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Let $C$ be the category of commutative Banach algebras and let $U : C \to \text{Set}$ be the usual forgetful functor. The holomorphic functional calculus guarantees that every holomorphic function $f : \mathbb{C}^n \to \mathbb{C}$ defines a natural transformation $U^n \to U$ (in a way which is compatible with composition, etc.). Does every natural transformation $U^n \to U$ have this form?

Similarly, let $C$ be the category of commutative C*-algebras and let $U : C \to \text{Set}$ be the usual forgetful functor. The continuous functional calculus guarantees that every continuous function $f : \mathbb{C}^n \to \mathbb{C}$ defines a natural transformation $U^n \to U$. Does every natural transformation $U^n \to U$ have this form?

In both cases the first thing to try would be to check if $U$ is representable, and in both cases I think it can't be.

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  • $\begingroup$ I'm away from my books and MathSciNet right now, but the name you might wish to look up is William Zame $\endgroup$
    – Yemon Choi
    Commented Jun 5, 2013 at 6:45
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    $\begingroup$ For the first, have you try to use the fact that the algebra of function holomorphic on a closed disk (its completion of course). Should represente the functor which send an algebra $B$ on the set of element of norme smaller or equal to k. I think this should solve your problem at least if one suppose some conditions of "continuity" or "boundeness" of you natural transformation... $\endgroup$
    – Simon Henry
    Commented Jun 5, 2013 at 7:41
  • $\begingroup$ @Simon: this could work. No continuity or boundedness hypotheses should be necessary; this sort of thing is enforced by naturality (see the argument below). $\endgroup$
    – Qiaochu Yuan
    Commented Jun 5, 2013 at 7:55

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I'll just consider a natural map $\phi:U\to U$; the multivariate case should be similar.

Let $D_n$ be the closed disk of radius $n$ centred at the origin, and let $A_n$ be the Banach algebra of functions that are continuous on $D_n$ and holomorphic on the interior. Let $\iota_n\in A_n$ be the identity map. By naturality for the restriction maps $A_{n+1}\to A_n$, the functions $\phi(\iota_n)\in A_n$ will fit together to give an entire function $f$. Now consider an arbitrary Banach algebra $B$ and an element $b\in B$. Choose $n>\|b\|$, so functional calculus gives a homomorphism $\beta:A_n\to B$ with $\beta(\iota_n)=b$. (Here we can use the simplest version of functional calculus using convergent power series; we do not need need anything about the spectrum of $b$, or $B$-valued contour integrals.) Naturality of $\phi$ with respect to $\beta$ tells us that $\phi(b)=f(b)$, as required.

I think that $A_n$ represents the functor $U_n(B)=\{b\in B:\|b\|\leq n\}$ (or should that be $\rho(b)\leq n$?), and $U$ is the colimit of these.

UPDATE:

I didn't read Simon Henry's comment properly before; now I see that the argument above is more or less what he proposed.

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  • $\begingroup$ Cool. I think basically the same argument works for C*-algebras as well (using the C*-algebras of continuous functions on $D_n$ instead). $\endgroup$
    – Qiaochu Yuan
    Commented Jun 5, 2013 at 21:39
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    $\begingroup$ I think that in the definition of $U_n$ there definitely should be spectral radius, since Banach algebra homomorphisms are not necessarily contractive but they shrink spectra (of course, one may assume that morphisms are contractive, but the spectral radius seems to be more appropriate for working with functional calculus). Moreover, there is some subtlety connected with the fact that there can be a problem with defining functional calculus if spectrum touches the boundary of $D_n$; I don't have any particular example at hand but von Neumann's inequality wouldn't be famous, if it were true. $\endgroup$
    – Mateusz Wasilewski
    Commented Jun 6, 2013 at 6:57
  • $\begingroup$ It then suggests that probably there should be strict inequality in the definition of $U_n$ instead. However, there are Banach algebras for which this extended functional calculus is valid and for them $U_n(B)$ would be "too small" to be isomorphic to $Hom(A_n,B)$. Category theory is completely outside my comfort zone, so I hope it makes sense. $\endgroup$
    – Mateusz Wasilewski
    Commented Jun 6, 2013 at 7:03
  • $\begingroup$ @Mateusz: yes, the question of what happens when $\rho(b)=n$ is delicate. It may be that $A_n$ does not actually represent $U_n$, but that is close to the truth and motivates the line of argument anyway. $\endgroup$
    – Neil Strickland
    Commented Jun 6, 2013 at 7:24
  • $\begingroup$ Yes, I think it's true in any case that $U(-)$ is the colimit over $\text{Hom}(A_n, -)$. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 12, 2013 at 19:23

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