# Question regarding the definition of the action of $sl_2$ on a vector space

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 $$(a e) \cdot (b x) := ab (e \cdot x)?$$

• 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. – user45765 Mar 9 at 16:57
• What is $U(sl_2)$? – Johnny T. Mar 9 at 17:02
• 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 $x,y\in sl_2$. – user45765 Mar 9 at 17:14
• "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. – Dietrich Burde Mar 9 at 17:14
• @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..? – Johnny T. Mar 9 at 17:59