1
$\begingroup$

In J H H Chalk's paper "Algebraic Lattices", he defines the notion of a lattice in ultrametric space as the following:

Given a non-archimedean field $K$ with ring of integers $\mathcal{O}_K$ and completion $\tilde{K}$,

  1. The integer lattice is defined $\Lambda_0 = \mathcal{O}_K^n \subset V= \tilde{K}^n$,

  2. A lattice $\Lambda$ in $V$ is the image of $\Lambda_0$ under an invertible $\tilde{K}$-linear function $\lambda$ that sends $V$ to itself. The determinant $d(\lambda) = |\det{\lambda}|$,

  3. $\|\mathbf{x} - \mathbf{y}\|=\max_{1\leq i \leq n}|x_i - y_i|$ for all $\mathbf{x},\mathbf{y} \in V$ with entries $x_i, y_i$.

Define $K = \mathcal{Q}_p$ so its ring of integers is the $p$-adic integers. How can we define an inner product on this space that satisfies the properties of an inner product, e.g. to find the dual of a lattice?

$\endgroup$
2
  • 2
    $\begingroup$ What exactly are "the properties of an inner product" here, given that the field $\Bbb{Q}_p$ cannot be ordered (i.e. there is no notion of "$\ge 0$" for elements in $\Bbb{Q}_p$)? $\endgroup$ Jan 16, 2018 at 6:14
  • 2
    $\begingroup$ For the purposes of defining the dual lattice Isn't enough that the bilinear form is non-degenerate? What Torsten said is, of course, true, but do we really need ordering to get dual lattices? Won't $$\Lambda^*=\{x\in K^n\mid (x,y)\in\mathcal{O}_K\ \text{for all $y\in\Lambda$}\}$$ do nicely? $\endgroup$ Jan 16, 2018 at 12:04

1 Answer 1

1
$\begingroup$

On the complete metric space $\mathbb{Q}_p^n$ we introduce the inner product as usual: $$ (x,y)=x_1y_1+\cdots +x_ny_n $$ for $x=(x_1,\ldots ,x_n)$ and $y=(y_1,\ldots ,y_n)$. This satisfies the inequality $$ |(x,y)|_p\le |x|_p\cdot |y|_p. $$

Edit: As $K=\mathbb{Q}_p$ is not ordered, the usual third axiom $(x,x)> 0$ for $x\neq 0$ is commonly replaced by the axiom $(x,x)\neq 0$ whenever $x\neq 0$.

$\endgroup$
5
  • 1
    $\begingroup$ thank you! for clarification, how is the positive definiteness of this inner product proven? $\endgroup$
    – Chris
    Jan 14, 2018 at 22:35
  • 2
    $\begingroup$ Sorry, I don't understand the last comment. $\mathbb{Q}_p$ cannot be ordered, so a property like "$(x,x) \ge 0$" does not even make sense here. (Also, e.g. there is $x_1 \in \Bbb{Q}_2$ with $x_1^2 = -7$; so for the vector $x = (x_1,1,1,1,1,1,1,1) \in \Bbb{Q}_2^8$, we have $(x,x) = 0$). Maybe there is a different defintion of "positive definite" here that I'm not aware of? $\endgroup$ Jan 16, 2018 at 6:09
  • $\begingroup$ @TorstenSchoeneberg Yes, you are right, but for the non-archimedian case an "inner" product is defined accordingly, e.g., see here, page $23$. $\endgroup$ Jan 16, 2018 at 9:43
  • 1
    $\begingroup$ OK, but except for very small dimension $n$, that product will not even satisfy $(x,x) \neq 0$, see above example. Actually, it is a classical result (due to Witt?) that every quadratic form in $n \ge 5$ variables over a $p$-adic field is isotropic. However, of course your form is non-degenerate, and I think that should actually be good enough as per Jyrki's comment. $\endgroup$ Jan 17, 2018 at 3:21
  • 1
    $\begingroup$ For $n=2$ if $p \equiv 1\bmod 4$ then $( (1,i), (1,i)) = 0$ and for $n \ge 3$ there is always a non-zero solution to $x_1^2+x_2^2+x_3^2=0$. In $\Bbb{Q}_3^2$, we have $(x,x) = 0 \implies x=0$ and the bilinear form is invariant under $O_2(\Bbb{Q}_3)$. $\endgroup$
    – reuns
    Jun 13, 2019 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.