Lie algebra cohomology

Let $\mathfrak{g}$ be a simple complex Lie algebra of $rank(\mathfrak{g})\geq2$ and dimension $d$. Fix a (non-zero) invariant bilinear form $(\cdot,\cdot)$ on $\mathfrak{g}$ and let $\{x_i\}_{1\leq i\leq d}$ be an orthogonal basis of $\mathfrak{g}$ with respect to $(\cdot,\cdot)$ satisfying $$[x_i,x_j] = f_{ij}^{\;\;k} x_k ,$$ where $f_{ij}^{\;\;k}$ are the structure constants of $\mathfrak{g}$ and the Einstein summation rule of the dummy indices is assumed. Denote $\kappa_{ij}=(x_i,x_j)$ and set $\kappa^{ij}=(\kappa_{ij})^{-1}$ such that $\kappa_{il}\kappa^{lj}=\delta_{i}^{\,j}$. Moreover set $f_k^{\;ji} = \kappa^{jl} f_{kl}^{\;\;i}$. Then $$f_k^{\;ji} [x_i,x_j] = c\,x_k ,$$ where $c$ is the the eigenvalue of the quadratic Casimir operator $C=\kappa^{ij} x_i x_j$ in the adjoint representation.

Set $\mathfrak{g}^{n}=\mathfrak{g}\wedge\cdots\wedge\mathfrak{g}$ to be the $n^{th}$ antisymmetric product of $\mathfrak{g}$ and set $V_m$ to be a vector space spanned by the totally symmetric polynomials $$\{x_{i_1},x_{i_2},\ldots,x_{i_m}\}=\frac{1}{m!}\sum_{\pi} x_{\pi(i_1)},x_{\pi(i_2)},\ldots,x_{\pi(i_m)}$$ of order $m$; here the sum is over all permutations $\pi$. Consider a category of maps $Hom(\mathfrak{g}^{n},V_m)$.

Q1) Have such maps been classified? Does anyone know a good reference for this? I'm particularly interested in the maps in $Hom(\mathfrak{g}^{3},V_5)$.

Q2) Let $n=m=3$. Then $\beta_{abc}^{ijk}=f_a^{\;il} f_b^{\;jm} f_c^{\;kn} f_{lmn}$ is a map $$\beta_{abc}^{ijk} : x_a\wedge x_b\wedge x_c \mapsto \{x_i,x_j,x_k\}.$$ This map appears in the r.h.s. of the Drinfeld Yangian terrific relation (see e.g. Theorem 12.1.1 in A Guide to Quantum Groups by Chari-Pressley). Is the map $\beta_{abc}^{ijk}$ the unique map in $Hom(\mathfrak{g}^{3},V_3)$? Here by uniqueness I mean that any other map in $Hom(\mathfrak{g}^{3},V_3)$ is equivalent to $\beta_{abc}^{ijk}$ up to a constant.

Q3) A hypothesis: let $n$ be odd and $m$ be even or on the contrary. Then $Hom(\mathfrak{g}^{n},V_m)$ is trivial. Could this be true?

• When you write Hom, do you mean linear maps without any restriction, or do you also impose equivariance? If the latter, you are asking about equivariant maps $\Lambda^n\mathfrak{g} \to \operatorname{Sym}^m\mathfrak{g}$. I don’t know of any general results, but once you fix $\mathfrak{g}$, $m$ and $n$, you can simply decompose $\Lambda^n\mathfrak{g}$ and $\operatorname{Sym}^m\mathfrak{g}$ into irreducible modules (e.g., using LiE) and see whether they have any irreducible modules in common. – José Figueroa-O'Farrill Dec 1 '13 at 12:48