Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $F$ be a local nonarchimedean field. Let $n$ be a positive integer for which the group $\mu_n(F)$ of $n^{th}$ roots of unity in $F$ has order $n$. Let $\epsilon: \mu_n(F) \rightarrow C^\times$ be a faithful character.

The $n^{th}$ order Hilbert symbol, composed with $\epsilon$ will be denoted by $(x,y)_n$ for all $x,y \in F^\times$.

Somewhat remarkably, when $n = 2$, there is a gadget called the Weil index (depending on a nontrivial additive character of $F$), which is a function $w: F^\times \rightarrow \mu_4(C)$ satisfying $$(x,y)_2 = w(xy) w(x)^{-1} w(y)^{-1}.$$ In other words, the quadratic Hilbert symbol, viewed as a cocycle in $Z^2(F^\times, \mu_2)$ becomes a coboundary after applying the unique injective map on coefficients $\mu_2 \rightarrow \mu_4$.

The Weil index has many lovely properties, but I'm interested merely in the extent to which it generalizes to $n > 2$. Namely, does there exist a "higher-order Weil index" $w_n: F^\times \rightarrow \mu_{2n}(C)$ satisfying $$(x,y)_n = w_n(xy) w_n(x)^{-1} w_n(y)^{-1}?$$ Or perhaps does one need to pass to a larger group than $\mu_{2n}(C)$ for the "Hilbert cocycle" to become a coboundary?

The main difficulties occur when $n$ is a multiple of 4, I think. Any references or known results?

------------EDIT BELOW----------------

Oh no - I missed a dumb point. $(x,y)_n$ is skew-symmetric. So when $n$ is a multiple of $4$, we have $(x,y) \neq (y,x)$ quite often. Thus the existence of such a "higher-order Weil index" is not possible when $n$ is a multiple of $4$. Well... maybe this question will keep others from making similar mistakes.

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.