most recent 30 from http://mathoverflow.net 2013-06-19T00:23:06Z http://mathoverflow.net/feeds/question/70762 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70762/hilbert-symbol-and-weil-index-beyond-the-quadratic-case Marty 2011-07-19T18:23:57Z 2011-07-19T21:44:24Z

<p>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.</p> <p>The $n^{th}$ order Hilbert symbol, composed with $\epsilon$ will be denoted by $(x,y)_n$ for all $x,y \in F^\times$.</p> <p>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 $\mu_2 \rightarrow \mu_4$.</p> <p>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?</p> <p>The main difficulties occur when $n$ is a multiple of 4, I think. Any references or known results?</p> <p>------------EDIT BELOW----------------</p> <p>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.</p>