## § Specht module construction

#### § $A[\lambda][t]$, and its image

Define
$A[\lambda][t](x) \equiv \sum_{\pi \in C[t]} sgn(\pi) \pi(x).$
That is, $A[\lambda][t]$ creates a signed linear combination of $x$ by creating signed orbits of $x$ under the column stablizier of $t$. First consider
$A[\lambda][t](t) = \sum_{\pi \in C[t]} sgn(\pi) \pi(t).$
We claim that $A[\lambda][t]$ is a projection operator which projects onto the subspace spanned by $A[\lambda][t](t)$. To show this, let's consider the action of $A[\lambda][t]$ on some other tabloid $s$. There is a predicate we are interested in that determines whether $A[\lambda][t](x)$ is $0$ or $\pm t$: If $t$ has two elements $a, b$ that are in the same column of $t$, which are in the same row of $x$. If such elements $a, b$ exist, then the action of $\texttt{swap}(a, b)$ is trivial on $x$, as tabloids are invariant under row permutations. Furthermore, $\texttt{swap}(a, b)$ is in the column stabilizer $C[t]$, since $a, b$ are in the same column of $t$. Exploiting this, we write the group $C[t]$ as cosets of the subgroup $H \equiv \{ id, \texttt{swap}(a, b) \}$. Now the magic happens: the action on $x$ via $H$ turns out to be zero:
\begin{aligned} &\sum_{\pi \in H} sgn{\pi} \pi(x) \\ &= id(s) - \texttt{swap}(a, b)(x) \\ &= x - x = 0 \end{aligned}
Since $C[t]$ partitions as cosets of $H$, the entire action of $C[t]$ on $x$ becomes zero:
\begin{aligned} &\sum_{\pi \in C[t]} sgn{\pi} \pi(x) \\ &= \sum_{\pi \in C[t]/H} (\pi \cdot id)(x) - (\pi \cdot \texttt{swap}(a, b))(x) \\ &= \sum_{\pi \in C[t]/H} (\pi \cdot id) (x) - (\pi \cdot \texttt{swap}(a, b))(x) \\ &= \sum_{\pi \in C[t]/H} \pi \cdot (id(x) - \texttt{swap}(a, b)(x)) \\ &= \sum_{\pi \in C[t]/H} \pi (id(x) - \texttt{swap}(a, b)(x)) \\ &= \sum_{\pi \in C[t]/H} \pi (x - x) \\ &= \sum_{\pi \in C[t]/H} \pi (0) = 0 \end{aligned}
On the other hand, let us assume that elements in the same column of $t$ are always in different rows $x$ [ if they are in the same row, then the action is zero as we saw before. ] Let us focus on the $c$th column of $t$: say the elements in this column are $t[c], t[c], \dots, t[n][c]$. These elements will be in different rows of $x$. Since we can freely permute rows, we can move these elements $t[:][c]$ to the $c$th column of $x$. This makes column $x[:][c]$ of $x$ be a permutation of the $t[:][c]$ column of $t$. Now, there is a unique permutation which permutes every column of $x$ to be like the columns of $t$. Thus, there is a unique permutation $\pi \in C[x]$ such that $\pi(x) = t$. We can invert this, to find a permutation $\pi' \equiv \pi^{-1} \in C[t]$ such that $\pi'(t) = x$. This will force the value of $A[\lambda][t](x)$ to be equal to $\pm A[\lambda][t](t)$, since $x$ differs from $t$ by a permutation:
\begin{aligned} &A[\lambda][t](x) = \\ &= A[\lambda][t](\pi' t) \\ &= \sum_{\sigma \in C[t]} sgn(\sigma) \sigma(\pi' t) \\ &= \sum_{\sigma \in C[t]} sgn(\sigma) (\sigma \circ \pi') t) \\ &= \sum_{\sigma \circ \pi \in C[t]} sgn(\sigma) sgn(\pi') sgn(\pi') (\sigma \circ \pi') t) \\ &= sgn(\pi') \sum_{\sigma \circ \pi \in C[t]} sgn(\sigma \circ \pi) (\sigma \circ \pi') t) \\ &= sgn(\pi') \sum_{\sigma' \in C[t]} sgn(\sigma') \sigma'(t) \\ &= sgn(\pi') A[\lambda][t](t) \end{aligned}
Thus, we find that when $A[\lambda][t]$ acts on a tableaux $x$, the result is either $0$ [when $x$ cannot be obtained by a column permutation of $t$], or is $\pm A[\lambda][t](t)$ [when $x$ can be ontained by a column permutation of $t$]. Thus, the image of $A[\lambda][t]$ is a 1-dimensional subspace spanned by $A[\lambda][t](t)$. So the important property that we have uncovered is that $A[\lambda][t](x)$ is non-zero iff $t$'s columns can be permuted to produce $x$: written formally, we have:
\begin{aligned} &A[\lambda][t](x) \neq 0 \iff \exists \pi \in C[t], \pi(t) = x \\ &A[\lambda][t](x) = 0 \iff \not \exists \pi \in C[t], \pi(t) = x \\ \end{aligned}

#### § Inner product and $A[\lambda][t]$ is self adjoint

We impose the "canonical" inner product on the space of vectors spanned by tabloids, given by making all non-equal basis tabloids orthogonal:
$\langle \{t\} | \{t'\} \rangle \equiv \begin{cases} 1 & \{t \} \simeq \{ t' \} \\ 0 & \text{otherwise} \end{cases}$
Under this inner product, we claim that $A[\lambda][t]$ is self-adjoint: we have that $\langle A[\lambda][t](x), y \rangle = \langle x, A[\lambda][t](y) \rangle$. The key idea is that $A[\lambda][t](y)$ is made up of permutations which are unitary, since they simply permute the orthogonal basis vectors, and these permutations are arranged in $A[\lambda][t]$ such that the $A[\lambda][t]$ operator is self-adjoint:
\begin{aligned} &\langle A[\lambda][t](x) | y \rangle \\ &= \langle \sum_{\pi \in C[t]} sgn \pi \pi(x) | y \rangle \\ &= \sum_{\pi \in C[t]} sgn \pi \langle \pi(x) | y \rangle \\ &\text{(\pi^{-1} is a permutation of orthonormal basis, hence orthogonal)} \\ &\text{(\pi^{-1} preserves inner produce as orthogonal):} \\ &= \sum_{\pi \in C[t]} sgn \pi \langle \pi^{-1} \pi(x) | \pi^{-1} y \rangle \\ &= \sum_{\pi \in C[t]} sgn \pi \langle \pi^{-1} \pi(x) | \pi^{-1} y \rangle \\ &= \sum_{\pi \in C[t]} sgn \pi \langle x | \pi^{-1} y \rangle \\ &\text{(Sum over \pi^{-1}, is an automorphism:)} \\ &= \sum_{\pi^{-1} \in C[t]} sgn \pi^{-1} \langle x | \pi^{-1} y \rangle \\ &= \langle x | \sum_{\pi^{-1} \in C[t]} sgn \pi^{-1} \pi^{-1} y \rangle \\ &= \langle x | A[\lambda][t](y) \rangle \\ \end{aligned}

#### § $S[\lambda]$ is a irreducible subspace of $M[\lambda]$

• Define the subspace spanned by $\{ A[\lambda][t] : t \in \texttt{tabloid}(\lambda) \}$ as $S[\lambda]$ (for Specht). Thus, the $A[\lambda][t]$ span $S[\lambda]$.
• $S[\lambda]$ is invariant under $S[n]$, since the action of $\pi \in S[n]$ on $A[\lambda][t]$ sends $A[\lambda][t]$ to $A[\lambda][\pi(t)]$. Also, the full space $M[\lambda]$ is invariant under $S[n]$ by construction.
• The orbit of any $A[\lambda][t]$ under $S_n$ gives us the full set $\{ A[\lambda][t'] : t' \in \texttt{tabloid}$, since we can produce $A[t']$ from $A[t]$ by the action that permutes $t$ into $t'$.
• For all invariant subspace $U$, $U$ is either disjoint from $S[\lambda]$ or $U$ contains $S[\lambda]$. So it is impossible to reduce $S[\lambda]$ into a smaller invariant subspace $U$.
• Consider some invariant subsepace $U$. If it is disjoint from $S[\lambda]$, then we are done.
• Otherwise, assume there is some $x \in S[\lambda] \cap U$.
• As $x \in S[\lambda]$ and $S[\lambda]$ is spanned by $\{ A[\lambda][t](t) : t \in \texttt{tabloid} \}$, there must be some $t'$ along which $x$ has a component: $\langle x | A[\lambda][t'](t') \rangle \neq 0$.
• Since $A[\lambda][t']$ is symmetric, I can write the above as $\langle A[\lambda][t'](x) | t' \rangle \neq 0$. Now since the image of $A[\lambda][t']$ is the subspace spanned by $t'$, since $U$ is invariant under $A[\lambda][t']$, and since $\langle A[\lambda][t'](x) | t' \rangle \neq 0$, we can say that $A[\lambda][t'](x) = \alpha t' \in U$ for $\alpha \neq 0$. This tells us that we have the vector $t' \in U$.
• Once we have a single $t' \in U$, we win, since all the other $t$'s are obtained as permutations of $t'$, and $U$ is an invariant subspace of these permutations.
• TLDR: if we havs some common vector $x \in S[\lambda] \cap U$, then $\langle x | A[\lambda][t](t) \rangle \neq 0$. By self-adjoint, we get $\langle A[\lambda][t](x) | t \rangle \neq 0$. But $A[\lambda][t](x) = k_{t, x} A[\lambda][t](t)$, hence $k_{t, x} \neq 0$. Further, $A[\lambda][t](x) \in U$ since $U$ is invariant and $x \in U$, hence $k_{t, x} A[\lambda][t](t) \in U$ for $k_{t, x} \neq 0$ hence $A[\lambda][t](t) \in U$. This forces all of $S[\lambda] \in U$, since $U$ is invariant and $S[\lambda]$ is generated by the various $\{ A[\lambda][t](t) : t \texttt{tabloid} \}$, which are obtained by permutation of of $A[\lambda][t](t)$ for a given $t$.

#### § The argument, in the abstract

Let $V$ be a finite dimensional real vector space with inner product $\langle \cdot | \cdot \rangle$. Let $H: V \rightarrow V$ be a symmetric operator with rank 1 image, eigenvector $h \in V$. For simplicity, say that the eigenvalue of $h$ is $1$, so $H h= h$. ( $H$ Hermitian is defined as $\langle Hx | y \rangle = \langle x | H y \rangle$) Let $\mathcal O$ be a group of orthogonal matrices. Define a subspace $S$ of $V$ given by the $\mathcal O$-span of the image of $H$: $S \equiv span(\{ O h : O \in \mathcal O\})$. We wish to show that $S$ is an irreducible $\mathcal O, H$-invariant subspace. By construction, $S$ is $\mathcal O, H$-invariant, since it takes the subspace spanned by $h$ and makes it invariant under $\mathcal O$. To show that this is irreducible, suppose we have some $\mathcal O, H$ invariant subspace $W$. We wish to show that if $W$ contains a single vector from $S$, then it contains all of $S$: $W \cap S \neq \emptyset \implies S \subset W$.
• Suppose that $x \in W \cap S$. Since $S$ is spanned the various $\mathcal O h$, there must be some $O \in \mathcal O$such that $\langle x | O h \rangle \neq 0$.
• Since $O$ is orthogonal, we can shift the rotation towards $x$ by rotating the entire frame by $O^{-1}$, giving us $\langle O^{-1} x | h \rangle \neq 0$.
• Since $h$ is an eigenvector, we replace $h$ by $H h$ giving us $\langle O^{-1} x | H h \rangle \neq 0$.
• Since $H$ is hermitian, I rewrite the above as $\langle H O^{-1} x | h \rangle \neq 0$.
• Since $x \in W$ and $W$ is invariant under $\mathcal O$ and $H$, we have that $H O^{-1} x \in W$.
• Also, since the image of $H$ lies entirely along $h$, we have that $H O^{-1} x = \alpha_x h$. Combining with $\langle H O^{-1} x | h \rangle \neq 0$ gives us $\langle \alpha_x h | h \rangle \neq 0$, or $\alpha \neq 0$.
• Thus, the non-zero vector $\alpha_x h \in W$ (non-zero as $\alpha_x \neq 0$). Hence, the vector $h \in W$. Since $W$ is closed under $\mathcal O$ and $S$ is generated as $\mathcal O h$, we have that $S \subseteq W$.

#### § Showing that $H$ as thesigned linear combination of $\mathcal O$ is Hermitian

In the abstract, we define $H \equiv \sum_{O \in \mathcal O} |O| O$, which specializes to $H_t \equiv \sum_{\pi \in C_t } \pi sgn(\pi)$ in the tableaux theory. Now consider $H^T = \sum_{O \in \mathcal O} |O| O^T$ Since $O$ is orthogonal, $O^T = O^{-1}$. Furthermore, we have that $|O| = |O|^{-1}$ since:
\begin{aligned} &|O| = \pm 1 \\ |O^{-1}| = |O|^{-1} \\ &= (\pm 1)^{-1} = \pm 1 \end{aligned}
Combined, this tells us that $H^{T} = \sum_{O \in \mathcal O} |O|^{-1} O^{-1}$. Since $\mathcal O$ is a subgroup, the sum can be re-indexed to be written as $H^{T} = \sum_{O' \in \mathcal O} |O'| O'$, which is equal to $H$. Hence, we find that $H^T = H$, or $H$ defined in this way is hermitian.

#### § Showing that $H$ is rank 1

In the symmetric group case, we consider:
$A_x \equiv \sum_{\sigma \in C_t} sgn(\sigma) \sigma$
Now say we have some other $y$. The two cases are:
• $y \in Orb(x, C_x)$. We have $y = \pi x$ for $\pi \in C_x$ In this case, the expression for $A_x y$ can be written as $A_x (\pi x)$which is equal to $sgn(\pi) A_x(x)$. So this belongs to the subspace of $A_x(x)$.
• $y \not \in Orb(x, C_x)$. This means that we cannot rearrange the columns of tabloid $x$ to get tabloid $y$ (upto row permutation).
• That is, we have:
xa -> 1
xb -> 1
xc -> 2

• where two elements in the same column of $x$ (xa, xb) want to go to the same row of $y$. If all elements in the same column of $x$ (xa, xb, xc) wanted to go to different rows of $y$ (3, 1, 2), we could have permuted $x$ in a unique way as (xb, xc, xa) to match the rows. This tells us how to convert $x$ into $y$, for this column. If we can do this for all columns, we are done.
• The only obstruction to the above process is that we have two elements in the same column of $x$ (xa, xb) that want to go to the same row of $y$. Said differently, there is a permutation $p$ that swaps xa <-> xb that is in $C_x$ (since (xa, xb) are in the same column), whose action leaves $y$ unchanged (since $y$ a tabloid has these elements in the same row; tabloid invariant under row permutation).
• Thus, we can write $C_t$ as cosets of the subgroup $\{e, p\}$ whose action of $y$ will be:
\begin{aligned} &(sgn(e) e +sgn(p) p)(y) \\ &(e - p)(y) \\ &y - y = 0 \end{aligned}
• Thus, the action of the full $C_t$, written as cosets of $\{ e, p\}$ cancels out entirely and becomes zero, since every coset is of the form $h\{e, p\}$, ie $\{h, hp\}$. And the action of this will be:
\begin{aligned} &(sgn(h)h + sgn(hp)hp)y \\ &=sgn(h) hy + sgn(h)(-1) hp y \\ &=sgn(h) hy - sgn(h) hp y \\ &=sgn(h) hy - sgn(h) hy \\ &=0 \end{aligned}
• Thus, either an element $y$ is in the orbit $C_x$ or not. If it's in the orbit, we get answer $\pm A_x x$. If it's not, we get zero.

#### § Have we found all the irreps?

Recall that the number of irreps is upper bounded by the number of conjugacy classes of the group. This follows from character theory: (1) the characters of irreps are orthogonal in the space of class functions, and (2) the dimension of the space of class functions is is equal to the number of conjugacy classes, since there are those many degrees of freedom for a class function --- it must take on a different value per conjugacy class [TODO: finish my character theory notes ]. In our case, we have found one irrep per conjugacy class, since conjugacy classes of $S_n$ is determined by cycle type, and the shape of a diagram encodes the cycle type of a permutation. If we show that the irreps of different shapes/diagrams are inequivalent, we are done.

#### § Characterizing Maps $S[\lambda]$ to $S[\mu]$

We wish to prove the key lemma, which is that if we have a non-zero map $f: M[\lambda] \rightarrow M[\mu]$, then $\lambda \trianglerighteq \mu$. Let's consider the extreme cases with 3 elements:
λ = (1 1 1):
* * *

μ = 3:
#
#
#

• Let $l$ be a $\lambda$ tableau, $m$ be a $\mu$ tableau.
• Let's consider $A[l](m)$ and $A[m](l)$.
• For $A_l(m)$ to be non-zero, we need a way to send elements of $m$ in the same column ( #; #; #) to correct rows in $l$ ( * * *)But see that $l$has only one row, and $m$ has no choice: it must send all its elements in all columns to that single row of $l$. Thus, the $C_l$ [WRONG ] don't hinder us from doing the only thing we possibly can.
• For $A_m(l)$ to be non-zero, we need a way to send elements of $l$ in the same column, of which there are three columns, *, *, *, to different rows of $m$. But if $l$ were feeling stubborn, it could say that it wants each of its *'s to end up in the first row of $m$. $m$ will be overcrowded, so this leads to the map becoming zero.
• In general, if $\lambda \triangleright \mu$, then the map $A[\lambda](\mu)$ can be nonzero, since we need to send elements in the same column of $\mu$ to different rows of $\lambda$, and $\lambda$ is "bigger", [WRONG?! ]
Thus, we have found ALL irreps, since as argued before, there can be at most as many irreps as there are shapes/diagrams of $n$, and we've shown that each irrep that corresponds to a shape is distinct.

#### § All the $S[\lambda]$ are distinct irreps of $S_n$ by Schur's lemma

Suppose that $S[\lambda] \simeq S[\mu]$. Thus we have an invertible intertwining map $T: S[\lambda] \rightarrow S[\mu]$. By Schur's lemma, since we know that $S[\lambda]$ and $S[\mu]$ are irreps, we know that $T$ is a scalar multiple of the identity map. Let $m$ be a tabloid of shape $\mu$. We know that $A[\mu][m](m) \in S[\mu]$. Now consider $T^{-1}(A[\mu][m](m))$. This must be equal to $A[\mu][m](T^{-1}(m))$. This means that $T^{-1}(m)$ is not zero when acted upon by $A[\mu][m](m)$, thus $T^{-1}(m)$, of shape $\lambda$ must dominate shape $\mu$ [Argue why this is the case by adapting the proof seen before about spaces ]. Ruunning the argument is reverse, we get both directions of $\lambda \trianglerighteq \mu$ and $\mu \trianglerighteq \lambda$, there by establishing $\lambda = \mu$.

#### § Tabloid(3)

There's only one tabloid of shape 3, which is {1 2 3}. Thus we get a 1D complex vector space with basis vector b{1, 2, 3}. Every permutation maps b{1, 2, 3} onto itself, so we get the trivial representation where each element of S3 is the identity map.

#### § Tabloid(2, 1)

There are three tabloids of shape (2, 1), one for each unique value at the bottom. The top row can be permuted freely, so the only choice is in how we choose the bottom. We get the tableaux {1 2}{3} = [1 2] = [2 1], drawn as:
[1 2] = [2 1] = {1 2}
     [3      {3}

And similarly we get {1 3}{2} and {1 2}{3}. So we have a three dimensional vector space. Now let's look at the action of the A operator A: Tableaux -> GL(V(Tabloid(mu)). First of all, we see that the A operator uses tableaux and not tabloids (because we need to know which elements are in the same column). Recall that the action of A(t) on a tabloid x is to sum up linear combinations of $sgn(\pi)\pi(x)$, where $\pi$ is from the column stabilizer of t.
$A(t)(x) \equiv \sum_{\pi \in \texttt{col-stab}(t)} sgn(\pi) \pi(x)$
So let's find the action! The tableaux [1 2], ie:
[1 2]


has as column stabilizers the identity permutation, and the permutation (1 3) obtained by swapping the elements of the columns [1..] Thus, the action of A([1 2]) on a tabloid {k l}{m} is the signed linear combination of the action of the identity and the swap on {k l}{m}:
$A([1 2])(\{ k l \}\{m \}) = 1 \cdot \{k l\}\{m\} + (-1) \cdot {m l}{k}$
Recall that the basis of the Specht module is given by A([t])({t}), where we have the tableaux t act on its own tabloid. In the case where t = [1 2] we get the output
A([1 2])({1 2}{3}) = {1 2}{3} - {3 1}{2}

Similarly, we tabulate all of the actions of A(x)({x}) below, where we pick the equivalence class representative of tabloids as the tabloid whose row entries are in ascending order.
A([1 2])({1 2}{3})
= (id - (1, 3))({1 2}{3})
= {1 2}{3} - {3 1}{2}
= {1 2}{3} - {1 3}{2}

A([2 1])({2 1}{3})
= A([2 1])({2 1}{3})
= A([2 1])({1 2}{3})
= (id - (2, 3))({1 2}{3})
= {1 2}{3} - {1 3}{2}

A([1 3])({1 3}{2})
= (id - (1, 2))({1 3}{2})
= {1 3}{2} - {2 3}{1}

A([3 1])({3 1}{2})
= (id - (3, 2))({3 1}{2})
= (id - (3, 2))({1 3}{2})
= {1 3}{2} - {1 2}{3}

A([1 2])({1 2}{3})
= (id - (1, 3))({1 2}{3})
= {1 2}{3} - {3 2}{1}
= {1 2}{3} - {2 3}{1}

A([2 1])({2 1}{3})
= (id - (2, 3))({2 1}{3})
= (id - (2, 3))({1, 2}{3})
= {1 2}{3} - {1 3}{2}

If we now label the vector as {2 3}{1} = a, {1 3}{2} = b, {1 2}{3} = c, written in ascending order of the element of their final row, we find that A(x)(x) gave us the vectors:
A([1 2])({1 2}{3})
= {1 2}{3} - {1 3}{2} = c - b
A([2 1])({2 1}{3})
= {1 2}{3} - {1 3}{2} = c - b
A([1 3])({1 3}{2})
= {1 3}{2} - {2 3}{1} = b - a
A([3 1])({3 1}{2})
= {1 3}{2} - {1 2}{3} = b - c = -(c-b)
A([1 2])({1 2}{3})
= {1 2}{3} - {2 3}{1} = c - a
A([2 1])({2 1}{3})
= {1 2}{3} - {1 3}{2} = c - b

where the subspace spanned by the vectors (a-b), (b-c), (c-a) is two dimensional, because there a one-dimensional redundancy (a-b) + (b-c) + (c-a) = 0 between them. Furthermore, the basis vectors (a - b), (b - c), (c - a) are invariant under all swaps, and are thus invariant under all permutations, since all permutations can be written as a composition of swaps. So we have found a two-subspace of a three-dimensional representation of S3. To see that this subspace is irreducible, notice that given any permutation of the form k - l, we can swap the letters k, l and the third letter m to obtain the entire basis. Hence, this subspace is indeed irreducible, and the representation of Sn that we have is indeed an irreducible representation.

#### § Tabloid(1, 1, 1)

There are 6 tabloids of shape (1, 1, 1), given by the permutations of the numbers {1, 2, 3}. If we write them down, they're going to be (a) {1}{2}{3}, (b) {1}{3}{2}, (c) {2}{1}{3}, (d) {2}{3}{1}, (e) {3}{1}{2}, (f) {3}{2}{1}. This gives us a 6 dimensional vector space spanned by these basis vectors. Let's now find out the value of A()({1}{2}{3}) recall that we need to act on {1}{2}{3} with all column stabilizers of A().

#### § A on tabloid instead of tableaux

I claim that the different A_t and A_s for {t} = {s} differ only by sign [Why? Because we can reorder the elments of t and s to suffer a sign ]. Thus, we can directly define A_{t} on the tabloids , by defining it as first sorting the rows of t and then using A_t.