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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? –  16278263789 Nov 1 '11 at 23:00
    
(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 $\{v_i:i\geq j\}$. –  Mariano Suárez-Alvarez Nov 1 '11 at 23:11
    
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. –  16278263789 Nov 1 '11 at 23:25
    
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. –  Mariano Suárez-Alvarez Nov 1 '11 at 23:32
    
This is exercise 7b in Humphreys. he hasn't talked about Verma modules. I think I'm supposed to do it without Verma modules. –  16278263789 Nov 2 '11 at 0:42
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