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I learned that the Warning rank of a homogeneous polynomial $h\in \mathbb{C}[x_1, \cdots, x_n]_d$ of degree $d$ is defined by the smallest number of summands such that $h$ can be expressed as a sum of $d$-th powers of linear polynomials. For example, $XY=(\frac{X}{2}+\frac{Y}{2})^2+(i\frac{X}{2}-i\frac{Y}{2})^2$ so the Warning rank of $XY$ is $2$.

But how do we know that any homogeneous polynomial of degree $d$ can be writte as a sum of $d$-th powers of linear polynomials? I read a proof for the case of elementary symmetric polynomials, but I have no idea how to get the generalized result for arbitrary homogeneous polynomials.

I know that the dimension of $\mathbb{C}[x_1, \cdots, x_n]_d$ is $\binom{n+d-1}{d}$. I tried to find $\binom{n+d-1}{d}$ linearly independent polynomials each of which is a $d$-th power of a linear polynomial, but there is no progress so far.

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(Answer for the case $n=2$)

Put $ s = \dim \mathbb{C}[x, y]_d$ and let $0<a_1 < a_2 < \cdots < a_s$ be positive real numbers. Consider $L_i=(x+a_i y)^d$ for $1 \leq i \leq s$. We can show that $L_1, \cdots, L_s$ are linearly independent, as follows.

Observe that $L_i = \sum_{r=0}^d \binom{d}{r} a_i^r x^{d-r} y^r$. So it sufficient to show that determinant of the following matrix is nonzero: (Note that here $s=d+1$)

\begin{bmatrix} 1 & \binom{d}{1}a_1 & \binom{d}{2}a_1^2 & \cdots & \binom{d}{d}a_1^d \\ 1 & \binom{d}{1}a_2 & \binom{d}{2}a_2^2 & \cdots & \binom{d}{d}a_2^d \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & \binom{d}{1}a_s & \binom{d}{2}a_s^2 & \cdots & \binom{d}{d}a_s^d \end{bmatrix}

The determinant is given by $ (\det A) \prod_{r=0}^d \binom{d}{r} $, where $A= \left( a_i^{j-1} \right)_{ij}$ is the Vandermonde matrix. Since the Vandermonde determinant is given by $\prod_{1 \leq i < j \leq s } (a_j-a_i) \neq 0$, we are done.


(Answer for the general case)

Let $s = \dim \mathbb{C}[x_1, \cdots, x_n]_d$ and $0<a_1 < a_2 < \cdots < a_s$ be positive real numbers. Choose positive integers $m_1, \cdots, m_n$ such that $m_j > s(m_{j-1}+\cdots+m_1)$ for all $1<j\leq n$. Define $L_i = \left( \sum_{j=1}^{n} a_j^{m_j} x_j \right)^d$ for $1\leq i \leq s$. We will show $L_1, \cdots, L_s$ are linearly independent by calculating determinant as follows:

First, give an anti-lexicographical order on the set of all monomials $\mathfrak{B}=\{\prod_{j=1}^{n} x_j^{r_j} \mid \sum r_j = d \}$, a basis for $\mathbb{C}[x_1, \cdots, x_n]_d$. Observe that $L_i= \sum \binom{d}{r_1, r_2, \cdots, r_n} a_i^{\sum_{j=1}^{n}m_j r_j} \prod_{j=1}^{n} x_j^{r_j}$, where the indices run over all nonnegative integers $r_1, \cdots, r_n$ such that $r_1 + \cdots + r_n = d$. Here $\prod_{j=1}^{n} x_j^{n_j} < \prod_{j=1}^{n} x_j^{r_j}$ implies $\sum_{j=1}^{n} m_j n_j < \sum_{j=1}^{n} m_j r_j$, by our choice of $m_1, \cdots, m_n$.

Now it suffices to show that determinant of the following matrix $A$ is nonzero. Note that multinomial coefficients can be ignored and the order of columns can be changed since the determinant is alternating multilinear in the columns.

$$A= \begin{bmatrix} a_1^{i_1} & a_1^{i_2} & a_1^{i_3} & \cdots & a_1^{i_s} \\ a_2^{i_1} & a_2^{i_2} & a_2^{i_3} & \cdots & a_2^{i_s} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_s^{i_1} & a_s^{i_2} & a_s^{i_3} & \cdots & a_s^{i_s} \end{bmatrix} $$ Here $i_1>i_2 > \cdots > i_s$ are positive integers.

By the bialternant formula, $ \det A = s_\lambda(a_1, \cdots, a_s)\prod_{1\leq j<k\leq s} (a_j-a_k)$ holds, where $s_\lambda(x_1, \cdots, x_s)$ is the Schur polynomial associated to the partition $\lambda = (\lambda_1, \cdots, \lambda_s)$ with $\lambda_j = i_j - (s-j)$. Recall that the Schur polynomial is defined by the sum of monomials, $$s_\lambda (x_1, \cdots, x_s) = \sum_T x^T = \sum_T x_1^{t_1} \cdots x_s^{t_s}$$ where the summation is over all semistandard Young tableaux $T$ of shape $λ$. The exponents $t_1, \cdots , t_s$ give the weight of $T$, in other words each $t_i$ counts the occurrences of the number $i$ in $T$.

Since $a_1, \cdots, a_s$ are distinct positive real numbers, $s_\lambda(a_1, \cdots, a_s) \neq 0$. Therefore, $ \det A = s_\lambda(a_1, \cdots, a_s)\prod_{1\leq j<k\leq s} (a_j-a_k)$ is nonzero. Hence the claim follows.


Alternatively, we can prove the result by induction; the method is essentially same to the case $n=2$, as suggested in the exercise 23.5 in Introduction to Lie algebras and Representation Theory written by J.Humphreys. For details, see this answer.

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  • $\begingroup$ Great answer but here are a few comments: the "anti-lexicographic" order seems not to have been used (at least there is no need to define a sum or a product. Maybe to choose an (ordered) basis before writing an explicit matrix?). In the same paragraph $L_i= \sum \binom{d}{r_1, r_2, \cdots, r_n} a_i^{\sum_{j=1}^{n}m_j r_j} \prod_{j=1}^{n} x_j^{r_j}$ should simply be $L_i= \sum \binom{d}{r_1, r_2, \cdots, r_n} \prod_{j=1}^{n} (a_j^{m_j} x_j)^{r_j}$ and finally I do not get the business with $\prod_{j=1}^{n} x_j^{n_j} < \prod_{j=1}^{n} x_j^{r_j}$. Do we evaluate at some precise $(x_1, ... ,x_n)$ $\endgroup$ – Noix07 Jan 23 '20 at 16:08
  • $\begingroup$ and anyway it seems again not to be used afterwards $\endgroup$ – Noix07 Jan 23 '20 at 16:10
  • $\begingroup$ @Noix07 I agree that the proof could be done without the monomial order. It was intended to show all $i_1, i_2, \dots, i_s$ are distinct, i.e. $\sum m_j r_j$ are all distinct. Otherwise the determinant of $A$ turns to be zero. $\endgroup$ – Luxerhia Jan 24 '20 at 6:29
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This is too long to be a comment, but is also not a complete solution.

If you fix a monomial order such that the monomials are well ordered (which happens, for example, in lexicographic ordering, although since you are dealing with homogeneous polynomials, you will only have a finite number of monomials of degree d, so this isn't exactly necessary, but it would be for some similar problems), then it suffices to show that you can find a linear combination of $d$th powers of a linear polynomials that has the same leading term as your polynomial. Subtracting off those powers, you will then have a polynomial with a smaller leading term to consider. Because the set of monomials of degree $d$ is well ordered, this process has to end after a finite number of steps.

From here, it suffices to find a linear combinations of $d$th powers of linear whose leading term is the desired leading term. I don't immediately see how to do this, but this at least reduces the problem to something slightly more manageable.

For example, if $x>y$, then the monomials of degree $2$ are ordered $x^2>xy>y^2$, and suppose we have $ax^2+bxy+cy^2$. First, we subtract off a multiple of $x^2$ and we have something whose leading monomial is $xy$. Then we subtract off a multiple of $(x+y)^2+(ix)^2=2xy+y^2$ and we will be left with something whose leading monomial is (less than or equal to) $y^2$.

So we don't need to be able to produce a linear combination of $d$th powers equal to every given monomial, just having a given leading monomial.

For polynomials where the leading term is larger that all of its symmetries (e.g., $x^2y > xy^2$), you could use your result on symmetric polynomials. For other monomials, I do not have an immediate answer. If I think of one, I will update this answer.

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