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Let $B:V\times V \rightarrow K$ be a (weak!) bilinear form where $K$ is a finite field with base field $F$ and $V$ a vector space over $K$. Let $u,v \in V$ and $\lambda \in F (!)$

$B(u + v, w) = B(u, w) + B(v, w)$

$B(u, v + w) = B(u, v) + B(u, w)$

$B(λu, v) = B(u, λv) = λB(u, v)$ (actually this condition is not necessary because of above conditions.)

Alternating: $B(v, v) = 0$ holds for all $v ∈ V$, and

Nondegenerate: when $B(u, v) = 0$ holds regardless of $v, u$ must be zero.

My question is whether $V$ "can have" a symplictic basis or not. (I think the answer is yes. But i am not sure.)+(i don't know there is a definition of "weak bilinear form", i just defined it for not misunderstanding.)

bilinear form

symplectic vector space

symplectic basis

Edited part: has --> can have

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The answer is negative. Below is a slightly cheating counterexample. I say cheating, because you are probably looking for something more interesting.

Let $\overline{B}:V\times V\to F$ be a usual (non-degenerate) symplectic form over $F$ (among other things it thus has a symplectic basis over $F$), and let $e\in K\setminus F$ be a fixed element. Then the formula $$ B(u,v)=e \overline{B}(u,v) $$ gives another function $B:V\times V\to K$. This function inherits all the listed properties from $\overline{B}$.

But it cannot have a symplectic basis, because $B(u,v)\neq 1_K$ for all $u,v$.

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A concrete toy example. Let $K=\Bbb{F}_4=\{0,1,a,a+1=a^2\}$. Let $V=K$. We view $V$ as a vector space over $F=\{0,1\}\subset K$ with basis $\mathcal{B}=\{1,a\}$. Let $\overline{B}$ be the bilinear mapping gotten by extending $\overline{B}(1,1)=0=\overline{B}(a,a)$, $\overline{B}(1,a)=1=\overline{B}(a,1)$ bilinearly over $F$. Then the function $$ B(x,y)=a \overline{B}(x,y) $$ is a weakly bilinear function. There is no symplectic basis w.r.t. $B$, because for all $x,y\in V$ we have $B(x,y)\in\{0,a\}$.

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  • $\begingroup$ If we assume $[K:F]$ is odd. If we use trace map $Tr \: o \: B = \: \bar B$. Then $Tr(\bar B(u_i,v_j))=Tr(\delta_{i,j})=\delta_{i,j}$. Cant we construct a basis for $B$? $\endgroup$
    – vudu vucu
    Mar 24, 2015 at 11:56
  • $\begingroup$ Cant I choose $e=1 \in F$? $\endgroup$
    – vudu vucu
    Mar 24, 2015 at 11:56
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    $\begingroup$ You can, but the point was that if we select $e\notin F$, then the values of $B$ are in the set $Fe$ that intersects $F$ only at $0$ and misses $1$ entirely. The "dirty" trick is that with my counterexample the function $B$ is not surjective. That is just a way of exploiting the fact that "weak" bilinear is not bilinear over $K$, hence its image does not need to be a vector space over $K$. $\endgroup$ Mar 24, 2015 at 12:00

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