Take a closed disk \(D\) inside

$$ \frac12<\Re s<1, $$

with no zeta zero on its boundary. Remove a boundary arc of length \(\ell\). Now ask how often a vertical shift of zeta approximates zeta itself on the rest of the boundary.

If \(D\) is zero-free, ordinary universality on the full disk gives a positive-density set of shifts that works for every slit. The density has a floor independent of \(\ell\).

If \(D\) contains a zero, the answer collapses:

$$ \boxed{ \overline d(\mathcal A_\ell) \le C_D\exp(-\kappa_D/\ell^2). } $$

The exponent \(1/\ell^2\) comes from one tiny endpoint functional. That is the part I keep staring at.

The slit must carry the missing winding#

Let \(F\) be zero-free on \(D\) and close to zeta on the retained boundary. On the boundary put

$$ h=\frac F\zeta. $$

If \(D\) contains \(m\ge1\) zeta zeros, then

$$ \operatorname{wind}(h(\partial D),0)=-m. $$

On the retained arc, \(h\) stays in a small disk around \(1\). All of the missing winding must therefore occur across the slit. The slit may shrink, but the required argument change does not.

Equivalently, a global analytic logarithm \(L_F\) must satisfy

$$ \boxed{ |L_F(b_\ell)-L_F(a_\ell)|\ge c_D, } $$

where \(a_\ell,b_\ell\) are the slit endpoints.

Prime coordinates make the cost exact#

For the prime-linear Hilbert space

$$ g(s)=\sum_p c_p p^{-s}, \qquad \|g\|^2=\sum_p|c_p|^2, $$

consider the endpoint functional

$$ \Lambda_{a,b}(g)=g(b)-g(a). $$

Its exact squared dual norm is

$$ \boxed{ \|\Lambda_{a,b}\|_*^2 = \sum_p|p^{-b}-p^{-a}|^2. } $$

As the endpoints approach one another along a smooth arc,

$$ \boxed{ \|\Lambda_{a_\ell,b_\ell}\|_*^2 \sim \ell^2 \sum_p(\log p)^2p^{-2\Re s_*}. } $$

So imposing a fixed endpoint jump has minimum squared Hilbert cost

$$ \boxed{\Theta(\ell^{-2}).} $$

No vague entropy estimate is hiding here. It is the Riesz quotient of one explicit prime-log functional.

The random Euler law pays that cost#

Bagchi's weak limit for vertical zeta shifts is the random Euler product

$$ \zeta_\omega(s) = \prod_p(1-\omega(p)p^{-s})^{-1}. $$

Its logarithm splits into the prime-linear part and the higher prime powers. Across the slit, the higher powers move by only \(O(\ell)\). The prime-linear difference is

$$ \sum_p\omega(p)\bigl(p^{-b_\ell}-p^{-a_\ell}\bigr), $$

with variance \(\Theta(\ell^2)\). Hoeffding therefore charges a fixed jump with probability at most

$$ \exp(-c/\ell^2). $$

Portmanteau transfers the same upper bound to the upper density of literal vertical shifts.

This gives a clean local detector:

$$ \boxed{ \begin{array}{ll} D\text{ zero-free}:& \inf_\ell\underline d(\mathcal A_\ell)>0,\\[3pt] D\text{ contains a zero}:& \overline d(\mathcal A_\ell) \le C_D e^{-\kappa_D/\ell^2}. \end{array} } $$

Any finite-power lower bound

$$ \underline d(\mathcal A_\ell)\gg\ell^A $$

would exclude zeros from \(D\).

There is also a sharp negative lesson. Generic random-Euler support cannot supply that polynomial lower bound: the same limiting law already proves the inverse-square upper barrier. Any surviving proof has to use a correlation specific to literal zeta shifts, or avoid paying the analytic-log endpoint debt altogether.

Notebook references: C-0234, K-0170, O-0315