I need some help about the problem below.


Let $d\geq 4$ and $f$ a symmetric polynomial, homogeneous of degree $d$, in $n$ variables $x\_1,\dots,x\_n$, with real coefficients. We set


$$ E(f):=\sum\_{j=1}^{n}x\_{j}\frac{\partial^2 f}{\partial x\_{j}^2}. $$



> 
> Question: If $f\neq (x\_1+\ldots+x\_n)^d$ then the dimension of the real span of the set below
> $$ \big\{\frac{\partial f}{\partial x\_1},\ldots,\frac{\partial f}{\partial x\_n},E(f)\big\} $$
> is at least $n$. In symbols:
> $$ \dim\_{\mathbb{R}}\left(\mathrm{span}\big\{\frac{\partial f}{\partial x\_1},\ldots,\frac{\partial f}{\partial x\_n},E(f)\big\}\right)\geq n. $$
> 
> 
> 


**For degree 2 and 3 this assertion is already settled.**


Have any one any idea to prove this statement in general ?


Two considerations:


1- Recall that an equivalent condition for f to be homogeneous of degree d is that f must satisfy Euler's identity below
$$ \sum\_{j=1}^{n}x\_{j}\frac{\partial f}{\partial x\_{j}}=d\cdot f. $$


2- Another thing that can be useful to remember here is that any homogeneous polynomial $f$ of degree $d$ is a linear combination of $k$, $d$-th powers of linear forms, for some $k$.


This question was posted also on Math.SE.



Dimension of the span of all partial derivatives of a given homogeneous symmetric polynomial $f$ and the polynomial $E(f)$
If we're lucky, the $f\_j$, $1\le j\le n$ (I use index notation for partial derivatives) are already linearly independent. If not, then $v\cdot\nabla f=0$ for some vector $v\not= 0$, but then also $Pv\cdot\nabla f=0$ for all permutations of the components of $v$. I believe there are three possibilities, though I haven't proved this carefully: the $Pv$ will typically span $\mathbb R^n$ (then $f=0$), or they can span an $(n-1)$ dimensional subspace, so that $f$ can then only depend on $w\cdot x$; this corresponds to the exceptional example you mentioned. The only way to avoid such a large span is to have specifically $v=(1,1,\ldots , 1)$.


In this case, we can write
$$
f(x\_1,x\_2,\ldots ,x\_n)=f(0,x\_2-x\_1, \ldots, x\_n-x\_1) = g(x\_2-x\_1,\ldots, x\_n-x\_1) ,
$$
and $g(t\_2,\ldots , t\_n)=f(0,t\_2,\ldots ,t\_n)$ is still a symmetric homogeneous polynomial of its arguments.


Now suppose we have an additional linear relation $\sum a\_jf\_j=\sum x\_j f\_{jj}$. We can rewrite this as
$$
\sum\_{j\ge 2} b\_j g\_j(t\_2,\ldots ,t\_n) - \sum\_{j\ge 2} t\_j g\_{jj}(t\_2,\ldots ,t\_n) = x\_1\Delta f(0,t\_2,\ldots ,t\_n) ,
$$
with $t\_j=x\_j-x\_1$ and $b\_j=a\_j-a\_1$. Since $x\_1$ and the $t\_j$'s can be varied independently, both sides must be (identically) equal to zero. Now $g$ is symmetric, so we can as above manipulate for example the first term as follows:
$$
\left( \sum b\_j D\_j\right) g(t\_2,\ldots, t\_n) = \left( \sum b\_j D\_j\right) g(t\_{\pi 2},\ldots, t\_{\pi n}) = \sum b\_{\pi k} g\_k (t\_{\pi 2},\ldots ,t\_{\pi n})
$$
The same procedure applied to $\sum t\_j g\_{jj}$ produces
$$
\sum t\_{\pi k} g\_{kk}(t\_{\pi 2}, \ldots , t\_{\pi n}) ,
$$
so by comparing these we obtain that in fact $\sum b\_{\pi k} g\_k(t) -\sum t\_k g\_{kk}(t)=0$ for all permutations $\pi$ (this worked because the $b\_j$'s, unlike the $t\_j$'s, do not get reordered when we relabel the variables at the end). Thus we also have that
$$
\sum (b\_{\pi k}-b\_k) g\_k = 0 . \quad\quad\quad\quad (1)
$$


If (1) actually had non-zero coefficients, then, since the permutation $\pi$ is arbitrary, I again obtain $v\cdot\nabla g$ on an at least $(n-2)$ dimensional subspace of vectors $v$, so $g$ could only depend on $\sum c\_j t\_j$, and, being a symmetric function, it really would have to depend on $\sum t\_j$ only, but this isn't working.


So the conclusion is that we must have $b\_2=\ldots =b\_n$, and thus also $a\_2=\ldots =a\_n$. I can repeat the whole argument with another variable taking the role of $x\_1$ to see that in fact (if $n\ge 3$, but $n=2$ is easy to do directly) $a\_1=\ldots =a\_n$. So, to sum this up, the only possibility left open at this point is $E(f)=\Delta f=\sum f\_j=0$. I don't think there can be such an $f$, but it's apparently not quite as easy as I originally thought and I'm running out of steam now, so I'll leave it at that for now. (Perhaps the way to go is to exploit the extra symmetry of $g$ that comes from swapping $x\_1$ with another variable in $f$; this produces lots of identities.)




---


**Addendum:** I have an argument now, but it's very messy and there must be a better, more insightful way of doing this. Here's a very quick outline: as suggested above, swap $x\_1$ and $x\_2$, say. This shows that
$$
g(t\_2,\ldots, t\_n) = g(-t\_2,t\_3-t\_2,\ldots , t\_n-t\_2) .
$$
Use this on both sides of $\sum t\_j g\_{jj} = c\sum g\_j$ to produce identities for $g\_k$. After some fooling around with these and with the help of Euler's identity, I finally arrived at
$$
\left( |x|^2 - n\overline{x} \right)^2 f\_j = d\left( x\_j-\overline{x}\right) f ,
$$
with $\overline{x} = (x\_1+\ldots + x\_n)/n$, and $d$ is the degree of $f$ and $g$. Take the derivative wrt $x\_j$ on both sides and then sum over $j$. Use that $\Delta f=0$ and $\sum x\_j f\_j = df$. After a calculation, this shows that $f=0$.


