# The formula for a perhaps basic identity (move from stackexchange)

The following question is moved from math stackexchange. It seems that this is not a popular question, but I really want to know the answer so I moved it to here. The question reads as follows.

We know the expansion of the following product

$\displaystyle\prod_{k=1}^n(1+x+y_k)$

can be expressed by the formula

$\displaystyle\sum_{k=0}^n(1+x)^{n-k}e_k(y_1, \ldots, y_n),$

where the $e_k$'s are the elementary symmetric functions,

$e_k (x_1 , \ldots , x_n )=\displaystyle\sum_{1\le j_1 < j_2 < \dots < j_k \le n} x_{j_1} \dotsm x_{j_k}.$

My question is whether we have a nice formula for the expansion of the following product

$\displaystyle\prod_{1\leq k\leq n, 1\leq\ell\leq m}(1+x_\ell+y_k).$

Reference for the nice formula of the above expression will be highly appreciated. (It seems to me that it is related to generating functions, but I have no background in combinatorics.)

Thanks!~

• Just a thought: by the splitting principle, an expression you're seeking for would compute the Stiefel-Whitney/Chern/Pontrjagin classes of the tensor product of two bundles in terms of those of the factors. Obviously, lots of people thought about this but, AFAIK, no nice closed formula is known. – Alex Degtyarev Nov 1 '14 at 18:25
• math.stackexchange.com/questions/1000982/… – Will Jagy Nov 1 '14 at 18:33
• So in other words, this is $\prod_l\det(P_l)$, where $P_l$ is a diagonal matrix with entries $[1+x_l+y_i]_{i=1}^n$; i.e., $\det(P_1\cdots P_m)$. But expanding out the products seems to not be so cool; perhaps working with $\sum_{kl}\log ...$ may be more helpful. – Suvrit Nov 1 '14 at 18:41
• arxiv.org/abs/1012.0014v1 gives a sum expression using Schur functions and determinants. – darij grinberg Nov 1 '14 at 19:06
• Please do not quickly post questions on both MathOverflow and Math.StackExchange, as it is frowned upon by both communities. Please pick one of the sites, wait at least a few days for an answer and only then ask to migrate the question. – Ricardo Andrade Nov 1 '14 at 22:48

I guess the first product is expanded as $$\prod_{k=1}^n(1+x+y_k)=(1+x)^n\prod_k(1+y_k(1+x)^{-1})=\sum_{k\geq 0}e_k(y_1,\ldots,y_n)(1+x)^{-k+n}.$$ For the other products you can write $$\prod_{j=1}^m\prod_{i=1}^n(1+y_it_j)=\sum_\lambda e_\lambda(y_1,\ldots,y_n)m_\lambda(t_1,\ldots,t_m)=\sum_\lambda e_\lambda(\mathbf{y})m_\lambda(\mathbf{t})$$ where the sum is over $\lambda$ with $\ell(\lambda)\leq m$ and $m_\lambda$ is the monomial symmetric function. It seems that the product you are interested in looks something like $$\prod_{j=1}^m\left(\sum_k e_k(y)(1+x_j)^{n-k}\right)=\sum_\lambda e_\lambda(\mathbf{y})m_{\lambda-(n^m)}((\mathbf{1-x})^{-1})$$ In this formula, $\lambda-(n^m)=(\lambda_1-n,\ldots,\lambda_m-n)$ and you take the obvious definition of $m_\mu$ where $\mu$ is a partition with parts not necessarily positive.
$$\langle u,v\rangle = (u,\omega(v))$$
where $(\cdot,\cdot)$ is the standard bilinear form having Schur functions as an orthonormal basis and $\omega$ is the automorphism switching the elementary and homogeneous symmetric functions. In particular, I believe this product also can be expanded as $$\sum_{\lambda} s_{\lambda^t}(\mathbf{y})s_{\lambda-(n^m)}((\mathbf{1-x})^{-1}).$$