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Take the increasing event that a finite square torus graph contains some long horizontal or vertical rectangle crossing. Translations and a quarter turn make this a transitive increasing event on its bonds. A positive critical crossing bound and the Friedgut-Kalai sharp threshold theorem make its supercritical probability tend to one. A bounded family of shorter, thicker rectangles intercepts every possible long crossing. The square-root trick for positively associated events transfers the high probability to each fixed intercepting rectangle. Using every possible translate would lose the useful bounded exponent.

There is an absolute $C>0$ such that any nontrivial transitive increasing event on $n\ge 2$ independent bits satisfies ${\mu }_q(A)>1-\epsilon$ if ${\mu }_p(A)>\epsilon$, $0<\epsilon <1/2$, and the displayed bound holds with $p<q<1$. A uniform weighted Kahn-Kalai-Linial theorem gives one pivotal probability at least $c{\mu }_r(A)(1-{\mu }_r(A))\log (n)/n$. Transitivity makes the total influence at least $c{\mu }_r(A)(1-{\mu }_r(A))\log n$. The Margulis–Russo formula identifies this with the derivative of ${\mu }_r(A)$. Integration of the derivative of its log odds proves the threshold bound.

Partition $N=mw$ bits into $m$ tribes of size $w$, and let the tribes function be one when some tribe consists entirely of ones. Permuting tribes and permuting bits within a tribe acts transitively on the $N$ coordinates, so this is a transitive increasing event. Under independent bits of density $p$, its exact success probability isChoose $m=\lfloor (\log 2)2^w\rfloor$. Then $u(1/2)\to 1/2$, while $\log N=w\log 2+O(\log w)$. For fixed $0<x<1$, its unique $x$-quantile isPut $a_x=-\log (1-x)$. Expanding $1-e^{-a_x/m}=a_x/m+O_x(m^{-2})$ givesThus, for fixed $0<\epsilon <1/2$,This positive window prevents an improvement to $o(1/\log N)$ in the Friedgut-Kalai sharp threshold theorem. Moreover, as the fixed error parameter tends to zero, the displayed logarithmic coefficient is $\log (1/\epsilon )+\log \log (1/\epsilon )+o(1)$, so the theorem's logarithmic error dependence is also optimal up to constants. This last comparison takes $w\to \infty$ for each fixed $\epsilon$; the expansion is not being claimed uniformly for arbitrarily tiny $w$-dependent errors.

We prove the following precise form of the Friedgut-Kalai sharp threshold theorem. There is an absolute $C$ such that, for a nontrivial transitive increasing event $A\subseteq \{0,1{\}}^n$, $n\ge 2$, and $0<\epsilon <1/2$, its success probability increases from $\epsilon$ to $1-\epsilon$ over an interval of length at mostIn particular, ${\mu }_p(A)>\epsilon$ and $q-p\ge C\log ((1-\epsilon )/\epsilon )/\log n$, with $p<q<1$, imply ${\mu }_q(A)>1-\epsilon$. Here ${\mu }_p$ denotes independent bits equal to one with probability $p$. Symmetry means invariance under a permutation group acting transitively on the $n$ coordinates, not necessarily invariance under all coordinate permutations.

Use the allowed weighted-cube influence theorem in this exact form: an absolute $c>0$ exists such that, for every $n\ge 2$, $0<p<1$, and zero-one Boolean function of mean $u$ under ${\mu }_p$, some coordinate has pivotal probability at least $cu(1-u)\log n/n$. A coordinate's pivotal event is determined by the other coordinates and means that the two possible values of that coordinate give different outputs. This uniform form is a consequence of the weighted Kahn-Kalai-Linial theorem; it applies to nonsymmetric functions as well.

Write $u(p)={\mu }_p(A)$. Differentiating the finite product expression for $u(p)$, one coordinate at a time, givesThe last equality uses monotonicity, so each summand is the pivotal probability. This derives the Margulis–Russo formula in the normalization being used. Transitivity makes all the influences equal; applying the weighted influence theorem therefore yieldsNontriviality ensures $0<u(p)<1$ throughout $(0,1)$. Dividing and integrating the derivative of the log odds givesAt the two specified quantiles the left side is $2\log ((1-\epsilon )/\epsilon )$. Taking $C=2/c$ proves the window bound and, with a strict starting inequality, the asserted strict final inequality. This is the sharp threshold conclusion, rather than an assumption about a discontinuous finite-$n$ probability curve.

Let $T$ be the torus graph $(\mathbb{Z}/(5n)\mathbb{Z}{)}^2$, with independent bond density $p$. For $n$ large, all rectangles below are too small to identify any of their own vertices or edges. Let $E_n$ be the event that somewhere in $T$ there is a horizontal open crossing of a $4n$ by $n$ rectangle or a vertical open crossing of an $n$ by $4n$ rectangle. It is increasing and invariant under translations and a quarter turn. These symmetries act transitively on all $50n^2$ bonds, including both orientations. Thus it is a transitive increasing event.

A fixed $4n$ by $n$ rectangle has its ordinary planar distribution, so the supplied lower bound gives ${\mathbb{P}}_{1/2}^T(E_n)>c_4$. Choose any $0<\eta <\min (c_4,1/2)$. Since $p-1/2>0$ is fixed and $\log (50n^2)\to \infty$, the Friedgut-Kalai sharp threshold theorem implies ${\mathbb{P}}_p^T(E_n)>1-\eta$ for all sufficiently large $n$. As $\eta$ is arbitrary,It remains to transfer this symmetric statement to a fixed rectangle, without a growing union-bound loss. On each circular coordinate take the ten positions $a_j=\lfloor jn/2\rfloor$, $0\le j<10$. Consecutive circular gaps are at most $\lceil n/2\rceil$. Use the one hundred $3n$ by $2n$ rectangles starting at pairs $(a_i,a_j)$ and their one hundred quarter-turn images. Every crossed $4n$ by $n$ rectangle contains the full horizontal width of one of the first family, with its vertical span contained in that rectangle's height-$2n$ span: the permissible starting intervals in each coordinate have length $n$, so the chosen positions always supply a suitable start. The crossing, restricted between those two vertical sides, gives a crossing of the chosen $3n$ by $2n$ rectangle. The analogous assertion holds in the rotated family.

Consequently $E_n$ implies at least one of these two hundred increasing crossing events. Their common success probability is $a_n=h_p(3n,2n)$. The nth root trick givesOnly two hundred translates are used, independently of scale. This bounded exponent is the essential finite-torus crossing amplification step.

For a requested height $N$, set $s=\lfloor N/2\rfloor$. Glue $k$ width-$3s$, height-$2s$ rectangles, each successive pair overlapping in a square of side $2s$. Their horizontal crossings and the $k-1$ vertical overlap crossings imply a crossing of the union, which has width $(k+2)s$. An overlap square has crossing probability at least $a_s=h_p(3s,2s)$. Thus the Harris lemma givesChoose $k$ fixed and large enough that $(k+2)s\ge \lambda N$ for all sufficiently large $N$. Increasing height and shortening width can only help, so uniformly for $m\le \lambda N$,In particular it exceeds $1-\epsilon$ for every sufficiently large $N$, with the bound depending only on $p,\lambda ,\epsilon$.

For a sequence of nontrivial increasing events on finite product measures, write $p_x$ for the parameter at which the success probability is $x$. An additive sharp threshold means that $p_{1-\epsilon }-p_{\epsilon }\to 0$ for every fixed $0<\epsilon <1/2$. When the threshold parameter itself tends to zero, relative sharpness instead compares this window with $p_{1/2}$; additive sharpness alone is then a weaker assertion. The Friedgut-Kalai sharp threshold theorem provides an explicit additive bound for transitive increasing events.