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KLR grading gives integral multiplicity bookkeeping

Statement

Let $D$ be a KLR datum on a symmetrizable Cartan datum with no imaginary simple roots, over a field $k$ , in the conventions of The quiver Hecke (KLR) algebra of a symmetrizable Cartan datum. Let $R_{D}$ be the resulting graded KLR algebra, and let $K_{0} (R_{D})$ be the Grothendieck group of its graded projectives. Put $A = Z [q_{g r}, q_{g r}^{- 1}]$ , where $q_{g r}$ is the Laurent grading variable denoted by $q$ in the divided-power integral form, and let $A f_{D}$ be the integral negative quantum half. The canonical map $γ_{D} : A f_{D} \to K_{0} (R_{D})$ is an isomorphism of $A$ -algebras and graded twisted bialgebras; $q_{g r}$ records the grading shift. Assume the Axiom of Choice inherited through this supplier.

For each integer $n \geq 1$ , the irreducible highest-weight- $n$ module $V (n)$ defined by homogeneous polynomials of degree $n$ has $H$ -weights $n, n - 2, \dots, - n$ , each with multiplicity one. The diagonal ${s l}_{2}$ -module $V (1)^{\otimes n}$ has $H$ -weight $n - 2 k$ with multiplicity $(\binom{n}{k})$ for $0 \leq k \leq n$ . Here $H = 2 S^{z}$ .

For real $h$ and $β$ , define the zero-exchange finite matrix $H_{n}^{(0)} = - h S_{n}^{z}$ on $V (1)^{\otimes n}$ . Use $\exp (A) = \sum_{r = 0}^{\infty} A^{r} / r!$ for the finite-dimensional matrix exponential, convergent in operator norm, and let $Tr$ denote the basis-independent trace of a finite-dimensional endomorphism. Define its partition trace by $Z_{n} (β) = {Tr}_{V (1)^{\otimes n}} \exp (- β H_{n}^{(0)})$ . Then $Z_{n} (β) = (2 \cosh (β h / 2))^{n}, \cosh x : = \frac{e^{x} + e^{- x}}{2} .$ For a Gibbs interpretation, $h$ is an energy coefficient, $S_{n}^{z}$ is dimensionless, and $β$ has inverse-energy units. The KLR isomorphism supplies the negative-half categorification stated above; it supplies no identification with either finite-dimensional spin module, no tensor-product categorical realization, and no spin-chain energy spectrum.

Statement references

The Axiom of Choice

The quiver Hecke (KLR) algebra of a symmetrizable Cartan datum

Spin weight spaces and rank-one Hall realization

The KLR category categorifies the negative quantum half

Finite-dimensional matrix exponential

The basis-independent trace of an endomorphism of a finite-dimensional vector space

Khovanov–Lauda, Theorem 1.1 and §3.2

Khovanov–Lauda II, Theorem 8

Kleshchev, §9.1, Theorem 9.1

Facts & Assumptions

Given: An integer $n \geq 1$ , a KLR datum and ground field satisfying the stated imported theorem, and real scalars $h, β$ . For a physical interpretation of the Gibbs trace, $h$ is measured in joules and $β$ in inverse joules; the algebraic identity itself is dimensionally expressed by the product $β h$ .

[A1]

Adopt the Axiom of Choice (The Axiom of Choice). In the cited KLR supplier its exact use is the nondegeneracy of the form used in the proof of injectivity; no arbitrary family choice is used in the finite-dimensional calculations below.

[F1]

The KLR category categorifies the negative quantum half: for the stated KLR datum over a field, the canonical map from the divided-power integral negative half to the graded projective Grothendieck group is an isomorphism of $A$ -algebras and graded twisted bialgebras. Its proof uses [A1] through the nondegeneracy supplier for the injectivity step.

[F2]

The quiver Hecke (KLR) algebra of a symmetrizable Cartan datum: this KLR datum has a symmetrizable Cartan datum with no imaginary simple roots, a field $k$ , and polynomials $Q_{i j}$ satisfying the stated (Q1)–(Q4) conditions; $R_{D}$ is the graded algebra with the displayed idempotent, dot, and crossing generators and relations.

[F3]

Spin weight spaces and rank-one Hall realization: on $V = C^{2}$ the generators are $E, F, H$ as displayed there, $V (r) = C [x, y]_{r}$ with $E = x \partial_{y}$ , $F = y \partial_{x}$ , $H = x \partial_{x} - y \partial_{y}$ , and the tensor-product action is the sum over factors.

[F4]

Finite-dimensional matrix exponential: for every finite complex matrix $A$ , $\exp (A) = \sum_{r \geq 0} A^{r} / r!$ converges in operator norm.

[F5]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: for a finite-dimensional endomorphism, the trace is the matrix trace in any chosen basis and is independent of the basis.

Proof

1.1givenA1F1F2

Under [A1], apply [F1] to the KLR datum in the Given. Its hypotheses are the datum and field specified in [F2], and its conclusion is exactly the isomorphism $γ_{D} : A f_{D} \to K_{0} (R_{D})$ with the Laurent grading parameter $q_{g r}$ . This use makes no claim about a representation on the spin spaces.

1.2givenF3algebra

In $V (1)$ , $H$ has eigenvalue $1$ on $∣ ↑ ⟩$ and $- 1$ on $∣ ↓ ⟩$ . The elementary tensor basis of $V (1)^{\otimes n}$ therefore consists of $H$ -eigenvectors. A tensor with exactly $k$ down factors has eigenvalue $n - 2 k$ , and there are exactly $(\binom{n}{k})$ such tensors. Hence its weight character is $(z + z^{- 1})^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) z^{n - 2 k}$ .

1.3givenF3algebra

The monomials $x^{n - k} y^{k}$ form a basis of $V (n)$ . Applying $H = x \partial_{x} - y \partial_{y}$ gives $H (x^{n - k} y^{k}) = (n - 2 k) x^{n - k} y^{k}$ , so each listed weight space is one-dimensional. For $k > 0$ , $E (x^{n - k} y^{k}) = k x^{n - k + 1} y^{k - 1}$ is nonzero; for $k < n$ , $F (x^{n - k} y^{k}) = (n - k) x^{n - k - 1} y^{k + 1}$ is nonzero. Any nonzero submodule has a nonzero $H$ -weight vector, since the eigenvalues are distinct and the corresponding projections are polynomials in $H$ . Repeated $E$ then reaches $x^{n}$ , and repeated $F$ from $x^{n}$ reaches every basis monomial. Thus $V (n)$ is irreducible and its weights have multiplicity one.

2.1givenF3F4F5step 1.2algebra

On a tensor basis vector with $k$ down factors, $S_{n}^{z}$ has eigenvalue $(n - 2 k) / 2$ , so $H_{n}^{(0)}$ has eigenvalue $- h (n - 2 k) / 2$ . Summing the exponentials over the basis gives $Z_{n} (β) = \sum_{k = 0}^{n} (\binom{n}{k}) e^{β h (n - 2 k) / 2} = (e^{β h / 2} + e^{- β h / 2})^{n} = (2 \cosh (β h / 2))^{n} .$ The series [F4] acts diagonally on this eigenbasis, and [F5] identifies its trace with the sum of these diagonal entries. This is an exact finite-matrix trace identity for the stated operator.

3.1F1step 1.2step 1.3step 2.1∎

The isomorphism in [F1] identifies $A f_{D}$ with $K_{0} (R_{D})$ only. For $n \geq 2$ , the weight $n - 2$ has multiplicity $n$ in step 1.2 and multiplicity one in step 1.3, so the two characters differ; for $n = 1$ they coincide. The map in [F1] assigns neither module to its Grothendieck group. Likewise, [F1] has no Hamiltonian operator, while step 2.1 computes the trace only after $H_{n}^{(0)}$ is specified. A tensor-product categorical action or a spin-chain spectrum therefore requires additional data.

Remarks

The KLR Laurent variable $q_{g r}$ records internal grading shifts. It is distinct from the finite-field cardinality $q_{H}$ and from the Hall parameter $ν = \sqrt{q_{H}}$ in Spin weight spaces and rank-one Hall realization. The character and trace computations are finite-dimensional algebraic identities; their physical interpretation is conditional on the stated Hamiltonian convention.

KLR grading gives integral multiplicity bookkeeping

Statement

Statement references

Facts & Assumptions

Proof

Remarks

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