MHB Symmetric Polynomials Involving Discriminant Poly

mathjam0990
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Question: Let τ = (i, j) ∈ Sn with 1 ≤ i < j ≤ n. Prove: δ(rτ(1) , . . . ,rτ(n) ) = −δ(r1, . . . ,rn)

Note: Discriminant Polynomial δ(r1,r2,...,rn) = ∏ (ri - rj) for i<j

I am pretty confused on where to begin. Based on the note, does −δ(r1, . . . ,rn) then = (r1-r2)(r1-r2)·····(r1-rn)(r2-r3)(r2-r4)····(r2-rn)····(rn-1-rn) ?

Also, since τ = (i, j) is a transposition, does that suggest τ(1) = j (because i "goes to" j), then τ(2) = i (because i "goes to" j and j "goes to" i) ? Sorry I feel like I've got this all mixed up. Could anyone provide detailed insight please?

Thanks in advance!
 
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mathjam0990 said:
Question: Let τ = (i, j) ∈ Sn with 1 ≤ i < j ≤ n. Prove: δ(rτ(1) , . . . ,rτ(n) ) = −δ(r1, . . . ,rn)

Note: Discriminant Polynomial δ(r1,r2,...,rn) = ∏ (ri - rj) for i<j

I am pretty confused on where to begin. Based on the note, does −δ(r1, . . . ,rn) then = (r1-r2)(r1-r2)·····(r1-rn)(r2-r3)(r2-r4)····(r2-rn)····(rn-1-rn) ?

Also, since τ = (i, j) is a transposition, does that suggest τ(1) = j (because i "goes to" j), then τ(2) = i (because i "goes to" j and j "goes to" i) ? Sorry I feel like I've got this all mixed up. Could anyone provide detailed insight please?

Thanks in advance!

Let's just look at an example: take $n = 3$.

Then $\delta(x_1,x_2,x_3) = (x_1-x_2)(x_1-x_3)(x_2-x_3)$.

Suppose we examine the result of $\sigma = (1\ 2)$ applied to $\delta$, that is, we look at:

$(x_{\sigma(1)} - x_{\sigma(2)})(x_{\sigma(1)} - x_{\sigma(3)})(x_{\sigma(2)} - x_{\sigma(3)})$

We have:

$\sigma(1) = 2$
$\sigma(2) = 1$
$\sigma(3) = 3$, so we get:

$(x_2 - x_1)(x_2 - x_3)(x_1 - x_3) = (x_2 - x_1)(x_1 - x_3)(x_2 - x_3) = -\delta(x_1,x_2,x_3)$.

Note the only factor that changed sign in this example is $(x_1 - x_2)$. In general (for higher $n$) what will happen is that $(x_i - x_j)$ will change sign, factors that involve neither $x_i$ nor $x_j$ will remain the same, factors where $i,j < k$ and $k < i,j$ will keep the same sign, so the only factors that will change sign (besides ($x_i - x_j)$) are when $i < k < j$ where the two factors $(x_i - x_k),(x_k - x_j)$ will change sign together.
 
Deveno said:
$(x_2 - x_1)(x_2 - x_3)(x_1 - x_3) = (x_2 - x_1)(x_1 - x_3)(x_2 - x_3) = -\delta(x_1,x_2,x_3)$.

Note the only factor that changed sign in this example is $(x_1 - x_2)$. In general (for higher $n$) what will happen is that $(x_i - x_j)$ will change sign, factors that involve neither $x_i$ nor $x_j$ will remain the same, factors where $i,j < k$ and $k < i,j$ will keep the same sign, so the only factors that will change sign (besides ($x_i - x_j)$) are when $i < k < j$ where the two factors $(x_i - x_k),(x_k - x_j)$ will change sign together.

This is the part I wasn't seeing! Thanks a lot for breaking that down!
 
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