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Let $\mathcal{sl}_2$ be a Lie algebra over $\mathbb{C}$.

Each and every irreeducible representations of $\mathcal{sl}_2$ is uniquely determined by its maximal weight $n-1$ and is termed $V_n$. Then $\dim V_n = n$ and $V_n$ is the sum of one dimensional weight spaces each having weight $n-1-2i$ for $0 \leq i \leq n-1$. All finite dimensional representations of $\mathcal{sl}_2$ are sums of $V_n$ and can be thought of as $$ W = \bigoplus_j V_{n_j}^{\oplus a_j} $$

Let $W$ be a representation of $\mathcal{sl}_2$. A weight space, $W_i$ of weight $i$ is the eigenspace of $H$ each of whose vectors has eigenvalue $n_i$. Fixing a representation W, we reserve $d_i = \dim W_i$.

I don't understand the reason and I don't know to prove the following affirmative:

For any representation $W$ of $\mathcal{sl}_2$, the sequences $\{ d_{2i} \}_{i \in \mathbb{Z}}$ and $\{ d_{2i+1} \}_{i \in \mathbb{Z}}$ are unimodal and symmetric about $0$.

Reference: Representation Theory and Combinatorics sl2 and Applications

Vasco
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1 Answers1

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This is true because

1) every (finite-dimensional) representation of $sl_2$ is the direct sum of irreducible (fin.-dim.) representations (the last sentence in your highlighted paragraph)

2) it is true for irreducible (fin.-dim.) (sometimes called "simple") representations. These are explicitly described in the first two sentences of your first quote: For $V_n$, we have $$ d_{k}(V_n) = \begin{cases} 1 \text{ if } k=1-n, 1-n+2, ..., n-1-4, n-1-2, n-1 \\ 0 \text{ otherwise} \end{cases}$$

whose subsequences for even resp. odd $k$ have the described properties (actually, the odd (resp. even) one consists entirely of $0$'s if $n$ is odd (resp. even).)

Now you can check that for $W = \bigoplus_j V_{n_j}^{\oplus a_j} $, the $d_k$ for $W$ are given by $d_k(W) = \sum_j a_j d_k(V_{n_j})$, and both properties extend from the basic sequences $d_k(V_n)$ to such linear combinations.

For the same fact described in different words, compare the first part of my answer https://math.stackexchange.com/a/3288482/96384.

Added example:

The sequences of $d's$ look e.g. like this:

For $V(1)$:

$$...,0,0,\underbrace{0}_{k=-2},\underbrace{0}_{k=-1},\underbrace1_{k=0},\underbrace0_{k=1},\underbrace0_{k=2}, 0,0,...$$

For $V(3)$:

$$...,0,0,\underbrace{1}_{k=-2},\underbrace{0}_{k=-1},\underbrace1_{k=0},\underbrace0_{k=1},\underbrace1_{k=2}, 0,0, ...$$

For $V(4)$:

$$...,0,\underbrace1_{k=-3},\underbrace{0}_{k=-2},\underbrace{1}_{k=-1},\underbrace0_{k=0},\underbrace1_{k=1},\underbrace0_{k=2}, \underbrace1_{k=3},0, ...$$

So if e.g. you add three copies of $V(1)$, one of $V(3)$ and five of $V(4)$, we get that the sequence of $d$'s for $W= V(1)^3 \oplus V(3) \oplus V(4)^5$ is

$$...,0,\underbrace5_{k=-3},\underbrace{1}_{k=-2},\underbrace{5}_{k=-1},\underbrace4_{k=0},\underbrace5_{k=1},\underbrace1_{k=2}, \underbrace5_{k=3},0, ...$$

Now for this $W$, the sequence called $d_{2i}$ in your source takes just the even indices $k=2i$ of this, so it is

$$...,0,\underbrace{0}_{i=-2\\k=-4},\underbrace{1}_{i=-1\\k=-2},\underbrace4_{i=0\\k=0},\underbrace1_{i=1\\k=2},\underbrace0_{i=2\\k=4},0, ...$$

while the sequence called $d_{2i+1}$ singles out the odd indices and goes

$$...,0,\underbrace{5}_{i=-2\\k=-3},\underbrace{5}_{i=-1\\k=-1},\underbrace5_{i=0\\k=1},\underbrace5_{i=1\\k=3},0, ...$$

You will see that both highlighted sequences are unimodal and symmetric around $0$ (for the index $k$).

Torsten Schoeneberg
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  • Nice answer as always. But it might be worth mentioning that this all assumes the reps to be finite dimensional. – Tobias Kildetoft Sep 04 '19 at 20:29
  • The sequences of the numbers that have the described properties are the numbers $k$ and not $d_k =1 \text{ or }0$. Am I right? – Vasco Sep 05 '19 at 14:57
  • The sequences are the numbers $d$ with respect to the indices $k$ (a sequence consists of one number for each index). I've added an example. – Torsten Schoeneberg Sep 05 '19 at 17:10