Why complete symmetric polynomials and elementary symmetric polynomials are dual to each other? Here the definition of complete symmetric polynomial $h_{k}$ and elementary symmetric polynomial $e_{k}$ are:
$$
e_{k}=\sum_{1\le i_1<\cdots <i_k\le n}x_{i_1}\cdots x_{i_k}, h_{k}=\sum_{1\le i_1\le \cdots \le i_k\le n}x_{i_1}\cdots x_{i_k}
$$
I know that they are "dual" to each other in the symmetric function ring $\Lambda_{k}$, for the map $e_{k}\rightarrow h_{k}$ is an involution of the ring. But this does not explain some other beautiful dual relationship to me. 
For example in Pieri's formula we have
$$
s_{\lambda}e_{k}=\sum_{\mu\in \lambda\otimes 1^{k}}s_{\mu}, s_{\lambda}h_{k}=\sum_{\mu\in \lambda\otimes k}s_{\mu}
$$ 
And in the reverse side using Kostka numbers we have
$$
h_{\mu}=\sum K_{\lambda \mu}s_{\lambda}, e_{\mu}=\sum_{\lambda}K_{\lambda \mu}s_{\lambda^{*}}
$$
The Jacobi-Trudi formula claim that for $|\lambda|\le n$, we have
$$
s_{\lambda}=\det(h_{\lambda_i-i+j})_{1\le i,j\le n},s_{\lambda^{*}}=\det(e_{\lambda_i-i+j})_{1\le i,j\le n}
$$
To me, all these suggests that there some deeper relation underlying these dualizing relationships. The formulas are so stunningly beautiful that they cannot come from mere coincidence in computation. In particular if we consider their action on Schur polynomials using the Tableaux, we can visualize the dual relationship. I want to ask, is there any deep reason behind these dual relationships? The definition itself seems to reveal very little and I felt very puzzled by the unexpected beauty.  
 A: I think of all of the duality statements you wrote as a consequence of the fact that there is a ring involution of $\Lambda$ sending $e_k$ to $h_k$, so let me give a manifestation of that. First, the $e_k$ and $h_k$ are algebraically independent generators, so the existence of an automorphism given by $e_k \mapsto h_k$ or by $h_k \mapsto e_k$ is not that interesting, and the important fact is that these two maps are inverses of one another. That can be derived from the relation
$\displaystyle \sum_{i+j=n} (-1)^i e_i h_j = \delta_{0,n}$.
I think of this as coming from the Koszul complex: $e_i$ is the character of the $i$th exterior power functor and $h_j$ is the character of the $j$th symmetric power functor, and the Koszul complex is
$Sym(V) \gets V \otimes Sym(V) \gets \wedge^2(V) \otimes Sym(V) \gets \cdots$
which is graded and exact in positive degrees. The Euler characteristic of the degree $n$ piece is the identity I stated. This identity can be proven more directly without the Koszul complex, but hopefully this perspective is useful for you.
