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 TT’, 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 TT’ 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 TT’. ♦

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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 ij in the same row of T and same column of T’. Let t be the transposition (ij); 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 ij 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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