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The occupied spectral projector has no continuous extension through a zero gap

Example

For $n = 2, r = 1$ , let $H (t) = diag (- t, t)$ . For every $t \neq 0$ the lowest eigenvalue is simple, so its occupied projector is defined; at $t = 0$ the occupied/unoccupied gap vanishes and the lowest rank-one projector is not unique. The two one-sided limits of the defined projectors differ.

Facts & Assumptions

Given: The Hermitian path $H (t) = diag (- t, t)$ and the rank-one lowest-eigenvalue projector $Π_{1} (H (t))$ when the gap is positive.

[F1]

The occupied projector is uniquely defined only on a domain with a strict $r$ th gap (The occupied-projector response map on its strict-gap domain).

Verification

1.1givenF1algebra

If $t > 0$ , then $- t < t$ and the lowest eigenvector is $e_{1}$ , so $Π_{1} (H (t)) = e_{1} e_{1}^{*}$ . If $t < 0$ , then $t < - t$ and the lowest eigenvector is $e_{2}$ , so $Π_{1} (H (t)) = e_{2} e_{2}^{*}$ .

1.2givenF1algebra

At $t = 0$ , both eigenvalues equal $0$ , so the gap is $0$ and every rank-one orthogonal projector onto a line is a lowest-eigenspace projector; there is no unique $Π_{1} (H (0))$ under [F1].

2.1step 1.1step 1.2algebra∎

The limits as $t ↓ 0$ and $t ↑ 0$ are respectively $e_{1} e_{1}^{*}$ and $e_{2} e_{2}^{*}$ , which are distinct because their Frobenius distance is $\sqrt{2}$ . Thus no continuous extension of the uniquely gapped projector exists at $t = 0$ .

Depends on

The occupied-projector response map on its strict-gap domain

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Tosio Kato, Perturbation Theory for Linear Operators, 2nd ed., Chapter II §1.4–§1.6 (standard reference, not scraped)