A vector field $X$ on $\mathrm{GL}(n,\mathbb{R})$ with $\begin{cases} X.\mathrm{trace}=\mathrm{Det} \\X.\mathrm{Det}=-\mathrm{trace} \end{cases}$

Question

Is there a vector field $X$ on $\operatorname{M}_n(\mathbb{R})$ or $\operatorname{GL}(n,\mathbb{R})$ with the following condition: $$\begin{cases} X\cdot \operatorname{trace}=\operatorname{Det} \\X\cdot \operatorname{Det}=-\operatorname{trace} \end{cases}$$ where $\operatorname{Det}$ is determinant?

Answer 1

$\DeclareMathOperator\trace{trace}\DeclareMathOperator\Det{Det}$No. Let $\sum_{i,j}x_{ij}(M)\frac\partial{\partial m_{ij}}$ be this vector field. These conditions amount to writing $$\sum_ix_{ii}(M)=\det M,\qquad\sum_{ij}x_{ij}(M)\hat m_{ij}=-{\rm Tr}M,$$ where $\hat M$ is the cofactor matrix. Take $M=a I_n$, for which $\hat M=a^{n-1}I_n$. One must have $$\sum_ix_{ii}(a I_n)=a^n,\qquad a^{n-1}\sum_{i}x_{ii}(a I_n)=-na,$$ which is inconsistent when $a^{2n-1}\ne-an$.

Edit. Since both the trace and the determinant are homogeneous polynomial, you could be interested in a vector field in which the conditions are consistent with homogeneity. Then its coordinates would be themselves homogeneous polynomials in the entries of $M$. For instance, looking for a vector field $X$ such that $$X\cdot \trace=\Det,\qquad X\cdot \Det=-(\trace)^{2n-1}$$ makes sense. The last righ-hand side could be replaced by $- \trace^{n-1}\Det$.