Integral facet displacements with closed-alcove dominant representatives

In types $A_n,D_n,E_6,E_7$ there is a nonzero minuscule fundamental weight $\omega$ with $(\omega ,\theta )=1$. Here is the finite root-data check: the highest-root marks are respectively $(1,\dots ,1)$, $(1,2,\dots ,2,1,1)$, $(1,2,2,3,2,1)$ and $(2,2,3,4,3,2,1)$ in the standard diagram numbering. The $A,D$ vectors follow directly from $e_1-e_{n+1}$ and $e_1+e_2$ in their coordinate models. For $E_6,E_7$ use the first six or seven simple vectors of the $E_8$ realization specified in the Statement: summing with the stated marks gives respectively $(1,1,1,1,1,-1,-1,1)\mathrm{/}2$ and $-e_7+e_8$, actual roots in that realization, and pairing with the respective simple vectors is nonnegative. They are therefore the highest roots, since a simply-laced irreducible root system has a single dominant root in its root orbit. Every positive root is at most this highest root in the simple-root order, so a fundamental weight at a mark-one node has all positive-root pairings at most one, proving the minuscule assertion and $(\omega ,\theta )=1$.

For a finite facet $i$ form the full-rank finite root subsystem with base $\{-\theta \}\cup \{{\alpha }_j:j\mathrm{\ne }i\}$. Independence follows from $a_i>0$ in the highest-root relation; pairwise off-diagonal inner products are nonpositive, and these roots generate a finite reflection subgroup of $W$. Choose a vector $z$ pairing positively with all these independent roots, and minimize $(w\omega ,z)$ over their finite generated reflection subgroup. Call a minimizer ${\eta }_i$. If it paired positively with one of those roots, reflecting in that root would strictly lower its pairing with $z$, a contradiction. Thus $({\eta }_i,{\alpha }_j)\le 0$ for $j\mathrm{\ne }i$ and $({\eta }_i,\theta )\ge 0$. The remaining pairing $({\eta }_i,{\alpha }_i)$ is positive: if it were nonpositive, all simple pairings and their positive-mark sum would force every pairing zero, hence ${\eta }_i=0$, impossible. By [F2] it is exactly one. Set $r=m_i+1$ and $\lambda =\mu -r{\eta }_i$. Its $i$-th Dynkin coordinate is $-1$, all other finite coordinates are nonnegative, and its highest-root pairing has not increased. Therefore $\lambda +\rho$ lies on exactly the $i$-th finite facet, and $\lambda$ is in its relative interior. The dominant displacement is $r\omega$, whose highest-root pairing is $r\le k+1$.

For the upper facet in the same types put $r=k+1-(\mu ,\theta )>0$ and $\lambda =\mu +r\omega$. Every finite Dynkin coordinate stays nonnegative and the highest-root pairing becomes $k+1$, so $\lambda$ is in exactly that facet's relative interior. The dominant displacement is the dominant representative of $-r\omega$, whose highest-root pairing is again $r$ (the Weyl involution $-w_0$ preserves the highest root). Since $r\le k+1$, it lies in the required closed alcove.

It remains to handle $E_8$, where $h=30$, $k$ is positive odd, and the marks are $(2,3,4,6,5,4,3,2)$. Use its root model consisting of $\pm e_i\pm e_j$ and half sums with an even number of minus signs. Fix ${\alpha }_1=(e_1-e_2-e_3-e_4-e_5-e_6-e_7+e_8)\mathrm{/}2$, ${\alpha }_2=e_1+e_2$, ${\alpha }_3=e_2-e_1$, and ${\alpha }_j=e_{j-1}-e_{j-2}$ for $4\le j\le 8$. Their Cartan pairings give the $E_8$ diagram and the marked sum is $\theta =e_7+e_8$, whose Dynkin coordinates are $(0,0,0,0,0,0,0,1)$. For finite facets $i=1,\dots ,8$ take respectively the roots ${\eta }_1=-e_7+e_8,{\eta }_2=(e_1+e_2+e_3+e_4+e_5-e_6-e_7+e_8)\mathrm{/}2,{\eta }_3=(-e_1+e_2+e_3+e_4-e_5-e_6-e_7+e_8)\mathrm{/}2,{\eta }_4=-e_1-e_2,{\eta }_5=e_4+e_8,{\eta }_6=e_5+e_8,{\eta }_7=e_6+e_8,{\eta }_8=e_7+e_8\mathrm{.}$ Their Dynkin coordinate vectors, obtained by the displayed dot products, are respectively $(1,0,0,0,0,0,0,-1)$, $(0,1,0,0,0,0,-1,0)$, $(0,0,1,0,0,-1,0,0)$, $(0,-2,0,1,0,0,0,0)$, $(0,0,0,0,1,-1,0,0)$, $(0,0,0,0,0,1,-1,0)$, $(0,0,0,0,0,0,1,-1)$ and $(0,0,0,0,0,0,0,1)$. Thus each has pairing one at node $i$, nonpositive pairings elsewhere, and nonnegative highest-root pairing (respectively $0,0,0,0,1,1,1,2$). Set $r=m_i+1$ and $\lambda =\mu -r{\eta }_i$. The same coordinate inequalities as step 2.1 place $\lambda$ in the required finite facet's relative interior. Every root is Weyl conjugate to $\theta$, so the dominant displacement is $r\theta$. Because $a_i\ge 2$ and $k$ is odd, $m_i\le (k-1)\mathrm{/}2$, hence $(r\theta ,\theta )=2r\le k+1$.

For the $E_8$ upper facet let $r=k+1-(\mu ,\theta )$. If $r$ is even, take $\lambda =\mu +(r\mathrm{/}2)\theta$. It is dominant, is in that facet's relative interior, and its dominant displacement has highest-root pairing $r\le k+1$. If $r$ is odd, $(\mu ,\theta )$ is odd since $k+1$ is even, so $m_j>0$ for some odd-mark node $j\in \{2,5,7\}$. Choose respectively ${\beta }_2=-e_1+e_8$, ${\beta }_5=e_3+e_8$, ${\beta }_7=e_5+e_8$. Their Dynkin coordinate vectors are $(0,-1,1,0,0,0,0,0)$, $(0,0,0,1,-1,0,0,0)$ and $(0,0,0,0,0,1,-1,0)$, and each has highest-root pairing one. With $u=(r-1)\mathrm{/}2$ put $\lambda =\mu +u\theta +{\beta }_j$. Its only potentially negative coordinate is repaired by $m_j\ge 1$, and its highest-root pairing increases by $2u+1=r$, placing it exactly in the upper facet's relative interior. For every root $\alpha$, $\mathrm{\mid }(u\theta +{\beta }_j,\alpha )\mathrm{\mid }\le 2u+2=r+1$, since all root pairings have absolute value at most two. The highest-root pairing of the dominant displacement is the maximum of its pairings with roots, hence at most $r+1$. Oddness of $r$ gives $r\le k$, so this is at most $k+1$. This proves the upper-facet assertion too.

For an arbitrary affine alcove write $\mu =w\cdot {\mu }_0$ with ${\mu }_0$ in the bottom alcove and transport the constructed facet point by the same dot action. The difference of the two transported points is the finite linear Weyl part of $w$ applied to their original difference, so its ordinary dominant representative is unchanged. If the transported facet is inside the strict shifted dominant chamber, each shifted simple pairing at its relative-interior integral point is a positive integer, and subtracting $\rho$ leaves nonnegative Dynkin coordinates. Thus the transported point is dominant exactly as required. No tensor decomposition conclusion has been used or asserted.