Let \(V\) be a complex vector space, \(J\) a conjugation on \(V\), \(W\) the set of \(\alpha\) in \(V\) such that \(J\alpha = \alpha\), and \(f\) an inner product on \(W\). Show that:
There is a unique inner product \(g\) on \(V\) such that \(g(\alpha, \beta) = f(\alpha, \beta)\) for all \(\alpha, \beta\) in \(W\),
\(g(J\alpha, J\beta) = g(\beta, \alpha)\) for all \(\alpha, \beta\) in \(V\).
What does part (a) say about the relation between the standard inner products on \(\RR^1\) and \(C^1\), or on \(\RR^n\) and \(C^n\)?
Solution
1. Real decomposition \(V = W \oplus iW\):
1.1. A conjugation \(J\) on \(V\) satisfies \(J(c\alpha) = \bar{c}J(\alpha)\), \(J(\alpha + \beta) = J(\alpha) + J(\beta)\), and \(J^2 = I\).
definition of conjugation on a complex vector space.
1.2. For any \(\alpha \in V\), define \(\alpha_1 = \frac{\alpha + J\alpha}{2}\) and \(\alpha_2 = \frac{\alpha - J\alpha}{2i}\).
algebraic definition.
1.3. Check \(J\alpha_1 = \frac{J\alpha + J^2\alpha}{2} = \frac{J\alpha + \alpha}{2} = \alpha_1\) and \(J\alpha_2 = \frac{J\alpha - J^2\alpha}{-2i} = \frac{J\alpha - \alpha}{-2i} = \frac{\alpha - J\alpha}{2i} = \alpha_2\).Thus \(\alpha_1, \alpha_2 \in W\).
\(J(i v) = -i J(v)\) and \(J^2 = I\).
1.4. Moreover, \(\alpha = \alpha_1 + i\alpha_2\) uniquely, so \(V = W \oplus iW\) as real vector spaces.
if \(\alpha_1 + i\alpha_2 = 0\) with \(\alpha_1, \alpha_2 \in W\), then \(J(\alpha_1 + i\alpha_2) = \alpha_1 - i\alpha_2 = 0 \implies \alpha_1 = \alpha_2 = 0\).
2. Part (a): Uniqueness and existence of the complex inner product \(g\):
2.1. Uniqueness: If \(g\) is a complex inner product on \(V\) extending \(f\) on \(W\), then for \(\alpha = \alpha_1 + i\alpha_2\) and \(\beta = \beta_1 + i\beta_2\) with \(\alpha_j, \beta_j \in W\):\[ g(\alpha, \beta) = g(\alpha_1 + i\alpha_2, \beta_1 + i\beta_2) = g(\alpha_1, \beta_1) - i g(\alpha_1, \beta_2) + i g(\alpha_2, \beta_1) + g(\alpha_2, \beta_2). \]Since \(g|_{W \times W} = f\), we must have:\[ g(\alpha, \beta) = \big(f(\alpha_1, \beta_1) + f(\alpha_2, \beta_2)\big) + i \big(f(\alpha_2, \beta_1) - f(\alpha_1, \beta_2)\big). \]This uniquely determines \(g(\alpha, \beta)\) in terms of \(f\).
sesquilinearity of complex inner products.
2.2. Existence: Define \(g: V \times V \to \mathbb{C}\) by the formula in
2.1.
definition of \(g\).
2.3. Sesquilinearity: \(g\) is \(\mathbb{C}\)-linear in the first argument and conjugate-linear in the second argument.
linearity of \(f\) and real coordinates.
2.4. Conjugate symmetry: Using the symmetry of the real inner product \(f\):\[ \overline{g(\beta, \alpha)} = \big(f(\beta_1, \alpha_1) + f(\beta_2, \alpha_2)\big) - i \big(f(\beta_2, \alpha_1) - f(\beta_1, \alpha_2)\big) = \big(f(\alpha_1, \beta_1) + f(\alpha_2, \beta_2)\big) + i \big(f(\alpha_2, \beta_1) - f(\alpha_1, \beta_2)\big) = g(\alpha, \beta). \]
\(f(x, y) = f(y, x)\) for real inner products.
2.5. Positive-definiteness: For \(\alpha = \alpha_1 + i\alpha_2\):\[ g(\alpha, \alpha) = f(\alpha_1, \alpha_1) + f(\alpha_2, \alpha_2) \ge 0. \]If \(g(\alpha, \alpha) = 0\), then \(f(\alpha_1, \alpha_1) = 0\) and \(f(\alpha_2, \alpha_2) = 0\), so \(\alpha_1 = 0\) and \(\alpha_2 = 0\), whence \(\alpha = 0\).
positive-definiteness of \(f\) on \(W\).
3. Part (b): Show \(g(J\alpha, J\beta) = g(\beta, \alpha)\):
3.1. Since \(J\alpha = \alpha_1 - i\alpha_2\) and \(J\beta = \beta_1 - i\beta_2\):\[ g(J\alpha, J\beta) = \big(f(\alpha_1, \beta_1) + f(-\alpha_2, -\beta_2)\big) + i \big(f(-\alpha_2, \beta_1) - f(\alpha_1, -\beta_2)\big) = \big(f(\alpha_1, \beta_1) + f(\alpha_2, \beta_2)\big) - i \big(f(\alpha_2, \beta_1) - f(\alpha_1, \beta_2)\big). \]
applying the formula for \(g\).
3.2. Thus \(g(J\alpha, J\beta) = \overline{g(\alpha, \beta)} = g(\beta, \alpha)\).
4 of 2.
4. Relation to \(\mathbb{R}^n\) and \(\mathbb{C}^n\):
4.1. On \(V = \mathbb{C}^n\), let \(J(z) = \bar{z}\) be standard complex conjugation. Then \(W = \mathbb{R}^n\).
fixed points of complex conjugation in \(\mathbb{C}^n\) are \(\mathbb{R}^n\).
4.2. The standard real inner product on \(\mathbb{R}^n\) is \(f(x, y) = \sum_{k=1}^n x_k y_k\).Its unique complex inner product extension \(g\) on \(\mathbb{C}^n\) is the standard Hermitian inner product:\[ g(z, w) = \sum_{k=1}^n z_k \bar{w}_k. \]
direct application of Part (a).
5. Conclusion:The unique extension exists, satisfies \(g(J\alpha, J\beta) = g(\beta, \alpha)\), and standardly relates the Euclidean inner product on \(\mathbb{R}^n\) to the Hermitian inner product on \(\mathbb{C}^n\). Q.E.D.
1 through 4.