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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!

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  • 1
    $\begingroup$ (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. $\endgroup$ – Qiaochu Yuan Sep 16 at 3:35
  • $\begingroup$ @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! $\endgroup$ – eightc Sep 16 at 6:16
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    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Sep 16 at 6:20
  • $\begingroup$ 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. $\endgroup$ – eightc Sep 16 at 6:23

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