Questions tagged [gelfand-duality]

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Representing measurable map to compact space as a continuous map

Let $\Omega$ be a measurable space equipped with a $\sigma$-ideal $\mathcal{N}$ (though of as the "null sets"). Define the compact Hausdorff space $$ \tilde{\Omega} := \mathrm{Spec}(L^\infty(...
user avatar
2 votes
1 answer
170 views

Unifying categorical equivalences and dualities for the functors : Gelfand spectrum and Zariski spectrum, Structural sheaf and Continuous sections

I would be very grateful for any references I might be led to, from a categorical point of view for the functors: $\textsf{Spec}_{\mathscr{Z}\textrm{arisky}}(-)$, related to $\mathcal{O}(-)$, which ...
Julien Dalpayrat-Glutron's user avatar
1 vote
1 answer
191 views

Adjunction via Gelfand duality

$\DeclareMathOperator\Hom{Hom}$For which unital $C^{\ast}$-algebras $A$ does it hold that for all compact Hausdorff $S$ we have the bijection: \begin{align*} \Hom(A, C(S)) \cong \Hom(S, \Hom (A, \...
Luiz Felipe Garcia's user avatar
4 votes
0 answers
162 views

Gelfand's transform for noncommutative $C^*$-algebras

Please excuse me if this is well-known, I am not very familiar with the general theory of $C^*$-algebras. Let $A$ be a unital separable liminal $C^*$-algebra (in the case I am interested in, ...
Antoine Labelle's user avatar
3 votes
0 answers
160 views

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 ...
Mirco A. Mannucci's user avatar
14 votes
3 answers
1k 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 ...
Mirco A. Mannucci's user avatar
0 votes
1 answer
90 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?
Prakash's user avatar
11 votes
2 answers
260 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 ...
Igor Khavkine's user avatar
5 votes
1 answer
461 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 ...
Nathanael Skrepek's user avatar
4 votes
0 answers
82 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 ...
Lviv Scottish Book's user avatar
5 votes
1 answer
174 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 ...
Tomasz Kania's user avatar
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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 ...
Daron's user avatar
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8 votes
1 answer
707 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 ...
Alex M.'s user avatar
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2 votes
1 answer
291 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 $C$, ...
Daron's user avatar
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5 votes
2 answers
973 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 ...
Ilan Barnea's user avatar
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5 votes
1 answer
864 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 ...
Ali Taghavi's user avatar
1 vote
0 answers
305 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{...
compmath's user avatar
2 votes
1 answer
223 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 ...
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0 votes
0 answers
205 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}\...
Ali Taghavi's user avatar
3 votes
0 answers
276 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 ...
Adam Hughes's user avatar
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0 votes
1 answer
267 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 ...
Dave's user avatar
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10 votes
5 answers
2k 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 ...
Joshua Seaton's user avatar
0 votes
1 answer
382 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, ...
roger123's user avatar
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37 votes
5 answers
4k 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 ...
Dmitri Pavlov's user avatar