Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 13 (Probabilistic Combinatorics)

The Lusternik-Schnirelmann-Borsuk theorem has the following two equivalent covering formulations. For every integer $d\ge 0$, a cover of the sphere $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The same assertion holds with open sets in place of closed sets. Thus $d+1$ antipodal-pair-free open sets, or $d+1$ antipodal-pair-free closed sets, cannot cover $S^d$. The dimension-zero case simply says that one set covering the two-point sphere contains both points.

Here is why the two versions agree. A finite open cover of a compact metric space admits a closed shrinking that still covers: sufficiently small closed balls subordinate to the open cover can be grouped according to their containing open member. Applying the closed version to that shrinking proves the open version. Conversely, if a nonempty closed subset $C$ of $S^d$ avoids antipodal pairs, compactness gives positive distance between $C$ and $-C$. A sufficiently small open neighbourhood of $C$ still avoids antipodal pairs. Enlarge each member of a hypothetical closed counterexample in this way; the open version rules it out. Empty members cause no difficulty.

Another common equivalent formulation is the Borsuk-Ulam theorem: every continuous $f:S^d\to {\mathbb{R}}^d$ has $f(x)=f(-x)$ for some $x$, or, equivalently, every continuous odd function $g:S^d\to {\mathbb{R}}^d$ has a zero. For example, if $d+1$ antipodal-pair-free open sets covered $S^d$, a subordinate partition of unity $({\varphi }_1,\dots ,{\varphi }_{d+1})$ would give the odd functionIt cannot vanish, since some ${\varphi }_i(x)>0$, whereas antipodal-pair-freeness forces ${\varphi }_i(-x)=0$. In the other direction, if an odd function $g:S^d\to {\mathbb{R}}^d$ never vanishes, choose $d+1$ vectors $u_i$ forming a regular simplex centred at the origin in ${\mathbb{R}}^d$. For $d\ge 1$, the open sets $\{x:u_i\cdot g(x)>0\}$ cover $S^d$ and none contains an antipodal pair, contradicting the covering theorem.

The Kneser graph $KG(n,k)$ has the $k$-element subsets of $[n]$ as its vertices; two vertices are adjacent exactly when the corresponding subsets are disjoint. The Lovász theorem on Kneser graphs determines its chromatic number. A graph colouring is therefore a partition of the $k$-sets into intersecting families.

First construct a graph colouring with $d+2=n-2k+2$ colours. If a $k$-set meets $[d+1]$, give it the colour of its least element. Give every remaining $k$-set colour $d+2$. Two sets with one of the first $d+1$ colours share that colour's element. The last colour consists of $k$-sets in a $(2k-1)$-element ground set, so it too is an intersecting family. Thus $\chi (KG(n,k))\le d+2$.

For the converse we establish the Gale hemisphere lemma explicitly, using a signed moment curve. Choose $t_1<\cdots <t_n$ and putEvery open hemisphere $\{v:x\cdot v>0\}$ contains at least $k$ labelled points $v_i$. To see this, put $p(t)=x_0+x_{1}t+\cdots +x_d{t}^d$. It is a nonzero polynomial of polynomial degree at most $d$. First suppose none of the $p(t_i)$ vanishes. Write $b_i=\mathrm{sgn}((-1{)}^{i}p(t_i))$. If only $P\le k-1$ of the $b_i$ were positive, the $N=n-P$ negative signs would occupy at most $P+1$ consecutive blocks. There would be at leastadjacent pairs with both $b_i,b_{i+1}$ negative. Each such pair forces $p(t_i)$ and $p(t_{i+1})$ to have opposite signs. The intermediate value theorem would give $d+1$ distinct roots of a polynomial, a contradiction.

If $p$ vanishes at $z$ of the sample points, then $z\le d$. A Lagrange interpolation polynomial $q$ of polynomial degree at most $z-1$ can be chosen with $(-1{)}^{i}q(t_i)<0$ at all those points. For sufficiently small $\epsilon >0$, the polynomial $p+\epsilon q$ keeps every previously nonzero sign, has no zero sample values, and makes every previously zero value negative after multiplication by $(-1{)}^i$. Its positive count is exactly the positive count of $p$, so the preceding argument proves the open hemisphere assertion also in this case. For $d=0$, the labelled points alternate between the two points of $S^0$; distinct labels, rather than distinct positions, are what is needed.

Suppose now that $KG(n,k)$ had a graph colouring with at most $d+1$ colours, padding the palette with unused colours if necessary. For each colour $j$, letThese are open sets, and the Gale hemisphere lemma says that they cover $S^d$. If both $x$ and $-x$ belonged to $U_j$, two $k$-sets of colour $j$ would lie in opposite open hemispheres. They would be disjoint, hence adjacent in the Kneser graph, contradicting the graph colouring. The Lusternik-Schnirelmann-Borsuk theorem excludes this cover. Therefore the lower bound matches the explicit colouring:

Use the identification of two-element subsets of $[n]$ with the edges of the complete graph $K_n$. A colour class in the Kneser graph $KG(n,2)$ is an intersecting two-element set family, so its edges must pairwise meet.

Such a class is contained either in a star graph or in a triangle in a graph. Indeed, if not all edges have a common vertex, take two meeting edges $ab,ac$. An edge not containing $a$ must then be $bc$. Any further edge meeting all three of $ab,ac,bc$ is one of these three. Classes with at most two edges already lie in a star graph.

Assume that a graph colouring used $c\le n-3$ colours. Classify $s$ classes as stars and the remaining $t=c-s$ classes as triangles. Delete a chosen centre of each star class, leaving $m\ge n-s\ge t+3$ vertices. Every edge between those remaining vertices must belong to a triangle class, of which each covers at most three edges. HenceButa contradiction. Thus $\chi (KG(n,2))\ge n-2$. For the matching upper bound, assign a two-set the colour of its smaller element if that element is at most $n-3$, and otherwise give it colour $n-2$. The last class comprises the three pairs on the last three elements, an intersecting family. Consequently an entirely combinatorial argument gives

Take information entropy in bits and put $H(X_{\varnothing })=0$. Write $C=A\cap B$, $U=A\setminus B$, and $V=B\setminus A$. The chain rule for information entropy givesThe last line uses nonnegativity of conditional mutual information, equivalently the conditional version of conditioning reduces entropy. All random variables are finite-valued, so every conditional entropy here is finite. Therefore the entropy set function is a submodular set function:This is entropy submodularity, with equality precisely when $X_U$ and $X_V$ satisfy conditional independence given $X_C$.

An elementary union-intersection compression replaces two occurrences $A_1,A_2$ in a multiset by $A_1\cup A_2,A_1\cap A_2$. By entropy submodularity, this changes the sum of information entropies byA compression of an entropy sum is obtained by iterating these elementary union-intersection compressions. Summing the inequalities over the sequence, with repetitions counted according to their multiset multiplicities, proves the required monotonicity:Neither the number of occurrences of an individual coordinate nor the total number of sets changes under union-intersection compression.

The empty family satisfies the bound, so assume $\mathcal{F}$ is nonempty and choose a graph according to the uniform distribution on a finite set $F\in \mathcal{F}$. Let $Y$ be its random vector of edge indicators, indexed by $E=\left(\genfrac{}{}{0ex}{}{[n]}{2}\right)$. Then its information entropy is $H(Y)={\log }_2|\mathcal{F}|$.

For each vertex $v$, put $S_v=\{e\in E:v\in e\}$, so $Y_{S_v}$ records the graph neighbourhood of $v$. The possible graph neighbourhoods form an intersecting family of subsets of $[n]\setminus \{v\}$: the graph intersection of any two members of $\mathcal{F}$ has no isolated vertex at $v$. A subset and its complement cannot both occur. Pairing the $2^{n-1}$ subsets into complementary pairs gives at most $2^{n-2}$ possible graph neighbourhoods. The maximum entropy on a finite alphabet therefore givesThis argument applies for $n\ge 2$. For $n=1$ the assumed nonempty family cannot exist, since its only possible graph has an isolated vertex.

Each edge belongs to exactly two of the sets $S_v$. To obtain the needed instance of Shearer inequality directly from the preceding compression argument, repeatedly apply a union-intersection compression to incomparable members of the multiset $(S_v{)}_{v=1}^n$. Each such step strictly increases ${\sum }_S|S{|}^2$, byThe potential is bounded and integer-valued, so the process terminates in a chain under inclusion. Coordinate multiplicities stay equal to two, so, for $E\ne \varnothing$, the terminal chain consists of two copies of $E$ and $n-2$ empty sets. The compression of an entropy sum now givesExponentiating proves the entropy bound for graphs with isolated-vertex-free intersections:

Throughout this question the binomial random graph has $p=1/2$, as fixed in the introduction. Let $L={\log }_{2}n$ and $s=\lceil 2L\rceil$. Count the $s$-vertex independent sets by a random variable $I_s$. Linearity of expectation givesFor the second inequality, $s\ge 2L$ implies $sL-s(s-1)/2\le s/2\le L+1/2$. For the limit, $s!\ge (s/2{)}^{\lfloor s/2\rfloor }$ grows faster than $n$. By the first moment method, with high probability there is no such independent set; any larger independent set would contain one. Thus the independence number satisfies $\alpha (G)\le s-1<2L$ with high probability.

Every class of a proper graph coloring is an independent set, giving the claimed chromatic lower bound:

For the clique count in a binomial random graph, writeFix $0<\epsilon <1$. At $r_{-}=\lfloor (2-\epsilon )L\rfloor$, the estimates $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\ge (n/r{)}^r$ and $r=O(L)$ giveThis eventually exceeds $(9/5)L$. At $r_{+}=\lceil (2+\epsilon )L\rceil$, the upper bound $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\le n^r$ instead givesso ${\mu }_{r_{+}}<n^{9/5}$. Moreoverso no later clique size can regain the threshold. The maximum defining $r_0$ exists for sufficiently large $n$, since ${\mu }_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)/2\ge n^{9/5}$ then. We have $r_{-}\le r_0<r_{+}$. Letting $\epsilon$ decrease to zero proves the asymptotic size:

We need independent sets in every sufficiently large remaining induced subgraph, rather than just one in the initial random graph. The useful concentration variable is the maximum edge-disjoint clique packing in the complement graph; changing one edge changes that variable by at most one. This avoids the much larger sensitivity of the ordinary clique count in a binomial random graph.

First work in a binomial random graph $H\sim G(m,1/2)$, take $r=r_0(m)$, and let $X$ count its $r$-cliques. Put $\mu =\mathbb{E}X\ge m^{9/5}$ and let $Z$ be its maximum edge-disjoint clique packing size. Form the clique conflict graph whose vertices are these cliques, adjacent when they share an edge. If $Q$ counts unordered conflicting pairs, the Caro-Wei bound gives, for each realization,To verify the bound, randomly order the vertices of a finite graph and retain each vertex preceding all its graph neighbours. These retained vertices are an independent set, with expected size ${\sum }_v(1+\mathrm{deg}v{)}^{-1}\ge X^2/(X+2Q)$ by the Cauchy-Schwarz inequality. Interpret the ratio as zero when $X=0$. A second application of the Cauchy-Schwarz inequality, now to expectations, givesCounting ordered pairs of $r$-sets with $j$ common vertices, including identical sets when $j=r$, yieldsThe shared edges account for the factor $2^{\left(\genfrac{}{}{0ex}{}{j}{2}\right)}$. We claimHere are the estimates, including the large overlaps. For $2\le j\le \lfloor r/2\rfloor$, the probability that a random $r$-set contains any specified $j$ elements is at most $(r/(m-r){)}^j$. Summing over their choices inside the first $r$-set givesThe logarithm of $T_j$ is a convex function, a quadratic in $j$, so the maximum on this interval is at an endpoint. We have $T_2=O(r^4/m^2)$. As $r=(2+o(1)){\log }_{2}m$, the other endpoint satisfiesand is smaller than $T_2$ for sufficiently large $m$. Summing at most $r$ terms gives $O(r^5/m^2)$.

For the other half of the overlaps, write $j=r-s$, where $0\le s\le \lceil r/2\rceil$. The exact identity and an upper bound areThe quantity in parentheses is $m^{-1/2+o(1)}$, which tends to zero. The geometric series is thus at most $2/\mu$ for sufficiently large $m$. This proves the claim. In fact $r^5/m^2=o(m^{-9/5})$ and $1/\mu \le m^{-9/5}$, so eventuallyNow expose the $N=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edge indicators one at a time. The edge-exposure martingale $M_i=\mathbb{E}[Z\mid \text{first }i\text{ indicators}]$ starts at $\mathbb{E}Z$ and ends at $Z$. Deleting one edge destroys at most one member of any edge-disjoint clique packing. Hence changing one indicator changes $Z$ by at most one; coupling the remaining indicators shows $|M_i-M_{i-1}|\le 1$.

Precisely, the Azuma-Hoeffding inequality says that if $(M_i{)}_{i=0}^N$ is a martingale and $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then, for $t>0$,If all $c_i=0$, the martingale is constant. Applying the lower-tail inequality with $c_i=1$ and $t=\mathbb{E}Z$ givesfor sufficiently large $m$. This is the only martingale concentration inequality needed.

Return to the original binomial random graph $G\sim G(n,1/2)$ and set $m=\lceil n/({\log }_{2}n{)}^2\rceil$. For each fixed $m$-element vertex set $U$, the complement graph of $G[U]$ has distribution $G(m,1/2)$. The union bound therefore givessince $m^{8/5}/n\to \infty$. Thus with high probability every vertex set of size at least $m$ contains an independent set of sizeApply greedy colouring by removing independent sets: remove such an independent set, give it a fresh colour, and repeat while at least $m$ vertices remain. Give each of the fewer than $m$ final vertices its own colour. This uses at mostcolours, since $m=o(n/{\log }_{2}n)$. Together with the lower bound, this gives the chromatic number of the half-density binomial random graph:

For each element $x$ of the product set $A\cdot B$, let $s_x$ count its representations as $ab$ with $(a,b)\in A\times B$. The ratio equality $a/b=c/d$ is equivalent to $ad=cb$. Swapping the two $B$ coordinates is a bijection between the ratio-equality quadruples and the equal-product quadruples. Therefore the multiplicative energy satisfiesThe Cauchy-Schwarz inequality gives the energy lower bound:For the upper bound, place the Cartesian product $P=A\times B$ in the strictly positive quadrant. For every occupied Euclidean ray from the origin let $P_{\lambda }$ be its points and $r_{\lambda }=|P_{\lambda }|$. The slope is $\lambda =b/a$, soTake $\log$ to be the natural logarithm and put $L=\lceil \log |B|\rceil \ge 1$, using the non-triviality assumption $|B|\ge 2$. Partition the possible occupancies into $L$ classes: $[e^j,e^{j+1})$ for $0\le j<L-1$, and $[e^{L-1},|B|]$ for the last class. This last closed endpoint ensures that the partition works even when the top endpoint is attained. Within each class the ratio of any two occupancies is at most $e$.

Fix one class and order its occupied Euclidean rays by increasing slope, with occupancies $r_1,\dots ,r_t$. Write $S=|A+A||B+B|$, the cardinality of $(A+A)\times (B+B)$. If $t=1$, then $r_1^2\le |A||B|\le S$, since adding any fixed element gives an injection of $A$ into its sumset $A+A$, and similarly for $B$.

If $t\ge 2$, the sums $P_i+P_{i+1}$ contain exactly $r_i{r}_{i+1}$ different points, by injectivity of sums on two distinct rays. Indeed, the two direction vectors are linearly independent, so the coefficients of a sum uniquely recover its two summands. Positivity puts every sum strictly inside the open planar sector between those two Euclidean rays. The sectors between successive selected Euclidean rays are disjoint, even if there are additional unselected rays between them. All these sums belong to $(A+A)\times (B+B)$, and henceFor neighbouring occupancies their ratio lies between $e^{-1}$ and $e$, soSumming covers every $r_i^2$ at least once, and gives ${\sum }_i{r}_i^2\le 4S$. The same bound holds for empty or singleton classes by the preceding observations. Adding over the $L$ classes proves the multiplicative energy sumset bound $E(A,B)\le 4LS$. Combining both bounds yields the sum-product conclusion:If instead $\log$ is interpreted as base two, use the same $L=\lceil {\log }_2|B|\rceil$ occupancy classes with $2$ in place of $e$; $2+2^{-1}<4$ proves that convention as well. Positivity is essential to the sector argument.