I found this topic in a book ['Metric Affine Geometry' by Ernst Snapper and Robert J. Troyer](https://books.google.ru/books/about/Metric_Affine_Geometry.html?id=PVbvAAAAMAAJ&redir_esc=y).
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.



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


