Suppose we have $\mathbb{Z}_n$ the group of residues modulo $n$ and $\mathbb{F}_q$ a Galois finite field with $q$ elements where $q=p^m$ with $p$ prime and $m\in\mathbb{N}$ and suppose $n\vert q-1$.

Suppose we consider the set $ZF = \{f:\mathbb{Z}_n\to\mathbb{F}_q\vert\, f\text{ function}\}$. We see that $\#ZF=q^n$ and that $ZF$ is a vector space. (a simple basis could be $\{e_n(m) = \delta_{nm} \vert n,m\in\mathbb{Z}_n\}$)

If we define the inner product in $ZF$ as $\langle f,g\rangle = \sum_{k\in\mathbb{Z}_n}f(k)g(-k)$ then we can show that it is in fact an inner product.

It is linear in the first argument ($\langle \alpha(f+h),g\rangle=\alpha\langle f,g\rangle + \alpha\langle h,g\rangle$)

It is symmetric since if we have $f(m)g(-m)$ for some $m$ then in a finite group we define $-m:=h$ so that $h+m\equiv 0\mod n$ but then $f(h)g(-h)=f(-m)g(m)$, and since we are summing over all $\mathbb{Z}_n$ the resulting sum is finite and then not dependent on the order of the elements hence the same for every $f(m)g(-m)$ and every $f(-m)g(m)$.

It is defined positive, in fact for the inner product to be zero $f(m)f(-m)$ must be zero for every m in $\mathbb{Z}_n$ (we are multiplying over $F_q$, there aren't two non-zero elements so that $\alpha\cdot\beta=0$ since it's a field and so an intergrity domain).

But now the dilemma, i knew that pre-Hilbert spaces are defined over vector spaces over $\mathbb{R}$ or $\mathbb{C}$. Can i say that $ZF$ is a pre-Hilbert space nonetheless? What's the catch, what did i do wrong?

In that case, how does the completeness (or closure) works in a finite case? Is it free from the finiteness (in the sense that a finite pre-Hilbert space is automatically a Hilbert space)?

EDIT: As pointed out in the comments i can't hope for definite positiveness. So it's not a pre-Hilbert space. What other property can i ask that's weaker and makes thing somewhat work?

  • $\begingroup$ Say $p=5$ , $n=2$. Take $f(0)=2$, and $f(1)=1$. Then $\langle f, f\rangle = 2^2 +1 = 0$. $\endgroup$ – peter a g Feb 20 at 14:56
  • $\begingroup$ Ok, asking for definite positiveness could be a bit too much then, already going from $\mathbb{R}$ to $\mathbb{C}$ we have to lose the symmetry. So maybe we have to lose that in finite fields, what other property can i use? $\endgroup$ – WhiteEyeTree Feb 20 at 15:23

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