Let \(f\) be the bilinear form on \(C^2\) defined by \[ f((x_1, x_2), (y_1, y_2)) = x_1y_2 - x_2y_1. \] Show that
if \(T\) is a linear operator on \(C^2\), then \(f(T\alpha, T\beta) = (\det T)f(\alpha, \beta)\) for all \(\alpha, \beta\) in \(C^2\).
\(T\) preserves \(f\) if and only if \(\det T = +1\).
What does (b) say about the group of \(2 \times 2\) matrices \(M\) such that \(M^tAM = A\) where \[ A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}? \]
Solution
For column vectors \(\alpha,\beta\in\mathbb C^2\), let \[ [\alpha\ \beta] \] denote the \(2\times2\) matrix having these vectors as columns. Then \[ f(\alpha,\beta)=\det[\alpha\ \beta]. \]
1. Compute the effect of an arbitrary linear operator.
Let \(M\) be the matrix of \(T\) in the standard basis. Then \[ [T\alpha\ T\beta]=M[\alpha\ \beta]. \] Therefore \[ \begin{aligned} f(T\alpha,T\beta) &=\det(M[\alpha\ \beta])\\ &=(\det M)\det[\alpha\ \beta]\\ &=(\det T)f(\alpha,\beta). \end{aligned} \] Thus \[ \boxed{f(T\alpha,T\beta)=(\det T)f(\alpha,\beta).} \]
2. Characterize the operators preserving \(f\).
If \[ \det T=1, \] step 1 immediately gives \[ f(T\alpha,T\beta)=f(\alpha,\beta) \] for all \(\alpha,\beta\), so \(T\) preserves \(f\).
Conversely, if \(T\) preserves \(f\), take the standard basis vectors \(e_1,e_2\). Since \[ f(e_1,e_2)=1, \] step 1 gives \[ 1=f(Te_1,Te_2)=(\det T)f(e_1,e_2)=\det T. \] Hence \[ \boxed{T\text{ preserves }f\iff\det T=1.} \]
3. Identify the matrix group preserving \(A\).
The form \(f\) has matrix \[ A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \] so the equation \[ M^tAM=A \] is exactly the statement that the linear operator with matrix \(M\) preserves \(f\). By step 2, this happens exactly when \[ \det M=1. \] Therefore \[ \boxed{ \{M\in GL_2(\mathbb C):M^tAM=A\}=SL_2(\mathbb C).} \] In particular, the complex symplectic group in dimension two is \(SL_2(\mathbb C)\).