$A[\lambda][t](x) \equiv \sum_{\pi \in C[t]} sgn(\pi) \pi(x).$

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 $t[1][c], t[2][c], \dots, t[n][c]$. 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 $\begin{aligned}
&A[\lambda][t](x) = \\
&= A[\lambda][t](\pi' t) \\
&= \sum_{\sigma \in C[t]} sgn(\sigma) \sigma(\pi' t) \\
&= \sum_{\sigma \in C[t]} sgn(\sigma) (\sigma \circ \pi') t) \\
&= \sum_{\sigma \circ \pi \in C[t]} sgn(\sigma) sgn(\pi') sgn(\pi') (\sigma \circ \pi') t) \\
&= sgn(\pi') \sum_{\sigma \circ \pi \in C[t]} sgn(\sigma \circ \pi) (\sigma \circ \pi') t) \\
&= sgn(\pi') \sum_{\sigma' \in C[t]} sgn(\sigma') \sigma'(t) \\
&= sgn(\pi') A[\lambda][t](t)
\end{aligned}$

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:
$\begin{aligned}
&A[\lambda][t](x) \neq 0 \iff \exists \pi \in C[t], \pi(t) = x \\
&A[\lambda][t](x) = 0 \iff \not \exists \pi \in C[t], \pi(t) = x \\
\end{aligned}$

$\langle \{t\} | \{t'\} \rangle \equiv
\begin{cases}
1 & \{t \} \simeq \{ t' \} \\
0 & \text{otherwise}
\end{cases}$

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}$

- 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 $t$.

- 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
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 $S \subseteq W$.*non-zero*

$\begin{aligned}
&|O| = \pm 1 \\
|O^{-1}| = |O|^{-1} \\
&= (\pm 1)^{-1} = \pm 1
\end{aligned}$

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.
$A_x \equiv \sum_{\sigma \in C_t} sgn(\sigma) \sigma$

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.

```
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?! ]

`Tabloid(3)`

`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)`

`(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][3]`

= `[2 1][3]`

, drawn as:
```
[1 2] = [2 1] = {1 2}
[3] [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 `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][3]`

, ie:
```
[1 2]
[3]
```

has as column stabilizers the identity permutation, and the
permutation `(1 3)`

obtained by swapping the elements of the columns `[1..][3]`

Thus, the action of `A([1 2][3])`

on a tabloid `{k l}{m}`

is the signed linear combination of the action
of the identity and the swap on `{k l}{m}`

:
$A([1 2][3])(\{ k l \}\{m \}) = 1 \cdot \{k l\}\{m\} + (-1) \cdot {m l}{k}$

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][3]`

we get the output
```
A([1 2][3])({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][3])({1 2}{3})
= (id - (1, 3))({1 2}{3})
= {1 2}{3} - {3 1}{2}
= {1 2}{3} - {1 3}{2}
```

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

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

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

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

```
A([2 1][3])({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][3])({1 2}{3})
= {1 2}{3} - {1 3}{2} = c - b
A([2 1][3])({2 1}{3})
= {1 2}{3} - {1 3}{2} = c - b
A([1 3][2])({1 3}{2})
= {1 3}{2} - {2 3}{1} = b - a
A([3 1][2])({3 1}{2})
= {1 3}{2} - {1 2}{3} = b - c = -(c-b)
A([1 2][3])({1 2}{3})
= {1 2}{3} - {2 3}{1} = c - a
A([2 1][3])({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)`

`(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])({1}{2}{3})`

recall that we need to act
on `{1}{2}{3}`

with all column stabilizers of `A([1][2][3])`

.
`A`

on tabloid instead of tableaux `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 `t`

and then using `A_t`

.