# A vanishing condition for cup products in Galois cohomology

Let $k$ be a field of characteristic $\neq 2$. For a non-zero element $a \in k^*$, let us write $[a] \in H^1(k,\mathbb{Z}/2)$ for the Galois cohomology class corresponding to the quadratic extension $k(\sqrt{a})/k$. If $a,b \in k^*$ are non-zero elements then the class $[a] \cup [b] \in H^2(k,\mathbb{Z}/2)$ is trivial if and only if $b$ is a norm from $k(\sqrt{a})/k$. In particular, if $ab = -1$ then $b=-a$ is a norm from $k(\sqrt{a})/k$ and hence $[a] \cup [b] = 0$. I have a reason to believe the following generalization is true:

Claim: Let $a_1,...,a_n \in k^*$ be such that $\prod_i a_i = -1$. Then $[a_1] \cup [a_2] \cup ... \cup [a_n] = 0$.

Is this claim true?

One way to approach this problem is via quadratic forms. Let us denote by $\left<b_1,...,b_n\right>$ the isomorphism class of the quadratic form $\sum_i b_i x_i^2$. The Witt ring $W(k)$ of $k$ is the ring generated by isomorphism classes of non-degenerate quadratic forms over $k$, modulu the relation $\left<1,-1\right> \sim 0$ (more precisely, the addition in $W(k)$ is determined by the direct sum operation and multiplication by tensor product). Let $I \subseteq W(k)$ be the ideal generated by forms of even rank. A deep result in the theory of quadratic forms (closely related to Milnor's conjecture), is that there is a natural isomorphism of graded rings $$\oplus_n I^n/I^{n+1} \stackrel{\cong}{\longrightarrow} H^*(k,\mathbb{Z}/2) .$$ In particular, the class of $\left<1,-a\right> \in I$ (mod $I^2$) is sent to the class $[a] \in H^1(k,\mathbb{Z}/2)$, and hence the class of $\left<1,-a_1\right> \otimes ... \otimes \left<1,-a_n\right>$ is sent to $[a_1] \cup ... \cup [a_n]$. A possible strategy is then to show that if $\prod_i a_i = -1$ then the quadratic form $\left<1,-a_1\right> \otimes ... \otimes \left<1,-a_n\right>$ (which has rank $2^n$) is trivial (in the sense that it is isomorphic to a direct sum of $2^n$ copies of $\left<1,-1\right>$).

• Pernickety remark : you should add $n\geq 2$ in the claim. Oct 1 '15 at 12:25
• And in fact, over $\mathbf R$, the quadratic form $<1,-a>^{\otimes s}$ is never hyperbolic ... for $a<0$, in particular for $a=-1$ and $s$ odd ... Oct 1 '15 at 12:58
• Yes, over $\mathbb R$, one has $\prod_{i=1}^3-1=-1$ and $[-1]\cup[-1]\cup[-1]\ne 0$ in $H^3(\mathbb R,\mathbb Z/2)$. Oct 1 '15 at 13:20
• On the other hand, over any field $k$ one has $[x_1]\cup [x_2]\cup\dotsb\cup [x_{2m-1}]\cup[-x_1\dotsb x_{2m-1}]=$ $[x_1]\cup\dotsb\cup [x_{2m-1}]\cup[-x_1\cdot -x_2\cdot\dotsc\cdot-x_{2m-1}]=$ $[x_1]\cup \dotsb\cup [x_{2m-1}]\cup[-x_1]+\dotsb+[x_1]\cup\dotsb\cup[x_{2m-1}]\cup[-x_{2m-1}]=$ $0+\dotsb+0=$ $0$. So your assertion is true for $n$ even and false for $n$ odd. Oct 1 '15 at 13:47
• And for the same reason, one has $[a_1]\cup\dotsb\cup[a_n]=0$ for any $a_i$ such that $\prod_ia_i=1$ when $n$ is odd. Oct 1 '15 at 13:53

For each $l$, let $\mu_l$ denote the Galois module of $l$-roots of unity. In particular, let us use $\mu_2 = \{1,-1\}$ instead of $\mathbb{Z}/2$. For each $n$ we have an inclusion of Galois modules $\mu_l \longrightarrow \mathbb{G}_m$. Now indeed it is not true that if $\prod_i a_i = -1$ then $[a_i] \cup ... \cup [a_i] = 0$. However, it is true that in this case the image of $[a_1] \cup ... \cup [a_n]$ in $H^n(k,\mathbb{G}_m)$ is $0$ (which happens to be sufficient for my original application).
To see this, consider the short exact sequence of Galois modules $$0 \longrightarrow \mu_2 \longrightarrow \mu_4 \longrightarrow \mu_2 \longrightarrow 0 .$$ We then obtain a long exact sequence in homology groups, and in particular a boundary map $$H^{n-1}(k,\mu_2) \longrightarrow H^n(k,\mu_2) .$$ Unwinding the definitions, one can check that this boundary map sends a class $\alpha \in H^{n-1}(k,\mu_2)$ to the class $\alpha \cup [-1] \in H^n(k,\mu_2)$. Now, as explained in the answer above, it is sufficient to consider the case where $n > 1$ is odd. Let $a_1,...,a_n$ be such that $\prod_i a_i = -1$. Then, again as explained above, we have $$[a_1] \cup ... \cup [a_n] = [a_1] \cup ... \cup [a_{n-1}] \cup [-1]$$ and so $[a_1] \cup ... \cup [a_n]$ is in the image of the boundary map $H^{n-1}(k,\mu_2) \longrightarrow H^n(k,\mu_2)$. It follows that the image of $[a_1] \cup ... \cup [a_n]$ in $H^n(k,\mu_4)$ is $0$, and hence the image in $H^n(k,\mathbb{G}_m)$ is $0$ as well.
• It's a bit more complicated, actually. The Bockstein operator (boundary map) $H^{n-1}(k,\mu_2)\to H^n(k,\mu_2)$ corresponding to the short exact sequence $0\to\mu_2\to\mu_4\to\mu_2\to0$ always differs from the boundary map corresponding to the short exact sequence $0\to\mathbb Z/2\to\mathbb Z/4\to\mathbb Z/2\to0$ by the operator of multiplication with $[-1]$. However, which of them is which depends on whether $n$ is odd or even. Oct 1 '15 at 21:11
• When $n$ is odd, the former boundary map (which you are interested in) is the multiplication with $[-1]$ and the latter one vanishes. When $n$ is even, the former one vanishes and the latter one is the multiplication with $[-1]$. Oct 1 '15 at 21:19
• So this extension is neither $0\to\mathbb Z/2\to\mathbb Z/4\to\mathbb Z/2\to 0$, nor $0\to\mu_2\to\mu_4\to\mu_2\to0$, but, in fact, their difference in the group of extensions in the abelian category of discrete $G_k$-modules over $\mathbb Z$. This difference is indeed a short exact sequence of $\mathbb Z/2$-vector spaces, as the $\mathbb Z/4$ and the $\mu_4$-sequences represent the same extension in the category of abelian groups (when you forget the action of $G_k$), so their difference must split as an exact sequence of abelian groups, hence its middle term is a $\mathbb Z/2$-vector space. Oct 2 '15 at 19:58
• This proves that the difference between the two extensions with the middle terms $\mathbb Z/4$ and $\mu_4$ corresponds to a certain class in $H^1(k,\mathbb Z/2)$, and the difference between the two Bockstein operators is the multiplication with this cohomology class. Computing this degree-one cohomology class as $[-1]$ is the "unwinding the definitions" part of the story. Oct 2 '15 at 20:02