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In Nestruev's (2000) Smooth Manifolds and Observables, the authors define an $\mathbb{R}$-algebra as a commutative, associative algebra with unit (p. 21). A natural generalization of this definition would drop the requirement of a unit. (For example, any self-adjoint, commutative, and non-unital C*-algebra defines such a "non-unital $\mathbb{R}$-algebra".) I am interested in how many of Nestruev's constructions carry over to the non-unital case. For example: is the notion of a smooth envelope of a geometric $\mathbb{R}$-algebra $\mathcal{F}$ well-defined if $\mathcal{F}$ lacks a unit?

P.S. Apologies if this question is too preliminary; I am new to posting on this site. Many thanks for reading :)

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    $\begingroup$ This is a pretty wide question. Without looking at the book, it looks like you are asking how much of a whole book can be reproduced by dropping an assumption. I suggest picking one single result you are interested in, asking about that. $\endgroup$
    – David Roberts ♦
    Commented Sep 9, 2021 at 7:24
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    $\begingroup$ What is the motivation behind such a question? A nonunital R-algebra can be turned into a unital R-algebra using the unitization functor. This reduces the study of nonunital algebras to unital algebras. In geometric language, this corresponds to studying pointed spaces instead of spaces. $\endgroup$
    – Dmitri Pavlov
    Commented Sep 9, 2021 at 13:35
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    $\begingroup$ Very crudely speaking, one can argue that encoding a non-compact LCH space $X$ in terms of the non-unital $C^\ast$-algebra $C_0(X)$ is just the price you pay for working with an honest $C^\ast$-algebra of globally bounded functions without invoking any particular choice of compactification. Nestruev’s commutative-algebraic approach doesn’t require anything to be globally bounded, so the only non-unital algebras they ever need invoke are prime ideals of unital algebras. $\endgroup$
    – Branimir Ćaćić
    Commented Sep 9, 2021 at 14:23
  • $\begingroup$ David, thank you for the suggestion; I've edited the question to ask just about smooth envelopes. YCor, thanks for making the title more specific. Dimitri, my motivation is exactly the intuition that Branimir describes (i.e., that I want to work with an "honest" $C^*$-algebra of globally bounded functions like $C_0 (X)$, or a dense subspace of it containing only smooth functions). $\endgroup$
    – supergeneric
    Commented Sep 9, 2021 at 18:42
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    $\begingroup$ @DmitriPavlov: the existence of unitizations of Cstar algebras does not really reduce the non-unital theory to the unital theory, and indeed there are many instances where experience rather than categorical dictat indicates that one should work with the multiplier algebra not the unitization. $\endgroup$
    – Yemon Choi
    Commented Sep 10, 2021 at 1:33

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For example: is the notion of a smooth envelope of a geometric R-algebra F well-defined if F lacks a unit?

Yes. Recall the construction: $F$ is geometric if the Gelfand homomorphism $$\def\Map{\mathop{\rm Map}} \def\Spec{\mathop{\rm Spec}} \def\R{{\bf R}} \def\Hom{\mathop{\rm Hom}} G\colon F → \Map(\Spec(F),\R)$$ is injective, where $\Spec(F)=\Hom(F,\R)$ is the set of homomorphisms $F→\R$ and $\Map$ denotes the set of maps of sets $\Spec(F)→\R$.

The smooth envelope of $F$ is then constructed as the real algebra $E$ generated by the image of $G$ inside $\Map(\Spec(F),\R)$ and closed under smooth compositions: if $f_1,…,f_n∈E$ and $g\colon \R^n→\R$ is smooth, then $g(f_1,…,f_n)∈E$.

What changes if $F$ is nonunital? Homomorphisms $F→\R$ are defined in the same manner as before, and in fact coincide with unital homomorphisms $\def\hF{{\hat F}} \hF→\R$, where $F→\hF$ is the unitization of $F$.

The same definition of the Gelfand homomorphism continues to work.

The resulting smooth envelope is a unital algebra even if $F$ is not: indeed, the unit is produced by taking $n=0$ and $g=1$ in the construction of the smooth envelope of $F$. Even if we exclude $n=0$, we can always take $g=1$ for any $n$, which still gives us the unit.

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