# What are the fields in which $-1$ is not a square and over which every finite-dimensional vector space has a nice inner product?

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.

• Property 2' always holds: you can just define $\tilde{\rho}$ separately on every 1-dimensional subspace of $V$. – Eric Wofsey Sep 13 at 20:52
• 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. – Malkoun Sep 13 at 20:57
• 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? – Qiaochu Yuan Sep 13 at 21:26

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.
• 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? – Malkoun Sep 13 at 18:25