Three cases on the ordinary exceptional plane

Proof. The degree-one diagonal space contains $$1,t,a,b,t^2,ta,tb,ab,tQ_2,bQ_2,aQ_3,Q_2Q_3, D:=t(Q_3-Q_2).$$ Their leading exponents, in coordinate order $(t,a,b)$, are the thirteen vectors $p_0,\ldots,p_{12}$ below; each occurs once. Put $R=\beta b^2+\gamma bt-\alpha\kappa t^2$. The following degree-two sections give two more certified leading monomials: $$\begin{aligned} G_5&=(tb)D-\alpha t^2(tQ_2)=t^3R, &\operatorname{in}(G_5)&=\beta t^3b^2,\\ H_6&=D(bQ_2+\alpha tQ_2)-\alpha t^2(Q_2Q_3)=t^2Q_2R, &\operatorname{in}(H_6)&=\beta t^2ab^3. \end{aligned}$$ Take the convex hull $P$ of the following normalized vectors (the last two are divided by degree two): $$\begin{array}{c|rrrrrrrrrrrrrrr} i&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\\\hline t&0&0&0&0&0&0&0&1&1&1&1&2&2&1&3/2\\ a&0&0&1&1&1&2&2&0&0&1&1&0&1&1/2&0\\ b&0&1&0&1&2&1&2&0&1&0&1&0&0&3/2&1 \end{array}$$ An exact facet description is $$\begin{gathered} t,a,b\ge0,\qquad |a-b|\le1,\\ t+2a\le4,\quad 2t+b\le4,\quad t+2b\le4,\\ 5t+2a+4b\le12,\qquad4t+a+3b\le9. \end{gathered}$$ For a reproducible rational integration certificate, take $c=(1/2,1/2,1/2)$ and cone the following boundary triangles to $c$: $$\begin{array}{c|rc|r} (i,j,k)&\operatorname{vol}[c,p_i,p_j,p_k] &(i,j,k)&\operatorname{vol}[c,p_i,p_j,p_k]\\\hline (12,2,5)&1/3 &(12,6,5)&5/12\\ (14,12,11)&5/24 &(13,6,4)&5/24\\ (13,12,6)&13/24 &(13,14,12)&5/24\\ (1,2,0)&1/12 &(1,6,5)&1/6\\ (1,6,4)&1/12 &(1,2,5)&1/6\\ (12,2,0)&1/6 &(12,11,0)&1/6\\ (14,1,0)&1/8 &(14,11,0)&1/6\\ (13,1,4)&1/6 &(13,14,1)&1/8 \end{array}$$ Each tetrahedron has moment equal to its volume times $(c+p_i+p_j+p_k)/4$. Summation gives $$\operatorname{vol}(P)=\frac{10}{3},\qquad \int_P(t,a,b)\,dV=\left(\frac73,\frac{253}{96},\frac{81}{32}\right).$$ The semigroup generated by these fifteen certified monomials has full lattice: it contains $(1,0,0,0)$ and its translates by each coordinate unit vector. Consequently its degree-$m$ cardinality is $\operatorname{vol}(P)m^3+O(m^2)$ and its exponent sums have leading term $m^4\int_P(t,a,b)\,dV$. These conclusions follow either from the conductor of an affine semigroup in its saturation or from lattice-point counting. The full initial space has dimension $N_m=(10/3)m^3+O(m^2)$, so at most $O(m^2)$ initial monomials are missing from this generated semigroup. All exponents are bounded by $4m$ because the original generators have ordinary degree at most two in each factor. Their contribution to exponent sums is therefore $O(m^3)$. This proves the asserted asymptotic means without asserting a complete SAGBI basis. ◻

Proof. Normalize as before, obtaining $Q_2=ab+\kappa t^2$, $Q_3=Q_2+t(\alpha a+\beta b+\gamma t)$. The case $\alpha\beta\ne0$ is proved above. If exactly one of $\alpha,\beta$ is nonzero, exchange the projection labels and the coordinates $a,b$ if necessary, and perform the triangular normalization again, so that $\alpha\ne0$, $\beta=0$.

Retain the thirteen degree-one monomials of the first lemma, and use its same explicit sections $G_5,H_6$. Now $R=\gamma bt-\alpha\kappa t^2$. If $\gamma\ne0$, their leading exponents are $(4,0,1)$ and $(3,1,2)$. If $\gamma=0$, coprimality forces $\kappa\ne0$: otherwise the bihomogeneous $q_2,q_3$ share their $x_3$ factor. In this case their leading exponents are $(5,0,0)$ and $(4,1,1)$.

Finally, if $\alpha=\beta=0$, coprimality forces $\gamma\ne0$. Then $D=\gamma t^3$ replaces the pivot $at^2$ by $t^3$; no degree-two pivots are needed. Exact rational triangulation gives the following volumes and centroids; the degree-two exponents in the middle column are divided by two when taking convex hulls. $$\begin{array}{c|c|c|c} \text{condition}&\text{additional exponents} &\operatorname{vol}(P)&(\mu_t,\mu_a,\mu_b)\\\hline \alpha\beta\ne0 &(3,0,2),(2,1,3)&10/3 &(7/10,253/320,243/320)\\ \alpha\ne0,\ \beta=0,\ \gamma\ne0 &(4,0,1),(3,1,2)&10/3 &(231/320,253/320,59/80)\\ \alpha\ne0,\ \beta=\gamma=0,\ \kappa\ne0 &(5,0,0),(4,1,1)&10/3 &(119/160,253/320,229/320)\\ \alpha=\beta=0,\ \gamma\ne0 &at^2\text{ replaced by }t^3&10/3 &(31/40,59/80,59/80) \end{array}$$ The vertices are the displayed thirteen degree-one vectors, with the two additional exponents divided by two in the first three rows. In the last row replace $(2,1,0)$ by $(3,0,0)$. Formula (34.1) computes each entry by coning the boundary facets; additional facet certificates are recorded just below. The same full-lattice and Hilbert-mass argument proves that these are the actual asymptotic lexicographic moments. Every displayed coordinate is at most $253/320$, and the monomial-degeneration lemma applies. ◻

Proof. Thirteen degree-one diagonal sections are $$1,t,a,b,t^2,ta,tb,ab,tQ_2,bQ_2,aQ_3,Q_2Q_3,D:=tQ_3.$$ Their leading exponents are distinct. In coordinate order $(t,a,b)$, the sorted list is $$\begin{aligned} p_0,\ldots,p_{12}=\big(& (0,0,0),(0,0,1),(0,1,0),(0,1,1), (0,1,2),(1,0,0),(1,0,1),\\ &(1,1,0),(1,1,1),(1,2,0),(1,2,1),(2,0,0),(2,1,0)\big). \end{aligned}$$ Set $R=Cb^2+\gamma bt-\kappa t^2$. Two degree-two diagonal sections are certified by the polynomial identities $$\begin{aligned} G&=(tb)D-t^2(tQ_2)=t^3R,\\ J&=(bQ_2)D-(tQ_2)^2=t^2Q_2R. \end{aligned}$$ Their leading exponents, divided by degree two, are as follows: $$\begin{array}{c|cc} \text{condition}&p_{13}=\tfrac12\exp\operatorname{in}(G) &p_{14}=\tfrac12\exp\operatorname{in}(J)\\\hline C\ne0&(3/2,0,1)&(1,1/2,3/2)\\ C=0,\ \gamma\ne0&(2,0,1/2)&(3/2,1/2,1)\\ C=\gamma=0,\ \kappa\ne0&(5/2,0,0)&(2,1/2,1/2). \end{array}$$ Let $P$ be the convex hull of $p_0,\ldots,p_{14}$. For completeness, all three polytopes have the common inequalities $$t,a,b\ge0,\qquad a-t\le1,\qquad b-a\le1,\qquad t+a\le3.$$ Their additional facet inequalities are respectively $$\begin{array}{c|l} C\ne0&2t+b\le4,\quad 2t+a+3b\le7,\quad4t+a+3b\le9;\\ C=0,\gamma\ne0&t-a+2b\le3,\quad t\le2,\quad3t+a+2b\le7, \quad3t+a+4b\le9;\\ C=\gamma=0,\kappa\ne0&2t-a+3b\le5,\quad2t+a+b\le5, \quad4t+a+5b\le11. \end{array}$$ Exact integration gives $$\begin{array}{c|c|ccc} \text{condition}&\operatorname{vol}(P)&\int_Pt&\int_Pa&\int_Pb\\\hline C\ne0&10/3&65/24&253/96&69/32\\ C=0,\gamma\ne0&10/3&89/32&253/96&25/12\\ C=\gamma=0,\kappa\ne0&10/3&137/48&253/96&193/96. \end{array}$$ The facets supply a rational integration certificate. Alternatively, cone each facet triangulation to $(1/2,1/2,1/2)$ and use $\operatorname{vol}[c,p_i,p_j,p_k]= |\det(p_i-c,p_j-c,p_k-c)|/6$; the first moment of this tetrahedron is its volume times $(c+p_i+p_j+p_k)/4$.

The generated affine semigroup has full lattice, since the degree-one list includes the origin and all three coordinate unit vectors. Its degree-$m$ cardinality is therefore $(10/3)m^3+O(m^2)$, and its exponent sums have leading term $m^4\int_P(t,a,b)$. All these semigroup elements occur as leading monomials of products of the fifteen displayed sections. The full initial space has the same leading Hilbert coefficient, $N_m=(10/3)m^3+O(m^2)$, so there are only $O(m^2)$ additional leading monomials. Every exponent is bounded by $4m$, and consequently these extra monomials contribute only $O(m^3)$ to exponent sums. Dividing the displayed moments by $10/3$ proves the asserted means. The stated threshold follows from the independently justified monomial degeneration lemma. ◻

Proof. Put $U=\beta a+b$, $V=a+\gamma b$, $P=bU$, and $R=aV$. The binary quadrics $P$ and $R$ are coprime: neither $a$ divides $P$ nor $b$ divides $R$, and the remaining factors $U,V$ are independent precisely when $\beta\gamma\ne1$.

Since $tQ_2,tQ_3$ are degree-one diagonal sections, their span contains $t^2a,t^2b$. If $\beta\ne0$, thirteen degree-one pivot exponents are $$\begin{aligned} p_0,\ldots,p_{12}=\big(& (0,0,0),(0,0,1),(0,1,0),(0,1,1), (1,0,0),(1,0,1),(1,1,0),\\ &(1,1,1),(1,2,0),(2,0,0),(2,0,1),(2,1,0),(2,2,0)\big). \end{aligned}$$ They come from $1,a,b,ab,t,ta,tb,t^2,t^2a,t^2b,bQ_2,aQ_3,Q_2Q_3$.

We next certify an entire sequence of sections rather than assume that a finite list forms a SAGBI basis. For every $m\ge1$, the degree-$m$ diagonal space contains $$t^m\,\langle a,b\rangle\operatorname{Sym}^{m-1}\langle P,R\rangle.$$ Indeed, for $0\le q\le m-1$, the section $t^m aR^{m-1-q}P^q$ has $$(r,s,h,i,k)=(1,m-q,q,q,m-1-q),$$ and the analogous section with $b$ has $$(r,s,h,i,k)=(1,m-1-q,q+1,q,m-1-q).$$ Both satisfy the defining inequalities.

These $2m$ binary forms are linearly independent. To see this, set $z=a/b$. The rational map $z\mapsto R/P$ has degree two because $R,P$ are coprime binary quadrics, so $[\mathbb C(z):\mathbb C(R/P)]=2$. A linear dependence between $aR^{m-1-q}P^q$ and $bR^{m-1-q}P^q$, after dividing by $bP^{m-1}$, would give $zA(R/P)+B(R/P)=0$. This forces $A=B=0$. Thus the displayed subspace is all binary forms of degree $2m-1$. In particular, $$t^m b^{2m-1}\in W_m \qquad\text{for every }m\ge1.$$ Its normalized exponent tends to $p_{13}=(1,0,2)$.

Let $P_\infty=\operatorname{conv}\{p_0,\ldots,p_{13}\}$. An exact facet description is $$0\le t\le2,\quad a,b\ge0,\quad a-t\le1,\quad b-t\le1, \quad a+b\le2,\quad2t+a+2b\le6.$$ Direct rational integration gives $$\operatorname{vol}(P_\infty)=10/3,\qquad \int_{P_\infty}(t,a,b)\,dV=(10/3,53/24,47/24).$$

Here is a precise volume argument allowing the limit point. For fixed $M$, use the thirteen degree-one sections and the certified section $t^Mb^{2M-1}$. Their finitely generated semigroup has full lattice, and its body is $$P_M=\operatorname{conv}\{p_0,\ldots,p_{12},(1,0,2-1/M)\}.$$ The bodies $P_M$ increase to $P_\infty$ up to closure. The degree-$m$ semigroup cardinality is $\operatorname{vol}(P_M)m^3+O_M(m^2)$, and its exponent sums have leading term $m^4\int_{P_M}(t,a,b)$. The full initial space has cardinality $(10/3)m^3+O(m^2)$. Since every exponent is at most $4m$, the difference between its exponent sums after division by $m^4$ and those of the fixed semigroup is at most $4(10/3-\operatorname{vol}(P_M))+o_M(1)$ in each coordinate. First let $m$ tend to infinity and then let $M$ tend to infinity. This proves existence of the full asymptotic means and gives $(1,53/80,47/80)$.

If $\beta=0$, the thirteen degree-one pivot exponents instead are $$\begin{aligned} \big(& (0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0), (1,0,1),(1,0,2),\\ &(1,1,0),(1,2,0),(2,0,0),(2,0,1),(2,1,0),(2,1,1)\big). \end{aligned}$$ Their convex hull already has volume $10/3$, with moments $(10/3,25/12,25/12)$. Its facets are $$0\le t\le2,\quad a,b\ge0,\quad a-t\le1,\quad b-t\le1, \quad a+b\le2,\quad t+a\le3,\quad t+b\le3.$$ The ordinary finite-semigroup volume argument proves the second row. This includes $\beta=\gamma=0$ without any additional argument. ◻