MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be a number field, which we assume for simplicity to be Galois and totally real. Set $\mathcal{O}_p=\mathcal{O}_F\otimes_{\mathbf{Z}}\mathbf{Z}_p$. The norm map on $\mathcal{O}_F$ extends uniquely to a multiplicative function $N: \mathcal{O}_p \to \mathbf{Z}_p$, and we set $\mathcal{O}_p^{1}=x\in \mathcal{O}_p \, \mathrm{with}\, Nx=1$. The famous Leopoldt conjecture asserts that the image of $\mathcal{O}_F^\times$ in $\mathcal{O}_p$, a priori contained in $\mathcal{O}_p^1$, is actually topologically dense in $\mathcal{O}_p^1$. Let us define the Leopoldt locus $\mathcal{L}$ as the topological closure of $\mathcal{O}_F^\times$ in $\mathcal{O}_p^1$.

Now, the Leopoldt conjecture is equivalent to other basic statements, for example the vanishing of $H^2_{\mathrm{cts}}(G_{F'},\mathbf{Q}_p)$ where $F'/F$ is the maximal extension unramified outside the primes dividing $p$ and $\infty$. In fact, the rank of this Galois cohomology group is equal to the codimension of $\mathcal{L}$ inside $\mathcal{O}_{p}^1$, so it makes sense to ask: can we start with a basis of $H^2(G_{F'}, \mathbf{Q}_p)$, i.e. a bunch of 2-cocycles, and use them to write down analytic functions on $\mathcal{O}_p^1$ which cut out $\mathcal{L}$ as their common zero locus?

I'm assuming the answer is "yes, and this is a baby version of some ideas of Minhyong Kim", but I don't know well enough to say.

share|cite|improve this question
Downvotes without an explanatory comment? For shame, for shame. – David Hansen Aug 12 '12 at 2:06

Your Answer


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

Browse other questions tagged or ask your own question.