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First K-types and ladder coefficients in I(epsilon, nu)

Example

Assume the Axiom of Choice (The Axiom of Choice). Tabulate the $K$ -types $f_{- 3}, \dots, f_{3}$ of $I_{0, ν}$ and of $I_{1, ν}$ and the values of the raising and lowering operators $L_{E_{\pm}} f_{n} = \frac{1 + ν \pm n}{2} f_{n \pm 2}$ on them, and check that for $ν = n \in W_{ε}$ the coefficient at the expected $K$ -type vanishes.

Facts & Assumptions

Given: AC, $ε \in {0, 1}$ , $ν \in C$ , and the K-type decomposition and ladder operators.

[F1]

$f_{n} (k_{θ}) = e^{i n θ}$ is a K-type exactly for $n \equiv ε (m o d 2)$ , and these are all K-types (K-type decomposition of the SL2(R) principal series).

[F2]

$L_{W} f_{n} = n f_{n}$ and $L_{E_{\pm}} f_{n} = \frac{1 + ν \pm n}{2} f_{n \pm 2}$ , with $L_{E_{-}} f_{n} = 0$ exactly when $ν = n - 1$ and $L_{E_{+}} f_{n} = 0$ exactly when $ν = - (n + 1)$ (Derived action and raising/lowering formulas in the compact picture).

[F3]

$W_{0}$ is the odd integers and $W_{1}$ is the even integers (The normalized principal series I(epsilon, nu)).

[A1]

AC is inherited through the principal-series and ladder suppliers; this explicit tabulation uses no additional choice (The Axiom of Choice).

Verification

technique · substitute each parity-allowed weight into [F2]

For the K-types in the requested range, the table is:

$ε$	$n$	$L_{W} f_{n}$	$L_{E_{+}} f_{n}$	$L_{E_{-}} f_{n}$
$0$	$- 2$	$- 2 f_{- 2}$	$\frac{ν - 1}{2} f_{0}$	$\frac{ν + 3}{2} f_{- 4}$
$0$	$0$	$0$	$\frac{1 + ν}{2} f_{2}$	$\frac{1 + ν}{2} f_{- 2}$
$0$	$2$	$2 f_{2}$	$\frac{ν + 3}{2} f_{4}$	$\frac{ν - 1}{2} f_{0}$
$1$	$- 3$	$- 3 f_{- 3}$	$\frac{ν - 2}{2} f_{- 1}$	$\frac{ν + 4}{2} f_{- 5}$
$1$	$- 1$	$- f_{- 1}$	$\frac{ν}{2} f_{1}$	$\frac{ν + 2}{2} f_{- 3}$
$1$	$1$	$f_{1}$	$\frac{ν + 2}{2} f_{3}$	$\frac{ν}{2} f_{- 1}$
$1$	$3$	$3 f_{3}$	$\frac{ν + 4}{2} f_{5}$	$\frac{ν - 2}{2} f_{1}$

1.1F1F2A1algebra

For $ε = 0$ , the only indices in ${- 3, - 2, \dots, 3}$ with the required parity are $- 2, 0, 2$ ; for $ε = 1$ they are $- 3, - 1, 1, 3$ . Applying the three formulas in [F2] to these indices gives every entry of the table, and [F1] shows that the table omits no K-type in the requested range.

2.1F2F3step 1.1algebra∎

Let $m \in W_{ε}$ with $m \geq 0$ ; $m = 0$ occurs only for odd parity. At $ν = m$ , [F2] gives $L_{E_{-}} f_{m + 1} = 0$ and $L_{E_{+}} f_{- m - 1} = 0$ , since their coefficients are respectively $(1 + m - (m + 1)) / 2$ and $(1 + m + (- m - 1)) / 2$ . At $ν = - m$ , it gives $L_{E_{-}} f_{1 - m} = 0$ and $L_{E_{+}} f_{m - 1} = 0$ , since both coefficients are zero. When $m = 0$ , these parameter cases coincide and the table shows $L_{E_{-}} f_{1} = L_{E_{+}} f_{- 1} = 0$ at $ε = 1, ν = 0$ . For the other exceptional values visible in the table, $m = 1$ in even parity and $m = 2$ in odd parity: the positive parameter zeros occur at $f_{\pm 2}$ and $f_{\pm 3}$ , respectively, and the negative parameter zeros occur at $f_{0}$ and $f_{\pm 1}$ . These are exactly the boundary arrows expected from the exceptional K-type strings; no parity class is identified with the other.

Remarks

Kerr's formula (2.6) gives the same raising and lowering coefficients in the right-translation basis used here. Etingof's §9.1 formulas (4)–(5) use an abstractly normalized weight basis, so they serve as a convention check rather than a literal coefficient-by-coefficient table source. The table above is computed directly from the local ladder formulas.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Matt Kerr, Notes on the Representation Theory of SL2(R) (CBMS workshop writeup) (standard reference, not scraped)
Pavel Etingof, Representations of Lie Groups (MIT 18.757 lecture notes, Fall 2023) (standard reference, not scraped)