### implicit function problem

From fakalin. If $$F(x, y, z)$$ is a function of 3 variables, and the relation $$F(x, y, z) = 0$$ defines each of the variables in terms of the other two, namely $$x = f(y, z)$$, $$y = g(x, z)$$, and $$z = h(x, y)$$, then show that:

$$\left(\frac{\partial x}{\partial y}\right) \left(\frac{\partial y}{\partial z}\right) \left(\frac{\partial z}{\partial x}\right) = -1$$.

One may obtain a vague feeling that e.g. “$$\frac{\partial x}{\partial y}$$” (more properly $$\frac{\partial f}{\partial y}$$) should be somewhat related to $$(\frac{\partial F}{\partial x})^{-1} \frac{\partial F}{\partial y}$$. Indeed, a version of the Implicit Function Theorem will provide that $$\frac{\partial f}{\partial y}(y,z) = (-1)\left(\frac{\partial F}{\partial x}(f(y,z),y,z)\right)^{-1} \frac{\partial F}{\partial y}(f(y,z),y,z)$$, and similarly for the other terms, thus immediately solving the problem. However, it does not give much insight into where the $$(-1)$$ term comes from.

Instead we look at the problem as two ways of defining a manifold, in this case a 2-manifold in 3-space. The explicit definitions, picking one w.l.o.g., $$\{(x,y,z): x = f(y,z), y = \textbf{id}(y), z = \textbf{id}(z)\}$$ we shall call its generator form (or perhaps, “tangent form”). The implicit definition $$\{(x,y,z): F(x,y,z)=c\}$$ we shall call its constraint form (or perhaps, “co-tangent” form).

Recall that a differential manifold has a tangent space at each point, which can be obtained from both definitions. Specialized to this problem, we linearize around $$(x,y,z)$$ by taking derivatives. Thus the generator form defines the tangent space directly by its bases:

$$\left[\begin{array}{c}dx\\dy\\dz\end{array}\right] = \left[\begin{array}{cc} \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \\ 1 & 0 \\ 0 & 1 \end{array} \right] \left[\begin{array}{c}dy\\dz\end{array}\right]$$

(analogously for $$g$$ and $$h$$), while the constraint form defines the tangent space by its normal:

$$\left[\begin{array}{ccc} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} & \frac{\partial F}{\partial z} \end{array} \right] \left[\begin{array}{c}dx\\dy\\dz\end{array}\right] = 0$$

From orthogonality, we obtain $$\frac{\partial F}{\partial x} \frac{\partial f}{\partial y} + \frac{\partial F}{\partial y} = 0$$ and $$\frac{\partial F}{\partial x}\frac{\partial f}{\partial z} + \frac{\partial F}{\partial z}= 0$$, and likewise for the other cases, so we see where the $$(-1)$$ term comes from.

What is more, the fact that we used the terms “generator” and “constraint” is evocative of linear codes, and tangent spaces are indeed linear subspace codes. Note that the generator form has the “generator matrix”

$$\left[\begin{array}{c} \nabla f \\ I_2\end{array}\right]$$

while the constraint form, if we divide through by $$\frac{\partial F}{\partial x}$$, has the “constraint matrix”

$$\left[\begin{array}{cc} I_1 & -\nabla f \end{array}\right]$$

which is exactly the relationship between systematic generator and parity-check matrices of such codes.