Which of the following functions \(f\), defined on vectors \(\alpha = (x_1, x_2)\) and \(\beta = (y_1, y_2)\) in \(C^2\), are (sesqui-linear) forms on \(C^2\)?
\(f(\alpha, \beta) = 1\).
\(f(\alpha, \beta) = (x_1 - \bar{y}_1)^2 + x_2\bar{y}_2\).
\(f(\alpha, \beta) = (x_1 + \overline{y}_1)^2 - (x_1 - \overline{y}_1)^2\).
\(f(\alpha, \beta) = x_1\overline{y}_2 - \overline{x}_2 y_1\).
Solution
A sesquilinear form on \(\mathbb C^2\) must be linear in \(\alpha\) and conjugate-linear in \(\beta\).
1. Part (a) is not a sesquilinear form.
The constant function \[ f(\alpha,\beta)=1 \] does not satisfy \(f(0,\beta)=0\), which every linear function of the first variable must satisfy.
2. Part (b) is not a sesquilinear form.
Expanding gives \[ f(\alpha,\beta) =x_1^2-2x_1\overline{y}_1+\overline{y}_1^{\,2} +x_2\overline{y}_2. \] The terms \(x_1^2\) and \(\overline{y}_1^{\,2}\) violate linearity in the first variable and conjugate-linearity in the second, respectively.
3. Part (c) is a sesquilinear form.
Using \((a+b)^2-(a-b)^2=4ab\), \[ \begin{aligned} f(\alpha,\beta) &=(x_1+\overline y_1)^2-(x_1-\overline y_1)^2\\ &=4x_1\overline y_1. \end{aligned} \] This is linear in \(\alpha\) and conjugate-linear in \(\beta\). Hence it is a sesquilinear form.
4. Part (d) is not a sesquilinear form.
The first term \[ x_1\overline y_2 \] has the correct sesquilinearity, but the second term \[ -\overline x_2y_1 \] is conjugate-linear in the first variable and linear in the second. For example, scaling \(\alpha\) by \(i\) changes this second term by \(-i\) rather than by \(i\). Thus the whole expression is not linear in the first variable.
Therefore the only sesquilinear form in the list is \[ \boxed{\text{(c)}}. \]