MHB Symmetric Polynomials s1,s2,s3

mathjam0990
Messages
28
Reaction score
0
Express r12+r22+...+rn2 as a polynomial in the elementary symmetric polynomials s1, s2, . . . ,sn.

I'm sure the equation we are dealing with is (r1+r2+...+rn)2 which is very large to factor out but should yield r12+r22+...+rn2+(other terms)

I believe s1=r1+r2+...+rn

s2=Σri1ri2 for 1≤i1≤i2≤n

s3=r1r2⋅⋅⋅⋅⋅⋅rn

So the answer should be r12+r22+...+rn2 = s12 - (something with s2,...sn) Sorry I am not sure what to employ here to break this all the way down.

If there is anyone who could provide an explanation, that would be amazing. Thank you!
 
Physics news on Phys.org
mathjam0990 said:
Express r12+r22+...+rn2 as a polynomial in the elementary symmetric polynomials s1, s2, . . . ,sn.

I'm sure the equation we are dealing with is (r1+r2+...+rn)2 which is very large to factor out but should yield r12+r22+...+rn2+(other terms)

I believe s1=r1+r2+...+rn

s2=Σri1ri2 for 1≤i1≤i2≤n

s3=r1r2⋅⋅⋅⋅⋅⋅rn

So the answer should be r12+r22+...+rn2 = s12 - (something with s2,...sn) Sorry I am not sure what to employ here to break this all the way down.

If there is anyone who could provide an explanation, that would be amazing. Thank you!
.

now $s_1 = r_1 + r_2 + r_3 + r_ 4 ...\cdots r_n$
because we need to evaluate $r_1^2 + r_2^2 + ..\cdots + r_n^2$ I am tempted to sqaure $S_1$ which shall give
square terms and additional ones

As we get $s_1^2 = r_1^2 + r_2^2 + r_3^2 + r_ 4 ...\cdots r_n^2 + r_1 r_2 + r_1 r_3 + ...$
$= r_1^2 + r_2^2 + r_3^2 + r_ 4 ...\cdots r_n^2 + 2 \sum_{p=1}^{m-1} \sum_{m=1}^{n} r_p r_m$ (one comes from $r_p r_m$ and another from $r_m r_p$ for $p \ne m$
$= r_1^2 + r_2^2 + r_3^2 + r_ 4 ...\cdots r_n^2 + 2 s_2$

hence given sum = $s_1^2 - 2s_2$
 
mathjam0990 said:
Express r12+r22+...+rn2 as a polynomial in the elementary symmetric polynomials s1, s2, . . . ,sn.

I'm sure the equation we are dealing with is (r1+r2+...+rn)2 which is very large to factor out but should yield r12+r22+...+rn2+(other terms)

I believe s1=r1+r2+...+rn

s2=Σri1ri2 for 1≤i1≤i2≤n

s3=r1r2⋅⋅⋅⋅⋅⋅rn

So the answer should be r12+r22+...+rn2 = s12 - (something with s2,...sn) Sorry I am not sure what to employ here to break this all the way down.

If there is anyone who could provide an explanation, that would be amazing. Thank you!

It's easier to illustrate what the $s_i$ are with a particular $n$. Let's use $n = 4$. Suppose our variables are $r_1,r_2,r_3,r_4$. Then our polynomials are in the polynomial ring:

$F[r_1,r_2,r_3,r_4]$ (where $F$ is our underlying field).

Given $f \in F[r_1,r_2,r_3,r_4]$, we can have $\sigma \in S_4$ operate on $F[r_1,r_2,r_3,r_4]$ by:

$\sigma(f(r_1,r_2,r_3,r_4)) = f(r_{\sigma(1)},r_{\sigma(2)},r_{\sigma(3)},r_{\sigma(4)})$.

We say a polynomial $f \in F[r_1,r_2,r_3,r_4]$ is *symmetric* if $\sigma(f) = f$.

The ELEMENTARY symmetric polynomials are the "basic" symmetric polynomials of any given degree > 0. Let's see how they are constructed, for $n = 4$, by seeing what they ought to be (note we don't care about constant terms, since $\sigma$ never affects them, so for simplicity's sake, we'll always assume all constant terms are 0).

For degree $1$, our polynomials are just linear combinations of the $r_i$. So the "simplest" symmetric combination would be:

$s_1 = r_1 + r_2 + r_3 + r_4$.

This can't be improved on, eliminating any term would break the symmetry.

for degree $2$, our polynomials would be linear combinations of $r_ir_j$ ($i = j$ might happen) + terms of lower degree. since we can capture symmetric terms of lower degree by the polynomial above, we will only look at symmetric combinations of $r_ir_j$.

At first, it might seem the best candidate would be:

$f = r_1^2 + r_1r_2 + r_1r_3 + r_1r_4 + r_2^2 + r_2r_3 + r_2r_4 + r_3^2 + r_3r_4 + r_4^2$ (all possible degree 2 terms summed).

However, note that:

$s_1^2 = (r_1 + r_2 + r_3 + r_4)^2 =$

$ r_1^2 + r_1r_2 + r_1r_3 + r_1r_4 + r_2r_1 + r_2^2 + r_2r_3 + r_2r_4 + r_3r_1 + r_3r_2r_3^2 + r_4r_1 + r_4r_2 + r_4r_3 + r_4^2$

$= r_1^2 + r_2^2 + r_3^2 + r_4^2 + 2(r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4)$

If we call the "cross-terms" $2g$, we have:

$f = s_1^2 - 2g + g$ where the $s_1^2 - 2g$ gives us just the degree 2 terms in $f$ that are squares.

Thus all we need to get $f$ is $s_1$ and $g$ (both of these are still symmetric) and $g$ is "more minimal" than $f$ (fewer terms). So we should pick $g$ as $s_2$:

$s_2 = r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4$

The same sort of logic applies to the higher terms (but the algebra is horrendous), so:

$s_3 = r_1r_2r_3 + r_1r_2r_4 + r_1r_3r_4 + r_2r_3r_4$

(Do you see the pattern? We pick $s_k = \sum\limits_{i_1 < i_2 < \cdots < i_k} r_{i_1}r_{i_2}\cdots r_{i_k}$)

and finally $s_4 = r_1r_2r_3r_4$.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

Similar threads

Replies
2
Views
3K
Replies
2
Views
4K
Back
Top