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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. –  user38268 Oct 1 '12 at 13:37
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1 Answer 1

up vote 1 down vote accepted

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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Thanks for your answer. However did you mean to talk of collecting terms into $2 \times 2$ determinants? –  user38268 Oct 1 '12 at 15:16
    
No, I mean that you gather your last three sums together (expanding $x_1^{l_1}/(1-x_1)$ in powers of $x_1$ and so on) then organize the terms of this big sum into an infinite sum of row expansions of 3x3 determinants like the one on the right hand side of your first equality. –  mt_ Oct 1 '12 at 15:21
    
I think I get the gist of what you are saying. Thanks for your answer, I think I can work out the rest. –  user38268 Oct 1 '12 at 15:30
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