Why commutator transforms like a vector

Question

Let's define \begin{equation} [X, Y]^i=X^j \frac{\partial}{\partial x^j} Y^i-Y^j \frac{\partial}{\partial x^j} X^i \end{equation} Then for coordinate transformation $x^{i\prime}(x^1,x^2,\ldots,x^n)$: \begin{equation} X^{j\prime}=\dfrac{\partial x^{j\prime}}{\partial x^i}X^i;\quad Y^{k\prime}=\dfrac{\partial x^{k\prime}}{\partial x^l}Y^l;\quad \dfrac{\partial}{\partial x^{m\prime}}=\dfrac{\partial x^n}{\partial x^{m\prime}}\dfrac{\partial}{\partial x^n} \end{equation} I'm getting: \begin{equation} [X,Y]^{k\prime}=\left(\dfrac{\partial x^{j\prime}}{\partial x^i}X^i\right)\left(\dfrac{\partial x^n}{\partial x^{j\prime}}\dfrac{\partial}{\partial x^n}\right)\left(\dfrac{\partial x^{k\prime}}{\partial x^l}Y^l\right)-\left(\dfrac{\partial x^{j\prime}}{\partial x^l}Y^l\right)\left(\dfrac{\partial x^n}{\partial x^{j\prime}}\dfrac{\partial}{\partial x^n}\right)\left(\dfrac{\partial x^{k\prime}}{\partial x^l}X^l\right) \end{equation} But if commutator is also vector, I should get something like: \begin{equation} [X,Y]^{k\prime}=\dfrac{\partial x^{k\prime}}{\partial x^i}[X, Y]^i \end{equation} It doesn't seem to appear here

Answer 1

Ok, I solved it: \begin{equation} [X,Y]^{i\prime}=\left(\dfrac{\partial x^{j\prime}}{\partial x^k}X^k\right)\left(\dfrac{\partial x^n}{\partial x^{j\prime}}\dfrac{\partial}{\partial x^n}\right)\left(\dfrac{\partial x^{i\prime}}{\partial x^l}Y^l\right)-\left(\dfrac{\partial x^{j\prime}}{\partial x^k}Y^k\right)\left(\dfrac{\partial x^n}{\partial x^{j\prime}}\dfrac{\partial}{\partial x^n}\right)\left(\dfrac{\partial x^{i\prime}}{\partial x^l}X^l\right)= \end{equation} \begin{equation} =\left(\delta^n_kX^k\right)\left(\dfrac{\partial}{\partial x^n}\right)\left(\dfrac{\partial x^{i\prime}}{\partial x^l}Y^l\right)-\left(\delta^n_kY^k\right)\left(\dfrac{\partial}{\partial x^n}\right)\left(\dfrac{\partial x^{i\prime}}{\partial x^l}X^l\right)= \end{equation} \begin{equation} =X^n\dfrac{\partial}{\partial x^n}\left(\dfrac{\partial x^{i\prime}}{\partial x^l}Y^l\right)-Y^n\dfrac{\partial}{\partial x^n}\left(\dfrac{\partial x^{i\prime}}{\partial x^l}X^l\right) \end{equation} \begin{equation} =\dfrac{\partial x^{i\prime}}{\partial x^l}\left(X^n\dfrac{\partial}{\partial x^n}Y^l-Y^n\dfrac{\partial}{\partial x^n}X^l\right)+\left(Y^lX^n\dfrac{\partial}{\partial x^n}\dfrac{\partial x^{i\prime}}{\partial x^l}-X^lY^n\dfrac{\partial}{\partial x^n}\dfrac{\partial x^{i\prime}}{\partial x^l}\right) \end{equation} Second bracket equals zero while the first equals: \begin{equation} \dfrac{\partial x^{i\prime}}{\partial x^l}[X,Y]^l \end{equation}