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Changing the equality case in the tensor rule breaks the partial-inverse axiom

Statement refuted

Claim refuted: on the product of two crystals one may exchange the two equality cases while retaining the comparison direction, namely

${\tilde{f}}_{i}^{'} (b_{1} \otimes b_{2}) = {\tilde{f}}_{i} b_{1} \otimes b_{2} if φ_{i} (b_{1}) \geq ε_{i} (b_{2}), {\tilde{e}}_{i}^{'} (b_{1} \otimes b_{2}) = {\tilde{e}}_{i} b_{1} \otimes b_{2} if φ_{i} (b_{1}) > ε_{i} (b_{2}),$

with the second factor used in the remaining cases and $0$ whenever the selected factor maps to $0$ ; that is, the two equality cases of The tensor product crystal $B_{1} \otimes B_{2}$ and its two-case convention are exchanged.

Counterexample: take $g = {s l}_{2}$ , the rank-one simples of highest weights $a = b = 2$ with divided bases $v_{j} = F_{1}^{(j)} v_{0}, w_{k} = F_{1}^{(k)} w_{0}$ , and the vertex $b = v_{1} \otimes w_{1}$ . Then $φ_{1} (v_{1}) = 1 = ε_{1} (w_{1})$ , so the modified rule uses the non-strict comparison and acts on the first factor: ${\tilde{f}}_{1}^{'} (v_{1} \otimes w_{1}) = v_{2} \otimes w_{1}$ . At the image $b^{'} = v_{2} \otimes w_{1}$ one has $φ_{1} (v_{2}) = 0 \leq ε_{1} (w_{1}) = 1$ , and the modified ${\tilde{e}}_{1}^{'}$ acts on the first factor only when $φ_{1} > ε_{1}$ , so it acts on the second factor: ${\tilde{e}}_{1}^{'} (v_{2} \otimes w_{1}) = v_{2} \otimes w_{0} \neq v_{1} \otimes w_{1}$ . Hence ${\tilde{f}}_{1}^{'} b = b^{'}$ while ${\tilde{e}}_{1}^{'} b^{'} \neq b$ : the partial-inverse axiom fails, so $(B (a) \otimes B (b), {\tilde{e}}^{'}, {\tilde{f}}^{'})$ is not a crystal, and a fortiori not the crystal $B (a) \otimes B (b)$ of The crystal basis of a tensor product and the two-case operator rule. The same failure occurs if one uses the strict comparison for both operators, or the non-strict comparison for both; by contrast the rank-one computation of Rank-one tensor strings and their crystal limit rule shows that with the convention of The tensor product crystal $B_{1} \otimes B_{2}$ and its two-case convention the string through $v_{0} \otimes w_{1}$ is $v_{0} \otimes w_{1}, v_{1} \otimes w_{1}, v_{1} \otimes w_{2}$ of length three, whereas the modified rule produces the four-element chain $v_{0} \otimes w_{1}, v_{1} \otimes w_{1}, v_{2} \otimes w_{1}, v_{2} \otimes w_{2}$ , so the modified rule has the wrong lowering-chain length and cannot be the crystal of $V (2) \otimes V (2) = V (4) \oplus V (2) \oplus V (0)$ .

Facts & Assumptions

Given: The rank-one case $g = {s l}_{2}$ , $i = 1$ , the simples of highest weights $a = b = 2$ with divided bases $v_{j} = F_{1}^{(j)} v_{0}$ , $w_{k} = F_{1}^{(k)} w_{0}$ ( $0 \leq j, k \leq 2$ ), the vertex $b = v_{1} \otimes w_{1}$ , and the modified pair of operators ${\tilde{f}}_{1}^{'}, {\tilde{e}}_{1}^{'}$ of the Statement refuted.

[F1]

The tensor product crystal (The tensor product crystal $B_{1} \otimes B_{2}$ and its two-case convention): the library convention uses the strict comparison for ${\tilde{f}}_{i}$ and the non-strict comparison for ${\tilde{e}}_{i}$ , i.e. ${\tilde{f}}_{i}$ acts on the first factor if and only if $φ_{i} (b_{1}) > ε_{i} (b_{2})$ and ${\tilde{e}}_{i}$ if and only if $φ_{i} (b_{1}) \geq ε_{i} (b_{2})$ , with value $0$ when the chosen factor maps to $0$ .

[F2]

The tensor rule theorem (The crystal basis of a tensor product and the two-case operator rule): $(L_{1} \otimes L_{2}, B_{1} \otimes B_{2})$ with the convention [F1] is a crystal basis of $M_{1} \otimes M_{2}$ ; in particular its operators satisfy the partial-inverse axiom ${\tilde{f}}_{i} b = b^{'} ⟺ b = {\tilde{e}}_{i} b^{'}$ .

[F3]

The rank-one tensor strings (Rank-one tensor strings and their crystal limit rule): for divided bases $v_{j} = F^{(j)} v_{0}$ , $w_{k} = F^{(k)} w_{0}$ of rank-one simples of highest weights $a, b \geq 0$ one has, modulo $q$ , $\tilde{f} (v_{j} \otimes w_{k}) = v_{j + 1} \otimes w_{k}$ if $a - j > k$ and $v_{j} \otimes w_{k + 1}$ otherwise, with zero outside the divided-basis ranges; in particular for $a = b = 2$ the string through $v_{0} \otimes w_{1}$ is $v_{0} \otimes w_{1} \to v_{1} \otimes w_{1} \to v_{1} \otimes w_{2}$ .

[F4]

Crystal graph invariants (The crystal graph, $ε_{i}$ , $φ_{i}$ and the weight function): on the rank-one chain $v_{0} \overset{1}{\to} v_{1} \overset{1}{\to} v_{2}$ one has $ε_{1} (v_{j}) = j$ and $φ_{1} (v_{j}) = 2 - j$ , and likewise for the $w_{k}$ .

[F5]

The partial-inverse axiom of a crystal basis (Crystal lattices, crystal bases and their crystals): for the crystal of a module one has ${\tilde{f}}_{i} b = b^{'}$ if and only if $b = {\tilde{e}}_{i} b^{'}$ ; a coloured graph for which this fails is not a crystal of a crystal basis.

[F6]

The $ε / φ$ bookkeeping of the tensor product (The weight, $ε_{i}$ and $φ_{i}$ formulas for the tensor product crystal) fixes which comparison governs each operator in [F1] and identifies the equality case.

Counterexample

technique · direct

1.1givenF4

Set $a = b = 2$ and index the divided bases by their string positions, so that $ε_{1} (v_{j}) = j$ , $φ_{1} (v_{j}) = 2 - j$ and similarly $ε_{1} (w_{k}) = k$ , $φ_{1} (w_{k}) = 2 - k$ by [F4]. The vertex $v_{1} \otimes w_{1}$ satisfies $φ_{1} (v_{1}) = 1 = ε_{1} (w_{1})$ ; this is the equality case in which the conventions differ.

2.1step 1.1

Under the modified rule ${\tilde{f}}_{1}^{'}$ acts on the first factor at this vertex, because it uses $φ_{1} \geq ε_{1}$ : ${\tilde{f}}_{1}^{'} (v_{1} \otimes w_{1}) = v_{2} \otimes w_{1}$ .

3.1step 2.1algebra

At $v_{2} \otimes w_{1}$ one has $φ_{1} (v_{2}) = 0$ and $ε_{1} (w_{1}) = 1$ , so both the modified and the original rule select a factor by the comparison of $0$ with $1$ ; the modified ${\tilde{e}}_{1}^{'}$ acts on the first factor only when $φ_{1} > ε_{1}$ , which fails, so it acts on the second: ${\tilde{e}}_{1}^{'} (v_{2} \otimes w_{1}) = v_{2} \otimes w_{0}$ .

4.1step 3.1step 1.1F4F5

Therefore ${\tilde{e}}_{1}^{'} {\tilde{f}}_{1}^{'} (v_{1} \otimes w_{1}) = v_{2} \otimes w_{0} \neq v_{1} \otimes w_{1}$ , contradicting [F5]. For the strict-for-both variant use $v_{0} \otimes w_{1}$ : lowering gives $v_{1} \otimes w_{1}$ , but strict raising at the tie acts on the second factor and gives $v_{1} \otimes w_{0}$ . For the non-strict-for-both variant use $v_{1} \otimes w_{1}$ : lowering gives $v_{2} \otimes w_{1}$ , and raising gives $v_{2} \otimes w_{0}$ . Thus each alternative fails partial inversion.

5.1step 4.1F3

Compare component sizes: the original rank-one tensor rule of [F3] gives ${\tilde{f}}_{1} (v_{j} \otimes w_{k}) = v_{j + 1} \otimes w_{k}$ if $2 - j > k$ and $v_{j} \otimes w_{k + 1}$ otherwise; through $v_{0} \otimes w_{1}$ this is $v_{0} \otimes w_{1} \to v_{1} \otimes w_{1} \to v_{1} \otimes w_{2}$ , a chain of three, which is the component $B (2)$ in the decomposition of $V (2) \otimes V (2) = V (4) \oplus V (2) \oplus V (0)$ . The modified rule instead produces $v_{0} \otimes w_{1} \to v_{1} \otimes w_{1} \to v_{2} \otimes w_{1} \to v_{2} \otimes w_{2} \to 0$ , a chain of four, which is not a component of the crystal of any module in that decomposition.

6.1step 4.1step 5.1F1F2F6∎

With the library comparison direction fixed, exchanged equality cases and either uniform comparison fail the partial-inverse axiom. The exchanged-equality lowering chain also has a length absent from the correct rank-one crystal, while the library convention satisfies both the axiom and the rank-one string computation.

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Sources

Kyeonghoon Jeong, Seok-Jin Kang and Masaki Kashiwara, Crystal Bases for Quantum Generalized Kac-Moody Algebras, arXiv:math/0305390 (standard reference, not scraped)
Mark Shimozono, Crystals for Dummies (lecture notes, 28 August 2005) (standard reference, not scraped)
Richard Borcherds, Mark Haiman, Theo Johnson-Freyd, Nicolai Reshetikhin and Vera Serganova, Berkeley Lectures on Lie Groups and Quantum Groups (book-length lecture notes, last updated 18 January 2024) (standard reference, not scraped)