Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $V$ be a $\mathbb{C}$-vector space of finite dimension. Denote its $d$-th symmetric power by $V^{\odot d}$. I am looking for a proof that $V^{\odot d}$ is generated by the elements $v^{\odot d}$ for $v\in V$.

A different way to look at it is the following: Consider the polynomial ring $R=\mathbb{C}[x_1,\ldots,x_n]$ and $f$ a homogeneous polynomial of degree $d$: Then, I want to show that there are linear polynomials $h_1,\ldots,h_k$ such that $f$ is a linear combination of the $d$-th powers $h_i^d$.

In the case $d=2$, this follows from $2xy = (x+y)^2 - x^2 - y^2$. For higher $d$, I recall seeing a proof involving multinomial coefficients once, but I do not remember the details. I have tried to work it out again, but it seems a bit cumbersome, so I am asking whether you know any textbook where this result is proved. If you know an easy proof, I'd be very happy if you could outline it, though.

share|improve this question
2  
Maybe en.wikipedia.org/wiki/Newton%27s_identities will help. –  KCd May 16 '12 at 8:58

2 Answers 2

up vote 2 down vote accepted

In my answer here I note that symmetric tensors, as multilinear functionals, descend to linear maps on the symmetric power of the underlying vector space. I then reason that if we could show that $\mathrm{Sym}^n V$ is generated by $n$th powers of elements from $V$ the question on tensors would then be answered in its general form decisively. I remark that this is formally equivalent to the elementary symmetric polynomials $e_n$ being expressible as sums of $n$th powers of homogeneous polynomials.

This was the subject of my question here, which received a correct answer (containing a proof of the claim) from user m_l. It was very combinatorial and indeed involved multinomial coefficients, though I'm not sure how related it is to what you've seen before. (Unfortunately, at this point in time I am the only person to have upvoted poor m_l.) It requires the characteristic of the base field be greater than the power $n$ in question (or zero, of course).

share|improve this answer

Let $W$ be the (finite-dimensional) vector space generated by $d$-th powers of linear functions in $x_1,\dots,x_n$. Let $h_1,\dots,h_d$ be such linear functions. Consider the polynomial map $f:\mathbb{C}^d\to W$ given by $f(t_1,\dots,t_d)=(t_1h_1+\dots+t_d h_d)^d$. As $$\frac{\partial}{\partial t_1}\cdots\frac{\partial}{\partial t_d}f=d!\;h_1\dots h_d,$$ we have $h_1\dots h_d\in W$. This shows that any degree-$d$ monomial is in $W$, and therefore also any homogeneous degree-$d$ polynomial.

share|improve this answer
    
I don't understand what the $h_i$'s are: surely there are more than $d$ $d$-th powers of linear functions? –  Georges Elencwajg May 16 '12 at 9:24
    
@GeorgesElencwajg: I edited my last sentence to make it clearer. –  user8268 May 16 '12 at 9:36
    
I don't quite understand your argument. Assuming that $f$ is holomorphic, which seems alright: How and why is $\partial/\partial t_1\cdots\partial/\partial t_d f$ an element of $W$? –  Jesko Hüttenhain May 16 '12 at 11:24
    
@rattle: well, if $f:\mathbb{C}^n\to W$, then $\partial f/\partial t_1$ is again a function $\mathbb{C}^n\to W$, hence so is $\partial^2 f/\partial t_1\partial t_2$, etc. –  user8268 May 16 '12 at 14:21
1  
Dear user8268, I certainly didn't vote you down and I think there is absolutely no reason why you should take a month off from this site: I hope you will not keep that promise! –  Georges Elencwajg May 16 '12 at 14:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.