# Presentation of Grothendieck-Witt group $GW(\mathbb{F})$ in terms of generators and relations.

Let $$\mathbb{F}$$ be a field, which for the sake of this discussion, is such that char $$\mathbb{F} \neq 2$$.

By Corollary 9.4 in Scharlau's Quadratic and Hermitian Forms, the Grothendieck-Witt group $$GW(\mathbb{F})$$ is generated by elements $$\langle \alpha \rangle, \alpha \in \mathbb{F}^{\times}$$, subject to the relations

1. $$\langle \alpha \rangle = \langle \alpha \beta^{2} \rangle$$ for all $$\alpha , \beta \in \mathbb{F}^{\times}$$.
2. $$\langle \alpha \rangle + \langle \beta \rangle = \langle \alpha + \beta \rangle + \langle (\alpha + \beta)\alpha \beta \rangle$$.

I understand the proof of this result as presented in Scharlau.

However, in Morel's $$\mathbb{A}^{1}$$ Algebraic topology over a field, lemma 2.9, he says that the second relation may be obtained from the first relation and the relation

$$\langle \alpha \rangle + \langle -\alpha \rangle = 1 + \langle -1\rangle$$.

I understand the motivation behind this relation (matrices of the form on the LHS are congruent to the hyperbolic plane when char $$\mathbb{F} \neq 2$$), but cannot formally derive the second relation using this and relation 1).

I am probably just missing a trick. Any help would be much appreciated!

Consider the case $$F=\mathbb{Q}_3$$. Let $$M$$ denote the free abelian group generated by $$\langle\alpha\rangle$$, $$\alpha\in \mathbb{Q}_3^\times$$. We have $$\mathbb{Q}_3^\times/(\mathbb{Q}_3^\times)^2=\{[1],[-1],[3],[-3]\}$$, so the quotient of $$M$$ by the first relation is a free abelian group with basis $$\langle 1 \rangle,\langle -1\rangle, \langle 3 \rangle,\langle -3\rangle$$. If Morel's claim is correct, then $$\operatorname{GW}(F)$$ must be the quotient of this group by the relation $$\langle 3\rangle + \langle -3\rangle = \langle 1\rangle + \langle -1\rangle.$$ However, we have $$\langle 3\rangle + \langle 3\rangle\neq \langle 6\rangle + \langle 54\rangle \quad( = \langle -3\rangle + \langle -3\rangle)$$ in this group, so the second relation does not hold.
On the other hand, it is true that the first and the second relation imply $$\langle \alpha\rangle + \langle -\alpha\rangle=\langle 1\rangle + \langle -1\rangle$$. Let $$x=\dfrac{\alpha+1}{2}$$ and $$y=\dfrac{\alpha-1}{2}$$, so that $$x^2-y^2=\alpha$$. By the second relation, we have $$\langle x^2\alpha\rangle + \langle -y^2\alpha\rangle = \langle (x^2-y^2)\alpha\rangle + \langle -x^2y^2(x^2-y^2)\alpha^3\rangle.$$ Using the first relation and $$x^2-y^2=\alpha$$, we can rewrite this as $$\langle \alpha\rangle + \langle -\alpha\rangle = \langle 1\rangle + \langle -1\rangle.$$ Perhaps this is what Morel wanted to note.
• Thanks, but could you spell out why in your counterexample that $\langle 3 \rangle + \langle 3 \rangle \neq \langle 6 \rangle + \langle 54 \rangle$ ? (I haven't really worked with $\mathbb{Q}_{3}$). Sep 30, 2021 at 10:09
• It is because $(\langle 3\rangle+\langle 3\rangle)-(\langle -3\rangle+\langle -3\rangle)$ is not contained in the subgroup generated by $(\langle 3\rangle+\langle -3\rangle) -(\langle 1\rangle+\langle -1\rangle)$ in the free abelian group with basis $\langle 1\rangle,\langle -1\rangle,\langle 3\rangle,\langle -3\rangle$. Sep 30, 2021 at 10:15