§ Specht module construction

§ A[λ][t]A[\lambda][t], and its image

Define
A[λ][t](x)πC[t]sgn(π)π(x). A[\lambda][t](x) \equiv \sum_{\pi \in C[t]} sgn(\pi) \pi(x).
That is, A[λ][t]A[\lambda][t] creates a signed linear combination of xx by creating signed orbits of xx under the column stablizier of tt. First consider
A[λ][t](t)=πC[t]sgn(π)π(t). A[\lambda][t](t) = \sum_{\pi \in C[t]} sgn(\pi) \pi(t).
We claim that A[λ][t]A[\lambda][t] is a projection operator which projects onto the subspace spanned by A[λ][t](t)A[\lambda][t](t). To show this, let's consider the action of A[λ][t]A[\lambda][t] on some other tabloid ss. There is a predicate we are interested in that determines whether A[λ][t](x)A[\lambda][t](x) is 00 or ±t\pm t: If tt has two elements a,ba, b that are in the same column of tt, which are in the same row of xx. If such elements a,ba, b exist, then the action of swap(a,b)\texttt{swap}(a, b) is trivial on xx, as tabloids are invariant under row permutations. Furthermore, swap(a,b)\texttt{swap}(a, b) is in the column stabilizer C[t]C[t], since a,ba, b are in the same column of tt. Exploiting this, we write the group C[t]C[t] as cosets of the subgroup H{id,swap(a,b)}H \equiv \{ id, \texttt{swap}(a, b) \}. Now the magic happens: the action on xx via HH turns out to be zero:
πHsgnππ(x)=id(s)swap(a,b)(x)=xx=0 \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]C[t] partitions as cosets of HH, the entire action of C[t]C[t] on xx becomes zero:
πC[t]sgnππ(x)=πC[t]/H(πid)(x)(πswap(a,b))(x)=πC[t]/H(πid)(x)(πswap(a,b))(x)=πC[t]/Hπ(id(x)swap(a,b)(x))=πC[t]/Hπ(id(x)swap(a,b)(x))=πC[t]/Hπ(xx)=πC[t]/Hπ(0)=0 \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 tt are always in different rows xx [ if they are in the same row, then the action is zero as we saw before.] Let us focus on the ccth column of tt: say the elements in this column are t[1][c],t[2][c],,t[n][c]t[1][c], t[2][c], \dots, t[n][c]. These elements will be in different rows of xx. Since we can freely permute rows, we can move these elements t[:][c]t[:][c] to the ccth column of xx. This makes column x[:][c]x[:][c] of xx be a permutation of the t[:][c]t[:][c] column of tt. Now, there is a unique permutation which permutes every column of xx to be like the columns of tt. Thus, there is a unique permutation πC[x]\pi \in C[x] such that π(x)=t\pi(x) = t. We can invert this, to find a permutation ππ1C[t]\pi' \equiv \pi^{-1} \in C[t] such that π(t)=x\pi'(t) = x. This will force the value of A[λ][t](x)A[\lambda][t](x) to be equal to ±A[λ][t](t)\pm A[\lambda][t](t), since xx differs from tt by a permutation:
A[λ][t](x)==A[λ][t](πt)=σC[t]sgn(σ)σ(πt)=σC[t]sgn(σ)(σπ)t)=σπC[t]sgn(σ)sgn(π)sgn(π)(σπ)t)=sgn(π)σπC[t]sgn(σπ)(σπ)t)=sgn(π)σC[t]sgn(σ)σ(t)=sgn(π)A[λ][t](t) \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[λ][t]A[\lambda][t] acts on a tableaux xx, the result is either 00 [when xx cannot be obtained by a column permutation of tt], or is ±A[λ][t](t)\pm A[\lambda][t](t) [when xx can be ontained by a column permutation of tt]. Thus, the image of A[λ][t]A[\lambda][t] is a 1-dimensional subspace spanned by A[λ][t](t)A[\lambda][t](t). So the important property that we have uncovered is that A[λ][t](x)A[\lambda][t](x) is non-zero iff tt's columns can be permuted to produce xx: written formally, we have:
A[λ][t](x)0    πC[t],π(t)=xA[λ][t](x)=0    ∄πC[t],π(t)=x \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[λ][t]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:
{t}{t}{1{t}{t}0otherwise \langle \{t\} | \{t'\} \rangle \equiv \begin{cases} 1 & \{t \} \simeq \{ t' \} \\ 0 & \text{otherwise} \end{cases}
Under this inner product, we claim that A[λ][t]A[\lambda][t] is self-adjoint: we have that A[λ][t](x),y=x,A[λ][t](y)\langle A[\lambda][t](x), y \rangle = \langle x, A[\lambda][t](y) \rangle. The key idea is that A[λ][t](y)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[λ][t]A[\lambda][t] such that the A[λ][t]A[\lambda][t] operator is self-adjoint:
A[λ][t](x)y=πC[t]sgnππ(x)y=πC[t]sgnππ(x)y(π1 is a permutation of orthonormal basis, hence orthogonal)(π1 preserves inner produce as orthogonal):=πC[t]sgnππ1π(x)π1y=πC[t]sgnππ1π(x)π1y=πC[t]sgnπxπ1y(Sum over π1, is an automorphism:)=π1C[t]sgnπ1xπ1y=xπ1C[t]sgnπ1π1y=xA[λ][t](y) \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[λ]S[\lambda] is a irreducible subspace of M[λ]M[\lambda]

  • Define the subspace spanned by {A[λ][t]:ttabloid(λ)}\{ A[\lambda][t] : t \in \texttt{tabloid}(\lambda) \} as S[λ]S[\lambda] (for Specht).Thus, the A[λ][t]A[\lambda][t] span S[λ]S[\lambda].
  • S[λ]S[\lambda] is invariant under S[n]S[n], since the action of πS[n]\pi \in S[n] on A[λ][t]A[\lambda][t] sends A[λ][t]A[\lambda][t] to A[λ][π(t)]A[\lambda][\pi(t)]. Also, the full space M[λ]M[\lambda] is invariant under S[n]S[n] by construction.
  • The orbit of any A[λ][t]A[\lambda][t] under SnS_n gives us the full set {A[λ][t]:ttabloid\{ A[\lambda][t'] : t' \in \texttt{tabloid}, since we can produceA[t]A[t'] from A[t]A[t] by the action that permutes tt into tt'.
  • For all invariant subspace UU, UU is either disjoint from S[λ]S[\lambda] or UU contains S[λ]S[\lambda]. So it is impossibleto reduce S[λ]S[\lambda] into a smaller invariant subspace UU.
  • Consider some invariant subsepace UU. If it is disjoint from S[λ]S[\lambda], then we are done.
  • Otherwise, assume there is some xS[λ]Ux \in S[\lambda] \cap U.
  • As xS[λ]x \in S[\lambda] and S[λ]S[\lambda] is spanned by {A[λ][t](t):ttabloid}\{ A[\lambda][t](t) : t \in \texttt{tabloid} \}, there must be some tt' along which xx has a component:xA[λ][t](t)0\langle x | A[\lambda][t'](t') \rangle \neq 0.
  • Since A[λ][t]A[\lambda][t'] is symmetric, I can write the above as A[λ][t](x)t0\langle A[\lambda][t'](x) | t' \rangle \neq 0.Now since the image of A[λ][t]A[\lambda][t'] is the subspace spanned by tt', since UU is invariant under A[λ][t]A[\lambda][t'],and since A[λ][t](x)t0\langle A[\lambda][t'](x) | t' \rangle \neq 0, we can say that A[λ][t](x)=αtUA[\lambda][t'](x) = \alpha t' \in U for α0\alpha \neq 0.This tells us that we have the vector tUt' \in U.
  • Once we have a single tUt' \in U, we win, since all the other tt's are obtained as permutations of tt', and UU is an invariantsubspace of these permutations.
  • TLDR: if we havs some common vector xS[λ]Ux \in S[\lambda] \cap U, then xA[λ][t](t)0\langle x | A[\lambda][t](t) \rangle \neq 0. Byself-adjoint, we get A[λ][t](x)t0\langle A[\lambda][t](x) | t \rangle \neq 0. But A[λ][t](x)=kt,xA[λ][t](t)A[\lambda][t](x) = k_{t, x} A[\lambda][t](t), hence kt,x0k_{t, x} \neq 0.Further, A[λ][t](x)UA[\lambda][t](x) \in U since UU is invariant and xUx \in U, hence kt,xA[λ][t](t)Uk_{t, x} A[\lambda][t](t) \in U for kt,x0k_{t, x} \neq 0 hence A[λ][t](t)UA[\lambda][t](t) \in U.This forces all of S[λ]US[\lambda] \in U, since UU is invariant and S[λ]S[\lambda] is generated by the various {A[λ][t](t):ttabloid}\{ A[\lambda][t](t) : t \texttt{tabloid} \},which are obtained by permutation of of A[λ][t](t)A[\lambda][t](t) for a given tt.

§ The argument, in the abstract

Let VV be a finite dimensional real vector space with inner product \langle \cdot | \cdot \rangle. Let H:VVH: V \rightarrow V be a symmetric operator with rank 1 image, eigenvector hVh \in V. For simplicity, say that the eigenvalue of hh is 11, so Hh=hH h= h. (HH Hermitian is defined as Hxy=xHy\langle Hx | y \rangle = \langle x | H y \rangle) Let O\mathcal O be a group of orthogonal matrices. Define a subspace SS of VV given by the O\mathcal O-span of the image of HH: Sspan({Oh:OO})S \equiv span(\{ O h : O \in \mathcal O\}). We wish to show that SS is an irreducible O,H\mathcal O, H-invariant subspace. By construction, SS is O,H\mathcal O, H-invariant, since it takes the subspace spanned by hh and makes it invariant under O\mathcal O. To show that this is irreducible, suppose we have some O,H\mathcal O, H invariant subspace WW. We wish to show that if WW contains a single vector from SS, then it contains all of SS: WS    SWW \cap S \neq \emptyset \implies S \subset W.
  • Suppose that xWSx \in W \cap S. Since SS is spanned the various Oh\mathcal O h, there must be some OOO \in \mathcal Osuch that xOh0\langle x | O h \rangle \neq 0.
  • Since OO is orthogonal, we can shift the rotation towards xx by rotating the entire frame by O1O^{-1},giving us O1xh0\langle O^{-1} x | h \rangle \neq 0.
  • Since hh is an eigenvector, we replace hh by HhH h giving us O1xHh0\langle O^{-1} x | H h \rangle \neq 0.
  • Since HH is hermitian, I rewrite the above as HO1xh0\langle H O^{-1} x | h \rangle \neq 0.
  • Since xWx \in W and WW is invariant under O\mathcal O and HH, we have that HO1xWH O^{-1} x \in W.
  • Also, since the image of HH lies entirely along hh, we have that HO1x=αxhH O^{-1} x = \alpha_x h. Combining with HO1xh0\langle H O^{-1} x | h \rangle \neq 0 gives us αxhh0\langle \alpha_x h | h \rangle \neq 0,or α0\alpha \neq 0.
  • Thus, the non-zero vector αxhW\alpha_x h \in W (non-zero as αx0\alpha_x \neq 0). Hence, the vector hWh \in W.Since WW is closed under O\mathcal O and SS is generated as Oh\mathcal O h, we have that SWS \subseteq W.

§ Showing that HH as thesigned linear combination of O\mathcal O is Hermitian

In the abstract, we define HOOOOH \equiv \sum_{O \in \mathcal O} |O| O, which specializes to HtπCtπsgn(π)H_t \equiv \sum_{\pi \in C_t } \pi sgn(\pi) in the tableaux theory. Now consider HT=OOOOTH^T = \sum_{O \in \mathcal O} |O| O^T Since OO is orthogonal, OT=O1O^T = O^{-1}. Furthermore, we have that O=O1|O| = |O|^{-1} since:
O=±1O1=O1=(±1)1=±1 \begin{aligned} &|O| = \pm 1 \\ |O^{-1}| = |O|^{-1} \\ &= (\pm 1)^{-1} = \pm 1 \end{aligned}
Combined, this tells us that HT=OOO1O1H^{T} = \sum_{O \in \mathcal O} |O|^{-1} O^{-1}. Since O\mathcal O is a subgroup, the sum can be re-indexed to be written as HT=OOOOH^{T} = \sum_{O' \in \mathcal O} |O'| O', which is equal to HH. Hence, we find that HT=HH^T = H, or HH defined in this way is hermitian.

§ Showing that HH is rank 1

In the symmetric group case, we consider:
AxσCtsgn(σ)σA_x \equiv \sum_{\sigma \in C_t} sgn(\sigma) \sigma
Now say we have some other yy. The two cases are:
  • yOrb(x,Cx)y \in Orb(x, C_x). We have y=πxy = \pi x for πCx\pi \in C_x In this case, the expression for AxyA_x y can be written as Ax(πx)A_x (\pi x)which is equal to sgn(π)Ax(x)sgn(\pi) A_x(x). So this belongs to the subspace of Ax(x)A_x(x).
  • y∉Orb(x,Cx)y \not \in Orb(x, C_x). This means that we cannot rearrange the columns of tabloid xx to get tabloid yy (upto row permutation).
  • That is, we have:
xa -> 1
xb -> 1
xc -> 2
  • where two elements in the same column of xx (xa, xb) want to go to the same row of yy. If all elementsin the same column of xx (xa, xb, xc) wanted to go to different rows of yy (3, 1, 2), we could havepermuted xx in a unique way as (xb, xc, xa) to match the rows. This tells us how to convert xx into yy,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 xx (xa, xb) thatwant to go to the same row of yy. Said differently, there is a permutation pp that swaps xa <-> xb that is inCxC_x (since (xa, xb) are in the same column), whose action leaves yy unchanged (since yy a tabloid hasthese elements in the same row; tabloid invariant under row permutation).
  • Thus, we can write CtC_t as cosets of the subgroup {e,p}\{e, p\} whose action of yy will be:
(sgn(e)e+sgn(p)p)(y)(ep)(y)yy=0 \begin{aligned} &(sgn(e) e +sgn(p) p)(y) \\ &(e - p)(y) \\ &y - y = 0 \end{aligned}
  • Thus, the action of the full CtC_t, written as cosets of {e,p}\{ e, p\} cancels out entirely and becomes zero, since every cosetis of the form h{e,p}h\{e, p\}, ie {h,hp}\{h, hp\}. And the action of this will be:
(sgn(h)h+sgn(hp)hp)y=sgn(h)hy+sgn(h)(1)hpy=sgn(h)hysgn(h)hpy=sgn(h)hysgn(h)hy=0 \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 yy is in the orbit CxC_x or not. If it's in the orbit, we get answer ±Axx\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 [WIP: finish my character theory notes]. In our case, we have found one irrep per conjugacy class, since conjugacy classes of SnS_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 M[λ]M[\lambda] to M[μ]M[\mu]

We wish to prove the key lemma, which is that if we have a non-zero map f:M[λ]M[μ]f: M[\lambda] \rightarrow M[\mu], then λμ\lambda \trianglerighteq \mu. Let's consider the extreme cases with 3 elements:
λ = (1 1 1):
* * *
μ = 3:
#
#
#
  • Let ll be a λ\lambda tableau, mm be a μ\mu tableau.
  • Let's consider A[l](m)A[l](m) and A[m](l)A[m](l).
  • For Al(m)A_l(m) to be non-zero, we need a way to send elements of mm in the same column (#; #; #) to correct rows in ll (* * *)But see that llhas only one row, and mm has no choice: it must send all its elements in all columns to that single row of ll. Thus, the ClC_l [WRONG]don't hinder us from doing the only thing we possibly can.
  • For Am(l)A_m(l) to be non-zero, we need a way to send elements of ll in the same column, of which there are three columns, *, *, *, to different rows ofmm. But if ll were feeling stubborn, it could say that it wants each of its *'s to end up in the first row of mm. mm will be overcrowded, so thisleads to the map becoming zero.
  • In general, if λμ\lambda \triangleright \mu, then the map A[λ](μ)A[\lambda](\mu) can be nonzero, since we need to send elements in the same column of μ\mu to differentrows of λ\lambda, and λ\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 nn, and we've shown that each irrep that corresponds to a shape is distinct.

§ All the S[λ]S[\lambda] are distinct irreps of SnS_n by Schur's lemma

Suppose that S[λ]S[μ]S[\lambda] \simeq S[\mu]. Thus we have an invertible intertwining map T:S[λ]S[μ]T: S[\lambda] \rightarrow S[\mu]. By Schur's lemma, since we know that S[λ]S[\lambda] and S[μ]S[\mu] are irreps, we know that TT is a scalar multiple of the identity map. Let mm be a tabloid of shape μ\mu. We know that A[μ][m](m)S[μ]A[\mu][m](m) \in S[\mu]. Now consider T1(A[μ][m](m))T^{-1}(A[\mu][m](m)). This must be equal to A[μ][m](T1(m))A[\mu][m](T^{-1}(m)). This means that T1(m)T^{-1}(m) is not zero when acted upon by A[μ][m](m)A[\mu][m](m), thus T1(m)T^{-1}(m), of shape λ\lambda must dominate shape μ\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 λμ\lambda \trianglerighteq \mu and μλ\mu \trianglerighteq \lambda, there by establishing λ=μ\lambda = \mu.

§ Working it out for S3

§ 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][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 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(π)π(x)sgn(\pi)\pi(x), where π\pi is from the column stabilizer of t.
A(t)(x)πcol-stab(t)sgn(π)π(x) 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([12][3])({kl}{m})=1{kl}{m}+(1)mlkA([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)

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])({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

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.