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$\mathbf{Corrected\ Question}$: How to express $$ \sum_{i,j\in\{0,1,2\}\atop i\neq j} x_ix_j^2 $$ by means of the elementary symmetric functions $$ \begin{array}{} e_1=x_0+x_1+x_2\\ e_2=x_0x_1+x_1x_2+x_0x_2\\ e_3=x_0x_1x_2 \end{array} $$ and $$ \begin{array}{l} p_2=x_0^2+x_1^2+x_2^2\\ p_3=x_0^3+x_1^3+x_2^3\ ? \end{array} $$ Thanks!

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  • $\begingroup$ Uhm... Last two edits were not symmetric. I suspect it was a problem. And I suspect this one is not symmetric either. $\endgroup$
    – user228113
    Aug 7, 2016 at 14:34
  • $\begingroup$ Thanks. Do you mean that the given function is not symmetric? $\endgroup$
    – boaz
    Aug 7, 2016 at 14:38
  • $\begingroup$ For instance, if you switch $x_0$ and $x_2$, you get $$ x_2x_1^2+x_1x_0^2+x_0x_2^2\ne x_0x_1^2+x_1x_2^2+x_2x_0^2$$ The monomial $x_0x_1^2$ has six "symmetric friends", not three. You need all of them. $\endgroup$
    – user228113
    Aug 7, 2016 at 14:38
  • $\begingroup$ Thanks @G.Saggetelli. I corrected the question. $\endgroup$
    – boaz
    Aug 7, 2016 at 14:51

2 Answers 2

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$$a b^2+a c^2+b a^2+b c^2+ c a^2+ c b^2=-3 a b c + (a+b+c)(a b + b c + c a)$$

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The five given polynomial are redundant, so there can be many ways to do it: for instance, since $$(x_0+x_1+x_2)^3=\\=x_0^3+x_1^3+x_2^3+6x_1x_2x_3+3x_0x_1^2+3x_1x_2^3+3x_1x_2^2+3x_2x_0^2+3x_2x_1^2+3x_1x_0^2$$

Your polynomial is $$\frac13 (x_0+x_1+x_2)^3-2x_0x_1x_2-\frac13 (x_0^3+x_1^3+x_2^3)$$

The polynomials $e_1,e_2,e_3$ are a tight choice, and Lozenges' answer covered them.

I'll illustrate an approach (with some trick that might turn useful in general) to find the polynomial in terms of $e_1,p_2,p_3$.

For an expression in terms of $e_1,p_2,p_3$, notice that, the polynomial being homogeneous of degree $3$, you want to find it as linear combination of $e_1^3$, $e_1p_2$, and $p_3$. $$ae^3_1+be_1p_2+cp_3=\text{your polynomial}$$

This is ultimately an exercise of linear algebra (finding the the coordinates of a vector with respect to the basis of a subspace that contains it). It could be written explicitly in matricial form using the standard coordinates of $\Bbb C_{3}^{\hom}[x_0,x_1,x_2]$, but let's cut through the details with tricks and exercises for the reader.

By confronting coefficients of $x_0^3+x_1^3+x_2^3$, you see that you want $a+b+c=0$

Since terms in the form $x_0x_1x_2$ come only from $e_1^3$, you want $a=0$.

You should be able to go on.

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