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.
 A: 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.

