trivial convergent sequences in the maximal ideal space of the measure algebra?

Consider the measures on the circle, $M(\mathbb T)$, endowed with the convolution product which makes it a unital Banach algebra under the total variation norm. Denote by $\Delta$ the maximal ideal ...

**0**

votes

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86 views

### Can we express separability of a ray-remainder in terms of the function algebra?

Let $X = [0, 1)$ be a ray and $C(X)$ the algebra of bounded continuous real functions. The spectrum of $C(X)$ is the Stone-Cech compactification $\beta [0,1) $ of the ray. It's easy to see the ...

**7**

votes

**1**answer

610 views

### Recovering a smooth manifold from its tensor fields

1) Consider the algebra $\mathcal T (M) = \bigoplus \limits _{p, q \ge 0} \mathcal T ^{p, q} (M)$, where $\mathcal T ^{p, q} (M)$ is the space of tensor fields of type $(p,q)$. I should endow it with ...

**2**

votes

**1**answer

249 views

### Can we Characterise Rings of Continuous Functions?

Suppose $K$ is some nice space, for example $\mathbb R$ or $\mathbb C$. Let $X$ be a set and $C$ a ring of functions $X \to K$. Is there any way to determine, from the algebraic structure of $C$, ...

**5**

votes

**2**answers

764 views

### Commutative von Neumann algebras and localizable measure spaces

This is not my subject so I apologize if my question is too obvious or understood from other pages.
I read some pages such as
Reference for the Gelfand-Neumark theorem for commutative von Neumann ...

**5**

votes

**1**answer

673 views

### Proper continuous image of metrizable space

Motivated by the following post, "Gelfand duality" and the fact that "a Hausdorff continuous image of a compact metric space is metrizable", we ask:
What is a counter example of two locally ...

**1**

vote

**0**answers

255 views

### Properties of a function from its pullback

Edit: I have now removed the duplication previously referred to. Thank you.
Let $M$ and $N$ be smooth manifolds and $T: M \to N$ be a smooth map. Let $ \mathcal{F}(M,\mathbb{R})$ (resp.$ \mathcal{...

**1**

vote

**1**answer

174 views

### Spectrum of a Banach algebra homeomorphic to the spectrum of one of its elements

In G. B. Folland - A Course in Abstract Harmonic Analysis we can read the following
" (1.15) Proposition. Let be $A$ a (complex) commutative unital Banach algebra with unit $e$, let $x_0 \in A$ and ...

**0**

votes

**0**answers

195 views

### A noncommutative analogy of the tube lemma

Assume that $A$ and $B$ are two unital commutative Banach algebras. Assume that $\phi \in \mathcal{M} (A)$ is an element of the maximal Ideal space. Define $\alpha: A\hat{\otimes} B \to \mathbb{C}\...

**4**

votes

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210 views

### Tauberian theorem from generalized Gelfand transform

Wiener's theorem gives the necessary and sufficient conditions for the set of translates of a set of functions to be dense in $L^1(\mathbb{R}^n)$, which translates algebraically into a statement about ...

**0**

votes

**1**answer

247 views

### Dual notion of a local homeomorphism between topological spaces for C*-algebras

Given two locally compact topological spaces $X$ and $Y$, and a local homeomorphism $f : X \to Y$, Gelfand duality gives us a homomorhism $Cf : C_0(Y) \to C_0(X)$. How does the fact that $f$ is a ...

**10**

votes

**5**answers

1k views

### Do subalgebras of C(X) admit a description in terms of the compact Hausdorff space X?

In light of the well-known theorem of Gelfand that, bluntly put, ends up saying that unital abelian C*-algebras are the 'same' as compact Hausdorff topological spaces, I tried to compile a dictionary ...

**0**

votes

**1**answer

348 views

### The functor of continuous functions from compact CW-spaces to the reals

The contravariant functor $C(-)$ given by
$$
\hom_{Top}(-,\mathbb{R}):cCW\to Rng
$$
where $cCW$ is the category of compact CW complexes is injective on objects. What is known about surjectivity, ...

**34**

votes

**5**answers

3k views

### Reference for the Gelfand duality theorem for commutative von Neumann algebras

The Gelfand duality theorem for commutative von Neumann algebras states that the following three categories are equivalent:
(1) The opposite category of the category of commutative von Neumann ...