Discussion
The quadratic form associated to the trace form \(\mathrm{Tr}_{\mathbf R}^{\mathbf C} : \mathbf C \times \mathbf C \to \mathbf R\), \((x, y) \mapsto \mathrm{Tr}(\lambda_{XY})\), where \(\lambda_{XY} : \mathbf{C} \to \mathbf{C}\) is the \(\mathbf{R}\)-linear map defined by \(\lambda_{XY}(z) = xyz\), is :
Answer: D
The quadratic form is \(X^{2} - Y^{2}\).
For \(w \in \mathbf{C}\), multiplication by \(w = a + ib\) is the \(\mathbf{R}\)-linear map with matrix \(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\) on the basis 1, \(i\), so its trace is \(2a = 2\,\mathrm{Re}\,w\). Hence the trace form is \(B(x, y) = 2\,\mathrm{Re}(xy)\).
Its quadratic form at \(z = X + iY\) uses \(z^{2} = X^{2} - Y^{2} + 2iXY\):
\[ B(z, z) = 2\,\mathrm{Re}(z^{2}) = 2(X^{2} - Y^{2}) \]
With the usual convention \(q(z) = \tfrac{1}{2}B(z, z)\), this is \(X^{2} - Y^{2}\); with \(q = B(z, z)\) it differs only by the factor 2. Either way it is indefinite, of signature (1, 1), which rules out \(X^{2}\), \(Y^{2}\) and the positive definite \(X^{2} + Y^{2}\).
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