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}.


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.


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. ♦


Simplify the following sum:

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


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)}.


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.


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.


  • \{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).

warningHowever, 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.


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.


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. ♦

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