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Verma modules for sl2

Example

Assume the Axiom of Choice. Let ${s l}_{2} = {s l}_{2} (C)$ have basis $e, f, h$ with $[e, f] = h$ , $[h, e] = 2 e$ , $[h, f] = - 2 f$ (The special linear Lie algebra sl_2), and for every linear functional $λ \in (C h)^{*}$ let $M (λ)$ be the induced module $U ({s l}_{2}) \otimes_{U (C h \oplus C e)} C_{λ}$ , realised concretely as $U ({s l}_{2}) / J_{λ}$ with $J_{λ}$ the left ideal generated by $e$ and $h - λ (h)$ (Universal enveloping algebra, Highest-weight vectors and modules). Then $M (λ)$ has basis $v_{k} = f^{k} \cdot v_{0}$ , $k \geq 0$ , on which $h \cdot v_{k} = (λ (h) - 2 k) v_{k}, f \cdot v_{k} = v_{k + 1}, e \cdot v_{k} = k (λ (h) - k + 1) v_{k - 1};$ it is infinite-dimensional, and it has a finite-dimensional simple quotient exactly when $λ (h)$ is a nonnegative integer. In that case the quotient is $V (n)$ of All irreducible finite-dimensional sl2 modules with $n = λ (h)$ .

Set $v_{- 1} = 0$ in the displayed action formula. The one-dimensional Borel module $C_{λ}$ has $h$ acting by $λ (h)$ and $e$ acting by zero.

Facts & Assumptions

Given: The Axiom of Choice, ${s l}_{2}$ with its basis, a functional $λ \in (C h)^{*}$ determined by the scalar $n_{λ} = λ (h)$ , the Borel subalgebra $C h \oplus C e$ , and the quotient module $M (λ)$ with generator $v_{0} = 1 + J_{λ}$ .

[A1]

The Axiom of Choice (The Axiom of Choice) is inherited from the cited general definitions of highest weight and dominant integral weight. The explicit rank-one calculations and supplied-basis PBW argument make no further use of choice.

[L1]

The monomials $f^{a} h^{b} e^{c}$ in the ordered basis $(f, h, e)$ form a basis of $U ({s l}_{2})$ ; moreover $U ({s l}_{2}) = U (C f) U (C h) U (C e)$ as a linear span (Poincaré–Birkhoff–Witt theorem).

[L2]

In $M (λ)$ one has $e \cdot v_{0} = 0$ and $h \cdot v_{0} = λ (h) v_{0}$ , and $f^{k} \cdot v_{0}$ is the class of $f^{k}$ ; the module is generated by $v_{0}$ (Highest-weight vectors and modules).

[L3]

In $U ({s l}_{2})$ the commutation identity $[e, f^{k}] = k f^{k - 1} (h - (k - 1))$ holds for every $k \geq 1$ , by induction from $[e, f] = h$ and $[h, f] = - 2 f$ . [L1]

[L4]

A finite-dimensional irreducible ${s l}_{2}$ -module of highest weight $m \geq 0$ is isomorphic to $V (m)$ of All irreducible finite-dimensional sl2 modules.

Verification

technique · direct

1.1

The relations $h \mapsto λ (h)$ and $e \mapsto 0$ define a character $χ_{λ} : U (C h \oplus C e) \to C$ . By PBW, multiplication is a linear isomorphism $U (C f) \otimes U (C h \oplus C e) ⟶ U ({s l}_{2}) .$ Tensoring this factorisation over $U (C h \oplus C e)$ with the one-dimensional module $C_{λ}$ identifies $M (λ)$ linearly with $U (C f)$ . Thus the classes of $f^{k}$ , $k \geq 0$ , are a basis of $M (λ)$ .

L1L2

2.1

Therefore $M (λ)$ is infinite-dimensional, and the action on $v_{k} = f^{k} \cdot v_{0}$ is $h \cdot v_{k} = (λ (h) - 2 k) v_{k}$ (eigenvalue computation), $f \cdot v_{k} = v_{k + 1}$ , and $e \cdot v_{k} = [e, f^{k}] \cdot v_{0} = k (λ (h) - k + 1) v_{k - 1}$ by [L3]; in particular $e \cdot v_{k} = 0$ exactly when $k = 0$ or $λ (h) = k - 1$ .

L2L3step 1.1

3.1

If $λ (h) \notin Z_{\geq 0}$ , then $e \cdot v_{k} \neq 0$ for every $k \geq 1$ . Given $0 \neq w = \sum_{k = 0}^{N} c_{k} v_{k}$ , choose the largest $K$ with $c_{K} \neq 0$ . For $k < K$ one has $e^{K} v_{k} = 0$ , whereas $e^{K} v_{K} = K! \prod_{j = 1}^{K} (λ (h) - j + 1) v_{0} \neq 0.$ Thus $e^{K} w$ is a nonzero multiple of $v_{0}$ . Hence $v_{0} \in U ({s l}_{2}) w$ , so every nonzero submodule is all of $M (λ)$ . The module is therefore simple and infinite-dimensional and has no nonzero finite-dimensional quotient.

step 2.1

3.2

If $λ (h) = n$ is a nonnegative integer, then $e \cdot v_{n + 1} = 0$ by step 2.1, and $K = \sum_{k \geq n + 1} C v_{k}$ is a nonzero proper submodule. Let $N$ be a submodule not contained in $K$ , and choose a finite nonzero sum $w = \sum c_{k} v_{k} \in N ∖ K$ . Since the $h$ -eigenvalues $n - 2 k$ are pairwise distinct, a polynomial in $h$ isolates from this finite sum a nonzero term $c_{j} v_{j} \in N$ with $j \leq n$ . Then $e^{j} v_{j} = j! \prod_{r = 1}^{j} (n - r + 1) v_{0} \neq 0$ , so $v_{0} \in N$ and $N = M (λ)$ . Consequently every proper submodule lies in $K$ ; hence $K$ is the unique maximal proper submodule and $M (λ) / K$ is the unique simple quotient.

step 2.1

4.1

On $M (λ) / K$ the classes of $v_{0}, \dots, v_{n}$ satisfy exactly the relations of $V (n)$ from All irreducible finite-dimensional sl2 modules, with $h$ -eigenvalues $n - 2 k$ and operators $f, e$ acting by $v_{k} \mapsto v_{k + 1}$ and $v_{k} \mapsto k (n - k + 1) v_{k - 1}$ ; hence the quotient is the finite-dimensional simple module $V (n)$ by [L4].

L4step 3.2

5.1

Collecting steps 1.1–4.1, $M (λ)$ is an infinite-dimensional highest weight module of highest weight $λ$ , and it has a finite-dimensional simple quotient exactly for $λ (h) \in Z_{\geq 0}$ , in which case that quotient is $V (λ (h))$ ; this proves all the assertions.

A1step 1.1step 2.1step 3.1step 3.2step 4.1∎

Depends on

Used by

A nondominant integral highest-weight module can be infinite-dimensional Counterexample

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alexander Kirillov Jr., An Introduction to Lie Groups and Lie Algebras (standard reference, not scraped)
Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (standard reference, not scraped)