Krein space

Remarks

Let $P_\pm$ be the projections of $K$ onto $K_\pm$ along the decomposition, and let $J = P_+ - P_-$. Then $\ip{x}{y}_J = [Jx, y]$ is a Hilbert space inner product on $K$, and $[x, y] = \ip{Jx}{y}_J$. The map $J$ is the fundamental symmetry; the decomposition is not unique, but all the norms $\ip{x}{x}_J^{1/2}$ are equivalent. If $K_-$ has finite dimension $\kappa$, the space is a Pontryagin space $\Pi_\kappa$.

Example: $K = \C^2$ with $[x, y] = x_1 \overline{y_1} - x_2 \overline{y_2}$, $K_+ = \C e_1$ and $K_- = \C e_2$. Clause 2 holds since $[e_1, e_2] = 0$, and $[x, x] = \abs{x_1}^2$ on $K_+$ and $-[x, x] = \abs{x_2}^2$ on $K_-$ are positive definite. The vector $e_1 + e_2$ satisfies $[e_1 + e_2, e_1 + e_2] = 0$, so the form is indefinite.

Non-example: $K = \C^2$ with $[x, y] = x_1 \overline{y_1}$. The vector $e_2$ satisfies $[e_2, y] = 0$ for every $y$. In a Krein space, write $e_2 = x_+ + x_-$; then $[x_+, x_+] = [e_2, x_+] = 0$ forces $x_+ = 0$, and in the same way $x_- = 0$. So the form of a Krein space is nondegenerate, and this degenerate form admits no fundamental decomposition.