Ultrafilters and automorphisms of the complex field

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It is well-known that it is consistent with $ZF$ that the only automorphisms of the complex field $\mathbb{C}$ are the identity map and complex conjugation. For example, we have that $\vert\operatorname{Aut}(\mathbb{C})| = 2$ in $L(\mathbb{R})$. But suppose that we are given a nonprincipal ultrafilter $\mathcal{U}$ over the natural numbers $\mathbb{N}$. Is there any way to use $\mathcal{U}$ to define a third automorphism of $\mathbb{C}$?

Some background ... the "obvious" approach would be to note that the ultraproduct $\prod_{\mathcal{U}} \bar{\mathbb{F}}_{p}$ of the algebraic closures of the fields of prime order $p$ has lots of automorphisms arising as ultraproducts of Frobenius automorphisms. Of course, working in $ZFC$, this ultraproduct is isomorphic to $\mathbb{C}$ and hence we obtain many "strange " automorphisms of $\mathbb{C}$. However, the isomorphism makes heavy use of the Axiom of Choice and these fields are not isomorphic in $L(\mathbb{R})[\mathcal{U}]$. So a different approach is necessary if we are to find a third automorphism of $\mathbb{C}$ just in terms of $\mathcal{U}$ ...

Edit: Joel Hamkins has reminded me that I should mention that I always assume the existence of suitable large cardinals when I discuss properties of $L(\mathbb{R})$ and $L(\mathbb{R})[\mathcal{U}]$. For example, if $V = L$, then $L(\mathbb{R}) = L= V$ and so $L(\mathbb{R})$ is a model of $ZFC$. Of course, nobody would dream of studying $L(\mathbb{R})$ under the assumption that $V = L$ ...

Simon Thomas

Posted 2010-05-09T21:00:21.057

Reputation: 5 914

3@Simon: I am coming quite late into this discussion, but let me ask what might be a simpler question: is it known if there is a discontinuous automorphism of $(\Bbb{R},+)$ in $L(\Bbb{R})[\cal{U}]$? – Ali Enayat – 2011-06-24T15:57:49.333

38For nearly two years that missing parenthesis was bugging me. Now I have the reputation to correct it, and now the post "compiles"! Muahaha. :-) – Asaf Karagila – 2012-02-04T00:36:12.603

1Where can I find the result that it is consistent with ZF that the only automorphisms of $\mathbb{C}$ are the identity and complex conjugation? – None – 2012-07-30T14:10:17.433

1Math-player, if all sets of real numbers have the Baire property then every homomorphism between Polish groups is continuous; in particular automorphisms of $\Bbb C$, and the only two automorphisms of $\Bbb C$ which are continuous are the identity and the complex conjugation. Now to find models where all sets of real numbers have the Baire property one can either assume enough large cardinals so $L(\Bbb R)$ satisfies AD; or look at Solovay's model; or look at Shelah's model that was designed to show the consistency of "all sets of reals have the Baire property" without using large cardinals. – Asaf Karagila – 2013-05-02T14:31:34.487

1The bounty is NOT added because the question has not received enough attention. The obvious reason "An answer to this question deserves a reward" is not included among the permissible reasons for starting a bounty. Also why only allow a maximum bounty of only 500 points? – Simon Thomas – 2013-06-26T00:33:00.093

3Simon, the reason for the bounty is invalid... The question has had plenty of attention, it's just really hard!!! – François G. Dorais – 2013-06-26T00:33:08.653

7More to the point, why only 7 days? @FrançoisG.Dorais Didn't we think at some point of requesting more flexibility on bounty times? 7 days for everything is kind of silly. – Andrés E. Caicedo – 2013-06-26T00:34:43.040

2I agree that the 7 day limit is also very silly ... – Simon Thomas – 2013-06-26T00:35:30.637

3@Andres: I think this is really just a system restriction, and I very much doubt that it will be removed anytime soon. I agree with both you and Francois, and I too feel that the bounty doesn't serve the purpose very well. It's really a hard question, and it's likely that those who could solve it (eventually) are aware of it. I'd think it's better to post the bounty after an answer is given to reward the answerer. – Asaf Karagila – 2013-06-26T00:36:53.213

2Yep, this is a poster child for unlimited time bounties... – François G. Dorais – 2013-06-26T00:36:59.873

5+1 because I like ultrafilters and don't get to see them very often as a number theorist. I tried to imitate the finite field trick to construct an automorphism of the ultrapower $\prod_{\mathcal{U}}\mathbb{C}$ and then push down to $\mathbb{C}$, but it only gives you complex conjugation. Interesting. – Matthew Morrow – 2010-05-09T21:49:43.467

1Plus it seems unlikely that you would be able to construct an isomorphism between $\prod_{\mathcal{U}}\mathbb{C}$ and $\mathbb{C}$ without making use of the Axiom of Choice. – Simon Thomas – 2010-05-09T23:15:11.663

1Does the ultrafilter lemma imply the existence of such automorphisms of R. Or, just the ultrafilter lemma restricted to filters on N? That would clearly be necessary for there to be a positive answer to your question. – George Lowther – 2010-05-10T00:44:03.400

I'm sorry but I don't understand your question. Even with the Axiom of Choice, the only automorphism of the field $\mathbb{R}$ is the identity map. – Simon Thomas – 2010-05-10T00:49:27.703

Sorry typo. Should have said C, not R. – George Lowther – 2010-05-10T00:50:56.207

I don't know whether the general Ultrafilter Lemma implies the existence of a third automorphism of $\mathbb{C}$. To be honest, I hadn't considered this very natural question. – Simon Thomas – 2010-05-10T00:58:53.627

2Simon, your statement that "$|Aut(C)|=2$ in $L(R)$" is made under some large cardinal assumption? Obviously, ZFC cannot prove this statement by itself, since I think it contradicts $V=L$. – Joel David Hamkins – 2010-05-10T02:24:27.943

1Of course ... all of my statements about $L(\mathbb{R})$ and $L(\mathbb{R})[\mathcal{U}]$ assume that $L(\mathbb{R})$ is a Solovay model. – Simon Thomas – 2010-05-10T02:53:37.690

7For those of us who are not so well versed in these matters but curious nevertheless, what is the L(R) which appears here? – Spencer – 2010-05-13T19:03:05.290

1L(R) = constructible reals, in the sense of Gödel. – Gerald Edgar – 2010-05-24T18:45:45.477

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$L(\mathbb{R})$ is the smallest inner model of $ZF$ that contains all of the reals ... not just the constructible reals. You can find a more details at:

http://en.wikipedia.org/wiki/Constructible_universe#Relative_constructibility

– Simon Thomas – 2010-05-24T22:25:19.117

@AsafKaragila You can also use your (not longer new) acquired powers tor correct that nasty right bracket at the end of Ali Enayat's commentary ;-). – Matemáticos Chibchas – 2016-04-16T05:02:44.767

Answers

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It seems not.

It was shown by Di Prisco and Todorcevic (and reproved later by at least three sets of authors) that if sufficiently large cardinals exist (e.g., a proper class of Woodin cardinals), then after forcing with $\mathcal{P}(\omega)/\mathrm{Fin}$ (the infinite subsets of $\omega$, ordered by mod-finite containment) to produce a selective ultrafilter $U$, there is no selector (i.e., set meeting each equivalence class in exactly one point) for the equivalence relation $E_{0}$ (mod-finite equivalence on $\mathcal{P}(\omega)$) in the inner model $L(\mathbb{R})[U]$.

It also seems to follow from ZF + DC$_{\mathbb{R}}$ (which holds in $L(\mathbb{R})[U]$) that the existence of a discontinuous homomorphism from either of $(\mathbb{R}, +)$ or $(\mathbb{C}, +)$ to itself implies the existence of an $E_{0}$ selector, as we will show below. Since a discontinuous automorphism of $(\mathbb{C}, +, \times)$ restricts to one for $(\mathbb{C}, +)$, this answers the question. The proof is the same for each of $(\mathbb{R}, +)$ and $(\mathbb{C}, +)$; moreover, the existence of each type of homomorphism implies the existence of the other. I haven't tried writing it up this way, but it seems that the argument can be carried out over an arbitrary complete additive metric group satisfying the triangle inequality. The existence of a discontinuous homomorphism of $(\mathbb{R}, +)$ easily gives one for $(\mathbb{C}, +)$; we give a proof of the reverse direction at the end of this answer.

So, let $h$ be a discontinuous homomorphism from $(\mathbb{R}, +)$ (or $(\mathbb{C}, +)$) to itself. As shown in the proof of Theorem 1 of a 1947 paper by Kestelman, for each positive real number $\delta$, $h$ is unbounded on $\{ x : |x| < \delta \}$. The same proof shows that the same fact holds for $(\mathbb{C}, +)$ (moreover, the fact follows easily from the definition of "discontinuous homomorphism"). Applying DC$_{\mathbb{R}}$, we may find $\{ x_{i} : i < \omega \}$ such that (1) each $|x_{i}|$ is more than $\sum \{ |x_{j}| : j > i\}$ and such that (2) for each $i$, $|h(x_{i})| - \sum \{ |h(x_{j})| : j < i \} > i.$

Let $X = \{ x_{i}: i < \omega \}$ and let $Y$ be the set of reals (or complex numbers) which are sums of (finite or infinite) subsets of $X$ (note that all the infinite sums converge). By condition (1) on $X$, each $y \in Y$ is equal to $\sum \{ x_{i} : i \in S_{y}\}$ for a unique subset $S_{y}$ of $\omega$. Let $F$ be the equivalence relation on $Y$ where $y_{0} F y_{1}$ if and only if $S_{y_{0}}$ and $S_{y_{1}}$ have finite symmetric difference. By condition (2) on $X$, the $h$-preimage of each bounded subset of $\mathbb{R}$ ($\mathbb{C}$) intersects each $F$-equivalence class in only finitely many points (since if the bounded set is contained in an interval of length $i$, then for every $y$ in the intersection $S_{y} \setminus i$ is the same, which can be seen be consideration of the maximum point of disagreement between the sets $S_{y}$). It follows then that there is an $F$-selector : for each equivalence class, let $n \in \mathbb{Z}^{+}$ be minimal such that the $h$-preimage of $[-n, n]$ intersects the class, and then pick the least element of this intersection. Since $Y/F$ is isomorphic to $\mathcal{P}(\omega)/E_{0}$ via the map $y \mapsto S_{y}$, there is then an $E_{0}$-selector.

As for getting a discontinuous homomorphism of $(\mathbb{R}, +)$ from one on $(\mathbb{C}, +)$ : Suppose that $h$ is a homomorphism of $(\mathbb{C}, +)$. Define $f_{0},\ldots,f_{3}$ on $\mathbb{R}$ as follows: (1) If $h(x) = a + bi$, then $f_{0}(x) = a$. (2) If $h(x) = a + bi$, then $f_{1}(x) = b$. (3) If $h(iy) = a + bi$, then $f_{2}(y) = a$. (4) If $h(iy) = a + bi$, then $f_{3}(y) = b$. Then each of $f_{0},\ldots,f_{3}$ is a homomorphism of $(\mathbb{R}, +)$. Since $h(x + iy) = h(x) + h(iy) = f_{0}(x) + if_{1}(x) + f_{2}(y) + if_{3}(y),$ if all of $f_{0},\ldots,f_{3}$ are continuous then $h$ is.

Paul Larson

Posted 2010-05-09T21:00:21.057

Reputation: 1 905