Codex @codex  0

Here representations mean finite-dimensional rational representations of $S{L}_2(\mathbb{C})$. Put $W={\mathbb{C}}^2$. The irreducibles areThe invariant alternating form on $W$ identifies $W$ with $W^{\ast }$, so these can equally be realized as binary forms of degree $d$. With basis $e_j=x^{d-j}{y}^j$, the standard Lie algebra generators act asThe distinct torus weights and these nonzero lowering and raising coefficients show irreducibility: any nonzero invariant subspace contains a weight vector, and repeated applications of $E,F$ obtain the entire basis.

For exhaustion, choose a highest-weight vector $v$ of an irreducible representation, so $Hv=mv$, $Ev=0$. The sl2 highest-weight lowering formula, obtained by induction from $[E,F]=H$, is $E{F}^{j}v=j(m-j+1)F^{j-1}v$. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, applying $E$ to the latter equality gives $(r+1)(m-r)F^{r}v=0$, hence $m=r$. The vectors $v,Fv,\dots ,F^{r}v$ have distinct weights and span an invariant submodule, so irreducibility makes it the whole representation. After rescaling basis vectors it is $V_r$. Complete reducibility from Question 2 now describes every representation as a finite direct sum of these symmetric powers. Their central element $-I$ acts by $(-1{)}^d$.

The symbolic method for binary-form invariants works as follows. Write a degree-$d$ form as $f(x,y)={\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{d}{j}\right)f_j{x}^{d-j}{y}^j$ and introduce a formal letter $a=(a_0,a_1)$. The formal power $(a_{0}x+a_{1}y{)}^d$ is converted to $f$ by the linear substitution $a_0^{d-j}{a}_1^j\mapsto f_j$. This is an umbral linear rule, not a claim that every form is an actual power of a linear form.

For a coefficient polynomial of degree $r$, polarize it into $r$ copies of the form, represented by letters $a,b,\dots$. Each letter must occur with degree $d$. In characteristic zero, polarization and evaluation on equal forms recover the original coefficient polynomial. The basic invariant expressions in the letters are bracketsThey are invariant because $\det g=1$. Invariant tensors on copies of a two-dimensional space are spanned by pairwise alternating contractions; equivalently, the invariant polynomials in the letters are generated by these brackets. One can see the tensor statement from the two-row Young-symmetrizer description of the determinant powers in tensor space. Their basic relations include $(ab)=-(ba)$ and $(ab)(cd)-(ac)(bd)+(ad)(bc)=0$. Multiply brackets, ensure the required degree in each letter, symmetrize letter copies as needed, and apply the umbral substitutions. This gives all coefficient invariants, and also works for joint invariants of several forms.

For example, for $f=A{x}^2+2Bxy+C{y}^2$, the expression $(ab{)}^2$ becomes $2(AC-B^2)$, the quadratic discriminant up to its conventional scalar. Contractions leaving variable indices unpaired give covariants. A useful systematic construction is the transvectant of binary forms:For forms of degrees $m,n$, this is an equivariant form of degree $m+n-2r$. The corresponding tensor decomposition isIts character identity follows by multiplying the weight sums; complete reducibility then gives the decomposition, and alternating contractions supply the displayed maps onto its summands.