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Apr 13, 2017 at 12:58 history edited CommunityBot
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Jan 30, 2015 at 10:06 vote accept Denis Serre
Jan 30, 2015 at 9:54 answer added Robert Bryant timeline score: 4
Jan 30, 2015 at 8:47 comment added Joonas Ilmavirta 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).
Jan 30, 2015 at 8:36 history edited Denis Serre CC BY-SA 3.0
added 63 characters in body
Jan 30, 2015 at 8:35 comment added Joonas Ilmavirta 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 $a_i^{j_1,\dots,j_d}$.
Jan 30, 2015 at 8:13 history asked Denis Serre CC BY-SA 3.0