Chain rule for partial derivatives of a function with vector arguments

Question

If I have two functions of vectors $x,y \in \mathbb{R}^n$ defined by \begin{equation} f(x + A y), \quad g(x - A y) \end{equation} for some matrix $A$ and I wish to compute the poisson bracket \begin{equation} \{ f, g\} = \sum_i \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial y_i} -\frac{\partial f}{\partial y_i} \frac{\partial g}{\partial x_i} . \end{equation} How would I go about doing this? Its clearly some application of the chain rule but I'm not entirely sure how to do it on vector valued functions like this.

Answer 1

The way your Poisson Bracket is written, it does not make sense that $f$ and $g$ are "vector valued". So I will assume scalar values, with vector inputs. $$f,g:\mathbb{R}^n\mapsto \mathbb{R}$$

Using index notation you have $$\begin{align} \frac{\partial f(x+Ay)}{\partial y_i} &= \frac{\partial f(x+Ay)}{\partial x_j}\frac{\partial (Ay)_j}{\partial y_i} \end{align} $$ then $$\frac{\partial (Ay)_j}{\partial y_i} = \partial_{y_i} (A_{jk}y_k) =A_{j,k}\delta_k^i = A_{j,i}$$ Therefore $$\frac{\partial f(x+Ay)}{\partial y_i} = A_{j,i}\partial_{x_j}f(x+Ay) = [A^T\nabla_xf(x+Ay)]_i $$ similarily $$\frac{\partial g(x-Ay)}{\partial y_i} = -[A^T\nabla_x g(x-Ay)]_i$$ Hence $$\begin{align} \{f,g\} &= -\nabla_x f\cdot A^T\nabla_xg - \nabla_x g\cdot A^T\nabla_x f \\ &=-\nabla_x f\cdot(A^T\nabla_x g + A\nabla_x g)|_{(x+Ay)} \end{align}$$