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I finally gave this statement its own entry, in order to be able to conveniently point to it:
embedding of smooth manifolds into formal duals of R-algebras
Such a statement is true for manifolds with finitely many or countably many connected components.
As Theo Johnson-Freyd once pointed out to me, and later expanded in this answer on MO: http://mathoverflow.net/a/91445 it is false for arbitrary paracompact Hausdorff manifolds, in particular, given two uncountable (discrete) sets S and T one can find a morphism of real algebras C^∞(T)→C^∞(S) that is not induced by a function S→T.
However, the construction is a very subtle set-theoretical argument that uses measurable cardinals.
Thanks. I have made the standard regularity assumptions explicit in the entry now and added pointer to this MO discussioon.
Ah, I was just wondering what sort of things break for uncountable disjoint unions of second countable manifolds.
EDIT: I was thinking continuum-many summands, which is still better behaved than for general uncountable coproducts.
added doi to
and
added pointer to Milnor’s original statement
and to these proofs for the case of isomorphisms:
Janusz Grabowski, Isomorphisms and ideals of the Lie algebras of vector fields, Inventiones mathematicae volume 50, pages 13–33 (1978) (doi:10.1007/BF01406466)
Jerrold Marsden, Ratiu, Abraham, Theorem 4.2.36 in: Manifolds, tensor analysis, and applications, Springer 2003 (ISBN:978-1-4612-1029-0)
Janusz Grabowski, Isomorphisms of algebras of smooth functions revisited, Arch. Math. 85 (2005), 190-196 (arXiv:math/0310295)
Added redirect: Milnor duality. To satisfy a link at duality between geometry and algebra.
Added:
The case of the category of smooth manifolds and diffeomorphisms is proved in
Interesting that such an early reference exists.
In trying to check it out on my phone, I only get to see the first 19 pages. Do you mean to say Pursell’s proof covers only diffeos/ring-isos, but not non-invertible maps?
Yes. There is also this announcement by Pursell’s PhD advisor M. E. Shanks: https://www.ams.org/journals/bull/1951-57-04/S0002-9904-1951-09521-X/S0002-9904-1951-09521-X.pdf, see page 295.
They never published any of this, except that they have a similar paper about Lie algebras: https://www.ams.org/journals/bull/1951-57-04/S0002-9904-1951-09521-X/S0002-9904-1951-09521-X.pdf.
added pointer to:
Added to the list of miracles:
And one more item:
Added a short simple proof of “Milnor’s exercise” (known before Milnor, it seems), since the proof in Michor et al. is too complicated:
\begin{proof} We provide a simplified proof using Hadamard’s lemma. Suppose $M$ is a smooth manifold and $\phi\colon C^\infty(M)\to\mathbf{R}$ is a homomorphism of real algebras.
If $M=\mathbf{R}^n$ for some $n\ge0$, then set $y_i=\phi(x_i)$, where $x_i\colon\mathbf{R}^n\to\mathbf{R}$ is the $i$th coordinate function. We have
$\phi(f)=\phi(f(y)+\sum_i (x_i-y_i)\cdot g_i)=f(y)+\sum_i (\phi(x_i)-y_i)\cdot \phi(g_i)=f(y),$where the functions $g_i$ are provided by Hadamard’s lemma.
For a general $M$, use Whitney’s theorem to embed $M$ into some $\mathbf{R}^n$ and consider the composition
$C^\infty(\mathbf{R}^n)\to C^\infty(M)\to \mathbf{R},$where the first homomorphism $\rho$ is given by restricting along the embedding. The composition is given by evaluating at some point $p\in\mathbf{R}^n$. We have $p\in M$ because every composition $\phi\rho$ must vanish on the kernel of $\rho$, but the evaluation homomorphism at $p\notin M$ does not. \end{proof}
That’s a good proof!
Interesting about this and similar proofs is that it manages to use the coordinate functions $x^i$ as if they were generators of $C^\infty(\mathbb{R}^n)$, even though they are not (not in the sense of $\mathbb{R}$-algebras). While it is the “$C^\infty$-ring“-structure on $C^\infty(\mathbb{R}^n)$ that does make the coordinate functions be actual generators, these kinds of proofs show that/why for many purposes this exctra structure is not actually necessary.
Re #18: Yes, and in fact, Hadamard’s lemma is key for the initial development of differential geometry.
It’s worth pointing out that analogous proofs work in the algebraic and complex holomorphic cases, where they show, in exactly the same manner, that the C-spectrum of the algebra of holomorphic functions on a Stein manifold $M$ is isomorphic to $M$. (Recall that Stein manifolds can be defined as complex manifolds admitting a proper holomorphic embedding into $\mathbf{C}^n$, which is precisely what we need for the proof.) Likewise, in the algebraic case we recover Nullstellensatz.
The statement could probably be generalized to algebras over Fermat theories.
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