Symmetrizers and Antisymmetrizers

\(\newcommand{\mb}{\mathbb}\)Let's talk about permutations.

Suppose we have 3 things, which we'll denote by \(a, b\) and \(c\). We can line them up in a row as \(abc\), or we can reorder them into, say, \(bca\), or \(cba\), and so on.
The permutations, not the orderings themselves but the movement from one order to another, form a group, in this case \(S_3\).
Every permutation can be achieved by repeatedly swapping two elements at a time. So we can get from \(abc\) to \(bca\) by staring with \(abc\) and first swapping \(a\) and \(b\) to get \(bac\) and then swapping \(a\) and \(c\) to get to \(bca\). For a permutation \(\sigma\), define \(l(\sigma)\) be the minimum number of two-element swaps needed to make \(\sigma\).

Now consider a tensor \(t_{abc}\). From it we can make a new tensor, \(t_{bca}\), or \(t_{cba}\), and so on.
We define a symmetrizer \(Sym_3\) to act on \(t_{abc}\) by sending it to
$$\frac{1}{3!}\sum_{\sigma \in S_3} t_{\sigma(a)\sigma(b)\sigma(c)} = \frac{1}{6}(t_{abc} + t_{bac} + t_{bca} + \cdots )$$ We also define an antisymmetrizer \(Asym_3\) to act on \(t_{abc}\) by sending it to
$$\frac{1}{3!}\sum_{\sigma \in S_3} (-1)^{l(\sigma)} t_{\sigma(a)\sigma(b)\sigma(c)} = \frac{1}{6}(t_{abc} - t_{bac} + t_{bca} - \cdots )$$ A few things to note about \(Sym_3\) and \(Asym_3\): they're idempotent, in other words \(Sym_3 \circ Sym_3 = Sym_3, Asym_3 \circ Asym_3 = Asym_3\). They're also self-adjoint, which I'm not going to explain here because it's not terribly interesting for my purposes. But it does mean that we can view \(Sym_3\) and \(Asym_3\) as projection operators, projecting from general \(3\)-index tensors to subspaces of symmetric and antisymmetric 3-index tensors.
Also, they're orthogonal, in that \(Asym_3 \circ Sym_3 = Sym_3 \circ Asym_3 = 0\); the two subspaces are disjoint.
We can, of course, talk about symmetrizers and antisymmetrizers for two index tensors, or four index tensors. For any \(n\), we can consider
$$Sym_n(t_{a_1\ldots a_n}) = \frac{1}{n!}\sum_{\sigma \in S_n} t_{\sigma(a_1)\ldots\sigma(a_n)}$$ $$Asym_n(t_{a_1\ldots a_n}) = \frac{1}{n!}\sum_{\sigma \in S_n} (-1)^{l(\sigma)}t_{\sigma(a_1)\ldots\sigma(a_n)}$$ and again we have that $$Sym_n \circ Sym_n = Sym_n, Asym_n \circ Asym_n = Asym_n$$ $$Asym_n \circ Sym_n = Sym_n \circ Asym_n = 0$$We can also talk about partial symmetrizers and partial antisymmetrizers, which act on only some of the indices of a tensor. We note, for instance, that applying a (partial) symmetrizer to some indices, and then another partial symmetrizer to a subset of those indices is the same as just applying the first symmetrizer, and that applying a partial symmetrizer and a partial antisymmetrizer can yield something other than 0, but only if they don't have more than one index in common.

So for any \(n\), we can look for projection operators on \(n\)-index tensors that we can build out of (partial) symmetrizers and antisymmetrizers, and indeed look for minimal projection operators, so that for any projection operator \(P\) in our collection, for any other projection operator \(Q\) on \(n\)-index tensors built from symmeterizers and antisymmetrizers, \(P \circ Q = Q \circ P\) and the result is either \(P\) or \(0\).
There's a well-known combinatorial setup for finding such sets of operators, usually via what are called Young Tableaux. Briefly, an \(n\)-box Young Tableau is a collection of \(n\) square boxes arranged in rows and columns so that each row is at least as long as the one below it and each column is at least as tall as the one to the right of it. If you take a Young Tableau and flip it along the diagonal, you get another Young Tableau which I'll call the conjugate Young Tableau.
We can associate \(n\)-box Young Tableaux with projections on \(n\)-index tensors by taking a Young Tableau and associating each box with an index. Then for each row, we symmetrize over the indices corresponding to the boxes in the row. Once we've finished with the rows, then for each column we antisymmetrize over the indices corresponding to the boxes in the column.
The result is a projection that is orthogonal to all other projections made from \(n\)-box Young Tableaux, and it is minimal amongst projections made from symmetrizers and antisymmetrizers.
And there you have, essentially, the representation theory of \(SL(k)\). To get irreducible representations of \(SL(k)\), you take \((\mb C^k)^{\otimes n}\) with the standard action of \(SL(n)\) on it, and then apply various projections built from \(n\)-box Young Tableaux.
This also gives you representations of \(S_n\), in that you take a tensor \(t_{a_1,\ldots,a_n}\) with no symmetry properties, and take the vector space of linear combinations of \(t_{\sigma(a_1),\ldots,\sigma(a_n)}\) for all permutations \(\sigma \in S_n\); this gives you an \(n!\)-dimensional vector space on which \(S_n\) acts, and applying the various projections made from \(n\)-box Young Tableaux gets you irreducible representations of \(S_n\).
There's a nice interplay of the actions of \(SL(k)\) and \(S_n\) on \((\mb C^k)^{\otimes n}\) via what is called Schur-Weyl duality, but I'm not going to go into that here.