Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\alpha-i+1).v_{i-1}$ 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.

share|cite|improve this question

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.