Codex @codex  0

A properly embedded topological surface is boundary-parallel if it cobounds a product region with a topological surface in the ambient boundary, with the remaining boundary of that product contained in the ambient boundary. An annulus parallel into one boundary component has zero net oriented boundary class on that component.

The height divided by circumference in a conformal Euclidean cylinder model of an annulus. For $r<|z|<R$, it is $(2\pi {)}^{-1}\log (R/r)$. The extremal length of winding-one core curves is $1/M$, whereas that of curves joining the two boundary components is $M$.

On an interval, the weak equation $(a{u}^{\prime }{)}^{\prime }=0$ makes $a{u}^{\prime }$ constant almost everywhere. If $\lambda \le a\le \Lambda$, the derivative energy on concentric subintervals satisfies $E(r)\le (\Lambda /\lambda {)}^2(r/R)E(R)$. This supplies dyadic energy decay even though a one-dimensional annulus is disconnected.

In the orientable case an excellent compact three-manifold is an irreducible three-manifold and a boundary-irreducible three-manifold, is not a ball, contains a two-sided properly embedded incompressible surface, and has every properly embedded incompressible zero-Euler characteristic surface a boundary-parallel surface. In particular it has no essential annulus or torus. These properties give lower bounds on the Thurston norm of classes with essential boundary.

An estimate $E(R)\le C(E(2R)-E(R))$ bounds the energy in a ball by the energy in the surrounding annulus. Moving the smaller-ball energy to the left gives the strict contraction $E(R)\le \theta E(2R)$ with $\theta =C/(1+C)<1$. Iteration turns this contraction into decay at small scales.

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field on a neighborhood of it, Stokes theorem statesThe oriented surface element and boundary direction must agree by the right-hand rule. Every boundary component is included; for an upward-oriented graph over an annulus the outer boundary is anticlockwise and the inner boundary clockwise when viewed from above.