# Are there only two smooth manifolds with field structure: real numbers and complex numbers?

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 $$\mathbb{C}$$?

• What about finite fields (considered as discrete 0-dimensional manifolds). Am I missing something? Oct 28, 2021 at 20:13
• @SamGunningham Yes, they are not connected. Oct 28, 2021 at 20:14
• Sorry, missed that asssumption! Oct 28, 2021 at 20:31
• 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 $\mathbb{Q}$) Oct 28, 2021 at 20:39
• @LoïcTeyssier Yep, a non-discrete finite topological space is not a real manifold. Yes, the question is about that. Oct 28, 2021 at 20:49

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".

• It's beautiful! Thanks! Oct 29, 2021 at 16:42
• 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. Nov 1, 2021 at 18:00
• @BenMcKay: nice observation that one can weaken the initial hypothesis on $\mathbb{F}$!
– M.G.
Nov 1, 2021 at 18:27
• 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. Nov 1, 2021 at 19:34
• Is the intermediate value theorem not topological enough? :-) Nov 1, 2021 at 19:44

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.

• Hmm, it seems p-adic numbers are also a locally compact field en.wikipedia.org/wiki/Locally_compact_field Oct 28, 2021 at 23:07
• Here is the complete classification jstage.jst.go.jp/article/jjm1924/19/2/19_2_189/_pdf Oct 28, 2021 at 23:11
• 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). Oct 28, 2021 at 23:26
• 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. Oct 28, 2021 at 23:27
• 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. Oct 29, 2021 at 0:35