This question has something to do with [that one](https://mathoverflow.net/q/194169).


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 $V\mapsto X\wedge V$.



Homogeneous polynomial vector fields tangent to the unit sphere
Isn't the answer just 
$$
D(n;d) = n{{n+d-1}\choose{d}}- {{n+d}\choose{d+1}}= d{{n+d-1}\choose{d+1}}\quad ?
$$


