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No nonzero finite-dimensional Weyl modules at nonzero central character in characteristic zero

Statement

Let $k$ be a field of characteristic zero, let $n \geq 1$ , and let $0 \neq ℏ \in k$ . Then no nonzero finite-dimensional $k$ -vector space $M$ admits a unital $k$ -algebra representation of $W_{n} (k; ℏ)$ .

Facts & Assumptions

Given: A field $k$ of characteristic zero, an integer $n \geq 1$ , a scalar $0 \neq ℏ \in k$ , and a nonzero finite-dimensional unital left $W_{n} (k; ℏ)$ -module $M$ .

[F1]

Polynomial Weyl algebra and its central parameter: the generators satisfy $p_{1} x_{1} - x_{1} p_{1} = ℏ 1$ .

[F2]

Unital left and right modules over a ring; unqualified module means left module: the module action satisfies $(a b) m = a (b m)$ and $1 m = m$ .

[F3]

The trace of a matrix commutator is zero: for square matrices over a commutative ring, the trace of their commutator is zero.

[F4]

The trace of a square matrix over a commutative ring: the trace of a square matrix is the sum of its diagonal entries.

[F5]

Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases: the columns of the matrix of a linear map are the coordinate columns of the images of the basis vectors.

[F6]

Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose: matrix products have entries $(A B)_{i j} = \sum_{r} a_{i r} b_{r j}$ , scalar multiplication is entrywise, and $I_{d}$ has diagonal entries $1$ and off-diagonal entries $0$ .

Proof

1.1givenF1F2F5F6choosealgebra

Choose a $k$ -basis $e_{1}, \dots, e_{d}$ of $M$ and let $D$ and $X$ be the matrices of the actions of $p_{1}$ and $x_{1}$ . Writing $X e_{j} = \sum_{r} x_{r j} e_{r}$ and $D e_{r} = \sum_{i} d_{i r} e_{i}$ , [F5] and the product formula [F6] show that the matrices of the composite actions $p_{1} x_{1}$ and $x_{1} p_{1}$ are $D X$ and $X D$ , respectively. The scalar copy of $k$ acts through the given vector-space structure, so the module axiom [F2] and relation [F1] give $D X - X D = ℏ I_{d}$ , where $d = \dim_{k} M \geq 1$ .

2.1step 1.1F3F4F6algebra

By [F3], $tr (D X - X D) = 0$ . By [F4] and [F6], $tr (ℏ I_{d}) = ℏ d$ , so $ℏ d = 0$ in $k$ .

3.1givenstep 2.1algebra∎

Characteristic zero makes the image of every positive integer $d$ nonzero in $k$ . Since $ℏ \neq 0$ and a field has no zero divisors, $ℏ d \neq 0$ , contradicting step 2.1. Hence no such nonzero module exists.

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Sources

François Dumas, Noncommutative Invariants, §2.1.1 (standard reference, not scraped)
Sheldon Axler, Linear Algebra Done Right, 4th ed., §8D (standard reference, not scraped)