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Let $\alpha$ be an arbitrary scalar in $\mathbb{C}$ and let $V(\alpha)$ be an infinite dimensional $\mathbb{C}$-vector space (with a countable basis). The formulas $h.v_i=(\alpha -2i).v_i$, $f.v_i=(i+1).v_{i+1}$ and $e.vi=(\alpha-i+1).v_{i-1}$ define an $sl(2,\mathbb{C})$-module structure.

If $\alpha+1=i$ is positive then $v_i$ is a maximal vector (it is easy to see this). Now I need to see that $v_0 \mapsto v_i$ induces a homomorphism $\phi: V(\alpha-2i) \rightarrow V(\alpha)$, that $\phi$ is injective, $Im(\phi)$ and $V(\alpha)/Im(\phi)$ is irreducible. I proved that $\phi$ is injective (I found $\phi(v_i)=(i+1)...(i+j)v_{i+j}$ )and I have an idea how to prove irreducibility but I don't exactly know what the image look like. Thx.

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The submodules of $V(\alpha)$ are all spanned, as vector spaces, by subsets of the basis $\{v_i:i\geq0\}$. This is easy to see using the fact that $h$ acts diagonally on the basis (and such classics as Vandermonde's determinant)

Moreover, by looking at the action of $e$ and $f$, it is also easy to see that a submodule if $V(\alpha)$ is in fact spanned by a contiguous subset of $\{v_i:i\geq0\}$. This means that there are not many options for the image of your map.

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    I am sorry I don't understand the use of the word contiguous. Is $Im(\phi)=${$(i+1)...(i+j)v_{i+j}: i\geq 0$}? Also is my formula for $\phi$ true?2011-11-01
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    (Please use a more sensible user name...) The space spanned by $\{(i+1)...(i+j)v_{i+j}: i\geq 0\}$ is the same as the space spanned by $\{v_i:i\geq j\}$.2011-11-01
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    What do the $V(\alpha)$ look like actually? What does it mean for an element to be in $V(\alpha)$? By the action of $e$ and $f$ it seems like we should just get the whole basis back.2011-11-01
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    Can you answer the same question for the finite dimensional simple modules of $\mathfrak{sl}_2$? The answer is very similar and in exactly the same spirit. If you do not know the finite dimensional version, it might be very useful for you to drop for a while the Verma modules and read a bit on it before.2011-11-01
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    This is exercise 7b in Humphreys. he hasn't talked about Verma modules. I think I'm supposed to do it without Verma modules.2011-11-02
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    @16278263789, $V(\alpha)$ *is* a Verma module! In any case, my suggestion does not imply at all that «you do it with Verma modules», whatever that may mean: I suggested that you try to understand better the finite dimensional modules before trying to do the infinite dimensional ones. For example, you asked «what dothe $V(\alpha)$ look like actually?» and the answer to that is «xactly like the finite dimensional modules but bigger». To make sense ofth answer you need to know how the finite dimensional modules look like, though...2011-11-02
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    thank you. I think I solved it now.2011-11-02
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    @16278263789, you should writ an answer to this question yourself explaining the solution and accept it, then :)2011-11-02