Degree and codegree of standard tableaux

Counting addable and removable nodes (1). Deleting the removable node $A$ of residue $i$ makes $A$ addable. Apart from this change, all corner changes occur immediately next to $A$ and have residues $i-1$ or $i+1$, which differ from $i$ since $e\ge 2$ (for $e=2$ the two neighbor residues coincide with each other). Hence the signed count of $i$-nodes strictly above $A$ is unchanged, as is the signed count strictly below it. There is no other addable or removable $i$-node in the row of $A$. Splitting the signed corner count of $\mu$ into its above and below parts and the removable node $A$ therefore gives $d_A(\mu )+d^A({\mu }^A)=d_i(\mu )+1$.

The Cartan formula (2). Write $u_j$ for the basis vector indexed by $j\in \mathbb{Z}\mathrm{/}e\mathbb{Z}$ in the free abelian group on residues, and choose $R$ greater than the number of nonzero rows of $\mu$. Put $b_r={\mu }_r-r$. The signed corner vector is ${\sum }_i{d}_i(\mu )u_i=u_{-R}+{\sum }_{r=1}^R(u_{b_r+1}-u_{b_r})$: adjacent row-boundary terms cancel precisely when the corresponding node is not a corner, and the tail term cancels the artificial bottom boundary. For the empty partition this telescopes to $u_0$, so $d_i(\mathrm{\varnothing })={\delta }_{i,0}=({\mathrm{\Lambda }}_0,{\alpha }_i)$. Adding a node of residue $j$ moves one $b_r=j-1$ to $j$ and changes this vector by $u_{j+1}-2u_j+u_{j-1}$. Thus each $d_i$ changes by $-a_{ij}$, including the double neighbor contribution when $e=2$. The vector with coordinates $({\mathrm{\Lambda }}_0-\alpha ,{\alpha }_i)$ has the same initial value and the same change when $\alpha$ increases by ${\alpha }_j$. Induction on the number of nodes proves (2).

Bricks in the first row (5)(i). All nodes carrying $m,\dots ,d$ lie consecutively in one row, by [F4] and standardness; since they are the largest entries, they form the rightmost segment of that row. At every addition in the codegree recursion for a brick in row one, there are no rows strictly above, so the increment is zero. Summing these increments proves $\mathrm{codeg}(T)=\mathrm{codeg}(T_{\le m-1})$.

The defect recursion (3). Expanding, $\mathrm{def}(\alpha )-\mathrm{def}(\alpha -{\alpha }_i)=({\mathrm{\Lambda }}_0,{\alpha }_i)-((\alpha ,\alpha )-(\alpha -{\alpha }_i,\alpha -{\alpha }_i))\mathrm{/}2=({\mathrm{\Lambda }}_0,{\alpha }_i)-(\alpha ,{\alpha }_i)+({\alpha }_i,{\alpha }_i)\mathrm{/}2=d_i(\mu )+1$ by (2) and $({\alpha }_i,{\alpha }_i)=2$, which is the displayed recursion [F3].

Degree plus codegree (4). For $d=0$ both definitions give $0=\mathrm{def}(0)$. For $d>0$ let $A$ be the node containing $d$ and $i=\mathrm{res}A$; by the induction hypothesis applied to $T_{\le d-1}$ and by step 2.1, $\mathrm{deg}(T)+\mathrm{codeg}(T)=\mathrm{deg}(T_{\le d-1})+\mathrm{codeg}(T_{\le d-1})+d_A(\mu )+d^A({\mu }^A)=\mathrm{def}(\alpha -{\alpha }_i)+d_i(\mu )+1=\mathrm{def}(\alpha )$, where the middle equality uses (1) [F2, F3].

Full residue bricks (5)(ii). Suppose the terminal brick has $e$ nodes in row $r$ and occupies columns $c+1,\dots ,c+e$. Put $\lambda =T_{\le m-1}$. If $r=1$, step 1.3 applies. If $r>1$, standardness implies that all nodes in rows above $r$ already occur in $\lambda$ and that ${\lambda }_{r-1}\ge c+e$: each entry above a brick node is smaller, and all entries outside the terminal brick are at most $m-1$. Immediately before each of the $e$ additions, row $r$ therefore has length strictly smaller than ${\lambda }_{r-1}$. The signed corner counts strictly above row $r$ stay fixed throughout these pre-addition shapes. Indeed, in the top $r-1$ rows each maximal plateau of equal row lengths contributes one addable node at the start of its first row and one removable node at the end of its last row, and the last such row remains removable because its length is greater than that of row $r$ before the addition. Thus the number of addable corners above the brick equals the number of removable corners above it. The brick residues $(c+1-r),\dots ,(c+e-r)$ run through $I$ once. Summing the codegree increment for these residues counts each of the fixed above addable corners once positively and each above removable corner once negatively, giving zero. Hence the codegrees before and after the brick agree. The boundary count, defect recursion and tableau induction establish all five clauses by finite computations and use no choice.