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.

  • $\begingroup$ Can you please provide a bit more background? Where does this problem appear? $\endgroup$ May 4, 2015 at 23:56
  • $\begingroup$ It comes from my PhD thesis $\endgroup$ May 4, 2015 at 23:59
  • 1
    $\begingroup$ This was also posted on Math.SE. Please indicate when you post in multiple forums. $\endgroup$
    – Pedro M.
    May 4, 2015 at 23:59
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    $\begingroup$ For more references and details please look at arxiv.org/abs/1504.07318 $\endgroup$ May 5, 2015 at 0:00
  • $\begingroup$ No, because the dimension is 1, when $f=(x_1+\cdots+x_n)^d$. Since any partial derivative gives us $$ \frac{\partial f}{\partial x_i}=d\cdot (x_1+\cdots+x_n)^{d-1}. $$ and $E(f)=d(d-1)\cdot (x_1+\cdots+x_n)^{d-1}$ we see that the desired dimension is always 1. $\endgroup$ May 5, 2015 at 0:13

1 Answer 1


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 $f=0$.

  • $\begingroup$ I remember that the Vandermonde determinant satiafies $E(f)=0$ $\endgroup$ May 5, 2015 at 18:36
  • $\begingroup$ @HectorBlandin: The Vandermonde determinant is anti-symmetric in its arguments, not symmetric, but it is true that this last point is not as clear as I originally thought. I'll update. $\endgroup$ May 5, 2015 at 22:39
  • $\begingroup$ By P you mean a permutation I think ? And $w\cdot x$ should be $P\cdot x$ where $x=x_1,\ldots,x_n$ ? $\endgroup$ May 6, 2015 at 14:43
  • $\begingroup$ @HectorBlandin: $w\cdot x$ is the usual dot product; $Px$ is a permutation (matrix, if you like) applied to $x$, so $Px = (x_{\pi 1},\ldots ,x_{\pi n})$. $\endgroup$ May 6, 2015 at 17:01
  • $\begingroup$ $w$ should be $P$ ? $\endgroup$ May 6, 2015 at 17:03

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