In the book Geometric Analysis on the Heisenberg Group and Its Generalizations proposition 1.3 says: Under the change of coordinates $$ y_1 = x_1 , \, y_2 = x_2, \, \tau = 4t -2x_1x_2,$$ the vector fields $$ X = \partial_{x_1}, \, Y = \partial_{x_2} +x_1 \partial_t, \, T = \partial_t $$ are transformed into $$ X = \partial_{y_1} - 2y_2\partial_{\tau}, \, Y = \partial_{y_2} +2y_1 \partial_{\tau} , \, T = 4\partial_{\tau}. $$ The author says that the proof follows from the following relationships $$ \partial_t = 4\partial_{\tau} , \, \partial_{x_2} = \partial_{y_2} -2y_1 \partial_{\tau} , \, \partial_{x_1} = \partial_{y_1} - 2y_2\partial_{\tau} . $$
I don't understand how this change of coordinates is done, nor how these relations in the demonstration were obtained, can anyone help me understand?
You must use the chain rule, because the two sets of coordinates are functions of each other. For instance, the derivative with respect to $t = t(y_1,y_2,\tau)$ is given by $$ \frac{\partial}{\partial t} = \frac{\partial y_1}{\partial t}\frac{\partial}{\partial y_1} + \frac{\partial y_2}{\partial t}\frac{\partial}{\partial y_2} + \frac{\partial \tau}{\partial t}\frac{\partial}{\partial \tau} = 0\cdot\frac{\partial}{\partial y_1} + 0\cdot\frac{\partial}{\partial y_2} + 4\cdot\frac{\partial}{\partial \tau}, $$ hence $\partial_t = 4\partial_\tau$. In order to speed up those computations, you can carry them out simultaneously with the help of the following relation : $$ \nabla_{(x_1,x_2,t)} = \frac{\partial(y_1,y_2,\tau)}{\partial(x_1,x_2,t)} \cdot \nabla_{(y_1,y_2,\tau)}, $$ which is nothing else than multidimensional chain rule, where $\nabla_{(x_1,x_2,t)} = (\partial_{x_1},\partial_{x_2},\partial_t)^T$ and $\nabla_{(y_1,y_2,\tau)} = (\partial_{y_1},\partial_{y_2},\partial_\tau)^T$ are the respective gradients and $$ \frac{\partial(y_1,y_2,\tau)}{\partial(x_1,x_2,t)} = \begin{pmatrix} 1 & 0 & -2x_2 \\ 0 & 1 & -2x_1 \\ 0 & 0 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 & -2y_2 \\ 0 & 1 & -2y_1 \\ 0 & 0 & 4 \end{pmatrix} $$ is the Jacobian matrix of the considered coordinate transformation, so that $$ \begin{pmatrix} \partial_{x_1} \\ \partial_{x_2} \\ \partial_t \end{pmatrix} = \begin{pmatrix} 1 & 0 & -2y_2 \\ 0 & 1 & -2y_1 \\ 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} \partial_{y_1} \\ \partial_{y_2} \\ \partial_\tau \end{pmatrix} = \begin{pmatrix} \partial_{y_1} - 2y_2\partial\tau \\ \partial_{y_2} - 2y_1\partial_\tau \\ 4\partial_\tau \end{pmatrix}. $$ Now you can substitute the new coordinates in the vector fields; for example, the second vector field becomes : $$ Y = \partial_{x_2} + x_1\partial_t = \left(\partial_{y_2} - 2y_1\partial_\tau\right) + y_1 \cdot 4\partial_\tau = \partial_{y_2} + 2y_1\partial_\tau $$