Polynomials and Representations XXXIV

Twisting

From the previous article, any irreducible polynomial representation of G= GL_n\mathbb{C} is of the form V(\lambda) for some \lambda \vdash d, l(\lambda) \le n such that \psi_{V(\lambda)} is the Schur polynomial s_\lambda(x_1, \ldots, x_n).

Now given any analytic representation V of G, we can twist it by taking V\otimes \det^k for an integer k. Then:

\displaystyle\psi_{V \otimes \det^k}= \psi_V \cdot \psi_{\det}^k = (x_1 \ldots x_n)^k \psi_V.

Twisting the irrep V(\lambda) with k\ge 0 gives us another irrep, necessarily of the form V(\mu). What is this \mu? Note that from \psi_{V(\lambda)} we can recover the partition \lambda by taking the minimal partition (with respect to \trianglelefteq). Hence from \psi_{V(\mu)} = (x_1\ldots x_n)^k\psi_{V(\lambda)} we must have \mu = \lambda + (k, \ldots, k). Thus:

Proposition. Any irreducible analytic representation of G can be uniquely written as:

\{\psi_{V(\lambda)} \otimes \det^k : l(\lambda) \le n-1, k\in\mathbb{Z}\}

where V(\lambda) is the polynomial representation satisfying:

\psi_{V(\lambda)} = s_\lambda(x_1, \ldots, x_n).

Since l(\lambda) \le n-1, s_\lambda(x_1, \ldots, x_n) is not divisible by x_n so the representation V(\lambda) \otimes \det^k is polynomial if and only if k\ge 0. Its degree is |\lambda| + kn.

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Dual of Irrep

The dual of the irrep V(\lambda) is also a rational irrep, so it is of the form V(\mu) \otimes \det^k for some partition \mu with l(\mu) \le n-1 and integer k. From:

\psi_{V(\lambda)^\vee}(x_1, \ldots, x_n) = \psi_{V(\lambda)}(x_1^{-1}, \ldots, x_n^{-1})

we take the term with the smallest exponent for x^\mu in lexicographical order. For large N, denoting \text{rev}(\alpha) for the reverse of a sequence \alpha, we have:

\lambda, \mu \vdash d, \lambda \trianglerighteq \mu \implies \text{rev}((N,\ldots, N) - \lambda) \trianglerighteq \text{rev}((N,\ldots, N) - \mu).

Hence \mu = (\lambda_1, \lambda_1 - \lambda_{n-1},  \lambda_1 - \lambda_{n-2}, \ldots, \lambda_1 - \lambda_2) and k = -\lambda_1. Pictorially we have:

partition_of_dual_representation

Weight Space Decomposition

By definition \psi_{V(\lambda)} is the character of V(\lambda) when acted upon by the torus group S. Since this polynomial is s_\lambda = \sum_{\mu} K_{\lambda\mu} m_\mu, as vector spaces we have:

\displaystyle V(\lambda) = \bigoplus_{\mu} \bigoplus_{\sigma} V(\lambda)_{\sigma(\mu)}

where:

  • \mu runs through all partitions with |\mu| = |\lambda| and l(\mu) \le n;
  • \sigma(\mu) runs through all permutations of \mu without repetition, e.g. if \mu = (5, 3, 3) we get 3 terms: (5, 3, 3), (3, 5, 3) and (3, 3, 5);
  • V(\lambda)_{\nu} is the space of all v\in V(\lambda) for which S acts with character x^\nu, i.e.

V(\lambda)_{\nu} = \{v \in V(\lambda) : D(x_1, \ldots, x_n) \in S \text{ takes } v\mapsto x^\nu v\}

and the dimension of V(\lambda)_{\nu} is K_{\lambda\nu}. This is called the weight space decomposition of V(\lambda). We will go through some explicit examples later.

Foreshadowing: SSYTs as a Basis

As noted above, the dimension of V(\lambda)_{\sigma(\mu)} is exactly the number of SSYT with shape \lambda and type \sigma(\mu). Thus in a somewhat ambiguous way, we can take, as a basis of V(\lambda), elements of the form \{v_T\} over all SSYT T of shape \lambda and entries from [n]={1,2,…,n}; each v_T lies in the space V(\lambda)_{\text{shape}(T)}.

However, such a description does not distinguish between distinct SSYT of the same type. For that, one needs a construction like the determinant modules (to be described later).

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Example: n=2

Consider G = GL_2\mathbb C. By the above proposition, each irreducible representation is given by V(m) \otimes \det^k where mk are integers and m\ge 0. To compute V(m), we need to find a polynomial representation of G such that

\psi_{V(m)} = s_m(x, y)= x^m + x^{m-1}y + \ldots + y^m

corresponding to the SSYT with shape (m) and entries comprising of only 1’s and 2’s. E.g. s_4(x,y) = x^4 + x^3 y + x^2 y^2 + xy^3 + y^4 from:

gl2_example_ssyt

Such a V(m) is easy to construct: take \text{Sym}^m V; if {ef} is a basis of V, then a corresponding basis of \text{Sym}^m V is given by \{e^i f^{4-i}\}_{i=0,\ldots,4}. If v_i := e^{2+i} f^{2-i}, then the diagonal matrix D(ab) takes v_i \mapsto a^{2+i} b^{2-i} v_i so its character is a^4 + a^3 b + a^2 b^2 + ab^3 + b^4 as desired.

The weight space decomposition thus gives:

V(m) = V(m)_{4,0} \oplus V(m)_{3,1} \oplus V(m)_{2,2} \oplus V(m)_{1,3} \oplus V(m)_{0,4}

where each V(m)_{i,j} is 1-dimensional and spanned by e^i f^{4-i}.

Example: d=2

Consider G = GL_n\mathbb{C}. We have:

\mathbb{C}^n \otimes_{\mathbb C} \mathbb{C}^n \cong \text{Sym}^2 \mathbb{C}^n \oplus \text{Alt}^2 \mathbb{C}^n,

where each component is G-invariant. As shown earlier, we have:

\begin{aligned}\psi_{\text{Sym}^2} &= \sum_{1\le i\le j \le n} x_i x_j = h_2(x_1, \ldots, x_n),\\ \psi_{\text{Alt}^2} &= \sum_{1 \le i < j \le n} x_i x_j = e_2(x_1, \ldots, x_n).\end{aligned}

Since the Schur polynomials are s_2 = h_2 and s_{11} = e_2, both \text{Alt}^2 and \text{Sym}^2 are irreps of G. The weight space decomposition of the two spaces are:

\displaystyle \begin{aligned}\text{Sym}^2\mathbb{C}^n &= \bigoplus_{1\le i\le j\le n}\mathbb{C}\cdot e_i e_j, \\ \text{Alt}^2\mathbb{C}^n &= \bigoplus_{1\le i<j \le n}\mathbb{C} \cdot (e_i \wedge e_j).\end{aligned}

Hence in their weight space decompositions, all components have dimension 1.

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Example: n=3

Now let us take G = GL_3\mathbb{C} and compute V(\lambda) where \lambda = (\lambda_1, \lambda_2) and \lambda_1 + \lambda_2 = d. To find s_\lambda(x,y,z) we need to compute K_{\lambda\mu} for all \mu\vdash d and l(\mu) \le 3. We will work in the plane X_1+ X_2+ X_3 = d; since partitions lie in the region X_1 \ge X_2 \ge X_3, we only consider the coloured region:

triangular_simplex

The point \lambda has \lambda_3 = 0. Assuming |\mu| = d, the condition \mu \trianglelefteq\lambda then reduces to a single inequality \mu_1 \le \lambda_1. Hence, \mu lies in the brown region below:

simplex_lambda_and_mu

To fix ideas, consider the case \lambda = (5,3). Calculating the Kostka coefficients gives us:

geometric_kostka_v2

Taking into account all coefficients then gives us a rather nice diagram for the weight space decomposition.

geometric_kostka_v1

E.g. we have \dim V(\lambda)_{2,3,3} =3 and \dim V(\lambda)_{4,3,1} = 2. These correspond to the following SSYT of shape (5,3):

weight_spaces_and_ssyt

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