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Let $L=\mathfrak{sl}(2,\mathbb{F})$ with the usual basis $(x, \ y, \ h)$ and $\text{char}\,\mathbb{F}=0$.
Let $Z(\lambda)$, $\lambda\in\mathbb{F}$ the infinite-dimensional $L$-module spanned by $\left(v_{0}, \ v_{1},\dots\right)$, $v_{i}=\frac{1}{i!}y^{i}\cdot v_{0}$ for which:
a) $h\cdot v_{i}=(\lambda-2i)v_{i}$,
b) $y\cdot v_{i}=(i+1)v_{i+1}$, and
c) $x\cdot v_{i}=(\lambda-i+1)v_{i-1}$ (where $v_{-1}=0$).

I was trying to proof the two statements:
I) Every submodule has a maximal vector.
II) If $\lambda+1$ is not a positive integer, then $Z(\lambda)$ is irreducible.

I tried the following:
I) My problem lies with the infinite-dimensional submodules:
First, let $U\subseteq Z(\lambda)$ be an infinite-dimensional submodule and $0\neq u\in U$ arbitrary. Write $u=\sum\nolimits_{i=0}^{t}\lambda_{k_{i}}v_{k_{i}}$ so that all $\lambda_{k_{i}}\neq 0$. It is $h\cdot u=\sum\nolimits_{i=0}^{t}\lambda_{k_{i}}(\lambda-2k_{i})v_{k_{i}}$ so $\color{red}{h\text{ acts diagonally on }U\text{ and therefore }v_{k_{i}}\in U}$. So $U$ is the direct sum of $\mathbb{F}v_{j}$, $j\in J\subseteq\mathbb{N}$. Let $r$ be minimal for $v_{r}\in U$, so $v_{r-1}\notin U$. But since $U$ is a submodule $x\cdot v_{r}$ has to be in $U$. This means $x\cdot v_{r}=0$ and $v_{r}$ is a maximal vector.
II) Suppose $Z(\lambda)$ is not irreducible, than a submodule $U\neq0$ exists. By the preceding paragraph, it has a maximal vector. $\color{red}{\text{It has the form }v_{k}}$, where $k>0$, since otherwise $U$ would be all of $Z(\lambda)$. That means it is: $0=x\cdot v_{k}=(\lambda-k+1)v_{k-1}$ since $v_{k-1}\neq 0$ it is $\lambda-k+1=0\iff \lambda+1=k$. This contradicts the choice of $\lambda$.
I have highlighted the parts that I think that could be problematic. For the first one I am not sure if it is that easy and for the second if the maximal vector really has to have that form.

Thank you for helping me.

Cheynne229
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Idun E.
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  • Please use "\cdot" instead of "." for the multiplication. Also, this is how you spell the following words: "module," "submodule," "positive," "infinite," and "direct." – Batominovski Jul 21 '15 at 09:13
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    I agree with Batominovski that you have not proved that $v_{k_i}\in U$. Here's a hint: Assume that $u$ has only two terms, say $u=a_1v_{k_1}+a_2v_{k_2}$ with $k_1\neq k_2$ and $a_1,a_2$ both non-zero. Can you then write $v_{k_1}$ and $v_{k_2}$ as linear combinations of $v$ and $h\cdot v$? Can you then generalize this to the case where $v$ has three terms, four ... by induction? – Jyrki Lahtonen Jul 21 '15 at 12:34
  • @ Batominovski: Thank you, my english is bit rusty and I forgot to turn of auto correct. I will proofread more carefully in the futur. – Idun E. Jul 21 '15 at 16:50
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    @ Jyrki Lahtonen: Thank you! I think I know, what to do, though I only got scalar multiple of $v_{k_{1}}$: As you said I started with $u=a_{1}v_{k_{1}}+a_{2}v_{k_{2}}$. Then $h\cdot u=a_{1}(\lambda-2k_{1})v_{k_{1}}+a_{2}(\lambda-2k_{2})v_{k_{2}}$ and $-2a_{1}(k_{1}-k_{2})v_{k_{1}}=h\cdot u - (\lambda-2k_{2})u$ is in $U$. For the induction step: $u=a_{1}v_{k_{1}}+\dots +a_{n}v_{k_{n}}$. $v=h\cdot u -(\lambda-2k_{1})$ eliminates the first summand and $v$ is a linear combination of $v_{k_{2}},\dots, v_{k_{n}}$. – Idun E. Jul 21 '15 at 17:10
  • The induction hypothesis yields that scalar multiples of $v_{k_{2}},\dots, v_{k_{n}}$ are in $U$. By adding $-a_{2}v_{k_{2}}-\dots - a_{n}v_{k_{n}}$ to $u$ one gets a scalar multiple of $v_{k_{1}}$ in $U$ and the hypothesis followes. – Idun E. Jul 21 '15 at 17:14
  • That's the idea. Tip: do no put space between the @-character and the username. It no longer works as a ping. I was just checking questions on Lie algebras and came back. – Jyrki Lahtonen Jul 24 '15 at 22:05
  • @JyrkiLahtonen: Thank you for the tip! I did not even know, it works that way! – Idun E. Jul 24 '15 at 23:24

1 Answers1

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I. I think you are trying to show that $U$ is also a weight $L$-module. However, I'm not sure if you understand how "$h$ acts diagonally on $U$" would imply $v_{k_i}\in U$. Do you know how to prove this? (I can give you hints, but I'm not sure you need them.) You are definitely right that $U$ is the direct sum of $\mathbb{F}v_j$ whenever $v_j$ occurs as a summand in an element of $U$.

In addition, I'm not sure what you meant by "But since $U$ is a submodule, $x\cdot v_r$ has to be." Can you clarify on that? Nonetheless, the conclusion is correct.

II. You might want to say "a proper submodule $U\neq 0$ exists." The red part is better written as "This maximal vector may be taken to be $v_k$ for some $k>0$." Since all of your weight spaces are either $0$- or $1$-dimensional, you are safe. The same cannot be said for, for example, representations of $\mathfrak{sl}_3$.

Non-Inductive Proof of the Red Text in I.

Let $\mu_i$ be the weight of $v_{k_i}$ (i.e., $\mu_i=\lambda-2k_i$). Then, consider the element $$h_i:=\left(h-\mu_1\right)\left(h-\mu_2\right)\cdots\left(h-\mu_{i-1}\right)\left(h-\mu_{i+1}\right)\cdots \left(h-\mu_t\right)$$ in the universal enveloping algebra $\mathfrak{U}(L)$ of $L$. Clearly, $h_i\cdot u \in U$. However, $$h_i\cdot u=\Bigg(\lambda_{k_i}\,\prod_{j\neq i}\,\left(\mu_i-\mu_j\right)\Bigg)\,v_{k_i}\,.$$ Therefore, $\Bigg(\lambda_{k_i}\,\prod_{j\neq i}\,\left(\mu_i-\mu_j\right)\Bigg)\,v_{k_i}\in U$. Since $\lambda_{k_i}\,\prod_{j\neq i}\,\left(\mu_i-\mu_j\right)\neq 0$, $v_{k_i}\in U$.

Batominovski
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  • Thank you for the answer! That was exactly the point I struggled with, but I think I got it now (see comment above). I meant, since $U$ is a submodule, $x$ has to send $v_{r}$ back to an element in $U$ and as the $v_{i}$ are linearly independent it has to be send to $0$. Thank you, that answers the question for II. – Idun E. Jul 21 '15 at 17:19
  • There is a non-inductive proof of your statement. See my edit above. – Batominovski Jul 21 '15 at 21:06