Bott–Samelson words related by a braid need not be isomorphic bimodules

The two words and their braid relation. In $S_3$ the products $s_1{s}_2{s}_1$ and $s_2{s}_1{s}_2$ both equal the longest element of $S_3$, the parabolic $W_{1,2}$, so the words $(1,2,1)$ and $(2,1,2)$ are related by the braid move and have the same product; their Bott–Samelson bimodules are $B_s{B}_t{B}_s$ and $B_t{B}_s{B}_t$.

The two decompositions. By [F1] applied to the adjacent pair $s,t$, $B_s{B}_t{B}_s\cong B_{1,2,1}\oplus B_s$ and $B_t{B}_s{B}_t\cong B_{1,2,1}\oplus B_t$ with degree-zero identifications, so the two bimodules have the common summand $B_{1,2,1}$ and the distinct extra summands $B_s$ and $B_t$.

The occurrence count of the graph $s_1$. By [F2] the number of occurrences of a graph $x$ in the $\mathrm{\Delta }$-flag of $B_{\underset{‾}{i}}$ equals the number of subexpressions of $\underset{‾}{i}$ whose product is $x$. For $\underset{‾}{i}=(1,2,1)$ the subexpressions with product $s_1$ are the one-term subexpressions on positions $1$ and $3$: choosing position $1$ alone gives $s_1$ and choosing position $3$ alone gives $s_1$, whereas $\{1,3\}$ gives $s_1{s}_1=e$ and $\{1,2,3\}$ gives the longest element, so there are exactly $2$ such subexpressions. For $\underset{‾}{i}=(2,1,2)$ the only subexpression with product $s_1$ is the one-term subexpression on position $2$, and again $\{1,3\}$ gives $s_2{s}_2=e$, so there is exactly $1$.

The two sums of multiplicities. By [F4] each occurrence of the graph $s_1$ in the $\mathrm{\Delta }$-flag of $B_{\underset{‾}{i}}$ is a quotient ${\mathrm{\Delta }}_{s_1}(d)$ for a unique degree $d$, and conversely every ${\mathrm{\Delta }}_{s_1}(d)$-quotient is an occurrence of the graph $s_1$; hence step 1.3 counts exactly the sum over $d$ of the multiplicities, giving ${\sum }_d(B_s{B}_t{B}_s:{\mathrm{\Delta }}_{s_1}(d))=2$ and ${\sum }_d(B_t{B}_s{B}_t:{\mathrm{\Delta }}_{s_1}(d))=1$.

The contradiction. Suppose $B_s{B}_t{B}_s\cong B_t{B}_s{B}_t$ as graded bimodules. By [F3] the multiplicity $(M:{\mathrm{\Delta }}_x(d))$ is intrinsic, so the two bimodules would have $(B_s{B}_t{B}_s:{\mathrm{\Delta }}_{s_1}(d))=(B_t{B}_s{B}_t:{\mathrm{\Delta }}_{s_1}(d))$ for every $d$, and summing over the finitely many degrees $d$ in which the flags of [F2] have quotients would give equal sums; but by step 2.1 the sums are $2$ and $1$.

Conclusion. The braid-related words $(1,2,1)$ and $(2,1,2)$ of $S_3$ have the same product in $S_3$ yet non-isomorphic Bott–Samelson bimodules $B_s{B}_t{B}_s\cong ̸B_t{B}_s{B}_t$, distinguished by the intrinsic multiplicities of the graph $s_1$: two occurrences in the first and one in the second, as counted by subexpressions in step 1.3. The common longest summand $B_{1,2,1}$ of step 1.2 therefore does not force the two words to give isomorphic bimodules, and the refuted claim is false. The argument uses only the finitely many subexpressions of the two words of length three and the displayed decompositions, so no choice principle is used. ∎