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This question has something to do with that one.

Let $n\ge1$ and $d\ge1$ be two given integers. Consider the polynomial vector fields $v=(v_1,\ldots,v_n)$ whose components $v_j$ are homogeneous of degree $d$ in the indeterminates $X_1,\ldots,X_n$.

What is the dimension $D(n;d)$of the subspace defined by the equation $$X_1v_1(X)+\cdots+X_nv_n(X)=0\quad ?$$

I computed this dimension for small dimensions: $D(1;d)=0$, $D(2;d)=d$ and $D(3;d)=d(d+2)$. I suspect that the problem has been solved a while ago and the formula is simple in terms of binomials. A solution might come by considering a the sequence of morphisms $$\cdots\rightarrow\Lambda_{n-2}({\mathbb R}^n)\otimes {\rm Hom}_n^{d-1}\rightarrow\Lambda_{n-1}({\mathbb R}^n)\otimes {\rm Hom}_n^{d}\rightarrow\Lambda_{n}({\mathbb R}^n)\otimes {\rm Hom}_n^{d+1},$$ where ${\rm Hom}_n^d$ denotes the space of homogeneous polynomials of degree $d$, and each arrow is of the form $V\mapsto X\wedge V$.

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    $\begingroup$ You also have $D(n;1)=\frac12n(n-1)$ because a linear vector field $v(X)=AX$ satisfies $X\cdot v(X)=0$ iff the matrix $A$ is skew-symmetric. I wonder if it would be useful in general to write $v_i(X)=\sum_{j_1,\dots,j_d}a_i^{j_1,\dots,j_d}X_{j_1}\cdots X_{j_d}$ and study the symmetries of $a_i^{j_1,\dots,j_d}$. $\endgroup$
    – Joonas Ilmavirta
    Commented Jan 30, 2015 at 8:35
  • $\begingroup$ A continuation to my previous comment: If we assume $a_i^{j_1,\dots,j_d}$ to be symmetric in the upper indices, we lose redundancy and have a parametrization of all $d$-homogeneous $n$-vector fields. We can then write $a_i^{j_1,\dots,j_d}$ as a sum of two parts, one of which is symmetric under swapping the lower index with an upper one and the other one antisymmetric. Now $X\cdot v(X)=0$ iff the symmetric part vanishes. The problem is then to find the dimension of the space of tensors with these symmetries (assuming my reasoning makes sense). $\endgroup$
    – Joonas Ilmavirta
    Commented Jan 30, 2015 at 8:47

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Isn't the answer just $$ D(n;d) = n{{n+d-1}\choose{d}}- {{n+d}\choose{d+1}}= d{{n+d-1}\choose{d+1}}\quad ? $$

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  • $\begingroup$ Yes indeed. I realized a bit late that it is enough to say that $v\mapsto x\wedge v\sim x\cdot v$ is onto $\Lambda_n({\mathbb R}^n)\otimes{\rm Hom}_n^{d+1}$. $\endgroup$
    – Denis Serre
    Commented Jan 30, 2015 at 10:08

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