2

Let $L=\mathfrak{sl}(2, \mathbb{F})$, $B$ a standard Borel subalgebra.
I am trying to solve exercise 20.4 from J.E. Humphreys "Introduction to Lie Algebras and Representation Theory", but I am stuck.
First of all $Z(\lambda)$ is constructed as $\mathfrak{U}(L)\otimes_{\mathfrak{U}(B)}D_{\lambda}$ by induced module construction. $1\otimes v^{+}$ is the maximal vector.
The exercise is to show that this is isomorphic to another definiton of $Z(\Lambda)$ given before. That means I have to finde a basis $(v_{0}, v_{1}, \dots )$ fulfilling:
a)$h.v_{i}=(\lambda-2i)v_{i}$
b)$y.v_{i}=(i+1)v_{i+1}$
c)$x.v_{i}=(\lambda-i+1)v_{i-1}$ with $v_{-1}=0$
for (x, h, y) the usual basis of $L$.
I don´t know exactly how $L$ acts on $Z(\lambda)$, but as $Z(\lambda)$ is an $\mathfrak{U}(L)$-module via the left multiplication, I thought it might be the same with $L$.
I tried the following:
$v_{0}=1\otimes v^{+}$
$v_{1}=y\otimes v^{+}$
$\dots$
$v_{i}=\frac{1}{i!} y^{i} \otimes v^{+}$
$\dots$
With this b) is obvious, but I don´t know how to continue with a) and c) or if this is even the correct idea.
Thank you for helping me.

Idun E.
  • 478
  • 1
    Could you verify that it worked for $v_0$ and $v_1$? It should just be a matter of remembering how a Lie algebra acts on a tensor product and remembering the relations of the universal enveloping algebra (i.e. the ones satisfied by the standard basis of the Lie algebra). – Tobias Kildetoft Jun 14 '15 at 18:51
  • Ok, if I try with $x$ acting on $v_{0}$ I get: $x.(1\otimes v^{+})=x.1\otimes v^{+} + 1\otimes x.v^{+} = x \otimes v^{+}$ I am not sure about the following, but as $x\in \mathfrak{U}(B)$ I would continue : $=1\otimes x.v^{+}=0$. But how would this with $y.v_{0}=y.1\otimes v^{+}+1\otimes y.v^{+}$ as $y.v^{+}$ is not defined? – Idun E. Jun 14 '15 at 19:40
  • Sorry, I forgot which tensor product this was. You only act on the left here, as that is how you turn this into a module for the larger algebra. – Tobias Kildetoft Jun 14 '15 at 19:44
  • So essentially I have to check how $x.y^{i}; \ y.y^{i}; \ h.y^{i} \in \mathfrak{U}(L)$ act? But how does $L$ act on $\mathfrak{U}(L)$? – Idun E. Jun 14 '15 at 19:51
  • 1
    You only have to check how $x$, $y$ and $h$ act. The Lie algebra is a subset of the enveloping algebra, so it just acts by multiplication. – Tobias Kildetoft Jun 14 '15 at 19:55
  • So b) should work as above. But aren´t $xy^{i}$, $hy^{i}$ and $y^{k}$ linearly independent and is it therefore even possible that a) and c) work with this basis? – Idun E. Jun 14 '15 at 20:09
  • 1
    I am not sure why those being linearly independent should matter. Try to sit down and write what happens for $v_0$ and $v_1$. Remember that in the enveloping algebra, you have for example $hy = yh - 2y$. – Tobias Kildetoft Jun 15 '15 at 06:56
  • Ok, I think I got it now: $x.v_{0}=x\otimes v^{+}= 1\otimes x.v^{+}=0$ and $x.v_{1}=xy\otimes v^{+}= yx\otimes v^{+} + h\otimes v^{+}= y\otimes x.v^{+} + 1\otimes h.v^{+}=\lambda(1\otimes v^{+})=\lambda v_{0}$ and similar with $h$. The rest should work via induction first for $h$ and then for $x$. – Idun E. Jun 15 '15 at 10:34
  • Yes, precisely. – Tobias Kildetoft Jun 15 '15 at 10:38

0 Answers0