# Showing a rep of $sl(2,\mathbb{K})$ is irreducible

Let $V$ be a $m+1$-dim $K$-vector space with char$K=0$. Let $(v_0,v_1,\dots,v_m)$ be a basis of $V(m)$. Now suppose I construct a representation of $sl(2,K)$ on this representation.

How do I show that this representation is irreducible?

A rep is called irreducible if the only representation subspace $U\subset V$ left invariant by the action of my algebra is $V$ itself and the zero subspace. But in general how does one show that this is indeed true for a given example?

For example, I have a representation. Now I can create a proper subspace $U$ by removing at least one basis vector from $V$, say $v_i$, so $U$ is spanned by $$(v_0,v_1,\dots,v_{i-1},v_{i+1},\dots, v_m).$$ Now I can show that this subspace is not closed under the action of my algebra. Then I can continue and remove all possible combinations of the basis vectors and find that none are invariant until I get to the zero space.

Is this a valid approach? Is there a more direct way?

EDIT: Here is a specific example. Let $X,Y,H$ be the basis of $sl(2,K)$ with $$X=\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right), \quad Y=\left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right), \quad H=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right).$$

I define $v_{-1}=0=v_{m+1}$ and $v_i=\frac{1}{i!}Y^iv_0$, where $v_0$ is a maximal vector. $\lambda$ is a highest weight of $H$. I represent the action of $X,Y,H$ on $V$ by $$X.v_i=(\lambda-i+1)v_{i-1}$$ $$Y.v_i=(i+1)v_{i+1}$$ $$H.v_i=(\lambda-2i)v_i$$ Now I can create proper subspaces by removing basis vectors from $V$. Remove $v_i$ and define $U$ to be the subspace spanned by $$(v_0,\dots,v_{i-1},v_{i+1},\dots,v_m)$$ Then it is clear that by acting with $X$ on $v_{i+1}$ I will obtain $v_i$ so this subspace is not closed. I can repeat this and remove all possible combinations of basis vectors from $V$, each time defining a new subspace. It is clear that unless I remove none (ie. start with $V$) or all (i.e start with zero space) then no subspace will be invariant, as I can always act with $X$ or $Y$ a sufficient number of times to get a vector not in this subspace.

Hence the representation is irreducible.

Is this correct?

• I think the action of $X$ and $Y$ should shift the index $i$ of $v_i$, in other words have $v_{i+1}$ or $v_{i-1}$ in the right hand side. – Marc van Leeuwen Jun 14 '16 at 20:45
• Yes absolutely, a typo on my behalf. – qftey Jun 14 '16 at 21:39

In the example a key fact is that the vectors $v_i$ already form a basis of diagonalisation for the action of $H$, and that all the eigenvalues are distinct. This implies that the only subspaces invariant for $H$ are those spanned by a subset of the vectors$~v_i$. If such a subspace contains some $v_i$, then in order to be stable for $X$ and $Y$ it must (with the proper formulas) also contain $v_{i-1}$ and $v_{i+1}$ if these exist. Then the only subsets of the $v_i$ that span a subspace invariant under $H,X,Y$ are the empty set and the set of all the $v_i$. This shows the representation is irreducible.