This is a follow-up of my this post. There is a nice way to represent and visualize partitions of $n$. The objects we will define will admit a natural action from $S_n$ and will permit us to think about the symmetric group in a very combinatorial way. Suppose $\lambda = \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_k$ is a partition of $n$. Then the Young diagram of shape $\lambda$ is an arrangement of $n$ boxes, there is $k$ rows and the $i$‘th row, starting from the top, contains $\lambda_i$ boxes. For example, here is the associated Young diagram for $n = 8$ and $\lambda = (3,2,2,1)$ :

If $\lambda \vdash n$, there is also the notion of a Young tableau of shape $\lambda$. This is simply a Young diagram of shape $\lambda$ filled with the numbers in $\{ 1, \ldots, n \}$ (with no repeats). A tableau of shape $\lambda$, or $\lambda-$tableau, is denoted by $t^{\lambda}$. For example, here are all the $\lambda-$tableaux for $\lambda = 2,1$ :

The entries in a tableau are designated as in a matrix, so the first tableau above has entries $t_{1,1} = 1, t_{1,2} = 2, t_{2,1} = 3$.

There is a natural action of $S_n$ on a tableau $t = (t_{ij})$ of shape $\lambda \vdash n$ defined as $\sigma \cdot t = (\sigma(t_{i,j}))$ for $\sigma \in S_n$. i.e. $\sigma$ just permutes the entries of the tableau. This action is clearly transitive.

The next step is to define an equivalence relation on the set of $\lambda-$tableaux which will be important for our construction of the irreducible representations of the symmetric group. If $t_1, t_2$ are two $\lambda-$tableaux, define them to be row equivalent if corresponding rows contain the same numbers and note this by $t_1 \sim t_2$. A $\lambda-$tabloid $[t]$ is then defined as an equivalence class of $\lambda-$tableaux for this relation. For example, here is a $(3,1)-$tableau:

The notation without the vertical bars is to suggest that the arrangement of numbers in different rows is irrelevant. Here is the set of all the $(2,1)-$tabloids:

The action of $S_n$ on tableaux induces an action of $S_n$ on tabloids. This gives, for every $\lambda \vdash n$, a representation of $S_n$ on the complex vector space $M^{\lambda} := \mathbb{C}\{ [t_1], \ldots, [t_k] \}$ generated by all $\lambda-$tableaux $[ t_1], \ldots, [ t_k ]$. Let’s look at some examples for different partitions of $n$:

If $\lambda = (n)$, then the associated diagram is just n horizontally aligned boxes so every $\lambda-$tableaux are row equivalent. In other words, there is only one $\lambda-$tabloid. We thus find $M^{(n)} \simeq \mathbb{C}$ and $S_n$ acts trivially. This is the trivial representation.

If $\lambda = (1^n)$, the associated diagram is n vertically aligned boxes so now the situation is that no two $\lambda-$tableaux are row equivalent. There is thus one different tabloid for every tableau. Also, every tableau can be identified with a permutation using one line notation and since this identification preserves the action of $S_n$, we find that $M^{(1^n)} \simeq \mathbb{C}S_n$ is the regular representation. In particular, the $M^{\lambda}$‘s are far from being irreducibles.

If $\lambda = (n-1,1)$, then a $\lambda-$ tabloid is uniquely determined by the number in its second row. So we can identity the tabloids with the set of numbers from $1$ to $n$. Clearly, $\sigma \in S_n$ will send the tabloid identified with $i$ to the tabloid identified with $j$ $(1 \leq i,j \leq n)$ if and only if $\sigma(i) = j$. This tells us that this identification is an equivalence of representations and so we have the defining representation $M^{(n-1,1)} \simeq \mathbb{C}\{e_1\ldots, e_n\}$.

In subsequent posts, we will see a way to construct, for each $\lambda \vdash n$, an irreducible submodule of $M^{\lambda}$ in a way that gives us all irreducible representations of $S_n$.