Polynomials and Representations X

Cauchy’s Identity

In this article, our primary focus is the ring $\Lambda_n$ of symmetric polynomials in $x_1, \ldots, x_n.$

Theorem (Cauchy’s Identity). Consider polynomials $h_\lambda, m_\lambda \in \Lambda_n$ over all partitions $\lambda.$ [Recall that $m_\lambda = 0$ if $l(\lambda) > n.$ ] We have an equality of formal power series:

$\displaystyle \sum_{\lambda} h_\lambda(x_1, \ldots, x_n) m_\lambda(y_1, \ldots, y_n) = \prod_{1\le i,j\le n} \frac 1 {1 - x_i y_j}.$

Note.

For convenience, we will use $x$ and $y$ to denote $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$ respectively. Also $x^\lambda$ means $x_1^{\lambda_1} x_2^{\lambda_2} \ldots.$

Note that the LHS sum is well-defined since for each $d\ge 0$ , there are only finitely many partitions of $d,$ so we get a finite sum in $\Lambda_n^{(d)}$ . The eventual sum lies in $\prod_{d\ge 0} \Lambda_n^{(d)}$ , a formal power series in $x_1, \ldots, x_n, y_1, \ldots, y_n.$

Proof

If $d = |\lambda|$ , we have $h_\lambda = \sum_{\mu\vdash d} M_{\lambda\mu} m_\mu$ . Thus we need to show:

$\displaystyle \sum_{d\ge 0}\sum_{\lambda,\mu \vdash d} M_{\lambda\mu} m_\mu(x) m_\lambda(y) = \prod_{1\le i,j\le n} \frac 1 {1 - x_i y_j}.$

The proof is combinatorial: each term on the RHS is

$(1 - x_i y_j)^{-1} = 1 + (x_i y_j) + (x_i y_j)^2 + \ldots.$

Now we compute the coefficient of $x^\mu y^\lambda$ in the expanded product. E.g. for $\lambda = (3,2,2)$ and $\mu = (4,3)$ , to obtain the term $x_1^3 x_2^2 x_3^2 \cdot y_1^4 y_2^3$ we can multiply:

$\displaystyle (x_1 y_1)^2 \cdot (x_3 y_1)^2 \cdot (x_1 y_2)^1 \cdot (x_2 y_2)^2 \implies \begin{array}{|c|ccc|}\hline & 3 & 2 & 2 \\ \hline 4 & 2 & 0 & 2 \\ 3 & 1 & 2 & 0 \\ \hline\end{array}.$

Hence each occurrence of $x^\mu y^\lambda$ in the expansion corresponds to a matrix of non-negative integers with row sum $\mu_i$ and column sum $\lambda_j$ , and the coefficient of $x^\lambda y^\mu$ on the RHS is exactly $M_{\lambda\mu}.$ This means RHS = LHS. ♦

Exercise

Simplify the following sum:

$\displaystyle \sum_\lambda e_\lambda(x_1, \ldots, x_n) m_\lambda(y_1, \ldots, y_n).$

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Hall Inner Product for $\Lambda^{(d)}_n$

Now $\{h_\lambda\}$ and $\{m_\lambda\}$ are dual bases under the Hall inner product and they satisfy Cauchy’s identity. We would like to prove the converse that Cauchy’s identity imply dual bases. However, the problem is that our results for the Hall inner product were proven for $\Lambda$ and not $\Lambda_n.$ To obtain similar results for $\Lambda_n$ , we need to be a bit more careful.

We already know that

$\{h_\lambda : \lambda \vdash d, \lambda_1 \le n\}$

form a basis of $\Lambda_n^{(d)}$ . It turns out we also have:

Proposition. Let $Z_d := \{\lambda : \lambda \vdash d, l(\lambda) \le n\}.$ Then the sets

$\{ h_\lambda : \lambda \in Z_d\}, \quad \{ s_\lambda : \lambda \in Z_d\}$

are $\mathbb{Z}$ -bases of $\Lambda_n^{(d)}.$

Proof

Consider the surjective map $\pi:\Lambda^{(d)} \to \Lambda_n^{(d)}$ which takes $x_i \mapsto 0$ for all $i>n.$ Its kernel has a basis given by the monomial symmetric polynomials $\{ m_\lambda : \lambda \vdash d, l(\lambda) > n\}.$ Now if $l(\lambda) > n$ and $\mu \trianglelefteq \lambda,$ we have $l(\mu) \ge l(\lambda) > n$ as well. Thus:

$l(\lambda) > n \implies s_\lambda = \sum_{\mu\trianglelefteq \lambda} K_{\lambda\mu} m_\mu \in \text{ker}(\pi).$

Since $\{s_\lambda : \lambda \vdash d\}$ is a basis of $\Lambda^{(d)},$ the set $\{s_\lambda : \lambda\vdash d, l(\lambda)\le n\}$ thus forms a basis of $\Lambda_n^{(d)}$ as desired.

Finally since $h_\lambda = \sum_{\mu\trianglerighteq\lambda} K_{\mu\lambda} s_\mu$ and the matrix $(K_{\mu\lambda})_{\mu, \lambda \in Z_d}$ is invertible, we see that $\{h_\lambda : \lambda\in Z_d\}$ also forms a basis of $\Lambda_n^{(d)}.$ ♦

Summary. The following are $\mathbb{Z}$ -bases of $\Lambda_n^{(d)}$ .

$\begin{aligned} \{ m_\lambda : \lambda \vdash d, l(\lambda) \le n\}, &\qquad \{ e_\lambda : \lambda \vdash d, \lambda_1 \le n\}, \\ \{h_\lambda : \lambda \vdash d, l(\lambda) \le n\}, &\qquad \{h_\lambda : \lambda\vdash d, \lambda_1 \le n\}, \\ \{s_\lambda : \lambda \vdash d, l(\lambda) \le n\}.&\end{aligned}$

For example, when d=4 and n=3, we obtain the following, where each row gives a $\mathbb{Z}$ -basis of the abelian group.

basis_of_formal_and_polynomial

Note: from this, it is clear that $\omega(s_{\lambda})\ne s_{\overline\lambda}$ when we are confined to $\Lambda_n^{(d)}.$ E.g. in the above diagram $\omega(s_4) \ne s_{1111} = 0.$

Now we define the Hall inner product for $\Lambda_n^{(d)}$ by letting $\{h_\lambda\}_{\lambda \in Z_d}$ and $\{m_\lambda\}_{\lambda \in Z_d}$ to be dual bases. As before, we have the following.

Proposition.

$\{s_\lambda\}_{\lambda \in Z_d}$ is an orthonormal basis.

The Hall inner product on $\Lambda_n^{(d)}$ is symmetric.

Proof. Same as before (check carefully).

warning However, the involution $\omega : \Lambda_n^{(d)} \to \Lambda_n^{(d)}$ is no longer unitary. This can be checked for n=2, d=3. Here, we get $s_3 = m_3 + m_{21}$ and $s_{21} = m_{21}= e_{21}.$ A bit of computation then shows that $\omega(s_{21}) = h_{21} = s_3 + s_{21}$ which is not a unit vector. The reader should examine the proof to see why it does not carry over.

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Criterion for Orthonormal Basis

Now let us prove the desired result.

Theorem. Suppose we have a ollection $f_\lambda, g_\lambda \in \Lambda^{(d)}_n$ indexed by $\lambda$ for all $\lambda \in Z_d$ and $d\ge 0.$ If Cauchy’s identity holds for them:

$\displaystyle \sum_{d\ge 0,\lambda \in Z_d} f_\lambda(x) g_\lambda(y) = \prod_{i,j\ge 1} (1 - x_i y_j)^{-1},$

then $f_\lambda$ and $g_\lambda$ are dual bases with respect to the Hall inner product.

Proof

We know that $\{h_\lambda\}_{\lambda\in Z_d}$ and $\{m_\lambda\}_{\lambda \in Z_d}$ both satisfy Cauchy’s identity and are dual. Taking the degree-d component in x or y:

$\displaystyle \sum_{\lambda \in Z_d} f_\lambda(x) g_\lambda(y) = \sum_{\lambda\in Z_d} m_\lambda(x) h_\lambda(y).$

Now, let us first show that $\{f_\lambda\}_{\lambda\in Z_d}$ is a basis of $\Lambda^{(d)}_n \otimes_\mathbb{Z} \mathbb{Q}$ ; comparing dimensions, it suffices to show that the set spans this space.

Suppose not; then there exists $h \in \Lambda_n^{(d)} - \{0\}$ such that $\left< f_\lambda, h\right> =0$ for all $\lambda \in Z_d.$
Write $h(x) = \sum_{\lambda\in Z_d} c_\lambda h_\lambda(x)$ , where the $c_\lambda$ are not all zero.
The two Cauchy identities give us (note: $\left< a(x,y), b(x)\right>$ is computed term-wise by taking $a(x,y)$ as a power series in x with coefficients in $\mathbb{Z}[\![ y ]\!]$ ):

$\displaystyle 0 = \left< \sum_{\lambda\in Z_d} f_\lambda(x) g_\lambda(y), h(x)\right> =\left< \sum_{\lambda\in Z_d} m_\lambda(x) h_\lambda(y), \sum_{\mu\in Z_d} c_\mu h_\mu(x)\right> = \sum_{\lambda\in Z_d} c_\lambda h_\lambda(y).$

Hence $h=0$ as desired.

Now it remains to show that $f_\lambda^\vee = g_\lambda$ , where the dual is taken over $\mathbb{Q}.$ We have:

$\displaystyle \left< m_\lambda(x), \sum_{\mu\in Z_d} h_\mu(x) m_\mu(y) \right> = m_\lambda(y)$

for any $\lambda \in Z_d$ ; by linearity we can replace $m_\lambda$ with any $f\in \Lambda_n^{(d)}.$ In particular

$\displaystyle f_\lambda^\vee(y) = \left< f_\lambda^\vee(x), \sum_{\mu\in Z_d} h_\mu(x) m_\mu(y) \right> =\left< f_\lambda^\vee(x), \sum_{\mu\in Z_d} f_\mu(x) g_\mu(y)\right> = g_\lambda(y)$

which is what we wanted. ♦