Finite gapped adiabatic estimate

[The projection is $C^2$ without assuming a smooth eigenbasis] Fix $s_0$ and list the finitely many distinct eigenvalues of $H(s_0)$. Around each distinct eigenvalue belonging to the chosen cluster take a positively oriented small circle containing no other eigenvalue of $H(s_0)$, with disjoint closed discs. A finite choice of radii suffices; one can take the first sufficiently small positive rational radius for each center. Let $\mathrm{\Gamma }$ be their finite union. On $\mathrm{\Gamma }$, $z-H(s_0)$ is invertible. Compactness and the Neumann formula show it remains invertible for all $s$ sufficiently near $s_0$. Define $P_{\mathrm{l}\mathrm{o}\mathrm{c}}(s)=\frac{1}{2\pi i}{\oint }_{\mathrm{\Gamma }}(z-H(s){)}^{-1}dz\mathrm{.}$ Finite diagonalization at each fixed $s$ and the scalar circle integral show that this is the orthogonal spectral projection onto the eigenvalues inside those discs. The inverse equations give $R^{\mathrm{\prime }}=R{H}^{\mathrm{\prime }}R$ and $R^{\mathrm{\prime }\mathrm{\prime }}=R{H}^{\mathrm{\prime }\mathrm{\prime }}R+2R{H}^{\mathrm{\prime }}R{H}^{\mathrm{\prime }}R$. The local contour is fixed, so continuous differentiation under its finite integrals gives $P_{\mathrm{l}\mathrm{o}\mathrm{c}}\in C^2$. Both $P(s)$ and $P_{\mathrm{l}\mathrm{o}\mathrm{c}}(s)$ are spectral projections of the same matrix and therefore commute. At $s_0$ they coincide. By the assumed continuity and the just-proved local continuity, their difference has norm less than $1$ on a smaller neighborhood. Distinct commuting orthogonal projections have difference norm $1$: simultaneous decomposition into their intersection and the two difference subspaces gives a unit vector with difference equal to $\pm 1$ if they differ. Hence $P=P_{\mathrm{l}\mathrm{o}\mathrm{c}}$ there. This proves $P\in C^2$ locally everywhere, and thus on $[0,1]$ with one-sided endpoint derivatives. It also shows that its rank is constant, without imposing a new rank hypothesis. If the cluster is empty or full, continuity makes that condition locally persistent by the same projection-distance argument; the projection is identically $0$ or $I$ on the connected interval, and the subsequent proof reduces to $K=X=0$. This handles any original empty-cluster convention without a nonempty-cluster restriction.

[Projection identities and elementary ODE construction] Differentiate $P^2=P$ to obtain $P^{\mathrm{\prime }}=P^{\mathrm{\prime }}P+P{P}^{\mathrm{\prime }}$ and $P{P}^{\mathrm{\prime }}P=0$. Because $P=P^{\ast }$, also $P^{\mathrm{\prime }}=P^{\mathrm{\prime }\ast }$. Thus $K=P^{\mathrm{\prime }}P-P{P}^{\mathrm{\prime }}$ is $C^1$, $K^{\ast }=-K$, and $[K,P]=P^{\mathrm{\prime }}P+P{P}^{\mathrm{\prime }}=P^{\mathrm{\prime }}\mathrm{.}$ For any continuous matrix coefficient $B(s)$, define $V_0(s)=I$ and $V_{j+1}(s)={\int }_0^{s}B(r)V_j(r)dr$. If $M=\sup \mathrm{\parallel }B\mathrm{\parallel }$, induction gives $\mathrm{\parallel }{V}_j(s)\mathrm{\parallel }\le (Ms{)}^j\mathrm{/}j!$. The uniformly convergent sum $V={\sum }_j{V}_j$ solves $V=I+{\int }_0^{s}BV$ and hence $V^{\mathrm{\prime }}=BV$ by the fundamental theorem of calculus. For two solutions with the same initial value, iterating the integral difference bound gives $\mathrm{\parallel }V-\tilde{V}\mathrm{\parallel }\le M^j\mathrm{/}j!\sup \mathrm{\parallel }V-\tilde{V}\mathrm{\parallel }$ for every $j$, proving uniqueness as $j\to \mathrm{\infty }$. The same recursion works for any initial matrix. For anti-Hermitian $B$, differentiation gives $(V^{\ast }V{)}^{\mathrm{\prime }}=V^{\ast }(B^{\ast }+B)V=0$, so $V^{\ast }V=I$; in finite dimension injectivity implies surjectivity, giving $V{V}^{\ast }=I$. Thus $V$ is unitary. This local construction avoids both the incorrect generic ordered-integral formula and every autonomous spectral/group theorem. Apply it to $B=K$ to obtain $W^{\mathrm{\prime }}=KW$, $W(0)=I$. Since $K\in C^1$, the equation makes $W\in C^2$. Differentiating $W^{\ast }PW$ yields $\frac{d}{ds}(W^{\ast }PW)=W^{\ast }(P^{\mathrm{\prime }}-[K,P])W=0,$ so $W^{\ast }P(s)W=P(0)=:P_0$. This is the correct Kato transport, with $(W^{\ast }{)}^{\mathrm{\prime }}=-W^{\ast }K$. No choice of instantaneous eigenvectors is involved.

[Smooth Hermitian commutator solution for any separated cluster] Put $\tilde{H}=W^{\ast }HW$ and $\tilde{K}=W^{\ast }KW$. Then $\tilde{H}\in C^2$, $\tilde{K}\in C^1$, and $[\tilde{H},P_0]=0$. The matrix $\tilde{K}$ is anti-Hermitian and has only cross blocks with respect to $P_0$, because $PKP=(I-P)K(I-P)=0$. Let $\mathcal{V}$ be the fixed finite-dimensional complex vector space of matrices $Y$ with $P_{0}Y{P}_0=(I-P_0)Y(I-P_0)=0$. The linear map $L_s:\mathcal{V}\to \mathcal{V}$, $L_s(Y)=[\tilde{H}(s),Y]$, is $C^2$ as a matrix-valued map on this fixed space. To prove invertibility at each $s$, diagonalize the two Hermitian diagonal blocks of $\tilde{H}(s)$ at that one time. In these finite bases every cross entry is multiplied by $\lambda -\mu$, where $\lambda$ belongs to the cluster and $\mu$ to its complement, or the reverse. The given full complementary spectral gap gives $\mathrm{\mid }\lambda -\mu \mathrm{\mid }\ge g$. Thus all cross coefficients are invertible and $\mathrm{\parallel }{L}_s^{-1}Y{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\le g^{-1}\mathrm{\parallel }Y{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\mathrm{.}$ This diagonalization is only a pointwise verification of a uniquely determined inverse; no simultaneous choice of bases over an uncountable interval is made. In fixed coordinates on $\mathcal{V}$, the Neumann inverse or adjugate formula proves $L_s^{-1}$ is $C^2$; its derivative is $-L_s^{-1}{L}_s^{\mathrm{\prime }}{L}_s^{-1}$. Define $\tilde{X}=L_s^{-1}\tilde{K}$, which is $C^1$. Since ${\tilde{K}}^{\ast }=-\tilde{K}$ and $[\tilde{H},Y{]}^{\ast }=-[\tilde{H},Y^{\ast }]$, taking adjoints of the equation shows $[\tilde{H},{\tilde{X}}^{\ast }]=\tilde{K}$. Both $\tilde{X}$ and ${\tilde{X}}^{\ast }$ lie in $\mathcal{V}$, so uniqueness gives ${\tilde{X}}^{\ast }=\tilde{X}$. Set $X=W\tilde{X}{W}^{\ast }$. It is $C^1$, Hermitian, and $[H,X]=K$. If $\mathcal{V}=\{0\}$, use its unique zero inverse solution $X=0$; the proof has no division by a nonexistent cross entry.

[Physical and modified finite propagators and intertwining] Apply the matrix ODE construction of step 2.1 to $B_T=-iTH$ and $B_A=-iTH+K$. Both coefficients are anti-Hermitian and continuous. It supplies unique $C^1$ unitary matrices $U_T,U_A$ with initial value $I$. Multiplication of the second differential equation by $i$ gives exactly the original self-adjoint-generator equation $i{U}_A^{\mathrm{\prime }}=(TH+iK)U_A$. Because $[H,P]=0$, product differentiation gives $\frac{d}{ds}(U_A^{\ast }P{U}_A)=U_A^{\ast }(P^{\mathrm{\prime }}+iT[H,P]-[K,P])U_A=0.$ Thus $U_A^{\ast }P(s)U_A=P(0)$, or $P(s)U_A(s)=U_A(s)P(0)$. This establishes the original transport conclusion for every $T>0$.

[Exact integration identity and the unchanged error constant] Set $D=U_A^{\ast }{U}_T$ and $F=U_A^{\ast }X{U}_T$. The equations give $(U_A^{\ast }{)}^{\mathrm{\prime }}=U_A^{\ast }(iTH-K)$ and $U_T^{\mathrm{\prime }}=-iTH{U}_T$. Therefore $D^{\mathrm{\prime }}=-U_A^{\ast }K{U}_T,F^{\mathrm{\prime }}=U_A^{\ast }(X^{\mathrm{\prime }}-KX+iT[H,X])U_T=U_A^{\ast }(X^{\mathrm{\prime }}-KX+iTK)U_T\mathrm{.}$ Substitution, with all signs retained, gives $iT{D}^{\mathrm{\prime }}=-F^{\mathrm{\prime }}+U_A^{\ast }(X^{\mathrm{\prime }}-KX)U_T\mathrm{.}$ Integrating from $0$ to $s$, and using $D(0)=I$, $F(0)=X(0)$, yields $iT(D(s)-I)=-F(s)+X(0)+{\int }_0^s{U}_A(r{)}^{\ast }(X^{\mathrm{\prime }}(r)-K(r)X(r))U_T(r)dr\mathrm{.}$ The unitaries have norm $1$, $X\in C^1$, and the interval is compact. Hence $\mathrm{\parallel }D(s)-I\mathrm{\parallel }\le \frac{2{\sup }_{r\in [0,1]}\mathrm{\parallel }X(r)\mathrm{\parallel }+{\int }_0^1(\mathrm{\parallel }{X}^{\mathrm{\prime }}(r)\mathrm{\parallel }+\mathrm{\parallel }K(r)\mathrm{\parallel }\mathrm{\parallel }X(r)\mathrm{\parallel })dr}{T}=\frac{C}{T}\mathrm{.}$ The constant is exactly the original public constant, finite and independent of $T$. Finally, $U_T=U_{A}D$ and transport gives $(I-P(s))U_A(s)P(0)=0$. Therefore $\mathrm{\parallel }(I-P(s))U_T(s)P(0)\mathrm{\parallel }=\mathrm{\parallel }(I-P(s))U_A(s)(D(s)-I)P(0)\mathrm{\parallel }\le \mathrm{\parallel }D(s)-I\mathrm{\parallel }\le C\mathrm{/}T\mathrm{.}$ This holds for every $T>0$ and $s\in [0,1]$, including zero leakage at $s=0$. No eigenvalue ordering within the cluster, rank-one hypothesis, stronger derivative hypothesis, or changed error constant has been used.