Propagation kernel and path-Laplacian gap

Set $V_0=I$ and $V_t=U_t\cdots U_1$. The block diagonal map $W={\sum }_{t=0}^T{V}_t\otimes \mathrm{\mid }{c}_t⟩⟨c_t\mathrm{\mid }$ is unitary because every $V_t$ is unitary and clock sectors are orthogonal. Conjugating an edge term by $W$ replaces $U_t$ in its off-diagonal block by $V_t^{\ast }{U}_t{V}_{t-1}=I$; its diagonal clock blocks remain unchanged. The ket-bra multiplications use the convention [F3]. Summing gives $W^{\ast }{H}_{prop}W=I_{comp}\otimes L_T$. Thus the kernel consists of vectors constant across clock time, and undoing $W$ gives exactly the history vectors ${\eta }_{\varphi }$, normalized when $\mathrm{\parallel }\varphi \mathrm{\parallel }=1$.

Put $M=T+1$. For $j=0,\dots ,M-1$, the vector with coordinates $v_j(r)=\cos (\pi j(r+1\mathrm{/}2)\mathrm{/}M)$, $r=0,\dots ,M-1$, is nonzero. The cosine addition identity verifies the interior eigenvalue equation for $L_T$ with eigenvalue $1-\cos (\pi j\mathrm{/}M)$; the same identity with the endpoint degree-one rows verifies both endpoint equations. These $M$ eigenvalues are distinct, so symmetry of $L_T$ makes the corresponding vectors orthogonal; by [F2] they form a full eigenbasis. The $j=0$ vector is constant and the smallest positive eigenvalue is the $j=1$ value.

For $x\in [0,\pi \mathrm{/}2]$, concavity of sine gives $\sin x\ge 2x\mathrm{/}\pi$. Hence $1-\cos (\pi \mathrm{/}M)=2{\sin }^2(\pi \mathrm{/}(2M))\ge 2\mathrm{/}{M}^2\mathrm{.}$ This proves the stated gap and includes $T=1$ ($M=2$).