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Let $\mathcal{C}$ be the class of all fields $k$ such that:

  1. $-1 \in k$ is not a square in $k$, and
  2. For any finite-dimensional vector space $V$ over $k$, there is a symmetric bilinear map:

$$g: V \times V \to k$$

such that for any nonzero $v \in V$, $g(v,v)$ is a nonzero square in $k$.

It is clear that $\mathcal{C}$ is non-empty, since $\mathbb{R} \in \mathcal{C}$. Could someone describe the class of such fields? Has it been studied before? Or maybe could someone name some more examples please? As a remark, if $k$ satisfies property 1 and if any sum of squares $a_1^2 + \cdots + a_n^2$ where the $a_i \in k$ and are not all $0$ is a nonzero square in $k$, then $k$ also satisfies property 2, and thus belongs to $\mathcal{C}$. Indeed, one could just let $g$ be the "Euclidean" inner product (whose definition is analogous to the Euclidean inner product, but with elements in $k$).

Edit: extra questions please. Let us consider the class $\mathcal{C}'$ of fields $k$ such that:

1'. there is a group homomorphism $\rho: k^* \to k^*$ such that $-1 \notin \operatorname{Im}(\rho)$ (Thank you @QiaochuYua for catching this typo), and

2'. for any finite-dimensional vector space $V$ over $k$, there is a map $\tilde{\rho}: V^* \to k^*$ whose restriction to $L^*:= L \setminus \{\mathbf{0}\}$ for any one-dimensional subspace $L \subseteq V$ can be identified with $\rho$ for any choice of basis vector of $L$. More precisely, given a one-dimensional subspace $L \subseteq V$ and $v \in L^*$ (i.e. $v$ is a nonzero element in $L$), we require $\tilde{\rho}$ and $\rho$ to be compatible in the following sense:

$$ \tilde{\rho}(a v) = \rho(a) \tilde{\rho}(v),$$

for any $a \in k^*$.

Note that $\mathcal{C}'$ is defined by weakening properties 1 and 2 of $\mathcal{C}$, in some sense by removing the restriction to quadratic maps. I am hoping that $\mathcal{C}'$ is strictly larger than $\mathcal{C}$.

As another remark, if $k$ is a formal real Pythagorean field (see the accepted answer), if $\rho$ is the square map, mapping $a \in k^*$ to $a^2$ and if $\tilde{\rho}$ on $k^n \setminus \{\mathbf{0}\}$ is the sum of squares of the components, we then see that $(k,\rho, \tilde{\rho})$ satisfy properties 1' and 2'. Thus, using @EricWofsey's answer, we see that $\mathcal{C} \subseteq \mathcal{C}'$. Is $\mathcal{C}'$ strictly bigger than $\mathcal{C}$ though? I think so, and hope so too.

Edit 2: following @QiaochuYuan's suggestion, I will tell you what I am trying to do. I am essentially trying to find an algebraic generalization of the Hopf map, that works over a large class of fields (assuming this can be done!). As you know, the Hopf map is a quadratic map from the three-dimensional unit sphere in $\mathbb{C}^2$ to the two-dimensional sphere in $\mathbb{R}^3$. Note that there are two fields here, $\mathbb{R}$ and $\mathbb{C}$. Note that both the domain and the target of the Hopf map can be enlarged, and the Hopf map is actually the restriction of a homogeneous quadratic map from $\mathbb{C}^2$ onto $\mathbb{R}^3$ which takes a nonzero vector in $\mathbb{C}^2$ to a nonzero image, and which is onto. Moreover, a preimage of a nonzero $v \in \mathbb{R}^3$ and a preimage of $-v$, under the Hopf map, are $\mathbb{C}$-linearly independent in $\mathbb{C}^2$.

So what I would like, is to consider two fields, $F$ and $k$, such that there is a homogeneous polynomial map, say $h$, from $F^m$ to $k^n$ of degree $d$, for some integers $m>1$ and $n>1$, which takes a nonzero $v \in F^m$ to a nonzero image in $k^n$, and which is onto. Moreover, such a map $h$ should have the property that for any nonzero $v \in k^n$, a preimage of $v$ and a preimage of $-v$ should be $F$-linearly independent. I was trying to abstract the kind of properties that I wanted $k$ to have, but I think, since it is not an easy task, that it is indeed better to tell you what I am trying to do.

I have a feeling that if $F$ is a finite extension of $k$, then one may be able to use the norm map of $F/k$ to define a "candidate Hopf map", but details remain to be checked.

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    $\begingroup$ Property 2' always holds: you can just define $\tilde{\rho}$ separately on every 1-dimensional subspace of $V$. $\endgroup$ – Eric Wofsey Sep 13 at 20:52
  • $\begingroup$ hmm, yes true... I am trying to avoid talking about continuity etc., in order to remain in an algebraic setting. But I guess I need to impose something, otherwise, as you wrote, 2' is always satisfied. $\endgroup$ – Malkoun Sep 13 at 20:57
  • $\begingroup$ Did you intend to write $-1 \not \in \text{im}(\rho)$ as property 1'? Otherwise $x \mapsto x^3$ works for property 1' as stated, for any field of characteristic $\neq 2$. Maybe you should just tell us what you want these fields for? $\endgroup$ – Qiaochu Yuan Sep 13 at 21:26
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Suppose $g$ is a symmetric bilinear form on a finite-dimensional vector space $V$ such that $g(v,v)$ is a nonzero square for any nonzero $v\in V$. Then $V$ has an orthonormal basis with respect to $g$. Indeed, you can start with any basis for $V$ and use the Gram-Schmidt process to orthonormalize it: the assumption that $g(v,v)$ is always a nonzero square is exactly what you need for Gram-Schmidt orthonormalization to work.

This means that if property (2) holds, then it actually holds whenver $g$ is just the usual inner product on $k^n$. So, property (2) is equivalent to saying that any sum of squares in $k$ which are not all zero is a nonzero square. Note that this implies that $-1$ is not a sum of squares in $k$ (and in particular implies (1)), since otherwise you could add $1$ to write $0$ as a sum of squares that are not all $0$. So, every such field $k$ can be made into an ordered field, and your $\mathcal{C}$ is the class of fields that can be made into ordered fields in which any sum of squares is a square. Such fields are known as formally real Pythagorean fields.

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  • $\begingroup$ Damn, beat me to it! As two simple examples, the real algebraic numbers are such a field (since they are even real closed), and the subfield of the real algebraic numbers obtained by repeatedly adjoining all square roots is also such a field. $\endgroup$ – Qiaochu Yuan Sep 13 at 18:20
  • $\begingroup$ Thank you! That was quite fast. By the way, how do you define the order? I guess you just define the positive elements in $k$ to be the sums of squares of elements that are not all $0$, right? $\endgroup$ – Malkoun Sep 13 at 18:25
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    $\begingroup$ @Malkoun: The order is generally not unique (and requires the axiom of choice to construct). The idea is you take a maximal subset that "looks like" the positive elements with respect to some ordering, and prove it can actually be the positive elements. $\endgroup$ – Eric Wofsey Sep 13 at 18:39
  • $\begingroup$ Please look at my edit. I tried to weaken conditions 1 and 2, to try to get more fields, while still having the properties I need (I think). $\endgroup$ – Malkoun Sep 13 at 19:07

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