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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?

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1 Answer 1

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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 $\overline V\times V$.

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