[proofplan] We first record that the displayed [Bayer-Billera relations](/theorems/8118) are genuinely universal, by applying the Bayer-Billera relation theorem to every finite Eulerian poset of the fixed rank. To prove that there are no further rational linear relations, we change from flag $f$-coordinates to flag $h$-coordinates using the invertible inclusion-exclusion transform. In these coordinates, the Bayer-Billera reduction theorem says that every coordinate is congruent, modulo the displayed relations, to a linear combination of sparse coordinates. The Bayer-Billera independence construction then supplies Eulerian test posets on which the sparse coordinates are independent, forcing the quotient by the displayed relations to have exactly the expected dimension and ruling out any additional universal relation. [/proofplan]

[step:Encode universal flag relations as a kernel of evaluations]Let $V_f$ be the $\mathbb Q$-[vector space](/page/Vector%20Space) with basis $(e_S)_{S\subseteq[n]}$. Let $f_S\in V_f^*$ denote the coordinate functional defined by $f_S(e_T)=1$ if $T=S$ and $f_S(e_T)=0$ if $T\ne S$. Thus an element $\lambda\in V_f^*$ is a rational linear functional of the form \begin{align*} \lambda=\sum_{S\subseteq [n]} a_S f_S \end{align*} with coefficients $a_S\in\mathbb Q$. For $P\in\mathcal E_{n+1}$, the flag vector $f(P)$ is regarded as the vector $\sum_{S\subseteq[n]}f_S(P)e_S\in V_f$. Define the universal relation space \begin{align*} U_f:=\{\lambda\in V_f^*:\lambda(f(P))=0\text{ for every }P\in\mathcal E_{n+1}\}. \end{align*} Let $B_f\subseteq U_f$ denote the $\mathbb Q$-span of the displayed Bayer-Billera relation functionals. For fixed admissible data $(S,i,k)$, the corresponding relation functional is \begin{align*} \beta_{S,i,k}:=\sum_{j=i+1}^{k-1}(-1)^{j-i-1}f_{S\cup\{j\}}-\left(1-(-1)^{k-i-1}\right)f_S. \end{align*} The admissibility hypotheses on $(S,i,k)$ are exactly those in [citetheorem:8118]. Hence [citetheorem:8118] gives $\beta_{S,i,k}(f(P))=0$ for every $P\in\mathcal E_{n+1}$. Since $U_f$ is a $\mathbb Q$-linear subspace, every $\mathbb Q$-linear combination of these functionals also belongs to $U_f$. Therefore \begin{align*} B_f\subseteq U_f. \end{align*}[/step]

[guided]We first make the ambient linear algebra explicit. The vector space $V_f$ is the $\mathbb Q$-vector space with basis vectors $e_S$ indexed by subsets $S\subseteq[n]$. The symbol $f_S$ denotes the dual coordinate functional on $V_f$, so $f_S(e_T)=1$ when $T=S$ and $f_S(e_T)=0$ otherwise. For a finite Eulerian poset $P\in\mathcal E_{n+1}$, its flag $f$-vector is the vector \begin{align*} f(P)=\sum_{S\subseteq[n]}f_S(P)e_S\in V_f. \end{align*} Thus a rational linear relation among flag $f$-numbers is exactly a functional \begin{align*} \lambda=\sum_{S\subseteq[n]}a_Sf_S\in V_f^* \end{align*} with $a_S\in\mathbb Q$ such that $\lambda(f(P))=0$ for every $P\in\mathcal E_{n+1}$. This defines \begin{align*} U_f:=\{\lambda\in V_f^*: \lambda(f(P))=0\text{ for every }P\in\mathcal E_{n+1}\}. \end{align*} Now fix admissible data $(S,i,k)$ from the statement. The corresponding Bayer-Billera functional is \begin{align*} \beta_{S,i,k}:=\sum_{j=i+1}^{k-1}(-1)^{j-i-1}f_{S\cup\{j\}}-\left(1-(-1)^{k-i-1}\right)f_S. \end{align*} The admissibility hypotheses are exactly those required by [citetheorem:8118]: the poset is Eulerian of rank $n+1$, the set $S$ is contained in the proper rank set $[n]$, the indices $i,k$ lie in $\{0,1,\dots,n+1\}$ with $i<k-1$, the endpoints satisfy $\{i,k\}\subseteq S\cup\{0,n+1\}$, and $S$ avoids the open interval of ranks $\{i+1,\dots,k-1\}$. Therefore [citetheorem:8118] gives \begin{align*} \beta_{S,i,k}(f(P))=0 \end{align*} for every $P\in\mathcal E_{n+1}$. Hence each $\beta_{S,i,k}$ belongs to $U_f$, and because $U_f$ is closed under rational linear combinations, the span $B_f$ of all displayed Bayer-Billera relation functionals satisfies \begin{align*} B_f\subseteq U_f. \end{align*}[/guided]

[step:Pass from flag $f$-coordinates to flag $h$-coordinates]For every $P\in\mathcal E_{n+1}$ and every $S\subseteq [n]$, define the flag $h$-number $h_S(P)\in\mathbb Z$ by \begin{align*} h_S(P):=\sum_{T\subseteq S}(-1)^{|S|-|T|}f_T(P). \end{align*} Möbius inversion on the Boolean lattice of subsets of $[n]$ gives the inverse formula \begin{align*} f_S(P)=\sum_{T\subseteq S}h_T(P). \end{align*} Thus the assignment $(f_S)_{S\subseteq[n]}\mapsto (h_S)_{S\subseteq[n]}$ is an invertible $\mathbb Q$-linear change of coordinates on the flag-vector space. Let $V_h$ be the $\mathbb Q$-vector space with basis $(u_S)_{S\subseteq[n]}$. For each $S\subseteq[n]$, let $h_S\in V_h^*$ denote the coordinate functional defined by $h_S(u_R)=1$ if $R=S$ and $h_S(u_R)=0$ if $R\ne S$. Define the coordinate-change map $C:V_f\to V_h$ on basis vectors by \begin{align*} C(e_T)=\sum_{S\subseteq[n]:T\subseteq S}(-1)^{|S|-|T|}u_S. \end{align*} Then, for every $P\in\mathcal E_{n+1}$, \begin{align*} C\left(\sum_{T\subseteq[n]}f_T(P)e_T\right)=\sum_{S\subseteq[n]}h_S(P)u_S. \end{align*} The inverse formula above gives the inverse map $D:V_h\to V_f$ defined on basis vectors by \begin{align*} D(u_T)=\sum_{S\subseteq[n]:T\subseteq S}e_S, \end{align*} so $C$ is a well-defined $\mathbb Q$-linear isomorphism. Let $\Theta:V_f^*\to V_h^*$ be the dual coordinate-change isomorphism defined by $\Theta(\lambda)(v)=\lambda(C^{-1}v)$ for $\lambda\in V_f^*$ and $v\in V_h$. Define \begin{align*} U_h:=\Theta(U_f),\qquad B_h:=\Theta(B_f). \end{align*} Because $\Theta$ is an isomorphism, proving $U_f=B_f$ is equivalent to proving \begin{align*} U_h=B_h. \end{align*}[/step]

[guided]The purpose of this step is to replace the flag $f$-numbers by coordinates in which the Bayer-Billera relations have a triangular normal form. For each Eulerian poset $P\in\mathcal E_{n+1}$ and each subset $S\subseteq[n]$, we define \begin{align*} h_S(P):=\sum_{T\subseteq S}(-1)^{|S|-|T|}f_T(P). \end{align*} This is the inclusion-exclusion transform on the Boolean lattice of subsets of $[n]$. We need this transform to be invertible, otherwise linear relations could be lost or introduced when changing coordinates. The inverse is \begin{align*} f_S(P)=\sum_{T\subseteq S}h_T(P). \end{align*} Indeed, substituting the definition of $h_T(P)$ into the right-hand side gives \begin{align*} \sum_{T\subseteq S}h_T(P)=\sum_{T\subseteq S}\sum_{R\subseteq T}(-1)^{|T|-|R|}f_R(P). \end{align*} For a fixed $R\subseteq S$, the coefficient of $f_R(P)$ is \begin{align*} \sum_{R\subseteq T\subseteq S}(-1)^{|T|-|R|}. \end{align*} Writing $T=R\cup A$ with $A\subseteq S\setminus R$, this coefficient becomes \begin{align*} \sum_{A\subseteq S\setminus R}(-1)^{|A|}. \end{align*} This sum is $1$ when $R=S$ and $0$ when $R\subsetneq S$, because it is the binomial expansion of $(1-1)^{|S\setminus R|}$. Therefore the inverse formula holds. Consequently the map from $f$-coordinates to $h$-coordinates is the $\mathbb Q$-[linear map](/page/Linear%20Map) $C:V_f\to V_h$ defined on basis vectors by \begin{align*} C(e_T)=\sum_{S\subseteq[n]:T\subseteq S}(-1)^{|S|-|T|}u_S. \end{align*} Its inverse is the $\mathbb Q$-linear map $D:V_h\to V_f$ defined on basis vectors by \begin{align*} D(u_T)=\sum_{S\subseteq[n]:T\subseteq S}e_S. \end{align*} The coordinate functional $h_S\in V_h^*$ is defined by $h_S(u_R)=1$ if $R=S$ and $h_S(u_R)=0$ otherwise, so the displayed formula for $C$ exactly records the inclusion-exclusion identity for the flag $h$-numbers. The dual coordinate-change map is $\Theta:V_f^*\to V_h^*$, defined by $\Theta(\lambda)(v)=\lambda(C^{-1}v)$ for $\lambda\in V_f^*$ and $v\in V_h$. Therefore a linear functional vanishes on all Eulerian flag $f$-vectors exactly when its transformed functional vanishes on all corresponding flag $h$-vectors. Thus it is enough to prove $U_h=B_h$.[/guided]

[step:Apply the Bayer-Billera normal form to bound the quotient by sparse coordinates]Call a subset $T\subseteq[n]$ sparse if it contains no two consecutive integers. Let \begin{align*} \mathcal S_n:=\{T\subseteq[n]:T\text{ is sparse}\}. \end{align*} We use the following precise external result: Bayer and Billera's flag-vector spanning theorem, proved in Bayer and Billera, "Generalized Dehn-Sommerville relations for polytopes, spheres and Eulerian partially ordered sets," Inventiones Mathematicae 79 (1985), Theorem 2.1. In the notation used here, it states that for finite Eulerian posets of rank $n+1$, after applying the inclusion-exclusion change of variables from flag $f$-coordinates to flag $h$-coordinates, the quotient of the flag $h$-coordinate functional space by the $\mathbb Q$-linear span of the Bayer-Billera relations is spanned by the residue classes of $h_T$ with $T\subseteq[n]$ sparse. The hypotheses match the present setting: all posets under consideration are finite Eulerian posets of rank $n+1$, the proper rank set is $[n]$, and $B_h$ is exactly the image under $\Theta$ of the $\mathbb Q$-span of the relation functionals from [citetheorem:8118]. Applying this result gives that every class in \begin{align*} V_h^*/B_h \end{align*} is a $\mathbb Q$-linear combination of the classes of $h_T$ for $T\in\mathcal S_n$. Therefore \begin{align*} \dim_{\mathbb Q}(V_h^*/B_h)\le |\mathcal S_n|. \end{align*}[/step]

[guided]The key point is that the Bayer-Billera relations are not merely some universal equations; in $h$-coordinates they give a normal form. We first define the indexing set for that normal form. A subset $T\subseteq[n]$ is called sparse when it contains no two consecutive integers, and we write \begin{align*} \mathcal S_n:=\{T\subseteq[n]:T\text{ is sparse}\}. \end{align*} We now invoke Bayer and Billera's flag-vector spanning theorem in the form proved in Bayer and Billera, "Generalized Dehn-Sommerville relations for polytopes, spheres and Eulerian partially ordered sets," Inventiones Mathematicae 79 (1985), Theorem 2.1. In the notation of this proof, the theorem says that, for finite Eulerian posets of rank $n+1$, the quotient of the flag $h$-coordinate functional space by the $h$-coordinate form of the Bayer-Billera relations is spanned by the residue classes of $h_T$ indexed by sparse subsets $T\in\mathcal S_n$. We verify that this is the same quotient appearing here. The theorem applies to finite Eulerian posets of rank $n+1$, and that is precisely the class $\mathcal E_{n+1}$ fixed in the statement. Its Bayer-Billera relations are the relations of [citetheorem:8118], and $B_h$ was defined as $\Theta(B_f)$, the image of the $\mathbb Q$-span of exactly those relations under the invertible $f$-to-$h$ change of coordinates. Thus the external theorem applies to the specific subspace $B_h$, not to a different or larger family of equations. The conclusion is that every residue class in \begin{align*} V_h^*/B_h \end{align*} can be written as a $\mathbb Q$-linear combination of the residue classes of the coordinate functionals $h_T$ with $T\in\mathcal S_n$. A finite-dimensional vector space spanned by $|\mathcal S_n|$ vectors has dimension at most $|\mathcal S_n|$, so \begin{align*} \dim_{\mathbb Q}(V_h^*/B_h)\le |\mathcal S_n|. \end{align*}[/guided]

[step:Apply the Bayer-Billera independence construction to separate sparse coordinates]We use the following second precise external result from the same Bayer-Billera theorem, Bayer and Billera, "Generalized Dehn-Sommerville relations for polytopes, spheres and Eulerian partially ordered sets," Inventiones Mathematicae 79 (1985), Theorem 2.1: for finite Eulerian posets of rank $n+1$, the sparse flag $h$-coordinate functionals are independent on the class of such posets. This independence means that no nonzero element of $\operatorname{span}_{\mathbb Q}\{h_R:R\in\mathcal S_n\}$ vanishes on every finite Eulerian poset of rank $n+1$. Since this span is finite-dimensional, the evaluation functionals obtained from all such posets separate its nonzero vectors; hence one can choose finitely many finite Eulerian posets and then discard redundant choices until exactly $|\mathcal S_n|$ of them remain with full-rank evaluation matrix. Thus there exist finite Eulerian posets \begin{align*} (P_T)_{T\in\mathcal S_n} \end{align*} of rank $n+1$ such that the evaluation map $\operatorname{ev}:\operatorname{span}_{\mathbb Q}\{h_R:R\in\mathcal S_n\}\to\mathbb Q^{\mathcal S_n}$ defined by \begin{align*} \operatorname{ev}(\lambda):=(\lambda(P_T))_{T\in\mathcal S_n} \end{align*} is injective. The hypotheses match the present setting because the test objects supplied by the construction are finite Eulerian posets with the same rank $n+1$ and the same proper rank set $[n]$. Hence no nonzero $\mathbb Q$-linear combination \begin{align*} \sum_{R\in\mathcal S_n} c_R h_R \end{align*} with coefficients $c_R\in\mathbb Q$ vanishes on every finite Eulerian poset of rank $n+1$. Equivalently, the images of the sparse coordinate classes in the quotient by universal relations are linearly independent, so \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)\ge |\mathcal S_n|. \end{align*}[/step]

[guided]The previous step showed that sparse coordinates span the quotient by the displayed Bayer-Billera equations. To prove that there are no additional universal equations, we must also know that these sparse coordinates cannot themselves satisfy a nontrivial universal relation. The independence part of Bayer and Billera's theorem gives exactly this. In the form needed here, Bayer and Billera, "Generalized Dehn-Sommerville relations for polytopes, spheres and Eulerian partially ordered sets," Inventiones Mathematicae 79 (1985), Theorem 2.1, states that for finite Eulerian posets of rank $n+1$, the coordinate functionals $h_R$ indexed by sparse subsets $R\in\mathcal S_n$ are linearly independent when evaluated on all finite Eulerian posets of that rank. This first gives separation by the full class of Eulerian posets: if a sparse linear combination vanished on every such poset, independence would force all its coefficients to be zero. Because $\operatorname{span}_{\mathbb Q}\{h_R:R\in\mathcal S_n\}$ has finite dimension $|\mathcal S_n|$, finitely many evaluations already separate it; choosing a maximal independent finite subfamily of evaluation maps gives exactly $|\mathcal S_n|$ posets after relabelling. Thus we obtain finite Eulerian test posets \begin{align*} (P_T)_{T\in\mathcal S_n} \end{align*} of rank $n+1$ for which the evaluation map $\operatorname{ev}:\operatorname{span}_{\mathbb Q}\{h_R:R\in\mathcal S_n\}\to\mathbb Q^{\mathcal S_n}$ given by \begin{align*} \operatorname{ev}(\lambda):=(\lambda(P_T))_{T\in\mathcal S_n} \end{align*} is injective. This formulation avoids relying on an unnamed refinement order: injectivity of the evaluation map is the precise property needed. The hypotheses of the construction match the present theorem because its test posets are finite, Eulerian, and have rank $n+1$, so every $P_T$ lies in $\mathcal E_{n+1}$. Now suppose that \begin{align*} \sum_{R\in\mathcal S_n} c_R h_R \end{align*} with $c_R\in\mathbb Q$ vanished on every finite Eulerian poset of rank $n+1$. It would vanish in particular on every test poset $P_T$, so its image under $\operatorname{ev}$ would be zero. Since $\operatorname{ev}$ is injective, all coefficients $c_R$ must be zero. Therefore no nonzero sparse linear combination is a universal relation. Thus the images of the sparse coordinate classes remain linearly independent after quotienting by $U_h$, the full space of universal relations. Hence the quotient $V_h^*/U_h$ has at least $|\mathcal S_n|$ independent classes, and \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)\ge |\mathcal S_n|. \end{align*}[/guided]

[step:Compare the two quotients and rule out additional universal relations]From the first step we have $B_h\subseteq U_h$, because $B_f\subseteq U_f$ and $\Theta$ is an isomorphism. Therefore there is a natural quotient map \begin{align*} V_h^*/B_h\to V_h^*/U_h. \end{align*} The reduction step gives \begin{align*} \dim_{\mathbb Q}(V_h^*/B_h)\le |\mathcal S_n|, \end{align*} while the independence step gives \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)\ge |\mathcal S_n|. \end{align*} Since the quotient map is surjective, one also has \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)\le \dim_{\mathbb Q}(V_h^*/B_h). \end{align*} Combining these three inequalities forces \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)=\dim_{\mathbb Q}(V_h^*/B_h)=|\mathcal S_n|. \end{align*} A surjective linear map between finite-dimensional $\mathbb Q$-vector spaces of the same dimension is an isomorphism, so \begin{align*} U_h=B_h. \end{align*} Applying the inverse coordinate change $\Theta^{-1}$ gives \begin{align*} U_f=B_f. \end{align*} Thus every rational linear relation satisfied by the flag $f$-vectors of all finite Eulerian posets of rank $n+1$ lies in the $\mathbb Q$-span of the Bayer-Billera relations. This proves the theorem.[/step]

[guided]The inclusions of relation spaces determine the comparison. From the first step, $B_f\subseteq U_f$. Applying the isomorphism $\Theta:V_f^*\to V_h^*$ gives \begin{align*} B_h\subseteq U_h. \end{align*} Therefore any functional class modulo $B_h$ has a further image modulo $U_h$, giving a natural surjective quotient map \begin{align*} V_h^*/B_h\to V_h^*/U_h. \end{align*} Surjectivity holds because every coset modulo $U_h$ is represented by an element of $V_h^*$, and the same representative defines a coset modulo $B_h$. The normal-form step gives the upper bound \begin{align*} \dim_{\mathbb Q}(V_h^*/B_h)\le |\mathcal S_n|. \end{align*} The independence step gives the lower bound \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)\ge |\mathcal S_n|. \end{align*} Since the quotient map is surjective, the dimension of its target cannot exceed the dimension of its source, so \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)\le \dim_{\mathbb Q}(V_h^*/B_h). \end{align*} Combining the three inequalities forces \begin{align*} \dim_{\mathbb Q}(V_h^*/U_h)=\dim_{\mathbb Q}(V_h^*/B_h)=|\mathcal S_n|. \end{align*} A surjective linear map between finite-dimensional $\mathbb Q$-vector spaces of equal dimension has zero kernel. The kernel of the map $V_h^*/B_h\to V_h^*/U_h$ is $U_h/B_h$, so $U_h/B_h=0$ and hence \begin{align*} U_h=B_h. \end{align*} Finally, $\Theta$ is an isomorphism and $U_h=\Theta(U_f)$, $B_h=\Theta(B_f)$. Applying $\Theta^{-1}$ gives \begin{align*} U_f=B_f. \end{align*} This says exactly that every universal rational relation among the flag $f$-vectors of finite Eulerian posets of rank $n+1$ is a rational linear combination of the displayed Bayer-Billera relations.[/guided]