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M2 Heavy Mode Schur Kernel Map

Example

Assume full AC. Let the protected and high subspaces each be $C^{1 ∣ 1}$ , with their standard even/odd gradings. For real coefficients $a, b, d$ , define the odd self-adjoint block operator by the even-to-odd matrix

Q_{+} = (\begin{array}{cc} a & b \\ b & d \end{array})

in the ordered bases $(P^{+}, H^{+})$ and $(P^{-}, H^{-})$ , and put $Q_{-} = Q_{+}^{*}$ . When $d = 2$ , $a = \frac{1}{2}$ , and $b = 1$ , the high block is invertible and the Schur complement is zero. Its kernel map sends $1 \in P^{+}$ to $(1, - \frac{1}{2}) \in P^{+} \oplus H^{+}$ , and similarly in odd parity. When $d = 0$ , $a = 0$ , $b = 1$ , the high block is not invertible and $Q_{+}$ is invertible, so the protected-block kernel does not map to a full-operator kernel. The Schur formula is undefined at the zero high eigenvalue. Both finite operators still have index zero, so this example demonstrates failure of the kernel comparison hypothesis, not an index mismatch. No physical membrane spectrum is asserted.

The block decomposition is encoded below; the proof uses the displayed matrices and verifies the cases directly.

Facts & Assumptions

Given: Full AC; the four one-dimensional graded summands, their standard Hermitian products, and the real block coefficients.

[F1]

The Axiom of Choice: full AC is the foundational assumption of the supplied block-index lemma.

[F2]

M2 Protected Block Index Comparison: for a finite-dimensional graded protected block and an invertible high block, the Schur complement and full operator have the same even-to-odd index as the protected block, with parity-preserving kernel map $u \mapsto (u, - D^{- 1} B^{*} u)$ .

Verification

1.1givenalgebra

All the displayed matrices act on their whole finite-dimensional source spaces and are closed. On $H = (P^{+} \oplus H^{+}) \oplus (P^{-} \oplus H^{-})$ , the full operator is $Q = (\begin{array}{cc} 0 & Q_{-} \\ Q_{+} & 0 \end{array})$ , so $Q (x_{+}, x_{-}) = (Q_{-} x_{-}, Q_{+} x_{+})$ . Since $Q_{-} = Q_{+}^{*}$ , block adjunction gives $Q^{*} = Q$ . For the grading $Γ (x_{+}, x_{-}) = (x_{+}, - x_{-})$ , direct substitution gives $Q Γ = - Γ Q$ , proving that $Q$ is odd and self-adjoint.

1.2F1F2givenalgebra

For $d = 2, a = \frac{1}{2}, b = 1$ , the protected block has coefficient $A_{+} = \frac{1}{2}$ , the high block has $D_{+} = 2$ , and the coupling is $B_{+} = 1$ . Thus $S_{+} = A_{+} - B_{+} D_{+}^{- 1} B_{+}^{*} = \frac{1}{2} - 1 \cdot \frac{1}{2} \cdot 1 = 0$ . The full even-to-odd matrix is $(\begin{array}{cc} \frac{1}{2} & 1 \\ 1 & 2 \end{array})$ , whose kernel is spanned by $(1, - \frac{1}{2})$ . The same real symmetric matrix acts in the odd direction, so it has the same one-dimensional odd kernel. The map in [F2] sends $1$ to $(1, - \frac{1}{2})$ because $- D^{- 1} B^{*} 1 = - \frac{1}{2}$ ; it preserves parity. The protected block $A_{+} = \frac{1}{2}$ is invertible, so its kernel and cokernel are zero; the Schur block $S_{+} = 0$ and the full matrix each have one-dimensional kernel and cokernel. Thus all three even-to-odd indices are zero.

2.1givenalgebra∎

For $d = 0, a = 0, b = 1$ , the high block $D_{+} = 0$ has no inverse, so $A_{+} - B_{+} D_{+}^{- 1} B_{+}^{*}$ is not defined. The protected block $A_{+} = 0$ has a one-dimensional even kernel, while the full matrix is $(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , whose square is the identity and whose kernel is zero; its odd adjoint matrix is identical. Hence the full operator has no kernel, although both protected vectors remain at zero. The index remains zero on each side: the full matrix is invertible, and the one-dimensional zero protected map has equal one-dimensional kernel and cokernel. Thus the Schur kernel formula fails precisely because the high inverse required for that formula does not exist; the example implies no physical spectral statement.

Depends on

Dependency tree · 2 levels

Sources

Atiyah and Singer, The Index of Elliptic Operators I, Annals of Mathematics 87 (1968)