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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 $V = L$ ...

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    $\begingroup$ +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. $\endgroup$ – Matthew Morrow May 9 '10 at 21:49
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    $\begingroup$ For those of us who are not so well versed in these matters but curious nevertheless, what is the L(R) which appears here? $\endgroup$ – Spencer May 13 '10 at 19:03
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    $\begingroup$ $L(\mathbb{R})$ 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: en.wikipedia.org/wiki/… $\endgroup$ – Simon Thomas May 24 '10 at 22:25
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    $\begingroup$ For nearly two years that missing parenthesis was bugging me. Now I have the reputation to correct it, and now the post "compiles"! Muahaha. :-) $\endgroup$ – Asaf Karagila Feb 4 '12 at 0:36
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    $\begingroup$ More 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. $\endgroup$ – Andrés E. Caicedo Jun 26 '13 at 0:34
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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.

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protected by Andrés E. Caicedo Oct 10 '13 at 1:49

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