Uniqueness of Jacobian

Question

Consider the simple stick-breaking function $$f : [0, 1]^2 \to S^3 : (v_1, v_2) \mapsto (x_1, x_2, x_3) = \big(v_1, (1 - v_1) v_2, (1 - v_1) (1 - v_2)\big),$$ where $S^N = \left\{(x_1, x_2, \ldots, x_N) \mid \sum_i x_i = 1, \forall i : x_i \geq 0\right\}$ is the $N$-dimensional probability simplex.

The inverse of this function should be $$f^{-1}(x_1, x_2, x_3) = \Big(x_1, \frac{x_2}{1 - x_1}\Big)$$ However, there is an equivalent formulation that makes use of the fact that $x_1 + x_2 + x_3 = 1$ $$f^{-1}_\mathrm{eq}(x_1, x_2, x_3) = \Big(x_1, \frac{x_2}{x_2 + x_3}\Big)$$

Both of these formulations should represent the same function, but if we compute the Jacobians for both functions, we get $$\begin{align*} \mathcal{J}_{f^{-1}}(x_1, x_2, x_3) &= \begin{pmatrix}1 & 0 & 0 \\ \frac{x_2}{(1 - x_1)^2} & \frac{1}{1 - x_1} & 0 \end{pmatrix} \\ \mathcal{J}_{f^{-1}_\mathrm{eq}}(x_1, x_2, x_3) &= \begin{pmatrix}1 & 0 & 0 \\ 0 & \frac{x_3}{(x_2 + x_3)^2} & \frac{-x_2}{(x_2 + x_3)^2} \end{pmatrix} = \begin{pmatrix}1 & 0 & 0 \\ 0 & \frac{1}{1 - x_1} - \frac{x_2}{(1 - x_1)^2} & \frac{-x_2}{(1 - x_1)^2} \end{pmatrix}, \end{align*}$$ which seems to suggest that there is no unique Jacobian for $f^{-1} = f^{-1}_\mathrm{eq}$.

Note that the difference between the second rows in the Jacobian is a constant. For higher-dimensional variants of these functions, I noticed that this constant offset is different for every row.

I am aware that constants are unique up to a constant term, but I thought derivatives would be unique (cf. this post with an answer for single-output functions). Therefore, I started wondering: are Jacobians actually unique? If yes, how should the example above be interpreted? If not, is there some specific notion of uniqueness like "unique up to ..."?

Answer 1

For simplicity let $g = f^{-1}$ and $h = f^{-1}_{eq}$

Now let $u\in \mathbb R^3$ such that $\sum_{i=1}^3 u_i = 0$

\begin{align} J(g)u &= \begin{bmatrix}u_1\\ u_1 \frac{x_2}{\left(1-x_1\right)^2} + u_2 \frac1{1-x_1}\end{bmatrix}\\ &= \begin{bmatrix}u_1\\ -(u_2 + u_3) \frac{x_2}{\left(1-x_1\right)^2} + u_2 \frac1{1-x_1}\end{bmatrix}\\ &= J(h)u \end{align}

Now why I choose $\sum_{i=1}^3 u_i = 0$ simply because I want $x + u\in S^3$.

Answer 2

You say that the two expressions represent the same function: but that is true only on $x_1+x_2+x_3=1$. There is no reason at all to suppose that their Jacobians coincide, even on that plane.

I find it helpful to consider a far simpler example.

The two functions $g_1(x)=x$ and $g_2(x)=1$ coincide at $x=1$: yet $g_1'(1)=1\ne 0=g_2'(1)$.