# Is there any definition of such semi-bilinear?

$K$ is a finite field which not equal to its base field $F_2$.

Let $f: V \rightarrow K$ be a function and $B(x,y)=f(x+y)+f(x)+f(y)$ such that $B(x+y,z)=B(x,z)+B(y,z)$ and $B(z,x+y)=B(z,x)+B(z,y)$ for every $x,y,z \in V$ (vector space over $K$) but $B(ax,y)$ is not necessary to equal to $aB(x,y)$ for every $a \in K$, same for $B(x,ay)$. Actually $B$ is linear over base field of $K$, but not linear over $K$.

Is there any such research area?

Example: $f(x)=Tr(x^9+x^5)$ where $Tr: F_{256} \rightarrow F_4$ is trace function.

• Is it exactly this situation, or not quite? Mar 27, 2015 at 15:58
• Thanks for this. I also did sth like that. I will add know an example. Mar 27, 2015 at 16:03
• While we're at it, the Hermitian inner product isn't really bilinear, but instead sesquilinear: it's linear in the first argument, but conjugate-linear in the second. If you're collecting examples; I can't shed any light on the general theory, unfortunately. Mar 27, 2015 at 16:09
• I am sorry, i wrote question wrong. But thanks for your link, they are really good for me. Mar 27, 2015 at 16:13
• If you are currently at UCD, you can ask Gary McGuire. He knows this stuff. Say "Hi" from me as well :-) Mar 27, 2015 at 18:27