Polynomials and Representations XXV

Properties of the Young Symmetrizer

Recall that for a filling $T$ , we have $R(T), C(T) \le S_d,$ the subgroup of elements which take an element of the i-th row (resp. column) of T to the i-th row (resp. column) of T. Then:

$a_T = \sum_{g\in R(T)} g,\quad b_T = \sum_{g\in C(T)} \chi(g)g,\quad c_T = a_T b_T,$

where $c_T$ is the Young symmetrizer. Recall the following results from earlier:

$g\in R(T) \implies a_T g = g a_T = a_T, \\ g\in C(T) \implies b_T g = g b_T = \chi(g)b_T.$

The following is obvious.

Lemma 1. If $g\in S_d$ , then $a_{g(T)} = g a_T g^{-1}$ and $b_{g(T)} = g b_T g^{-1}.$ Thus $c_{g(T)} = g c_T g^{-1}.$

Proof

This follows from $R(g(T)) = g R(T) g^{-1}$ and $C(g(T)) = g C(T)g^{-1}.$ E.g. the latter gives:

$b_{g(T)} = \sum_{x \in C(g(T))} \chi(x)x = \sum_{x\in C(T)} \chi(gxg^{-1}) gxg^{-1} = g b_T g^{-1}.$ ♦

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The following generalizes the Young symmetrizer.

Proposition 1. Consider the element $c = a_T b_{T'}$ for fillings $T, T'$ of shape $\lambda.$ Note that since $\mathbb{C}[G]c$ is contained in both $\mathbb{C}[G]a_T$ and $\mathbb{C}[G]b_{T'}$ it is either 0 or isomorphic to $V_\lambda.$

The following are equivalent.

$c \ne 0$ ;

no two distinct $i,j$ lie in the same row of $T$ and same column of $T'$ ;

there exist $g \in R(T)$ and $g' \in C(T')$ such that $g(T) = g'(T').$

Note

The third condition says: we can change T to T’ by permuting elements in each row, then elements in each column.

Proof

Suppose $c\ne 0$ but two distinct $i, j$ lie in the same row of T and same column of T’. Let $g\in S_d$ swap those two elements; then $g\in R(T) \cap C(T').$ Letting C run through a set of coset representatives of $R(T)/\left<g\right>,$ we have $a_T = (\sum_{x\in C} x)(1+g).$ But now we have $gb_{T'} = \chi(g)b_{T'} = -b_{T'}$ so $(1+g)b_{T'} = 0$ and we have c=0.

Suppose the second condition holds. Elements of the first row of T are in different columns of T’. Bringing those in T’ to the first row, there exist $g'_1\in C(T')$ and $g_1\in R(T)$ such that $g_1(T)$ and $g_1'(T')$ have identical first rows.

swapping_diagram_for_proof

Likewise, since the second row of $g_1(T)$ are in different columns of $g'_1(T')$ there exist $g'_2 \in C(T')$ and $g_2 \in R(T)$ such that $g_2 g_1(T)$ and $g'_2 g'_1(T')$ have the same first and second rows. Repeating, we eventually get the desired $g\in R(T)$ and $g'\in C(T').$

Finally, suppose the third condition holds; let $T'' = g(T) = g'(T')$ . By lemma 1,

$a_{T''} = g a_T g^{-1} = a_T, \quad b_{T''} = g' b_T g'^{-1} = b_{T'},$

So $c=a_T b_{T'} = a_{T''}b_{T''} = c_{T''}\ne 0.$ ♦

Lemma 2. If $T, T'$ are distinct SYT of the same shape, they do not satisfy the conditions of proposition 1.

Proof

Order all the fillings of shape $\lambda$ as follows: given T, T’, let k be the largest number occurring in different squares of T and T’; we write T’ > T if k occurs earlier in the word w(T’) than in w(T). For example, we have:

temp_ordering_in_proof

Note that for any SYT T we have:

$g \in R(T), g' \in C(T) \implies g(T) \ge T, g'(T) \le T$

since the largest entry of T which is moved by g must move to its left in w(T), and the largest entry moved by g’ moves to its right in w(T). Thus if T, T’ are SYT and g(T) = g’(T’) for some $g\in R(T), g'\in C(T')$ , we have $T \ge g(T) = g'(T') \ge T'$ so T = T’. ♦

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Lemma 3. For any $v\in \mathbb{C}[G]$ , $v$ is a multiple of $c_T$ if and only if:

for any $g\in R(T)$ and $h \in C(T)$ , we have $gvh = \chi(h)v.$

Proof

⇒ is left as an easy exercise, by proving $gv = v\chi(h)h = v.$

For ⇐, write $v= \sum_{x\in G} \alpha_x x$ where $\alpha_x \in \mathbb{C}.$ Then $\alpha_{gxh} = \chi(h)\alpha_x$ for any $x\in G$ , $g\in R(T)$ and $h\in C(T).$ Taking x=e, it suffices to show: if $x\in G$ is not of the form $gh$ for $g\in R(T), h\in C(T)$ , then $\alpha_x = 0.$

For that, consider the filling $T' := x(T)$ . We claim that $T, T'$ satisfy the conditions of proposition 1.

Indeed, if $g(T) = g'(T')$ for some $g\in R(T)$ and $g'\in C(T'),$ then

$x=g'^{-1}g, \quad g' \in C(T') = C(xT) = x C(T)x^{-1}.$

So $x^{-1} g' x = g^{-1} g' g \in C(T)$ and we have $x = g(g^{-1}g'g)^{-1} \in R(T)C(T)$ which is a contradiction.

Hence there exist distinct i, j in the same row of T and same column of T’. Let t be the transposition (i, j); then $t \in R(T)$ so $tv = v$ by the given condition. Since $t \in C(T') = x C(T) x^{-1}$ so $t' := x^{-1}tx$ satisfies $vt'=\chi(t') v = -v.$ Now consider the coefficient of tx in v. We have:

$\begin{aligned} tv = v&\implies \alpha_x = \alpha_{tx},\\ vt' = -v &\implies \alpha_{x} = -\alpha_{tx}.\end{aligned}$

Thus $\alpha_x = 0$ as desired. ♦

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Here is one immediate application of lemma 2.

Proposition 2. For any $w \in \mathbb{C}[G]$ , $c_T w c_T$ is a scalar multiple of $c_T.$

In the case $w=e$ we have $c_T^2 = nT$ where $n = \frac {d!}{\dim V_\lambda}.$

Proof

Indeed if $g\in R(T)$ and $h\in C(T)$ we have:

$g(c_T w c_T) h = (gc_T) w(c_T h) = (c_T)w(\chi(h) c_T) = \chi(h) c_T w c_T$

so by lemma 2, $c_T w c_T$ is a scalar multiple of $c_T.$

For $c_T^2$ , consider right-multiplication on $\mathbb{C}[G]$ by $c_T$ and let A be its trace. This map takes $g\mapsto gc_T$ where the coefficient of g is 1; taking the basis comprising of elements of G, we have $A = d!.$ On the other hand, the image of this map is $\mathbb{C}[G]c_T \cong V_\lambda$ and takes each $vc_T \mapsto vc_T^2 = nvc_T.$ Thus the map is a scalar map $n$ on $V_\lambda$ and we have $n\cdot \dim V_\lambda = d!.$ ♦

This gives the following.

Theorem. We have, as a direct sum of irreps:

$\displaystyle \mathbb{C}[S_d] = \bigoplus_{\lambda \vdash d} \bigoplus_{\substack{T = SYT \\ \text{of shape }\lambda}} \mathbb{C}[S_d] c_T.$

Proof

Let us show that $\sum_T \mathbb{C}[S_d] c_T$ is a direct sum, where T runs over all SYT of shape $\lambda.$ Indeed, if not we have:

$\overbrace{v_1 c_{T_1}}^{\ne 0} + v_2 c_{T_2} + \ldots + v_m c_{T_m} = 0,\quad (*)$

for some distinct SYT $T_1, \ldots, T_m$ and some $v_1, \ldots, v_m \in \mathbb{C}[G].$ Note that if $T\ne T'$ then by lemma 2, T and T’ do not satisfy the conditions in proposition 1, so we can find distinct i, j in the same row of T and same column of T’; using the same technique as the first paragraph of proof of proposition 1, we have $b_{T} a_{T'} =0$ and thus $c_T c_{T'} = 0.$

Right-multiplying (*) by $c_{T_1}$ thus gives $0 =v_1 c_{T_1}^2 = n v_1 c_{T_1}$ which gives $v_1 c_{T_1} = 0$ , a contradiction. Hence, the inner sum is a direct sum. The entire sum is a direct sum since there is no common irrep between distinct $\lambda.$

Hence, the RHS is a subspace of the LHS. Now we count the dimension: on LHS we get d!; on RHS we get $\sum_\lambda f_\lambda\dim V_\lambda = \sum_\lambda f_\lambda^2 = d!$ so the two sides are equal. ♦

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