I am trying to understand the following idea:
The Setup:
Consider the algebra $\mathfrak{sl}(2)$ with generators $\mathfrak{J}^0$,$\mathfrak{J}^{\pm}$. The algebra has the following commutation relations: $$ [\mathfrak{J}^0,\mathfrak{J}^{\pm}]= \pm \, 2 \, \mathfrak{J}^{\pm}, \quad[\mathfrak{J}^+,\mathfrak{J}^-]=\mathfrak{J}^0 $$ Then select the highest weight state $|k\rangle$ $$ \mathfrak{J}^+ |k\rangle =0, \quad \mathfrak{J}^0 |k\rangle = k|k\rangle $$ A module is spanned by the multiple application of lowering operators: $$ |k,n\rangle = (\mathfrak{J}^-)^n |k\rangle $$ Using the algebra relations one finds: $$ \mathfrak{J}^+|k,n\rangle = n(k-(n-1))|k,n-1\rangle $$ By construction $|k,0\rangle = |k\rangle$ is the highest weight.
The confusion
If $k \in \mathbb Z_{+}$ the state $|k,k+1\rangle$ is another highest weight with eigenvalue $k' = -2-k$ and therefore equivalent to $|k'\rangle$ $$ |k,k+1\rangle \hat{=} |-2-k\rangle $$ The eigenvalue $k'$ of this highest weight is negative and the lower representation therefore irreducible.
I don't understand the last part.
Edit: After the comments of Zachos and going through the calculation, I can see where the multiple highest weights come from. Just require $(k-(n-1))=0$ for highest weights.
Like that:
moving like -> with J+
|3,5> |3,4> |3,3> |3,2> |3,1> |3,0>
moving like <- with J-, you can move from |3,3> to |3,4> but the other way around is not possible.
Nevertheless, I thought the highest weights are unique. How is this compatible with the highest weight theorem?
The point of the paragraph is to introduce unitarity and multiplet shortening later.
To simplify notation, rename $\mathfrak{J}^0\mapsto J_0; ~~~\mathfrak{J}^{\pm}\mapsto J_{\pm}$, but move the 2 to the last commutator of charged generators, so the ladder ones move the eigenvalue of $J_0$ by one, not two!
Presumably you've learned that the operator $$ J^2\equiv J_-J_+ +J_0^2+ J_0= J_+J_- +J_0^2-J_0 $$ commutes with all tree Lie algebra elements above, so, it is simultaneously diagonalizable with them. One is interested in multiplets (modules) sharing eigenvalues of $J^2$, since the ladder operators cannot change those.
For simplicity, take $k=1$, and you may extend the following in the general case. You showed $$ J_+|1\rangle =0, \qquad J_0|1\rangle = |1\rangle \leadsto \\ J_-J_+|1\rangle =0 = (J^2 -1-1)|1\rangle ~~~\implies J^2|1\rangle = 2|1\rangle ,\leadsto \\ J^2|1,n\rangle =2|1,n\rangle. $$ Now, prove, by above, $$ J_0|1,1\rangle =0, \qquad J_0|1,2\rangle =-|1,2\rangle ,\leadsto \\ J_+|1,2\rangle =2|1,1\rangle, $$ non-vanishing.