CTA Gaussian-beam quasimodes and geodesic concentration

Embed the segment in a collar extension across $\mathrm{\partial }{M}_0$ and cover it by finitely many coordinate neighborhoods. On an embedded subsegment, parallel transport an orthonormal normal frame and set $F(t,y)={\exp }_{\gamma (t)}({\sum }_{\alpha }{y}_{\alpha }{E}_{\alpha }(t))$. The inverse function theorem gives local Fermi coordinates with $g_{jk}(t,0)={\delta }_{jk}$ and ${\mathrm{\partial }}_i{g}_{jk}(t,0)=0$. This construction is along an interval and uses no orientation of $M_0$.

In one such chart, seek $v_s=e^{is\mathrm{\Theta }}a$, $s=\tau +i\lambda$, with $\mathrm{\Theta }(t,0)=t$ and $\mathrm{Im}\mathrm{\Theta }(t,y)\ge c\mathrm{\mid }y{\mathrm{\mid }}^2$. Expand $\mathrm{\Theta }$ into homogeneous transverse Taylor polynomials. The quadratic polynomial is $\frac{1}{2}H(t)y\cdot y$; substituting into $g(d\mathrm{\Theta },d\mathrm{\Theta })=1$ gives the matrix Riccati equation $H^{\mathrm{\prime }}+H^2=F(t)$, where $F$ is determined by the second transverse Taylor coefficient of the inverse metric. An initial complex symmetric $H(t_0)$ with positive-definite imaginary part yields a smooth solution with positive imaginary part on a short subsegment. Each higher homogeneous phase term then solves a linear first-order ODE along $t$, because its newest coefficient occurs linearly in the eikonal equation.

Write $m=\dim M_0$, so the transverse dimension is $m-1$. Substituting $a={\tau }^{(m-1)\mathrm{/}4}{\sum }_{j=0}^N{s}^{-j}{a}_{-j}$ into $(-{\mathrm{\Delta }}_{g_0}-s^2)e^{is\mathrm{\Theta }}a$ cancels successive powers of $s$ by linear transport ODEs for the transverse Taylor coefficients of each $a_{-j}$. The leading coefficient solves $a_0(t,0)=c_0\exp (-\frac{1}{2}{\int }_{t_0}^t\mathrm{tr}H(r)dr)$. A finite Taylor realization to the order required for the fixed $K$ and a transverse cutoff produce a smooth local beam; Gaussian integration gives bounded $L^2$ norm and a residual that tends to zero faster than any prescribed power after increasing the Taylor order.

On an oriented transversal of dimension at least two, apply [F3] to complete the finite overlap matching and eliminate all distinct-branch cross terms. This gives the stated global family and limit; steps 1.1--2.1 describe its local ingredients, without claiming a new proof of the imported global estimates.

For a nonorientable transversal let $\pi :{\tilde{M}}_0\to M_0$ be its orientation double cover, with deck involution $\sigma$, and lift $\gamma$ to $\tilde{\gamma }$. Let $w_s$ be the oriented beam from [F3] along that lift. Define the descended function by ${\pi }^{\ast }{v}_s=w_s+{\sigma }^{\ast }{w}_s$. The residual and norm bounds follow by a finite sum and the covering integral formula. The two lifted segments have no common tangent branch: equality of position and velocity at different times would make the geodesic periodic, contrary to its nontangential boundary exit; opposite velocities at a repeated point would force a zero midpoint velocity by geodesic uniqueness. There are only finitely many crossing branches, by compactness and local uniqueness. The distinct-branch estimate in [F3], applied to these two lifts, therefore gives a zero mixed-term limit. For every continuous $\psi$ the covering formula is ${\int }_{M_0}\mathrm{\mid }{v}_s{\mathrm{\mid }}^2\psi dV={\int }_{{\tilde{M}}_0}\mathrm{\mid }{w}_s{\mathrm{\mid }}^2(\psi \circ \pi )dV+\mathrm{Re}{\int }_{{\tilde{M}}_0}{w}_s\overline{{\sigma }^{\ast }{w}_s}(\psi \circ \pi )dV\mathrm{.}$ The first term has precisely the required ray limit and the second tends to zero. In particular there is no extra factor two in the normalization. The same argument works if the cover is disconnected.

If the transversal has dimension one, the component containing the segment is an interval with arclength coordinate $t$ beginning at its incoming endpoint. Set $v_s=e^{ist}$ on this component and zero on all other components. The residual is zero and the asserted integral formula is exact. In dimension zero the given unit-speed segment cannot exist. This covers the dimensions not needed in the CTA application as well.