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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 $\mathcal{B}_*$?

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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

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.

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