GPT 6 Astra Ultra attempt on Math-6040
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Erdős Problem 986: off-diagonal Ramsey numbers
Statement and attribution
For each fixed $s\ge3$, prove $R(s,k)\gg_s k^{s-1}/(\log k)^{c(s)}$. We reconstruct Bradač's proof of the stronger bound \[ R(s,k)\gg_s \frac{k^{s-1}}{(\log k)^{2s-4}} \] from Section 2.5 of Off-diagonal Ramsey numbers, arXiv:2605.28793v3. That version credits an internal OpenAI model for the final improvement of the exponent to $s-1$. The reconstruction below includes the geometric graph, its spectrum, the counting argument, and the sampling step. No novelty is claimed.The projective polarity graph
Fix $t=s-1\ge2$ and a power $q$ of $2$. The vertices of $G$ are the one-dimensional subspaces of $\mathbb F_q^{t+1}$; join $a,b$ when $a\perp b$ for the usual nondegenerate dot product. Loops are allowed at this intermediate stage. The graph is regular with \[ n=\frac{q^{t+1}-1}{q-1}\asymp q^t,\qquad d=\frac{q^t-1}{q-1}\asymp q^{t-1}. \] Two different vertices have exactly $c=(q^{t-1}-1)/(q-1)$ common neighbors: their two independent linear equations cut out a $(t-1)$-dimensional vector subspace. If $A$ is the adjacency matrix, then $A^2=(d-c)I+cJ$, with $d-c=q^{t-1}$. Consequently, on the orthogonal complement of the constant vector, its eigenvalues have modulus $\lambda=q^{(t-1)/2}$. The usual decomposition of two indicator vectors into constant and orthogonal parts proves \[ \left|e_G(U,V)-\frac dn|U||V|\right| \le\lambda\sqrt{|U||V|}. \tag{1} \] Here edges between two sets are counted as ordered incidences, so $1$ also applies to overlapping sets.A digraph with no transitive $s$-tournament
Let $D$ have as vertices the ordered pairs $(a,b)$ with $a\perp b$. Draw an arc $(a,b)\to(a',b')$ exactly when $a\perp b'$ and $a'\not\perp b$. It has $nd\asymp q^{2t-1}$ vertices and has neither loops nor opposite arcs. Suppose $(a_i,b_i)$, $1\le i\le t+1$, form a transitive tournament in that order. Choose nonzero vector representatives $x_i,y_i$. We have $x_i\cdot y_j=0$ for $i\le j$, and $x_{i+1}\cdot y_i\ne0$. Choose also $x_{t+2}$ with $x_{t+2}\cdot y_{t+1}\ne0$. In a nontrivial relation $\sum_jc_jy_j=0$, let $i$ be its smallest nonzero coefficient. Taking the dot product with $x_{i+1}$ kills all later terms and leaves a contradiction. Thus the $t+1$ vectors $y_j$ form a basis. But $x_1$ is orthogonal to all of them, another contradiction.Counting ordered tuples with no forward arcs
Call a sequence $(a_1,b_1),\ldots,(a_m,b_m)$ consistent if $a_i\perp b_i$, and if $a_i\perp b_j$ for $i<j$, then $a_j\perp b_i$. This is exactly a tuple with no forward arc in $D$. Repetitions are allowed for this upper bound. For a consistent prefix $\sigma$, define \[ W_\sigma(y)=\operatorname{span}\{b_i:a_i\perp y\},\quad r_\sigma(y)=\dim W_\sigma(y),\quad U_r=\{y:r_\sigma(y)\le r\},\quad Z_r=\{y:r_\sigma(y)=r\}. \] An extension $(a,b)$ must have $a\perp W_\sigma(b)$ and $a\perp b$. If $r=r_\sigma(b)\le t$, there are at most $2q^{t-r}$ possibilities for $a$, even after dropping the latter condition; if $r=t+1$ there are none.For each nonempty $U_r$, $0\le r\le t$, choose $\ell\le r$ with $|Z_\ell|$ maximal. Then $|Z_\ell|\ge |U_r|/(t+1)$ and $|Z_r|\le|Z_\ell|$. Call $b\in Z_r$ popular if it lies in at least $|Z_\ell|/(16q)$ of the spaces $W_\sigma(y)$, $y\in Z_\ell$. Each such space contains at most $2q^{\ell-1}$ projective points (zero when $\ell=0$), so there are at most $32q^\ell$ popular $b$. Mark extensions using them. For this $r$ there are at most $32q^\ell\cdot2q^{t-r}\le64q^t$ marked extensions.
Call $a$ poor if $|N_G(a)\cap Z_\ell|\le |Z_\ell|/(8q)$, and write $P_r$ for these $a$. For sufficiently large $q$, $d/n\ge1/(4q)$. Combining the definition of poverty with $1$ gives \[ |P_r||Z_\ell|\ll q^{t+1}. \] Since $|Z_r|\le|Z_\ell|$, another application of $1$ gives \[ e_G(P_r,Z_r)\le\frac dn|P_r||Z_r|+ \lambda\sqrt{|P_r||Z_r|}\ll q^t. \] Mark all extensions with poor $a$, too. Summing over $r$ shows that each prefix has at most $h=C_tq^t$ marked children.
For an unmarked extension $(a,b)$ with $b\in Z_r$, $a$ is not poor and $b$ is not popular. At least $|Z_\ell|/(16q)$ points $y\in Z_\ell$ satisfy both $a\perp y$ and $b\notin W_\sigma(y)$. Their span dimension rises from $\ell$ to $\ell+1$. No dimension decreases. Thus this extension reduces $|U_\ell|$ by a factor at most \[ 1-\frac1{16(t+1)q}. \] Initially every $|U_\ell|$ is at most $n\ll q^t$. For each fixed $\ell$, only $O_t(q\log q)$ such reductions can occur while the integer $|U_\ell|$ remains positive; once it is zero it cannot be chosen again. There are only $t+1$ possibilities for $\ell$. Therefore every path of extensions has at most $w=C'_tq\log q$ unmarked steps.
The full number of children is at most $\Delta=nd\ll q^{2t-1}$. The number $F_k$ of consistent $k$-tuples is bounded by \[ F_k\le\sum_{i\le\min(w,k)}\binom{k}{i}\Delta^ih^{k-i} \le2^kh^k\max(1,\Delta/h)^w \le(C''_tq^t)^k \tag{2} \] provided $k\ge C'''_tq(\log q)^2$. The last inequality follows since the logarithm of the extra factor is $O_t(w\log q)=O_t(q(\log q)^2)$, which is absorbed by a fixed exponential in $k$.