I've stumbled in the german wikipedia over the question if a vector space over a finite field exists, that has a dot product.

Definition of dot product

A dot product over a $\mathbb{K}$-vector space V is a mapping $F : V \times V \rightarrow \mathbb{R}$ which is...

... bilinear: $\forall a, a_1, a_2, b, b_1, b_2 \in V \forall \lambda_1, \lambda_2, \mu_1, \mu_2: $ $\begin{align*} F(\lambda_1 \cdot a_1 + \lambda_2 \cdot a_2, b) &= \lambda_1 \cdot F(a_1, b) + \lambda_2 \cdot F(a_2, b)\\ F(a, \mu_1 \cdot b_1 + \mu_2 \cdot b_2) &= \mu_1 \cdot F(a, b_1) + \mu_2 \cdot F(a, b_2) \end{align*}$

... symmetric: $F(a, b) = F(b,a)$

... positive definite: $\forall a \in V: F(a,a) \geq 0 \land ( F(a,a) = 0 \Leftrightarrow a = 0)$

My try

Please correct my choice of words if you know what I mean and if you know how to write it in English. I am not used to write math texts in English.

$\mathbb{K} := \mathbb{Z} / 2 \mathbb{Z}$ Let $V$ be a $\mathbb{K}$-vector space over the set $\{0,1\}$.

$\mathbb{K}$ is a finite filed.

Define $\langle 0, 0 \rangle := 0$, $\langle 1, 0 \rangle := \langle 0,1 \rangle := 1$ and $\langle 1, 1 \rangle = 1$.


$\langle, \rangle$ is symmetric per definition

bilinear form

$f(1 + 1, 1) = f(0,1) = 1 \neq 0 = 1+1 = f(1,1) + f(1,1)$

Is it possible to define a dot product for a bigger vector space / finite field?

(If the answer is yes, please give me an example)

edit: argh - $f(a,a) = 0 \Leftrightarrow a = 0$, but $f(a,b)$ can be 0. I always forget that :-/

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    $\begingroup$ Should not $F$ take values in $\mathbb K$? Either way, the notion of positive is not defined in finite fields, so you'll have to do without that axiom. $\endgroup$ – user31373 Aug 22 '12 at 11:12
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    $\begingroup$ Indeed. It is in fact easy to prove that the prime field of an ordered field is isomorphic to the rationals. $\endgroup$ – kahen Aug 22 '12 at 11:19
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    $\begingroup$ See this question for more discussion, why an inner product in a vector space over a finite field does not work. Basically it is what LVK says: we cannot set aside a subset of positive elements. You can study quadratic forms (or related bilinear forms) though, if you are not afraid of vectors being orthogonal to themselves. $\endgroup$ – Jyrki Lahtonen Aug 22 '12 at 11:27
  • $\begingroup$ ... But if you define $\langle 0,1\rangle=\langle1,0\rangle=\langle0,0\rangle=0$ and $\langle1,1\rangle=1$ you do get something that seems to work. But only for the one-dimensional space - otherwise the problem of the set of "positive" elements not being closed under addition raises its ugly head. $\endgroup$ – Jyrki Lahtonen Aug 22 '12 at 11:31
  • $\begingroup$ @JyrkiLahtonen: You seem to be saying $\langle p,q\rangle=pq$. Are you suggesting to call the quadratic residus positive and the non-residus negative? That would be weird, but is the only notion I can think of where "positive" is closed under multiplication. $\endgroup$ – Marc van Leeuwen Aug 22 '12 at 11:59

Certainly the "obvious" inner product of $(v_1,\dots v_n)$ with $(w_1,\dots w_n)$ given by $\sum_n v_n w_n$ is defined for any (finite dimensional) vector space. It is symmetric and bilinear and fairly useful. It is used extensively, for example, when discussing linear codes over finite fields. (The codes are vector spaces!)

There isn't any reason to discouraged that the "positive definite" condition is missing. There are many interesting inner product spaces where the inner product is not positive definite... even over $\mathbb{R}$! The inner product of Minkowski space is not positive definite, for example, and this is the setting for special relativity.

Positive definiteness is kind of a "Euclidean ideal" that we would hope for, but the Minkowski space example shows that nature isn't always Euclidean, and that non-positive-definite inner products arise naturally.


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