# Is there a trigonometric field which is different enough from real numbers?

I found this topic in a book 'Metric Affine Geometry' by Ernst Snapper and Robert J. Troyer. I call a field $$k$$ trigonometric iff there is a quadratic form $$q$$ over $$k^2$$ such that every two lines through the origin in $$k^2$$ is isometric with respect to $$q$$. This condition is sufficient to introduce trigonometric functions over $$\mathbf{SO}(k^2,q)$$ in a geometric fashion. Hence, a name.

Obviously, $$\mathbb{R}$$ is trigonometric. I know, that to be trigonometric the field $$k$$ must be Pythagorean, that is for every finite sequence of values $$(\alpha_i)^n_{i=1} \in k^n$$ there is a $$\gamma \in k$$ such that $$\sum^n_{i=1} \alpha_i^2 = \gamma^2,$$

namely every sum of squares is a square. Secondly it must be a formally real field, which means that $$-1$$ is not a sum of squares. Hence, sadly $$\mathbb{Q},\mathbb{C},\mathbb{R}(x),\mathbb{Q}_p,\mathbb{F}_p$$ are all not trigonometric. Probably some extension of $$\mathbb{R}(x)$$ which allows square roots of formally positive functions may work. But I still doubt that it can be totally-ordered, and probably there are some clews in differential Galois theory. Maybe $$\hat{\mathbb{Q}} \cap \mathbb{R}$$, where $$\hat{\mathbb{Q}}$$ are algebraic numbers will work, or just adjoining enough real algebraic square roots to $$\mathbb{Q}$$ (call it a Pythagorean closure $$\overline{\mathbb{Q}}$$). At least it is Pythagorean and formally real. But I don't think it is interesting enough.

But I'm very a curious about finding an interesting example of trigonometric field different from $$\mathbb{R}$$. Trigonometric field $$k$$ different from $$\mathbb{R}$$ may mean formally that $$k$$ is not between $$\overline{\mathbb{Q}}$$ and $$\mathbb{R}$$. I would be very grateful if you could suggest one. If the result are negative, this would mean that class of all trigonometric fields has certain lower and upper bounds.

• I would look for: (1) Puiseux series or (2) Levi-Civita field or (3) Hahn series. These are real-closed ordered fields larger than $\mathbb R$. – Gerald Edgar Oct 25 '20 at 14:41
• Any real closed field (for example Puiseux series) is trigonometric, and likely to contain subfields which are not real closed but still trigonometric. I don't know whether they qualify as "interesting enough" though. – Reid Barton Oct 25 '20 at 14:41
• The reason I suggested real closed fields might not be that interesting is that maybe the easiest way to see that they are trigonometric is to observe that being a trigonometric field is a first-order condition in the language of fields, and real closed fields satisfy all the same such conditions as the reals do! So as far as first-order properties are concerned, nothing new happens in these real closed fields. – Reid Barton Oct 25 '20 at 14:52
• An ordered field in which every positive element has a square root is trigonometric (using the standard scalar product). This is usually far from real-closed (e.g., the subset of $\mathbf{R}$ generated from rational by field operations and taking square roots of positive elements). – YCor Oct 25 '20 at 18:43
• I didn't say they're more/less interesting. Real-closed field are the most trivial examples (since as said by Reid Barton it's a 1st order property), which is already sufficient to provide examples of arbitrary infinite cardinal. The question is maybe to just formulate a simple necessary and sufficient condition for a field (not in terms of quadratic forms!) to satisfy the property. – YCor Oct 25 '20 at 19:51

A field $$K$$ is trigonometric iff the sum of 2 squares is a square and $$-1$$ is not a square (equivalently, the set of nonzero squares is stable under addition), in which case the standard scalar product on $$K^2$$ satisfies the required condition.

Indeed, suppose that $$K$$ is trigonometric, so there is a nonzero quadratic form $$q$$ on $$K^2$$ such that $$\mathrm{O}(q)$$ is transitive on $$\mathbf{P}^1(K)$$ (if $$q=0$$ is allowed the condition is void! if nonzero, the kernel is invariant by the isometry group, so has to be reduced to $$\{0\}$$, i.e. $$q$$ is nondegenerate). Up tp rescale $$q$$ and change basis, we can suppose that $$q(x,y)=x^2+ty^2$$ for some $$t\in K$$. Transitivity of $$\mathrm{O}(q)$$ implies that $$q(x,y)$$ is nonzero for all nonzero $$(x,y)$$, and that $$q(x,y)/q(x',y')$$ is a square for all nonzero $$(x,y)$$ and $$(x',y')$$. Applying this to $$(1,0)$$ and $$(0,1)$$ already implies that $$t$$ is a square, and hence after again changing the basis, we can suppose that $$t=1$$.

Since $$q(x,y)/q(1,0)$$ is a square for all $$x,y$$, we obtain that $$x^2+y^2$$ is a square for all $$x,y$$. If $$-1$$ were a square, say $$i^2=-1$$, the element $$(1,i)$$ would have $$q(1,i)=0$$, contradiction.

Conversely (the converse is already in the comments), suppose that the conditions are satisfied, and fix the standard scalar product. Consider $$(a,b)\neq (0,0)$$. Since $$a^2+b^2$$ is a square, we can rescale it to assume $$a^2+b^2=1$$, and then it is in the orbit of $$(1,0)$$, using the rotation matrix $$\begin{pmatrix} a&-b\\b&a\end{pmatrix}$$.

(Note that the proof also implies that if a 2-dimensional quadratic form $$q$$ has $$\mathrm{O}(q)$$ transitive on $$\mathbf{P}^1(K)$$, then $$q$$ is equivalent to a scalar multiple of the usual scalar product.)

• more generally, fields for which $$x\le y$$ $$\Leftrightarrow$$ y-x is a square defines a total order. The smallest subfield of $$\mathbf{R}$$ among those stable under taking square roots of positive elements has this property, yet isn't real-closed;
• there are more examples, namely it can happen that there are elements $$y$$ such that neither $$y$$ nor $$-y$$ is a square. This is for instance the case for the ring $$\mathbf{R}(\!(t)\!)$$. Indeed, in this field, the nonzero squares are the elements of the form $$t^{2n}P$$ where $$P\in\mathbf{R}[\![t]\!]$$ has $$P(0)>0$$ and $$n\in\mathbf{Z}$$. In this field, none of $$\pm t$$ is a square.