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Compare the following two results:

Thm A) Let $A$ be a commutative $C^*$-algebra and let $X$ be its Gelfand spectrum. Gelfand duality says that there's a natural isometric $*$-isomorphism from $A$ to $C(X)$, the $C^*$-algebra of continuous functions on $X$.

Thm B) Let $A$ be a commutative ring and $X=\operatorname{Spec}A$. There's a natural sheaf of rings $\mathscr{O}$ on $X$ such that $A=\Gamma(X,\mathscr{O})$.

Both theorems are very close. I wonder if it's possible to obtain theorem A as a particular case of theorem B, or a variant thereof.

Gabriel
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    Affine varieties are not locally compact and Hausdorff, nor all rings are commutative C-star algebras, so no, but both results have the same flavour. – Fernando Muro Jan 13 '22 at 10:11
  • @FernandoMuro I'm sorry but I don't quite understand what you mean. I surely don't think that Thm B can be deduced from Thm A. I'm talking about the other direction. – Gabriel Jan 13 '22 at 10:16
  • The Gelfand spectrum doesn’t carry the Zariski topology, I believe. – Fernando Muro Jan 13 '22 at 10:22
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    This answer seems to suggest otherwise? https://mathoverflow.net/a/90810/131975 – Gabriel Jan 13 '22 at 10:29
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    The spectrum of a commutative C*-algebra is the space of maximal ideals (max spec). This is only the prime spectrum when the space X os totally disconnected. – Benjamin Steinberg Jan 13 '22 at 10:55
  • Chapters IV and V of Stone Spaces by Peter Johnstone have material about these things. – Paul Taylor Jan 13 '22 at 11:16
  • @BenjaminSteinberg I knew that but I don't know if its topology coincides with the Zariski topology. – Gabriel Jan 13 '22 at 11:46
  • The Zariski topology is quasi-compact, so the Gelfand spectrum of a non-unital C-star algebra won't carry the Zariski topology in general. I currently don't have an argument for the unital case, I'm also curious to know the answer, I would be surprised if both topologies agreed. – Fernando Muro Jan 13 '22 at 11:56
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    All Zarsiki open subsets of the sepctrum of a C-star algebra are open (complememt of vanishing locus of a continuous function is open). So the Zarsiki topology is coarser. If the spectrum of a unital algebra is compact and Hausdorff, then it is normal. (T4). In that case it seems that you could use the extension theorem to prove the that the Zarsiki topology is also finer. Namely, for any point and neighborhood of it, there is a continuous function that does not banish on the point, but vanishes on the complement of the neighborhood. – afh Jan 13 '22 at 13:13
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    If you are searching a common categorical generalization, you might start here: https://www.andrew.cmu.edu/user/awodey/preprints/lambek.pdf – NameNo Jan 13 '22 at 13:31
  • If you instead are only interested in commutative rings and in a spectrum between the prime and the maximal ones, look at this https://mahani.uk.ac.ir/documents/1000262/1220532/Sheaf%20Representation%20of%20Rings.pdf (unfortunately it gives no proofs, but I hope that a google scholar search might help) – NameNo Jan 13 '22 at 14:51
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    There is an obvious obstacle which needs to be addressed: Thm A involves objects with an involution and Thm B does not. (I insert my usual tired comment that commutative Banach algebras can look nothing like commutative Cstar algebras, and even the semisimple ones can behave differently.) So if you want to get Thm A from Thm B one has to work out what special properties are enjoyed by the underlying ring of a commutative Cstar algebra, and by that stage I feel like one may just as well develop classical Gelfand theory for Banach algebras anyway – Yemon Choi Jan 13 '22 at 16:09
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    @FernandoMuro I am not sure either, but the definition of Zariski topology always reminds me of the definition in Banach algebra world of the hull-kernel topology. It is known that the hull-kernel topology restricted to the max ideals of a commutative Cstar algebra coincides with the Gelfand topology (there are many "natural" examples of commutative Banach algebras where the restriction of the HK topology is much coarser) – Yemon Choi Jan 13 '22 at 16:12
  • Purely real commutative (identity as involution) C$^$-algebra are rings (order, real algebra, norm are uniquely defined from the ring). Their structure is also equivalent to (partially ordered abelian gropus with strong unit, which are then vector lattices used in Stone - Weierstrass) their "complexification" $A\oplus iA$ ("polarized" real $C^$-algebras, where a central antihermitian squre root of $-1$ is fixed; they are a sligthtly more precise structure than complex $C^*$-algebras, See https://mathoverflow.net/questions/160872/axiomatizing-orientation-in-the-complex-plane/161221#161221) – NameNo Jan 13 '22 at 17:09
  • This is a good question—so good that, I could swear, it's been asked before. Is there someone who has a sufficiently broader memory of MO arcana than mine to dig it up? – LSpice Jan 14 '22 at 00:20
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    @LSpice I have definitely seen variants on this question before, but I am less enthusiastic about this genre than you are, perhaps because I was trained in settings where the Gelfand transform is just an adjunction not a duality, and because the answers to these questions seem to get increasingly abstract without showing analysts like me whether Gelfand theory for CBAs is being superseded or just taken for granted. (No offence intended to the OP; it's a natural question, I just find that the answers lauded by everyone else are not my cup of tea) – Yemon Choi Jan 14 '22 at 00:34
  • For functional analysts: say you are interested in commutative radical Banach algebras https://www.deepdyve.com/lp/de-gruyter/universal-properties-of-some-commutative-radical-banach-algebras-isAccIcRHb?articleList=%2Fsearch%3Fquery%3Duniversal%2Bproperties%2Bof%2Bsome%2Bcommutative%2Bradical%2Bbanach%2Balgebras%26docNotFound%3Dtrue#bsSignUpModal You need more than the maximal spectrum, and the algebraic results apply, but you dislike the non-closed ideals you need. https://mathoverflow.net/questions/20268/did-gelfands-theory-of-commutative-banach-algebras-influence-algebraic-geometer – NameNo Jan 14 '22 at 13:41

2 Answers2

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Yes, both Theorem A and Theorem B are special cases of a more general construction.

Denote by $R$ the category of commutative unital C*-algebras or the category of commutative rings. Denote by $R'$ the full subcategory of $R$ given by reduced objects in $R$, meaning the only nilpotent element is zero. (All commutative unital C*-algebras are reduced.)

Given an object $X∈R$, we can consider its poset $\def\Spec{\mathop{\sf Spec}}\Spec X$ of quotient objects (= subobjects in the opposite category) that belong to $R'$.

One can show that $\Spec X$ is a locale, the (localic) Zariski/Gelfand spectrum of $X$. Furthermore, assuming the axiom of choice, this locale is spatial, so it corresponds to a topological space, namely, the traditional Zariski/Gelfand spectrum of $X$.

Given an open element $U$ of $\Spec X$ corresponding to a reduced quotient $X→Q$, its kernel $I⊂X$ is a radical ideal and we can consider the localization $X[S^{-1}]∈R$, defined using a universal property in the category $R$, where $S=\{a∈X\mid I⊂\sqrt{(a)}\}$.

The assignment $$U↦X[S^{-1}]$$ defines a sheaf on $\Spec X$ valued in $R$. This is precisely the structure sheaf of $\Spec X$ for both the Gelfand spectrum and the Zariski spectrum.

Of course, this construction is applicable to many other categories $R$:

  • finitely presented entire functional calculus algebras, recovering the Stein duality for globally finitely presented Stein spaces (i.e., complex geometry).
  • finitely generated germ-determined C^∞-rings, recovering the Dubuc duality for C^∞-loci (i.e., differential geometry);
  • other dualities for algebraic geometry, such as formal schemes, etc.
  • various versions of the above for derived geometry;
  • Boolean algebras, recovering the Stone duality for compact totally disconnected Hausdorff spaces;
  • complete Boolean algebras, recovering the Stonean duality for compact extremally disconnected Hausdorff spaces;
  • localizable Boolean algebras, recovering the Gelfand-type duality for compact strictly localizable enhanced measurable spaces.
Dmitri Pavlov
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    Do you happen to have "general conditions" on a(n abstract + maybe some structure ) category $R$ for such a construction to work and give similar results ? – Maxime Ramzi Jan 14 '22 at 08:45
  • Note that for Gelfand one uses only maximal i.e. prime and closed ideals (not all prime ideals). Se my first two comments to the question. The "commutative reduced = subdirect product of commutative domains", here glorified by a topology, is one of many kinds of (different) decompositions that become quite interesting for noncommutative (or Jordan) rings but all coincide for Boolean algebras. – NameNo Jan 14 '22 at 12:51
  • The special relation between prime and maximal ideals in Gelfand's case (every prime extends uniquely to a maximal; see 4.5 in this or the classical) explains why and how A) can be reduced to B) i.e. why the maximal spectrum (as subspace of the prime spectrum) is enough: https://core.ac.uk/download/pdf/82558373.pdf – NameNo Jan 14 '22 at 16:59
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    @MaximeRamzi: As far as I am aware, nobody has figured this out yet. I think Iskra's Really Modern Algebra describes how to recover R' from R in categorical terms (and makes many other interesting observations), and a recent question by John Baez makes some initial steps toward characterizing R, but there is a lot more to do. – Dmitri Pavlov Jan 14 '22 at 19:02
  • @NameNo: Ideals are automatically closed since we are looking at quotient objects in the category of commutative unital C*-algebras and the kernels of homomorphisms of such algebras are always closed. Maximal closed ideals are bijective with points in the spectrum and closed ideals are bijective with open subsets of the spectrum. – Dmitri Pavlov Jan 14 '22 at 20:54
  • @Dmitri Pavlov I was explaining the relation between your answer and my comments. You wrote correct things (including writing "maximal closed ideals" even if it's the same as "maximal ideals" https://math.stackexchange.com/questions/3496249/maximal-ideal-of-a-banach-algebra-is-closed ) – NameNo Jan 14 '22 at 21:35
  • @DmitriPavlov Can you give some references on this? – Luiz Felipe Garcia Dec 14 '23 at 20:46
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    References for most items can be found by following the hyperlinks in https://ncatlab.org/nlab/show/duality+between+algebra+and+geometry. For the last three items, see also https://arxiv.org/abs/2005.05284. I am not aware of a single unifying source. – Dmitri Pavlov Dec 15 '23 at 01:19
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As @PaulTaylor said, the book Stone Spaces by Peter Johnstone has some stuff about this. I'll explain below, in my words, the parts that I found relevant.

Let $A$ be a commutative (unital) $C^*$-algebra. Its Gelfand spectrum $\operatorname{sp}(A)$ is usually defined as the set of nontrivial multiplicative linear functionals with the weak-$*$ topology. It's a well-known result that the map $\operatorname{sp}(A)\to\operatorname{Specm}(A)$, sending $\varphi$ to $\ker\varphi$ is a bijection.

In the case of $C^*$-algebras over $\mathbb{R}$, this book affirms that this map is actually a homeomorphism. (The cited function $\hat{a}$ is the one that sends a maximal ideal $\mathfrak{m}$ to the image of $a$ through $A\to A/\mathfrak{m}\cong\mathbb{R}$.)

(I tried to prove that the same holds for complex $C^*$-algebras without much success.)

Answering @YemonChoi comments: the cited book proves (theorem IV.4.10) that the functor $\operatorname{Specm}$ defines a duality between the category of compact Hausdorff spaces and the full subcategory of $\mathsf{CRing}$ whose objects are real $C^*$-algebras. Moreover, he proves an analogous result to the Gelfand duality in section V.3.8 using only the prime spectrum. But I have to say that I found it somewhat unsatisfying.

Gabriel
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  • Yes with purely real commutative C$^$-algebras (involution is identity). But consider yourself what happens for general commutative real C$^$-algebras (equivalently complex commutative C$^*$-algebras with a fixed involution), for example taking two copies of the same compact T$_2$ space and the involution, on the algebra of continuous functions, that corresponds to the exchange of the two components. – NameNo Jan 14 '22 at 10:09
  • @NameNo thank you for your remark, I hadn't realized that. – Gabriel Jan 14 '22 at 10:17