# automorphisms of local rings vs local change of coordinates

Let $R$ be a local (commutative, associative) ring over a field of zero characteristic. (My typical examples are $k[[x_1,..,x_p]]/I$, $k\{x_1,.,x_p\}/I$, $C^\infty(\Bbb{R}^p,0)$. If it helps one can assume $R$ to be Henselian.)

I'd like to think of the ring automorphisms, $Aut(R)$, (those that act on the field as identity) as the local changes of coordinates, "$Aut(Spec(R))$". The two objects certainly coincide if $R$ is the localization of an affine ring.
More generally, let $S=k[x_1,..,x_p]/I$, let $S\subseteq R\subseteq \hat{S}$, the completion with respect to $(x_1,..,x_p)$. Then the two objects coincide for $R$.

But for $R=C^\infty(\Bbb{R}^p,0)$ there are endomorphisms which do not come from the local maps of coordinates. See Page 5. (In this particular example one has an endomorphism, not an automorphism. Still, it is not clear that here $Aut(R)=Aut(Spec(R))"=Aut(\Bbb{R}^p,0)$, the later is the group of germs of local diffeomorphisms).

1. Suppose $R$ is "geometric enough", so that one can speak of $Spec(R)$, its local coordinates, a local change of them. Does every local change of coordinates (that preserves the origin) extend to an automorphism of $R$?

2. For which "geometric" rings rings $Aut(R)=Aut(Spec(R))"$? (i.e. the group of all the automorphisms of $R$ vs the group of the local coordinate changes in $Spec(R)$.) what is the official notation for the "geometric" subgroup $Aut(Spec(R))"$ of $Aut(R)$? (The notation $Aut(Spec(R))$ is somewhat heavy/lengthy.)

Any paper/review on the state of the art in this direction?

• I think you haven't actually stated which two objects you want to coincide. – Damian Rössler Jan 5 '15 at 11:17
• What does $IC^{\infty}(...)$ stand for? – Qfwfq Jan 5 '15 at 12:52
• $I$ is separated by comma from the ring of germs of infinitely differentiable functions – Dmitry Kerner Jan 5 '15 at 13:36
• Oh, when I first read it my tablet was displaying it the wrong way. I can now see the comma, sorry :) – Qfwfq Jan 5 '15 at 17:00
• I still don't understand the question. Isn't Aut(R) = Aut(Spec R) an immediate consequence of what scheme theory is for? And again, "Spec R" makes sense for any commutative ring. Now that I've reread it for the fourth time it starts making sense, but I humbly think it could be written more clearly. – bananastack Jan 9 '15 at 9:37

About $C^\infty(\mathbb R^p,0)$, the ring of germs of smooth functions, I have the following remarks:

It's ideal of flat functions is notoriously ill behaved. In the final topology on this ring for the mapping $C^\infty(\mathbb R^p)\to C^\infty(\mathbb R^p,0)$ it is in the closure of zero. There are results available classifying closed ideals, see the book [Tougeron: Ideaux des functions differentiable, Springer 1972].

The exotic automorphism that you describe involves a non-continuous part. The Whitney extension theorem gives an extension operator from Whitney jets to functions, and describes when this extension operator can be chosen continuous. If I remember correctly, this is the case for closed sets which are the closures of their open interiors. A point is not of this class. This is an indication that the exotic automorphisms all come from discontinuous constructions.

If you do not insist on rings of germs but on the full rings $C^\infty(M)$ for smooth manifolds $M$, you have perfect duality between the category of manifolds and the these rings. See chapter 8 of this book. As spectrum you have to take the ideals of codimension 1. Ideals of finite codimension have interesting interpretations in terms of differential geometric constructions. See also the thorough treatment of $C^\infty$-rings in the first chapter of the book

• Moerdijk, Ieke; Reyes, Gonzalo E.: Models for smooth infinitesimal analysis. Springer-Verlag, New York, 1991. x+399 pp.

and the characterization of rings of smooth functions of manifolds in this paper.

• Thank you! Still, I do not know any place where a "non-geometric" automorphism of $C^\infty$ is constructed. (i.e. an automorphism which is identity on all the polynomials but not identity on the flat functions). Could you suggest smth? – Dmitry Kerner Jan 12 '15 at 6:48