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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!

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1 Answer 1

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I think Morel's claim is not correct.

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.

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  • $\begingroup$ 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 $\mathbb{Q}_{3}$). $\endgroup$
    – Sunny Sood
    Sep 30, 2021 at 10:09
  • $\begingroup$ 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 $\langle 1\rangle,\langle -1\rangle,\langle 3\rangle,\langle -3\rangle$. $\endgroup$ Sep 30, 2021 at 10:15
  • $\begingroup$ Ahh... I see. Thanks! $\endgroup$
    – Sunny Sood
    Sep 30, 2021 at 12:49

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