Is it true that in the category of connected smooth manifolds equipped with a compatible field structure (all six operations are smooth) there are only two objects (up to isomorphism) - $\mathbb{R}$ and $\mathbb{C}$?
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asked Oct 28, 2021 at 18:34
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2$\begingroup$ What about finite fields (considered as discrete 0-dimensional manifolds). Am I missing something? $\endgroup$– Sam GunninghamCommented Oct 28, 2021 at 20:13
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4$\begingroup$ @SamGunningham Yes, they are not connected. $\endgroup$– Arshak AivazianCommented Oct 28, 2021 at 20:14
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2$\begingroup$ Sorry, missed that asssumption! $\endgroup$– Sam GunninghamCommented Oct 28, 2021 at 20:31
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$\begingroup$ Well, they can be equipped with a topology that makes them connected... But in that case they are not manifolds modelled on $\mathbb{R}^n$, is that what you have in mind ? (I must confess I don't know where the theory of smooth manifolds breaks over $\mathbb{Q}$) $\endgroup$– Loïc TeyssierCommented Oct 28, 2021 at 20:39
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1$\begingroup$ @LoïcTeyssier Yep, a non-discrete finite topological space is not a real manifold. Yes, the question is about that. $\endgroup$– Arshak AivazianCommented Oct 28, 2021 at 20:49
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2 Answers
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Here is a series of standard arguments.
Let $(\mathbb{F},+,\star)$ be such a field. Then $(\mathbb{F},+)$ is a finite-dimensional (path-)connected abelian Lie group, hence $(\mathbb{F},+) \cong \mathbb{R}^n \times (\mathbb{S}^1)^m$ as Lie groups. Since $\mathbb{F}$ is path-connected, there is in particular a path $\gamma: [0,1] \to \mathbb{F}$ with $\gamma(0) = 0_{\mathbb{F}}$ and $\gamma(1) = 1_{\mathbb{F}}$. Now consider the homotopy $H: \mathbb{F} \times [0,1] \to \mathbb{F}$, $(x,t) \mapsto \gamma(t) \star x$. This gives a contraction of $\mathbb{F}$ and so we can exclude all the circle factors.
Now, fix $y_0 \in \mathbb{F}$ and consider the map $\widehat{y_0}: \mathbb{R}^n \to \mathbb{R}^n$, $x \mapsto x \star y_0$. Then $\widehat{y_0}$ is an additive map (but at the moment not necessarily linear with respect to the natural vector space structure on $\mathbb{R}^n$). It is not too difficult to see that by additivity we have $\forall q\in \mathbb{Q}: \widehat{y_0}(qx) = q \widehat{y_0}(x)$. Since $\widehat{y_0}$ is continuous (as being smooth), it now follows that it's actually $\mathbb{R}$-linear.
Thus $\mathbb{F}$ is an $\mathbb{R}$-algebra. From this point on one can finish either by the Frobenius theorem on the classification of finite-dimensional associative $\mathbb{R}$-algebras or invoke a theorem of Bott and Milnor from algebraic topology that $\mathbb{R}^n$ can be equipped with a bilinear form $\beta$ turning $(\mathbb{R}^n,\beta)$ into a division $\mathbb{R}$-algebra (not necessarily associative) only in the cases $n=1,2,4,8$.
EDIT: Another finishing topological argument is a theorem of Hopf saying that $\mathbb{R}$ and $\mathbb{C}$ are the only finite-dimensional commutative division $\mathbb{R}$-algebras. This is less of an overkill compared to invoking Frobenius or Bott–Milnor as the proof is a rather short and cute application of homology, see p.173, Thm. 2B.5 in Hatcher's "Algebraic Topology".
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3$\begingroup$ You don't really need connected, just positive dimensional with smooth multiplication. If you pick some $x\ne 0$ in the path component of $0$, and divide by $x$, you take the $x$ path component to the $1$ path component, but you fix $0$. So $0$ and $1$ are on the same path component. Then your homotopy above shows there is only one path component. $\endgroup$ Commented Nov 1, 2021 at 18:00
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1$\begingroup$ @BenMcKay: nice observation that one can weaken the initial hypothesis on $\mathbb{F}$! $\endgroup$– M.G.Commented Nov 1, 2021 at 18:27
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2$\begingroup$ One can have even less "overkill", because the question was specifically about fields and the only finite field extensions of $\mathbb{R}$ are $\mathbb{R}$ itself and $\mathbb{C}$ which is an easy consequence of the fact that all real polynomials of odd degree have a real root. $\endgroup$ Commented Nov 1, 2021 at 19:34
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2$\begingroup$ Is the intermediate value theorem not topological enough? :-) $\endgroup$ Commented Nov 1, 2021 at 19:44
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Finite-dimensional manifolds are locally compact, and the only non-discrete locally compact topological fields are the reals and the complex numbers. So the answer is yes.
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1$\begingroup$ Hmm, it seems p-adic numbers are also a locally compact field en.wikipedia.org/wiki/Locally_compact_field $\endgroup$ Commented Oct 28, 2021 at 23:07
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2$\begingroup$ Here is the complete classification jstage.jst.go.jp/article/jjm1924/19/2/19_2_189/_pdf $\endgroup$ Commented Oct 28, 2021 at 23:11
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4$\begingroup$ Of course, the $p$-adic numbers are not connected (in fact totally disconnected, though not discrete, so that the statement exactly as written is indeed false). $\endgroup$– LSpiceCommented Oct 28, 2021 at 23:26
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1$\begingroup$ If formal power series (I guess meaning Laurent polynomials?) over a finite field are given their topology coming from the valuation, then they, too, are totally disconnected. $\endgroup$– LSpiceCommented Oct 28, 2021 at 23:27
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3$\begingroup$ The formal power series over a field (a finite field or infinite field) form a commutative ring. If you pass to its fraction field you get the Laurent series over that field. It is strange (in English) to refer to that as a division ring when you don't have any actual noncommutative division rings under discussion too: it's better to call $\mathbf R$, $\mathbf Q_p$, $\mathbf F_p((t))$, and their finite extension fields rather than division rings. $\endgroup$– KConradCommented Oct 29, 2021 at 0:35