2
$\begingroup$

Let $a, b \in \mathbb{C}$ and let $V$ be an infinite dimensional vector space with basis $\{v_i | I \in \mathbb{Z}\}$. Define linear transformations $h$ and $f$ on $V$ by $$h(v_i) =(a+2i) v_i$$ and $$f(v_i) = v_{i-1} \forall i \in \mathbb{Z}$$.

Question 1: Set $e(v_0) = bv_1$. Show that there is a unique way to define $e(v_i)$. Show that there is a unique way to define $e(v_i)$ for all $i \in \mathbb{Z}$ in order to make V a module over $\mathcal{sl_2}$

Answer: I was able to figure it out by induction and I got $$e(v_i) = (b-i(a+i+1))v_{i+1}$$

Question 2: Show that any non-zero $sl_2$ sub-module $U$ of $V$ contains some $v_i$

Answer: I don't think I know what to do here but my understanding is to find a finite dimensional submodule $U$ of $V$ such that the linear transformation $h$ sends element in $U$ to some elements in $V$.

Question 3: Show that $V$ is reducible for if $$b= ja + j(j+1)$$ for some $j \in \mathbb{Z}$ and $V$ is irreducible if $b \neq ja + j(j+1)$ for all $j \in \mathbb{Z}$

Answer: if $b= ja + j(j+1)$ then $$e(v_j) =0$$ and showing that the

$$W = span \{.... \cdots....v_{j-1} , v_j\}$$ is a submodule of $V$ is enough. For the second part, I am guessing I have to show that V is generated by a single element say $v_{j+1}$ but I don't know how to show this.

P.S: I am pretty much new to Lie algebra. I will appreciate hints, suggestions, corrections and answers. Thanks.

$\endgroup$
3
$\begingroup$

You should state that $\text{sl}_2(\mathbb{C})$ is the $\mathbb{C}$-span of $h$, $e$, and $f$ with $[e,f]=h$, $[h,e]=2e$, and $[h,f]=-2f$. Not everybody can understand what $h$, $e$, and $f$ are.

More generally, what Question 2 asks you to do is to prove that, if $\mathfrak{h}$ is a Cartan subalgebra of a finite-dimensional semisimple Lie algebra $\mathfrak{g}$ and $M$ is an $\mathfrak{h}$-weight $\mathfrak{g}$-module, then any $\mathfrak{g}$-submodule of $M$ is an $\mathfrak{h}$-weight $\mathfrak{g}$-module. To prove the version in this question, suppose that $u$ is a nonzero element of a nonzero $\text{sl}_2(\mathbb{C})$-submodule $U$ of $V$. Then, $u=t_1\,v_{i_1}+\ldots+t_k\,v_{i_k}$ for some $t_1,\ldots,t_k\in\mathbb{C}\setminus\{0\}$ and for some integers $i_1<i_2<\ldots<i_k$. If $k=1$, we are done. Assume now that $k>1$.

First, observe that $$h^p\cdot u=\sum_{\mu=1}^k\,t_\mu\,\left(a+2i_\mu\right)^p\,v_{i_\mu}\,.$$ Take $P(X):=X-\left(a+2i_k\right)$ and $Q(X):=\prod_{\mu=1}^{k-1}\,\Big(X-\left(a+2i_\mu\right)\Big)$. Then, there exist polynomials $f(X),g(X)\in\mathbb{C}[X]$ such that $f(X)\,P(X)+g(X)\,Q(X)=1$. Now, we have $$g(h)\,Q(h)\cdot u=\sum_{\mu=1}^{k-1}\,t_i\,g(h)\,Q(h)\cdot v_{i_\mu}+t_k\,\big(1-f(h)\,P(h)\big)\cdot v_{i_k}\,.$$ Show that, for $\mu=1,2,\ldots,k-1$, $Q(h)\cdot v_{i_\mu}=0$ and $P(h)\cdot v_{i_k}=0$. Consequently, $t_k\,v_{i_k}=g(h)\,Q(h)\cdot u\in U$, whence $v_{i_k}\in U$. (To be rigorous, $h^p$, $f(h)$, $g(h)$, $P(h)$, and $Q(h)$ are viewed as elements of the universal enveloping algebra $\mathfrak{U}\big(\text{sl}_2(\mathbb{C})\big)$.)

For Question 3, use Question 2. Show that, if $b\neq j(a+j+1)$ for any $j\in\mathbb{Z}$, then $e^p\cdot v_i$ is a nonzero scalar multiple of $v_{i+p}$ for any $p$ and $i$.

$\endgroup$
  • $\begingroup$ Do you happen to know if the "preservation of Jordan normal form" theorem in Fulton and Harris holds for infinite dimensional representations? (If it does, then for question (2), $H$ would be diagonalisable when acting on $U$, so surely $U$ will have a basis of $v_i$'s?) $\endgroup$ – Kenny Wong Mar 7 '17 at 18:28
  • $\begingroup$ Can you state that theorem? I don't have access to the book at the moment. $\endgroup$ – Batominovski Mar 7 '17 at 18:31
  • $\begingroup$ Thank you Batominovski. I should be able to figure it out now. $\endgroup$ – Jaynot Mar 7 '17 at 18:37
  • $\begingroup$ "If $\rho : \mathfrak{g} \to \mathfrak{gl}(V)$ is any representation of a semisimple Lie algebra $\mathfrak{g}$, then $\rho(X_s)$ is the semisimple part of $\rho(X)$ and $\rho(X_n)$ is the nilpotent part of $\rho(X)$." (Here $X = X_s + X_n$ is the Jordan decomposition of $X$, viewed as a matrix acting on $\mathfrak{g}$ itself via the adjoint rep. Apparently $X_s$ and $X_n$ are themselves in $\mathfrak{g}$.) $\endgroup$ – Kenny Wong Mar 7 '17 at 18:37
  • $\begingroup$ It's not clear the representation $V$ in this theorem is assumed to be finite. $\endgroup$ – Kenny Wong Mar 7 '17 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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