# How does one define an inner product on the space $V=\mathbb{Q}_p^n$?

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?

• 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 $\Bbb{Q}_p$)? Jan 16, 2018 at 6:14
• 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? Jan 16, 2018 at 12:04

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 $x\neq 0$.
• 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? Jan 16, 2018 at 6:09
• @TorstenSchoeneberg Yes, you are right, but for the non-archimedian case an "inner" product is defined accordingly, e.g., see here, page $23$. Jan 16, 2018 at 9:43
• 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. Jan 17, 2018 at 3:21
• 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 $O_2(\Bbb{Q}_3)$. Jun 13, 2019 at 1:35