Von Neumann inertia, dense congruences, and coisotropic reduction

The positive spectral subspace of $J$ is strictly positive for its quadratic form: integrating the positive spectral variable gives a positive number for every nonzero vector there. If a closed module subspace $U$ is strictly positive, its projection into that positive spectral subspace is injective, because a vector with no positive spectral part has nonpositive quadratic form. Polar rank-nullity in [F1] therefore bounds ${\dim }_{A}U$ by the trace of the positive spectral projection. This proves the positive characterization, and applying it to $-J$ proves the negative one.

For two projections $p,q$ on a common finite module, rank-nullity for $(I-q):pH\to (I-q)H$ gives $\tau (p\wedge q)\ge \tau (p)-\tau (I-q)$; its kernel is exactly their closed range intersection. If $T$ is injective with dense image, let $q_{\epsilon }={\mathbf{1}}_{[\epsilon ,\mathrm{\infty })}(T{T}^{\ast })$. The traces $\tau (q_{\epsilon })$ tend to ${\dim }_A{H}_1$ by [F1], and $q_{\epsilon }{H}_1$ lies in the image of $T$ with a bounded module inverse on it. Intersect this range with a positive spectral subspace $p_{\delta }{H}_1={\mathbf{1}}_{[\delta ,\mathrm{\infty })}(J)H_1$. Its trace tends to $\tau (p_{\delta })$ as $\epsilon \downarrow 0$ by the intersection inequality. Pulling it back by the bounded inverse of $T$ gives a closed positive subspace for $T^{\ast }JT$ of the same dimension. Step 1.1 and then $\delta \downarrow 0$ show that the positive inertia of $T^{\ast }JT$ is at least that of $J$. Conversely, on every closed positive subspace for $T^{\ast }JT$, project $T$ into the positive spectral subspace of $J$; this map is injective, giving the reverse inequality by [F1]. Repeat with $-J$.

Assume $J$ is invertible and put $L=V^{{\perp }_J}\subseteq V$, $K=V\cap L^{\perp }$ and $Q=V^{\perp }$, where unmarked orthogonals use the underlying Hilbert product. These are closed module subspaces and $H=L\oplus K\oplus Q$. The identity $L=J^{-1}Q$ says that $J{\mathrm{\mid }}_L:L\to Q$ is a bounded isomorphism. In this decomposition the matrix of $J$ is $\left(\begin{array}{ccc}0& 0& B\\ 0& C& E\\ B^{\ast }& E^{\ast }& F\end{array}\right)$, where $B^{\ast }:L\to Q$ is invertible. Replace $k\in K$ by $k-(B^{\ast }{)}^{-1}{E}^{\ast }k$. This bounded invertible triangular change makes $K$ orthogonal, for the form, to both $L$ and $Q$, leaving its form $C$ unchanged. In the remaining $L\oplus Q$ block, replace $l$ by $l-\frac{1}{2}(B^{\ast }{)}^{-1}Fq$ to obtain $\left(\begin{array}{cc}0& B\\ B^{\ast }& 0\end{array}\right)$. Conjugation by $\mathrm{diag}(I_L,-I_Q)$ negates this last form; its positive and negative spectral projections consequently have the same trace, so its signature is zero. Step 2.1 applies to the bounded invertible changes just used, proving that the original signature is the signature of $C$.

The restriction of the form to $V=L\oplus K$ is $0\oplus C$, so it too has signature equal to that of $C$. This proves coisotropic reduction. For a map with a kernel, its pullback form vanishes on that kernel and has no cross terms with it; discard the kernel and use step 2.1 on its orthogonal complement whenever the remaining image is dense. For the trivial group, projection traces are ordinary complex dimensions; complexifying a real symmetric form preserves its real positive and negative dimensions by diagonalizing the finite real form. Thus the ordinary case is included.