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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.

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    $\begingroup$ I would look for: (1) Puiseux series or (2) Levi-Civita field or (3) Hahn series. These are real-closed ordered fields larger than $\mathbb R$. $\endgroup$ – Gerald Edgar Oct 25 '20 at 14:41
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    $\begingroup$ 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. $\endgroup$ – Reid Barton Oct 25 '20 at 14:41
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    $\begingroup$ 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. $\endgroup$ – Reid Barton Oct 25 '20 at 14:52
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    $\begingroup$ 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). $\endgroup$ – YCor Oct 25 '20 at 18:43
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    $\begingroup$ 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. $\endgroup$ – YCor Oct 25 '20 at 19:51
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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.)


(Added) Examples:

  • real-closed fields;
  • 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.
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  • $\begingroup$ @NikPronko yes, I should have said up to rescaling. It's fixed now; $\endgroup$ – YCor Oct 26 '20 at 6:46

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