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The lower shadow consists of the sets obtained by removing one element from a member:For a nonempty uniform set family, write its size in its unique decreasing binomial coefficient expansionThe Kruskal-Katona theorem states thatA colexicographic initial segment attains equality; thus this is an exact minimum, not just an asymptotic estimate.

Here is a direct proof of the requested Lovász shadow bound, avoiding any unproved numerical interpolation of the Kruskal-Katona theorem. The real binomial coefficient means $\left(\genfrac{}{}{0ex}{}{x}{r}\right)=x(x-1)\cdots (x-r+1)/r!$. It is strictly increasing for $x>r-1$. Since the set family is nonempty and its size is an integer, necessarily $x\ge r$.

Apply the usual left coordinate shifts of a set family until the set family is fixed by every shift. A shift preserves its size and does not increase its lower shadow; the witness argument is supplied in Solution 3. Termination follows because every nontrivial shift reduces ${\sum }_{A\in \mathcal{A}}{\sum }_{i\in A}i$. It suffices to prove the inequality for this shifted set family.

Induct on $r$, and, for each fixed $r$, on $|\mathcal{A}|$. The cases $r=1$ and $|\mathcal{A}|=1$ give, respectively, the single empty shadow member and the $r$ faces of an $r$-set. Split the set family at coordinate one, writing ${\mathcal{A}}_0$ for the members avoiding it and ${\mathcal{A}}_1$ for the sets obtained by deleting it from members containing it. Every $(r-1)$-subset of a member of ${\mathcal{A}}_0$ belongs to ${\mathcal{A}}_1$: replace the one omitted coordinate by coordinate one. ThereforeThe two terms count shadow members avoiding and containing coordinate one.

Put $b=|{\mathcal{A}}_1|$. This is positive. If $b<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity gives $|{\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r}\right)$. In this case $x>r$ and ${\mathcal{A}}_0$ is a smaller nonempty set family. The mathematical induction hypothesis on family size gives $|\partial {\mathcal{A}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, contradicting its containment in ${\mathcal{A}}_1$. Consequently $b\ge \left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$. Write $b=\left(\genfrac{}{}{0ex}{}{z}{r-1}\right)$, so $z\ge x-1$. The mathematical induction hypothesis on uniformity and the Pascal's identity now giveFor $r=2$ the degree-zero term is the constant one, which causes no difficulty.

For the graph application, identify a triangle-free graph with a subset of the $N$ possible edges. Removing an edge preserves being triangle-free, so $\partial {\mathcal{G}}_{m+1}\subseteq {\mathcal{G}}_m$. If ${\mathcal{G}}_{m+1}$ is nonempty, the Lovász shadow bound givesThus, for $1\le m<N$ with $p_m>0$,If the next set family is empty its probability is zero, so the same inequality holds without introducing $x_{m+1}$. If $p_m=0$, every later set family is empty; the divided expression is undefined and the correct assertion is this zero-probability conclusion.

Iterate the monotonicity of the effective ground-set parameter of a hereditary uniform layer. Whenever $p_{m+k}>0$, all the intermediate parameters are at least their layer sizes, and all factors below are nonnegative. We obtain the stronger estimateThe second inequality follows from $x_m\le N$ and the fact that $(x_m-t)/(N-t)$ decreases with $t$. If the target set family is empty, the requested bound follows directly when its right side is nonnegative. When $k>x_m+1$, the printed product can have signed factors and should instead be stopped at the first empty layer: its unrestricted signed form is not a meaningful probability bound. In the nonzero target range there is no such ambiguity. A uniformly safe extension, for $0\le k\le N-m$, isIt agrees with the printed expression whenever all its factors are nonnegative. In particular $0\le k\le m$ is always safe when $p_m>0$, and this includes the requested doubling case.

Taking $k=m$ in the nonzero range,If the $2m$-edge set family is empty the conclusion is immediate; if $2m>N$, interpreting the event as impossible gives the same conclusion. These qualifications distinguish the probability assertion from products or ratios outside their domains.

The Kruskal-Katona theorem states that a colexicographic initial segment minimizes the lower shadow of a uniform set family of prescribed size. Its numerical form is as follows. For $m>0$, write the unique combinatorial number system expansionEvery $\mathcal{F}\subseteq [n{]}^{(r)}$ of size $m$ satisfiesThe empty family has empty lower shadow. Equality is attained by the first $m$ $r$-sets in colexicographic order, where the largest differing element belongs to the later set. Iterating the Kruskal-Katona theorem shows that the same colexicographic initial segment minimizes every iterated lower shadow.

For $1\le r<n/2$, the Erdős-Ko-Rado theorem givesfor an intersecting family $\mathcal{F}\subseteq [n{]}^{(r)}$. All $r$-sets containing one prescribed point show that the bound is sharp.

For the proof from the Kruskal-Katona theorem, put $k=n-r$ and form the complement family $\mathcal{G}=\{[n]\setminus A:A\in \mathcal{F}\}\subseteq [n{]}^{(k)}$. Let $\mathcal{S}$ be its rank-$r$ iterated lower shadow. No member $B\in \mathcal{F}$ belongs to $\mathcal{S}$: containment in $[n]\setminus A$ would give $A\cap B=\varnothing$, impossible for an intersecting family of nonempty sets. HenceSuppose $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The first $\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$ members of the colexicographic order are all $k$-sets of $[n-1]$. Their rank-$r$ iterated lower shadow is all $r$-sets of $[n-1]$, since $r<k$. The next $k$-set contains $n$ and has an $r$-subset containing $n$, so taking even one more member strictly enlarges that iterated lower shadow. By the iterated Kruskal-Katona theorem, $|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. Together with $|\mathcal{F}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ this contradicts Pascal's identity and the preceding inequality. This proves the Erdős-Ko-Rado theorem.

For the Katona circle method proof, fix a cyclic ordering of $[n]$. A cyclic interval of length $r$ is specified by its final position. If no such interval belongs to $\mathcal{F}$, the bound below is automatic. Otherwise rotate one selected interval so that it ends at position $n$, and thus occupies positions $n-r+1,\dots ,n$. Intervals ending at positions $r,\dots ,n-r$ are disjoint from it and cannot be selected. Among the remaining endpoints, pair $j$ with $n-r+j$ for $1\le j\le r-1$. The corresponding intervals are disjoint when $n\ge 2r$, so at most one interval per pair is selected. Together with the interval ending at $n$, this gives the cyclic interval intersection bound of $r$ selected intervals.

Choose a uniformly random permutation and read its positions cyclically. A fixed $r$-set is a cyclic interval with probability $n/\left(\genfrac{}{}{0ex}{}{n}{r}\right)$: each cyclic position gives a uniformly distributed $r$-set, and for $r<n$ the $n$ intervals are distinct. Summing over $\mathcal{F}$, the expected value of the number of selected intervals is $n|\mathcal{F}|/(\genfrac{}{}{0ex}{}{n}{r})$. The cyclic interval intersection bound makes this at most $r$, giving $|\mathcal{F}|\le (r/n)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$.

Finally apply the Katona circle method to an arbitrary antichain $\mathcal{A}$. If it contains $\varnothing$ or $[n]$, it has exactly one member and the LYM inequality holds with equality. Otherwise all its members have sizes $1,\dots ,n-1$. In a fixed cyclic ordering, the cyclic intervals sharing a final position form a nested chain as their lengths increase. At most one of them can belong to the antichain. Summing over the $n$ final positions proves the cyclic interval antichain bound of $n$ intervals. Taking the expected value over a uniformly random permutation givesAfter division by $n$, this is the LYM inequality proved by cyclic intervals. The case $n=0$ consists only of the empty set and is immediate.

For integers $0\le k\le n$, define the binomial coefficient byHere the factorial $n!$ is the product of the positive integers up to $n$. This also counts the $k$-element subsets of an $n$-element set: an ordered choice has $n!/(n-k)!$ possibilities, and each subset is ordered in $k!$ ways.

Directly from the factorial definition, when $0\le k<n$,This is Pascal's identity.

The required hockey-stick identity follows by mathematical induction on $m$, with $n\ge 0$ fixed. For $m=0$, both sides are $1$. If it holds at $m$, adding the next binomial coefficient and using Pascal's identity givesTherefore for every $n,m\ge 0$,

Complexifying the Clifford algebra removes the signature dependence of its representation dimension. Pair the $2k$ gamma matrices into operators $a_i,a_i^{†}$ with $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. The $k$ commuting occupation operators have eigenvalues zero or one. Starting from a Clifford vacuum, applying every subset of the $k$ fermionic creation operators produces $2^k$ independent states. Conversely, these operators construct matrices on a $2^k$-dimensional space, and their products span its full matrix algebra. Thus an irreducible complex Clifford algebra module in $D=2k$ hasA general Clifford module is a direct sum of these; the statement concerns the minimal spinor.

The Lorentz group acts through even Clifford products, ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\Gamma }^{\mu },{\Gamma }^{\nu }]$. More precisely this is a representation of its double cover, the Spin group. The normalized product ${\Gamma }_{\ast }$ of all $2k$ matrices squares to one and anticommutes with each ${\Gamma }^{\mu }$, while commuting with every ${\Sigma }^{\mu \nu }$. Its eigenspaces are the two Weyl spinor representations, so the Dirac spinor is reducible under the connected spin group:Each summand is irreducible. The Clifford module itself remains irreducible because an individual ${\Gamma }^{\mu }$ interchanges the summands. In $D=2k+1$ one can use the old ${\Gamma }_{\ast }$ as the additional matrix, with an appropriate phase for its signature. No independent chirality matrix remains: the product of all $2k+1$ matrices is central and fixed to a scalar on an irreducible module. There are two choices of this central sign for the complex Clifford algebra, but each restricts to the same irreducible $2^k$-dimensional spinor representation of the connected spin group. An odd-dimensional spinor has no Weyl splitting. Majorana spinor reality conditions depend additionally on dimension and signature; they must be imposed separately from this complex dimension count.

For the massless field counts, use the rotational part $SO(D-2)$ of the little group and real physical polarizations. A massless p-form gauge field has only transverse, antisymmetric components after its gauge redundancy and equations of motion are imposed, giving $\left(\genfrac{}{}{0ex}{}{D-2}{p}\right)$ polarizations. A graviton has a symmetric traceless transverse tensor, giving $(D-2)(D-1)/2-1=D(D-3)/2$. A massless fermion obeys a Dirac equation which halves its off-shell spinor components. In ten dimensions a Majorana-Weyl spinor has sixteen real components off shell and eight physical components; in nine dimensions a Majorana spinor has sixteen off shell and eight physical components; in eleven dimensions a Majorana spinor has thirty-two off shell and sixteen physical components. A gravitino is a transverse vector-spinor with its gamma trace removed, leaving $(D-3)$ times the corresponding physical spinor dimension. Thus the ten-dimensional Majorana-Weyl gravitino has $7\times 8=56$ polarizations, the nine-dimensional Majorana gravitino has $6\times 8=48$, and the eleven-dimensional Majorana gravitino has $8\times 16=128$.

The common Neveu-Schwarz sector of type IIA supergravity and type IIB supergravity consists of a metric tensor, the Kalb–Ramond field $B_2$, and the dilaton $\varphi$. Their counts are respectively $35$, $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$, and $1$. The Ramond-Ramond sectors and fermions distinguish the theories. In type IIA supergravity the independent bosonic potentials and counts areThere are two Majorana-Weyl gravitini of opposite chirality, giving $56+56=112$, and two Majorana-Weyl dilatini of opposite chirality, giving $8+8=16$. Their total is $128$. With gravitini labelled ${\psi }^{+}$ and ${\psi }^{-}$, the corresponding dilatini have signs $-$ and $+$ respectively. Type IIA is nonchiral.

In type IIB supergravity, the independent bosonic potentials and counts areThe five-form field strength associated with $C_4$ is a self-dual differential form, so only half the seventy transverse four-form components are independent. In Lorentzian ten dimensions the Hodge star operator on five-forms squares to one, making a real self-duality condition possible. The two Majorana-Weyl gravitini have the same chirality and contribute $112$ states. Both Majorana-Weyl dilatini have the opposite chirality to those gravitini and contribute $16$ states. Type IIB is chiral, with $128+128$ bosonic and fermionic polarizations.

For dimensional reduction, retain the massless zero modes on a flat circle, with no flux, gauging or fermion twist. A metric splits into a lower-dimensional metric, one vector and one scalar. An ordinary $p$-form splits into a $p$-form and a $(p-1)$-form, according as it has no compact index or one. This preserves counts by Pascal's identity, $\left(\genfrac{}{}{0ex}{}{D-2}{p}\right)=\left(\genfrac{}{}{0ex}{}{D-3}{p}\right)+\left(\genfrac{}{}{0ex}{}{D-3}{p-1}\right)$.

Reducing type IIA supergravity to nine dimensions gives the following decompositions; each number is a count of physical states:Thus the bosonic maximal nine-dimensional supergravity multiplet contains one graviton, three massless vectors, two massless two-forms, one massless three-form and three real scalars, withEach ten-dimensional Majorana-Weyl gravitino becomes one nine-dimensional Majorana gravitino and one Majorana spin-one-half field, giving $56=48+8$. Each ten-dimensional dilatino gives one more Majorana spin-one-half field. The fermionic multiplet therefore has two gravitini and four spin-one-half fields, with $2(48)+4(8)=128$. The fields belong to one maximal gravity supermultiplet, not separate interacting matter multiplets.

Reducing type IIB supergravity gives $g_{\mu \nu },A_{\mu }^{\mathrm{metric}},\rho$ from the metric; one two-form and one vector each from $B_2$ and $C_2$; and the two scalars $\varphi ,C_0$. The four-form potential yields a nine-dimensional four-form and three-form, but the ten-dimensional self-dual differential form condition relates their field strengths. Retaining either one gives $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=\left(\genfrac{}{}{0ex}{}{7}{4}\right)=35$ independent polarizations, not seventy. The same bosonic count, three scalars, three vectors, two two-forms and one three-form, follows. The fermionic reduction likewise gives two gravitini and four spin-one-half fields. These are the same maximal nine-dimensional supergravity spectrum, as expected from T-duality. The spectrum is nonchiral in nine dimensions, since the odd-dimensional spinor representation has no independent Weyl chirality; the ten-dimensional distinction is lost on restriction to nine-dimensional Lorentz symmetry.

Eleven-dimensional supergravity has a metric with $11(8)/2=44$ states, a three-form with $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, and a Majorana gravitino with $128$. For direct reduction on a flat two-torus, write the internal indices as $i=1,2$. The metric gives $g_{\mu \nu }$, two vectors $g_{\mu i}$ and three symmetric components $g_{ij}$; the three-form gives $A_{\mu \nu \rho }$, two two-forms $A_{\mu \nu i}$ and one vector $A_{\mu 12}$. HenceThe eleven-dimensional Majorana spinor restricts to two nine-dimensional Majorana spinors. The vector-spinor consequently yields two gravitini and four spin-one-half fields, giving $128=2(48)+4(8)$. This is again the same $128+128$ maximal nine-dimensional multiplet.

For reduction of massless type IIA supergravity to four dimensions, use a flat six-torus and retain all zero modes, without flux or projections. Internal indices $i=1,\dots ,6$ giveThe multiplicities $21,15,20$ come respectively from a symmetric pair of six internal indices, $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)$. Before dualizing, this gives one graviton, $28$ vectors, $63$ scalars, seven two-forms and one three-form. The seven two-forms provide seven scalar polarizations, whereas the three-form has no local propagating polarization. Thus the final bosonic spectrum is one graviton, $28$ vectors and $70$ scalars, with $2+56+70=128$ states.

A ten-dimensional Majorana-Weyl spinor decomposes into four four-dimensional Majorana spinors. Each ten-dimensional gravitino supplies four four-dimensional gravitini and twenty-four spin-one-half fields, with $56=4(2)+24(2)$. Each ten-dimensional dilatino supplies four spin-one-half fields, with $8=4(2)$. Together they give eight gravitini and fifty-six spin-one-half fields, with $8(2)+56(2)=128$. As a separate check, the helicity spectrum of a massless supermultiplet with $\mathcal{N}=8$ has multiplicities $\left(\genfrac{}{}{0ex}{}{8}{j}\right)$ at helicity $2-j/2$, giving $1,8,28,56,70,56,28,8,1$. The reduced theory has the field content of four-dimensional N=8 supergravity, with $128+128$ physical states, exactly as in ten and eleven dimensions. Other compact manifolds or projections can reduce the number of preserved supercharges; the flat-torus assumption is essential to this spectrum.

Here is the local two-form scalar duality including its coupling dependence. For this calculation use signature $(-+++)$, $H_{\mu \nu \rho }=3{\partial }_{[\mu }{B}_{\nu \rho ]}$, and the contravariant volume tensor ${ϵ}^{0123}=+1$. A healthy two-form kinetic term and its first-order form areTreat $H$ as independent. Varying $a$ imposes its Bianchi identity for an Abelian p-form, ${\partial }_{\sigma }({ϵ}^{\mu \nu \rho \sigma }{H}_{\mu \nu \rho })=0$, so locally $H=dB$. Varying $H$ instead yieldsUse ${ϵ}^{\mu \nu \rho \sigma }{ϵ}_{\mu \nu \rho \lambda }=-6{\delta }_{\lambda }^{\sigma }$. The original kinetic term becomes $+(g_B^2/2)(\partial a{)}^2$, and the multiplier term becomes $-g_B^2(\partial a{)}^2$. Both terms must be substituted; replacing $H$ only in the original kinetic term would produce the wrong sign. ThusAt fixed normalization of the Bianchi multiplier, the dual kinetic coefficient is the inverse of the original coefficient: if the scalar convention is $-\frac{1}{2g_a^2}(\partial a{)}^2$, then $g_a=1/g_B$. A canonical scalar is $g_{B}a$ for constant $g_B$, but that rescaling hides the formal coupling inversion and changes any assigned scalar periodicity. For a scalar-dependent positive kinetic matrix ${\mathcal{G}}_{IJ}$ of several two-forms, the same calculation gives $-\frac{1}{2}({\mathcal{G}}^{-1}{)}^{IJ}\partial a_I\partial a_J$. Additional topological couplings modify the multiplier's derivative terms but do not change this basic inversion of the two-form kinetic matrix. The Hodge star operator exchanges the two-form Euler-Lagrange equations and Bianchi identities for an Abelian p-form with those of the dual scalars; the duality is local and global flux sectors require separate treatment.

A four-dimensional three-form instead has $F_4=d{C}_3=f{\mathrm{vol}}_4$. Its equation of motion, $d(g_3^{-2}\ast F_4)=0$, makes $f/g_3^2$ a spacetime constant. For constant coupling the first-order expression in this convention can be writtenHere the first-order construction includes a multiplier $q(F_4-d{C}_3)$: varying $C_3$ imposes $dq=0$, and the displayed density retains the term fixing the flux normalization. Eliminating $f$ gives $f=g_3^{2}q$ and ${\mathcal{L}}_{\mathrm{eff}}=-g_3^2{q}^2/2$. Thus it can encode a constant-flux contribution to the vacuum energy, but no local massless particle or scalar wave. On the zero-flux perturbative vacuum it contributes no state. The seven two-forms are scalar duals; the three-form is nondynamical, which completes the four-dimensional field count without discarding a possible global flux parameter.

On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.

Although the problem allows us to assume decay, it can be established independently. By compactness and uniform continuity, approximate $f$ uniformly by $h$ constant on residue classes modulo $p^r$. On this finite-dimensional space, $S^{p^r}=I$ andThe matrix of ${\Delta }^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\Vert {\Delta }^{j{p}^r+s}h{\Vert }_{\infty }\le p^{-j}\Vert h{\Vert }_{\infty }$ for $0\le s<p^r$. Thus $a_n(h)\to 0$. Since $|a_n(f)-a_n(h){|}_p\le \Vert f-h{\Vert }_{\infty }$, arbitrary uniform approximation proves automatic decay of Mahler coefficients.

For the last claim, construct the discrete antidifferentiation on the p-adic integersIts coefficient sequence is $0,a_0,a_1,\dots$, still tending to zero, so it is continuous. By Pascal's identity, $\Delta B_{n+1}=B_n$. The boundedness of $\Delta$ permits applying it to the uniform limit, giving $\Delta G=f$. For the stated linear map, invariance under translation by one now givesThus translation-invariant linear forms on p-adic continuous functions vanish:No continuity of $T$ has been assumed or used. In particular, one must not justify this by applying $T$ termwise to an infinite Mahler expansion; it is the existence of a continuous discrete antiderivative that makes the conclusion valid.

Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.

The diagonal Ramsey number $R(s)$ is the least positive integer $N$ such that every red-blue edge colouring of the complete graph $K_N$ contains a monochromatic copy of $K_s$.

For the existence and bound, write $R(r,s)$ for the least $N$ forcing either a red $K_r$ or a blue $K_s$. Given a vertex $v$ of $K_{R(r-1,s)+R(r,s-1)}$, the pigeonhole principle says that at least $R(r-1,s)$ incident edges are red or at least $R(r,s-1)$ are blue. In the first case those red neighbours contain a red $K_{r-1}$, which joins to $v$ to make a red $K_r$, or a blue $K_s$; the second case is symmetric. HenceStarting from $R(1,s)=R(r,1)=1$ and using Pascal's identity gives the binomial upper bound for a Ramsey numberThereforewhich proves both existence and the required estimate.

To determine $R(3)$, color a cycle graph $C_5$ red and its complement graph, also a $C_5$, yellow. This colors $K_5$ without a monochromatic triangle, so $R(3)>5$. At any vertex of a red-yellow $K_6$, at least three of the five incident edges have one color. If an edge among their three other endpoints has that color, it completes a monochromatic triangle with the chosen vertex; if none does, those three endpoints themselves form a triangle of the other color. Thus $R(3)\le 6$, and

The same local argument proves uniqueness on $K_5$. No vertex can have three incident edges of one color: their other endpoints would either contain an edge of that color or form a triangle of the other color. Every vertex therefore has vertex degree at most two in each color, and because the two degrees sum to four, each is exactly two. Each color class is consequently a regular graph of degree two on five vertices, hence a single $C_5$. Up to relabelling, the coloring is exactly the cycle-and-complement coloring above.

Finally, the monochromatic-triangle counting formula gives, for a coloring of $K_6$ with red degrees $d_v$,The bound is attained by coloring all edges of $K_{3,3}$ red and the two complementary copies of $K_3$ yellow: there are no red triangles and exactly two yellow triangles. The least requested number is therefore

Define the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ as the number of $r$-element subsets of an $n$-element set, equivalently as $n!/[r!(n-r)!]$. Partitioning the $r$-element subsets of $\{1,\dots ,n+1\}$ according to whether they contain $n+1$ proves Pascal's identityThus the forward difference operator satisfies

For $0\le k\le n$, the binomial coefficient isWhen $1\le k\le n-1$,This is Pascal's identity.

Let $D(a)$ denote the stated constant term and suppose every $a_i>0$. The rational identitycan be verified after clearing denominators, or by Lagrange interpolation. Multiplying it by the Dyson product and taking constant terms gives the recursionThe multinomial coefficientobeys the same recursion by the multinomial form of Pascal's identity.

If $a_k=0$, taking the constant term in $X_k$ forces the zero term from every factor involving $X_k$ and reduces the expression to the $(n-1)$-variable Dyson product with $a_k$ omitted. The same boundary reduction holds for $M(a)$. Finally $D(0,\dots ,0)=M(0,\dots ,0)=1$. Induction on $n$ and on $a_1+\cdots +a_n$ therefore proves the Dyson constant-term identity

Every positive integer $m$ has a unique binomial representationDefineThe Kruskal-Katona theorem says that every $r$-uniform family $\mathcal{A}$ of size $m$ hasand the initial segment of length $m$ in colexicographic order attains equality.

We prove the inequality by simultaneous mathematical induction on the ground-set size and on $r$. We use the elementary binomial-shadow arithmetic lemmaTo verify the lemma, greedily remove the largest term $\left(\genfrac{}{}{0ex}{}{q}{r}\right)$ from the binomial representation of $a+b$. If $a$ reaches that term first, apply the induction hypothesis to the remainders; otherwise transfer the excess to $b$. In the crossing case Pascal's identitygives exactly the two terms on the right. This induction also proves uniqueness of the greedy representation.

Choose the largest ground-set element $x$ and splitThe members of the lower shadow that avoid $x$ form $\partial {\mathcal{A}}_0\cup {\mathcal{A}}_1$, while those containing $x$ are the sets $B\cup \{x\}$ with $B\in \partial {\mathcal{A}}_1$. Hence, writing $a=|{\mathcal{A}}_0|$ and $b=|{\mathcal{A}}_1|$,This completes the induction.

Finally, the initial colex segment first contains all $r$-subsets of $[a_r]$, followed recursively by sets containing $a_r+1$ whose remaining elements form the appropriate initial segment at level $r-1$. Its shadow has the same recursive decomposition, and therefore has exactly $f_r(m)$ members. This proves sharpness.

Suppose $\mathcal{A},\mathcal{B}\subseteq [n{]}^{(r)}$ are cross-intersecting families. The iterated upper shadow ${\nabla }_{n-r}\mathcal{A}$ is disjoint frombecause $A\subseteq [n]\setminus B$ would mean $A\cap B=\varnothing$. HenceIf $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$, the upper-shadow form of the Kruskal-Katona theorem givesIt follows from Pascal's identity that $|\mathcal{B}|<\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$. Thus the two sizes cannot both exceed that number.

The Erdős-Ko-Rado theorem says that if $n\ge 2r$ and $\mathcal{A}\subseteq [n{]}^{(r)}$ is an intersecting family, thenThe star of all $r$-sets containing one fixed point attains equality.