# Irreducible representation of $sl_2(\mathbb{C})$: there is a unique irreducible representation for each non-negative $n$

I want to show that that for every non-negative $$n$$ there exists a unique irreducible representation $$V^{(n)}$$ of $$sl_2(\mathbb{C})$$ where $$\dim V^{(n)}=n+1$$. In other words, we need to show the following:

(1) There exists an irreducible representation $$V$$.

(2) Any two irreducible representation of the same dimension are isomorphic.

To show $$(1)$$, it's enough to define a vector space $$V$$ with linear operators $$H,X,Y$$ s.t. the basis of $$V$$ is given by $$\{v_0,v_1,\dots,v_n\}$$ where $$v_i=Y^iv_0$$ for $$i>0$$ and the operators satisfy the following equalities: $$Hv_i=(n-2i)v_i$$ $$Yv_i=v_{i+1}$$ $$Xv_i=i(n-i+1)v_{i-1} \text{ for all }v_i=Y^iv_0.$$ If such vector spaces exists, then it's enough to observe that the action defines a well-defined $$sl_2(\mathbb{C})$$-action on $$V$$ (for example, $$[X,Y]v=Hv$$ for all $$v\in V$$).

Finally, take $$V=\mathbb{R}^{n+1}$$, $$H= \begin{bmatrix} n & 0 & ... & 0\\ 0 & n-2 & ... & 0\\ ... & ... & ... & ...\\ 0 & 0 & ... & -n \end{bmatrix}, X= \begin{bmatrix} 0 & 1 & 0 & ... & 0\\ 0 & 0 & 1 & ... & 0\\ ... & ... & ... & ...\\ 0 & 0 & 0 & ... & 0 \end{bmatrix}, Y= \begin{bmatrix} 0 & 0 & 0 & ... & 0\\ 1 & 0 & 0 & ... & 0\\ 0 & 1 & 0 & ... & 0\\ ... & ... & ... & ...\\ 0 & 0 & 0 & ... & 0 \end{bmatrix}$$ and $$v_i=e_{i+1}$$ where $$e_{i+1}$$ is the standard basis vector. Let's show that $$V$$ is irreducible which will complete the proof of part $$(1)$$. We want to show if $$W\subset V$$ is $$sl_2\mathbb{C}$$-invariant subspace, then $$W=V$$. It's enough to show that $$e_i\in W$$ for some $$i$$. Indeed, using $$X$$ and $$Y$$, we can show that $$e_i\in W$$ for all $$i$$. From $$W\neq0$$ follows that we can take some $$x\neq0\in W$$ with $$x=a_1e_1+\dots+a_{n+1}e_{n+1}.$$ Then observe that $$Y^nx=a_1Y^ne_1=a_1e_{n+1}\in W$$. If $$a_1\neq0$$, then we are done. Otherwise, consider $$Y^{n-1}x$$. Since $$x\neq0$$, then there exists some $$a_i\neq0$$ which will assure that $$e_{n+2-i}\in W$$. So, the existence is proved if $$V$$ is irreducible.

Finally, let's show $$(2)$$. Let $$V$$ and $$W$$ be two irreducible representation of the same dimension. Then we have that $$V=\text{span}\{v_0,v_1,\dots,v_n\}$$ and $$W=\text{span}\{w_0, w_1,\dots,w_n\}$$ where $$v_i=Y^iv_0$$ and $$w_i=Y^iw_0$$ for $$i>0$$. Consider the map $$T:V\to W$$ which sends $$T(v_i)=w_i$$ and extend it linearly to a whole $$V$$. It's easy to check that $$T$$ is an isomorphism and $$sl_2(\mathbb{C})$$-invariant (i.e. $$T$$ is an intertwine). Therefore, $$V\cong W$$.

Are there another/faster arguments that can prove the statement? Thank you!

• (2) does not in any way follow from Schur's lemma, and is false for more general Lie algebras. You need to use specific facts about $\mathfrak{sl}_2$ to prove it. In 1) you haven't checked that the representation is irreducible. – Qiaochu Yuan Sep 16 at 3:35
• @QiaochuYuan thank you for your reply and your hints! I fixed my post. I should always check my assumptions lol Let me know if it looks better. Thanks! – eightc Sep 16 at 6:16
• 2) is closer but you should say more about how you know you can choose a basis that looks like that and how you know that $T$ is an intertwiner. 1) looks incomplete to me; you’ve maybe shown that $e_{n+1}$ is in $W$ but you’re not done showing you get everything else. – Qiaochu Yuan Sep 16 at 6:20
• Oh, I skipped those details in order not to make my post too long, and I don't know if you are supposed to post complete solutions here. – eightc Sep 16 at 6:23