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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?

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    $\begingroup$ 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 $\Bbb{Q}_p$)? $\endgroup$
    – Torsten Schoeneberg
    Jan 16, 2018 at 6:14
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    $\begingroup$ 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? $\endgroup$
    – Jyrki Lahtonen
    Jan 16, 2018 at 12:04

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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 $x\neq 0$.

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    $\begingroup$ thank you! for clarification, how is the positive definiteness of this inner product proven? $\endgroup$
    – Chris
    Jan 14, 2018 at 22:35
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    $\begingroup$ 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? $\endgroup$
    – Torsten Schoeneberg
    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 $23$. $\endgroup$
    – Dietrich Burde
    Jan 16, 2018 at 9:43
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    $\begingroup$ 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. $\endgroup$
    – Torsten Schoeneberg
    Jan 17, 2018 at 3:21
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    $\begingroup$ 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

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