The Soergel character grading alone does not prove category-O Koszulity

The interval. The word $s_2{s}_1{s}_3{s}_2$ has length $4$ and is reduced, since $3412$ has exactly the four inversions $(3,1),(3,2),(4,1),(4,2)$. Its reduced subwords give: $1234$ at length $0$; $2134,1324,1243$ at length $1$; $1342,1423,2314,3124,2143$ at length $2$; $1432,2413,3142,3214$ at length $3$; and $3412$ at length $4$. The four elements $1234,2134,1243,2143$ contain no reduced subword equal to $s_2=1324$, so they are not above $b$ by the reduced-subword criterion; the remaining ten elements are exactly $[b,w]$.

All gaps at most two have $P=1$. Suppose $y<z$ and $d:=\mathrm{\ell }(z)-\mathrm{\ell }(y)\le 2$. By [F3] the degree of $P_{y,z}$ is at most $(d-1)\mathrm{/}2\le 1\mathrm{/}2$, hence at most $0$; together with constant term $1$ this forces $P_{y,z}=1$. If $d=0$ then $y=z$ and $P_{y,y}=1$; if $y$ and $z$ are incomparable, $P_{y,z}=0$ by [F3]. In particular $P_{b,2413}=1$ and $P_{1234,1423}=1$, both pairs having gap $2$.

The recursion at $(\mathrm{i}\mathrm{d},2413)$. Write $\mathrm{i}\mathrm{d}=1234$ and use $2413=s_1{s}_3{s}_2$, a reduced word with left descent $s_1$. With $s=s_1$ one has $s2413=1423$ and $s\mathrm{i}\mathrm{d}=2134$, so $c=0$ in [F4] and $P_{\mathrm{i}\mathrm{d},2413}(q)=q{P}_{2134,1423}(q)+P_{\mathrm{i}\mathrm{d},1423}(q)-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\sum }_{\begin{array}{c}\mathrm{i}\mathrm{d}\le u\le 1423\\ s_{1}u<u,\mu (u,1423)\mathrm{\ne }0\end{array}}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\mu (u,1423)q^{(3-\mathrm{\ell }(u))\mathrm{/}2}{P}_{\mathrm{i}\mathrm{d},u}(q)\mathrm{.}$ The first summand vanishes: $1423=s_3{s}_2$ has no reduced subword $s_1$, so $2134\le ̸1423$ and $P_{2134,1423}=0$ by [F3] and [F5]. The correction sum is empty as well: $[\mathrm{i}\mathrm{d},1423]=\{1234,1324,1243,1423\}$ by step 1.1, and for each of these four elements left multiplication by $s_1$ swaps the values $1$ and $2$, raising the length from $0,1,1,2$ to $1,2,2,3$ respectively, so no element of the interval has the required left descent $s_1$. By step 1.2 the remaining term is $P_{\mathrm{i}\mathrm{d},1423}=1$, hence $P_{\mathrm{i}\mathrm{d},2413}=1$.

The main recursion. Apply [F4] to $b=1324$, $w=3412$ and the left descent $s=s_2$: $s_{2}w=2413$ and $s_{2}b=1234$, so $c=1$ because $s_{2}b$ has length $0<1=\mathrm{\ell }(b)$, and $P_{b,w}(q)=P_{\mathrm{i}\mathrm{d},2413}(q)+q{P}_{b,2413}(q)-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\sum }_{\begin{array}{c}b\le u\le 2413\\ s_{2}u<u,\mu (u,2413)\mathrm{\ne }0\end{array}}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\mu (u,2413)q^{(4-\mathrm{\ell }(u))\mathrm{/}2}{P}_{b,u}(q)\mathrm{.}$ By step 1.1 the interval $[b,2413]$ is $\{1324,1423,2314,2413\}$; left multiplication by $s_2$ swaps the values $2$ and $3$, turning these into $1234,1432,3214,3412$ of lengths $0,3,3,4$, so only $u=b$ has the required descent, and there $\mu (b,2413)=0$ because $\mathrm{\ell }(2413)-\mathrm{\ell }(b)=2$ is even by [F3]. The correction sum is therefore empty; step 2.1 gives $P_{\mathrm{i}\mathrm{d},2413}=1$ and step 1.2 gives $P_{b,2413}=1$, so $P_{b,w}=1+q$, with $\mu (b,w)=1$ and $\mathrm{\ell }(w)-\mathrm{\ell }(b)=3$.

The comparison. The polynomial $P_{b,w}(q)=1+q$ has leading coefficient $1$ (the coefficient of $q^1$, its degree being $1$) while its value at $q=1$ is $2$. The multiplicity theorem [F1] computes the ordinary Verma-flag multiplicity as $(P(w\cdot \lambda ):\mathrm{\Delta }(b\cdot \lambda ))=P_{b,w}(1)=2$, with $0$ dominant regular integral in ${\mathfrak{s}\mathfrak{l}}_4$ (indeed $⟨\rho ,{\alpha }_i^{\vee }⟩=1\mathrm{\ne }0$ for every simple root and $0$ is integral). Hence the leading coefficient $1$ of the Kazhdan–Lusztig polynomial differs from the multiplicity $2$: the refuted claim's coefficientwise clause is false, and neither individual coefficient of this polynomial equals the ordinary multiplicity; its total value at one does. This is a genuine failure of the claim and not an artefact of undefined quantities: $b\le w$, all polynomials are classical Kazhdan–Lusztig polynomials, and the multiplicity is that of the ordinary block ${\mathcal{O}}_0$.

The grading clause. The remaining structural clause of the refuted claim is not supplied by the cited results either: the multiplicity of [F1] is the ordinary (ungraded) Verma-flag multiplicity, and the projective dictionary of [F2] is the ungraded reduction $V(P(w\cdot \lambda ))={\bar{B}}_{w^{-1}}$, whose auxiliary grading lives on the polynomial lift and is not a grading of ${\mathcal{O}}_{\lambda }$. Thus the character identity at $q=1$ cannot establish that the block is graded or Koszul; the graded and $\mathrm{E}\mathrm{x}\mathrm{t}$ refinements belong to the separate Koszul-duality frontier, and step 4.1 refutes only the identification of an individual coefficient with an ordinary multiplicity. It does not refute graded decomposition numbers or the possibility of a Koszul grading: those require additional constructions and proofs, and graded multiplicities may sum to the ordinary multiplicity. The stated conjunction is refuted by its false leading-coefficient clause, with witness $(y,w)=(1324,3412)$ in ${\mathfrak{s}\mathfrak{l}}_4$.