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I'm trying exercise $A.39$ in Fulton and Harris. They suggest to first prove the formula

$|x_j^{l_i}| \prod_{j=1}^k(1-x_j)^{-1} = \sum |x_j^{m_i}| \hspace{1in} (\ast)$

where the sum on the right is over all $k$ - tuples $(m_1,\ldots,m_k)$ of non-negative integers with $m_1 \geq l_1\geq \ldots \geq m_k \geq l_k$. We induct on $k$:

Now the base case $k=1$ for $(\ast)$ is easy to see. So to see how my induction argument would work in general (using the inductive hypothesis to finish the problem) I looked at how the $k=3$ case can be reduced to the $k=2$ case. I then applied the "inductive hypothesis" to the $k = 2$ case. If I understand this, I am done. For $k = 3$, we have:

$\begin{eqnarray*} \frac{1}{(1-x_1)(1-x_2)(1-x_3)}\left|\begin{array}{ccc} x_1^{l_1} & x_2^{l_1} & x_3^{l_1} \\ x_1^{l_2}& x_2^{l_2} & x_3^{l_2} \\ x_1^{l_3} & x_2^{l_3} & x_3^{l_3} \end{array}\right| &=& \frac{x_1^{l_1} \left|\begin{array}{cc} x_2^{l_2} & x_3^{l_2} \\ x_2^{l_3} & x_3^{l_3} \end{array}\right| - x_2^{l_1}\left|\begin{array}{cc} x_1^{l_2} & x_3^{l_2} \\ x_1^{l_3} & x_3^{l_3} \end{array}\right| + x_3^{l_1}\left|\begin{array}{cc} x_1^{l_2} & x_2^{l_2} \\ x_1^{l_3} & x_2^{l_3} \end{array}\right| }{(1-x_1)(1-x_2)(1-x_3)} \\ &=& \frac{x_1^{l_1}}{(1-x_1)} \sum_{(m_2,m_3):m_2 \geq l_2 \geq m_3 \geq l_3} \left|\begin{array}{cc} x_2^{m_2} & x_3^{m_2} \\ x_2^{m_3} & x_3^{m_3} \end{array}\right| \\ &&- \frac{x_2^{l_1}}{(1-x_2)}\sum_{(m_2,m_3):m_2 \geq l_2 \geq m_3 \geq l_3} \left|\begin{array}{cc} x_1^{m_2} & x_3^{m_2} \\ x_1^{m_3} & x_3^{m_3} \end{array}\right| \\ &&+ \frac{x_3^{l_1}}{(1-x_3)}\sum_{(m_2,m_3):m_2 \geq l_2 \geq m_3 \geq l_3} \left|\begin{array}{cc} x_1^{m_2} & x_2^{m_2} \\ x_1^{m_3} & x_2^{m_3} \end{array}\right| \end{eqnarray*}$

where the last step was made by applying the "inductive hypothesis" to $k =2$ case. I can see that I am nearly done, however:

Problem: The three sums are taken for $(m_2,m_3) : m_2 \geq l_2 \geq m_3 \geq l_3$. This means that $m_2$ can possibly be unbounded. However to finish the problem I need that the sums be taken over those $m_2$ constrained to be less than $l_1$. How can I get around this?

Thanks.

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    I should add that I a know how to finish the proof of Pieri's formula from this identity.2012-10-01

1 Answers 1

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hint: When you collect terms into $3\times 3$ determinants, you will be able to find pairs of determinants related by a row swap which therefore cancel.

In detail: you have $ \sum (-1)^{1+r} x_r ^N | (x_j ^{m_i}) _{i \neq 1, j \neq r} | $

where the sum is over $r=1,2,3$, $N \geq l_1$ and $m_2 \geq l_2 > m_3 \geq l_3$. You crunch the sum over $r$ into a single $3 \times 3$ determinant; the top row is $(x_1^N, x_2^N, x_3^N)$. The second row is the same but with $m_2$ for $N$. Now if $X,Y \geq l_1$ you are seeing a determinant with $X$ as the power on the top row and $Y$ on the second row (from $N=X, m_2=Y$) and the other way round (from $N=Y, m_2=X$), so we may delete all such terms. This means we've thrown away all those terms with $m_2 \geq l_1$.

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    I think I get the gist of what you are saying. Thanks for your answer, I think I can work out the rest.2012-10-01