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Given a projective space $\mathbb{P}^n(\mathbb{R})$ and two points $x, y \in \mathbb{P}^n(\mathbb{R})$, the distance between $x$ and $y$ is defined as $$ d(x, y) = \frac{\|v_x \wedge v_y\|}{\|v_x\| \|v_y\|} $$ where $v_x$ and $v_y$ are non-zero vectors corresponding to $x$ and $y$, respectively. This expression is essentially the sine of the angle between the two lines in $\mathbb{R}^{n+1}$.

My question is: If we instead work over the function field over a finite field, such as $\mathbb{F}_q[T]$, where there is no natural notion of angle, how can we define a "distance" between points in projective space? As quadratic forms are well studied over the function field, I believe there might be some notion of distance. Any insight is very much appreciated.

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    $\begingroup$ It seems things make sense for every (complete?) normed field. Once a norm is fixed on $K^n$, there is not much than the choice of norm on $V\wedge V$ for $V=K^n$. $\endgroup$ Commented Nov 9, 2024 at 17:34
  • $\begingroup$ No, but I am asking in the real case, this $d(x,y)$ is nothing but $|\sin(x,y)|$, i.e., sine of the angle between two lines representing $x$ and $y$. In the function field, there is no such sense of angle; in fact the line also consists of just a bunch of points. Then what is the meaning of this distance intuitively? I guess $d(x,y)=|\sin(x,y)|$ does not make sense in this case right? $\endgroup$ Commented Nov 10, 2024 at 5:05
  • $\begingroup$ @YCor I am just not sure what does it signify as there is no notion of angle in the function field. If you can provide some insight. $\endgroup$ Commented Nov 15, 2024 at 5:25
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    $\begingroup$ Your definition of distance only depends on norms, not on any angle notion. If you have an ultrametric norm on $K$, you get, using the sup norm, a natural norm on $K^n$ and hence on $\bigwedge^2 K^n$. $\endgroup$ Commented Nov 15, 2024 at 10:58

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