Let $L$ be a simple $sl_2$-module of finite dimension over a field $K$ of characteristic $0$. Say $e,f,h$ is a basis of $sl_2$. In the notes I am reading it only defines the action of $e,f,h$ on the basis elements of $L$. From this I see that we get the action of $e,f,h$ on all elements of $L$.

I was wondering: Is the action of other elements of $sl_2$ determined by linearity? i.e. for example, given any $a, b \in K$ and $x \in L$, we define $$ (a e) \cdot (b x) := ab (e \cdot x)? $$

I was getting confused because it doesn't say anything about this. Thank you very much.

  • $\begingroup$ If $L$ is $sl_2$ module, then it is saying there is ring morphism $U(sl_2)\to End_K(L)$. Then the action is extending by linearity as $e,f,h$ spans $sl_2$ over $K$ via this ring homomorphism. $\endgroup$ – user45765 Mar 9 at 16:57
  • $\begingroup$ What is $U(sl_2)$? $\endgroup$ – Johnny T. Mar 9 at 17:02
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    $\begingroup$ If you have defined the module, then this is giving $sl_2\to End_K(L)$ morphism and this extends to tensor algebra but you also have $[x,y]-xy+yx$ inside the kernel of extended tensor algebra $T(sl_2)\to End_K(L)$. $U(sl_2)=T(sl_2)/([x,y]-xy+yx,...)$ where the quotient runs through all elements $x,y\in sl_2$. $\endgroup$ – user45765 Mar 9 at 17:14
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    $\begingroup$ "I was getting confused because it doesn't say anything about this." On the contrary, it says that $span\{e,f ,h\}$ is a vector space, and the module $L=V$ is also a vector space. $\endgroup$ – Dietrich Burde Mar 9 at 17:14
  • $\begingroup$ @user45765 When you say the $sl_2→End_K(L)$ morphism, this is a morphism of algebras? Also I guess the short answer to my original question is yes..? $\endgroup$ – Johnny T. Mar 9 at 17:59

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