Let $V$ be a complex vector space. A sesquilinear map (or conjugate-linear in the first variable and linear in the second) on a complex vector space $V$ is a map $f: V \times V \rightarrow \mathbb{C}$ that satisfies:
$f(v_1,cw_1) = cf(v_1,w_1)$, and $f(v_1, w_1 + w_2) = f(v_1, w_1) + f(v_1, w_2)$
$f(cv_1, w_1) = \bar{c}f(v_1,w_1)$, and $f(v_1 + v_2, w_1) = f(v_1,w_1) + f(v_2,w_1)$
for $c \in \mathbb{C}$ and $v_1, v_2, w_1, w_2 \in V$, $\bar{c}$ is the complex conjugate of $c$.
I read on the Wikipedia entry for "sesquilinear form" that a sesquilinear form can also be seen as a complex bilinear map
$\bar{V} \times V \rightarrow \mathbb{C}$
where $\bar{V}$ is the complex conjugate of $V$.
I wanted to ask:
This means that the map
$f: \bar{V} \times V \rightarrow \mathbb{C}$, on top of satisfying
$f(v_1, w_1 + w_2) = f(v_1, w_1) + f(v_1, w_2)$
$f(v_1 + v_2, w_1) = f(v_1,w_1) + f(v_2,w_1)$,
also satisfies:
$f(cv_1, w_1) = f(v_1,cw_1) = cf(v_1,w_1)$ ??
for $v_1, v_2 \in \bar{V}$, $w_1, w_2 \in V$ and $c \in \mathbb{C}$.
This last line I seem to get confused about, should it be
$cf(v_1,w_1)$? or $\bar{c}f(v_1,w_1)$ at the end?
The point is that the formula "$cv$" has different meaning in $\overline{V}$ and $V$. Saying that $\overline{V}$ is the complex conjugate of $V$ means that we define scalar multiplication in $\overline{V}$ as follows: $$c*v = \overline{c}\cdot v$$ where $*$ stands for scalar multiplication in $\overline{V}$, and $\cdot$ is scalar multiplication in $V$. (The convention to omit multiplication signs is really inconvenient here.)
So, if $v\in \overline V$ and $w\in V$, we have
$$f(c*v,w) = f(\bar c\cdot v,w) = \overline{\bar c} f(v,w) = cf(v,w)$$ This is what makes $f$ bilinear on $\overline V\times V$.