Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $d>0$ be even. Consider ${\mathbb{R}}[x_1,\ldots, x_n]_d$, i.e. polynomials of degree $d$.

Call a homogeneous polynomial $f$ of degree $d$ a polynomial in quadratic variables if it is of the form $f=p(y_1^2,\ldots, y_n^2)$ for some polynomial $p$. Here the $y_i=y_i(x_1,\ldots, x_n)$ are linear forms such that $\{y_1,\ldots, y_n\}$ is a basis of the degree 1 polynomials ${\mathbb{R}}[x_1,\ldots,x_n]_1$.

For example if we take $p(x,y)=5x^3y^4+(x-3y)^3$, and $x-y, x+y$ to be the linear polynomials, then this gives the example of $5(x-y)^6(x+y)^8+((x-y)^2-3(x+y)^2)^3$ $=5(x-y)^6(x+y)^8-(2x^2+8xy+2y^2)^3$ as an example of a polynomial in quadratic variables.

Is it true that polynomials in quadratic variables are dense in the vector space ${\mathbb{R}}[x_1,\ldots, x_n]_d$? In other words, it every polynomial of degree $d$ a limit of polynomials of degree $d$ in quadratic variables?

Edit: In response to Pete, Ewan's and Darij's comments, I will rephrase this question in terms of ring automorphisms. Firstly, let's ignore odd degree polynomials. In the language of linear change of variables, let $V$ be a real vector space space of dimension $n$. Consider the ring $\mathbb{R}[x_1^2,\ldots, x_n^2]$. Each choice of basis $\mathcal{B}=\{v_1,\ldots, v_n\}$ of the vector space $V$ induces a ring monomorphism $\mathcal{B}_*:\mathbb{R}[x_1^2,\ldots, x_n^2]\to Sym(V)$ to the symmetric algebra on $V$. This monomorphism is given by extending the map $x_i^2\mapsto v_i$.

Is every even degree element of $Sym(V)$ in the image of some induced monomorphism $\mathcal{B}_*$?

share|improve this question
I don't get it. The $y_i$ stay fixed? Do you really mean $p(y_1,...,y_n)$ and not $p(y_1^2,...,y_n^2)$ ? What about linear poylnomials?= –  darij grinberg Dec 16 '10 at 9:25
Something is wrong in your question, but I am not sure enough to edit it myself. It seems that $p$ should be $x^3y^4+(x-3y)^3$ and the quadratic forms be $(x-y)^2$ and $(x+y)^2$. Thus then they are not positive definite. –  Denis Serre Dec 16 '10 at 9:34
OK, but I still don't see how any non-even polynomial could be a limit of polynomials in quadratic variables. –  darij grinberg Dec 16 '10 at 10:42
Besides, I think you can WLOG set $y_1=x_1$, ..., $y_n=x_n$, because the polynomial ring $\mathbb R\left[x_1,...,x_n\right]$ is just one of many ways to coordinatize the symmetric algebra $\mathrm{Sym} V$ of an $n$-dimensional $\mathbb R$-vector space $V$, and the algebra doesn't change if you pass to different coordinates. –  darij grinberg Dec 16 '10 at 10:44
@Colin: you can't. $x^2+2xy$ is not invariant under $(x,y) \mapsto (x,-y)$, whereas any function of $x^2$ and $y^2$ is invariant under this automorphism. But what's your point? Darij is saying that you are only going to get polynomials which are even functions of your variables $y_1,\ldots,y_n$, so the answer to your question is "no". (He's also saying that the general case can be reduced to the special case $(y_1,\ldots,y_n) = (x_1,\ldots,x_n)$ via an automorphism of the ring, but why don't you address his first point first.) –  Pete L. Clark Dec 16 '10 at 11:49
show 6 more comments

1 Answer

The dimension of the linear space of homogenous polynomials of degree $d$ is $\binom{n+d-1}{d}$. The dimension of the space of homogenous polynomials of degree $d$ in squared linear variables is $$\binom{n+d/2-1}{d/2} \cdot (n^2-1),$$ where $n^2-1 = \dim GL_n(\mathbb R)$. One easily see that if $d$ and $n$ are sufficiently large, then the dimension of the set of squared polynomials is too small to exhaust all homogenous polynomials of degree $d$. Indeed,

$$\binom{n+d/2-1}{d/2} \cdot (n^2-1) \leq \frac{(n-1)\cdot n \cdots (n+d/2-1)}{1 \cdot 2 \cdots d/2} \cdot n^2$$ $$< \frac{(n-1)\cdot n \cdots (n+d-1)}{1 \cdot 2 \cdots d} = \binom{n+d-1}{d}$$ if

$$n^2 < \frac{(n+d/2)\cdots (n+d-1)}{(d/2+1) \cdots d} = \prod_{k=1}^{d/2} \frac{n-1+d/2+k}{d/2+k}.$$

But the RHS equals $$\prod_{k=1}^{d/2} \left( 1+\frac{n-1}{d/2+k} \right) \geq \left( 1+\frac{n-1}{d} \right)^{d/2} \sim \exp\left(\frac{n-1}2 \right)$$ for large $d$. So, if $n$ is large, we get $n^2 < \exp\left(\frac{n-1}2 \right) \leq \frac{(n+d/2)\cdots (n+d-1)}{(d/2+1) \cdots d}$ and this implies the claim.

share|improve this answer
add comment

Your Answer


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.