# Un-proving $1=-1$ in the development of the implicit mapping theorem. Edwards: Advanced Calculus of Several Variables.

The follow conundrum arose while attempting to translate to tensor notation the development of the implicit mapping theorem in C.H. Edwards, Jr.'s Advanced Calculus of Several Variables. I refer to pages 189 and 190.

As much as I would like to boast that I have proven the ever-elusive proposition that $1=-1$, I will assume that result is in error.

This is a preliminary observation: if $\mathbf{a}=\mathbf{a}[\mathbf{b}[\mathbf{a}]]$ then $\frac{d\mathbf{a}}{d\mathbf{a}}=\mathbf{I}=\frac{d\mathbf{a}}{d\mathbf{b}}\frac{d\mathbf{b}}{d\mathbf{a}}=\left\{ \frac{\partial a^{i}}{\partial b^{d}}\frac{\partial b^{d}}{\partial a^{j}}\right\}$. So the inverse matrix relationship is

$\left[\frac{d\mathbf{a}}{d\mathbf{b}}\right]^{-1}=\frac{d\mathbf{b}}{d\mathbf{a}}=\left\{ \frac{\partial a^{i}}{\partial b^{j}}\right\} ^{-1}=\left\{ \frac{\partial b^{j}}{\partial a^{i}}\right\}$.

The following manipulations in vector notation are fairly faithful to Edwards:

Assert $\mathbf{G}:\mathbb{R}^{m+n}\rightarrow\mathbb{R}^{n}$; $\mathbf{x}\in\mathbb{R}^{m}$; $\mathbf{y}:\mathbb{R}^{m}\rightarrow\mathbb{R}^{n}$ such that $\mathbf{G}[\mathbf{x},\mathbf{y}[\mathbf{x}]]=\vec{0}$. All mappings are assumed to be locally$\mathscr{C}^{1}$.

The differentiations are performed in some appropriate neighborhood of $\left\{ \mathbf{x}_{\alpha},\mathbf{y}_{\beta}\right\}$ where $\mathbf{G}[\mathbf{x}_{\alpha},\mathbf{y}_{\beta}]=\vec{0}$.

$\frac{d\mathbf{G}}{d\mathbf{x}}=\frac{\partial\mathbf{G}}{\partial\mathbf{x}}+\frac{\partial\mathbf{G}}{\partial\mathbf{y}}\frac{d\mathbf{y}}{d\mathbf{x}}=\left\{ 0\right\}$

$\frac{\partial\mathbf{G}}{\partial\mathbf{x}}=-\frac{\partial\mathbf{G}}{\partial\mathbf{y}}\frac{d\mathbf{y}}{d\mathbf{x}}$

$-\left[\frac{\partial\mathbf{G}}{\partial\mathbf{y}}\right]^{-1}\frac{\partial\mathbf{G}}{\partial\mathbf{x}}=\frac{d\mathbf{y}}{d\mathbf{x}}$ (assuming $\left|\frac{\partial\mathbf{G}}{\partial\mathbf{y}}\right|\ne0$).

I decided to write this out in the fromalism of tensor notation, employing Einstein summation convention on opposing raised and lowered indices.

Writing the previous expressions in component form results in

$\frac{dG^{i}}{dx^{j}}=\frac{\partial G^{i}}{\partial x^{j}}+\frac{\partial G^{i}}{\partial y^{b}}\frac{\partial y^{b}}{\partial x^{j}}=0$

$\frac{\partial G^{i}}{\partial x^{j}}=-\frac{\partial G^{i}}{\partial y^{b}}\frac{\partial y^{b}}{\partial x^{j}}$

$\left\{ \frac{\partial y^{j}}{\partial G^{i}}\right\} =\left[\frac{\partial\mathbf{G}}{\partial\mathbf{y}}\right]^{-1}$

$-\frac{\partial y^{i}}{\partial G^{b}}\frac{\partial G^{b}}{\partial x^{j}}=\frac{\partial y^{i}}{\partial G^{d}}\frac{\partial G^{d}}{\partial y^{b}}\frac{\partial y^{b}}{\partial x^{j}}$

$-\frac{\partial y^{i}}{\partial x^{j}}=\frac{\partial y^{i}}{\partial x^{j}}$,

Assuming $\left\{ \frac{\partial y^{i}}{\partial x^{j}}\right\} \ne\left\{ 0\right\}$, this implies $1=-1$!

The assumption that $\left|\frac{\partial\mathbf{G}}{\partial\mathbf{y}}\right|\ne0$ implies $\mathbf{y}=\mathbf{y}[\mathbf{G}]$ in some neighborhood of $\mathbf{G}[\mathbf{x}_{\alpha},\mathbf{y}_{\beta}]=\vec{0}$. The inverse relation appears to be $\mathbf{y}=\mathbf{y}[\mathbf{G}[\mathbf{x}_{\alpha},\mathbf{y}]]$. So I am inclined to write $\mathbf{G}_{\alpha}[\mathbf{y}]=\mathbf{G}[\mathbf{x}_{\alpha},\mathbf{y}]$, and $\mathbf{G}_{\beta}[\mathbf{x}]=\mathbf{G}[\mathbf{x},\mathbf{y}_{\beta}]$. Then $\frac{\partial\mathbf{G}}{\partial\mathbf{x}}=\frac{d\mathbf{G}_{\beta}}{d\mathbf{x}}$ and $\frac{\partial\mathbf{G}}{\partial\mathbf{y}}=\frac{d\mathbf{G}_{\alpha}}{d\mathbf{y}}$. Rewriting the previous expressions in this form renders

$\frac{d\mathbf{G_{\beta}}}{d\mathbf{x}}=-\frac{d\mathbf{G}_{\alpha}}{d\mathbf{y}}\frac{d\mathbf{y}}{d\mathbf{x}}$

$\left[\frac{\partial\mathbf{G}}{\partial\mathbf{y}}\right]^{-1}=\left[\frac{d\mathbf{G}\alpha}{d\mathbf{y}}\right]^{-1}=\left\{ \frac{\partial y^{j}}{\partial G_{\alpha}^{i}}\right\}$

$-\frac{\partial y^{i}}{\partial G_{\alpha}^{b}}\frac{\partial G_{\beta}^{b}}{\partial x^{j}}=\frac{\partial y^{i}}{\partial G_{\alpha}^{d}}\frac{\partial G_{\alpha}^{d}}{\partial y^{b}}\frac{\partial y^{b}}{\partial x^{j}}$.

$-\frac{\partial y^{i}}{\partial G_{\alpha}^{b}}\frac{\partial G_{\beta}^{b}}{\partial x^{j}}=\frac{\partial y^{i}}{\partial x^{j}}$.

Perhaps that remedies the situation, but it's not entirely clear to me why. By definition$\frac{\partial G_{\alpha}^{i}}{\partial y^{j}}=\frac{\partial G^{i}}{\partial y^{j}}$. So the offending assumption appears to be $\frac{\partial y^{j}}{\partial G_{\alpha}^{i}}=\frac{\partial y^{j}}{\partial G^{i}}$.

So my question is: have I correctly identified the error in my original development?

Assuming that to be the case, the follow-on question is: what does it mean?

I'm adding a graphic I created to illustrate the discussion of the implicit function theorem with $x\in\mathbb{R},y\in\mathbb{R},G:\mathbb{R}^2\rightarrow\mathbb{R}$.

The solid black curve is the solution set of $G[x,y[x]]=0$. The solid magenta curve is the set of points $\{x_*,y,G[x_*,y]\}$. The blue sheet is $\{x,y,G[x,y]\}$. The red arrow points in the $x$ direction, and the yellow sheet lies in the $x\times y$ plane.

• Judging from the accepted answer, you may find answers to questions such as this or this or this to be useful as further reading. – user14972 Jun 21 '17 at 9:47

The $y$ in $0=G(x,y(x))$ and $\tilde y=y(G(x_0,\tilde y))$ are different functions, which already is clear from the dimensions of the arguments, one has the dimension of $x$, the other the dimension of $G$ which is the dimension of $y$.

See also the Triple product rule $$\left({\frac {\partial x}{\partial y}}\right)_{z}\left({\frac {\partial y}{\partial z}}\right)_{x}\left({\frac {\partial z}{\partial x}}\right)_{y}=-1.$$ which is similar to your case.

Assuming that $x$, $y$ and $z=G(x,y)$ have the same dimensions, and that the relevant derivative matrices $\frac{∂G}{∂x}$ and $\frac{∂G}{∂y}$ are regular in $(x_0,y_0)$, setting $z_0=G(x_0,y_0)$. Then there exists locally an implicitly defined function $\psi(z)$ so that $z=G(x_0,ψ(z))$. For this we have $$I=\frac{∂G}{∂y}(x_0,y_0)\,\frac{∂ψ}{∂z}(z_0).$$ On the other hand one gets the solution $\phi(x)$ to $z_0=G(x,ϕ(x))$ with the derivative at the base point $$0=\frac{∂G}{∂x}(x_0,y_0)+\frac{∂G}{∂y}(x_0,y_0)\,\frac{∂ϕ}{∂x}(x_0)$$ Combining both to eliminate $\frac{∂G}{∂y}$ gives $$\frac{∂ϕ}{∂x}(x_0)=-\frac{∂ψ}{∂z}(z_0)\frac{∂G}{∂x}(x_0,y_0)$$ While one can combine the right side via the chain rule, the composite function is $x\mapsto ψ(G(x,y_0))$ which is not visibly related to $ϕ$ except by the fact that they have opposite derivatives in the base point $(x_0)$. To re-iterate \begin{align} &G(x,ϕ(x))&&=z_0\\ &G(x_0,ψ(G(x,y_0)))&&=G(x,y_0). \end{align}

• If I write $\frac{\partial y^{i}[\mathbf{G}]}{\partial G^{d}}\frac{\partial G^{d}[\mathbf{x},\mathbf{y}]}{\partial x^{j}}=-\frac{\partial y^{i}[\mathbf{G}]}{\partial G^{d}}\frac{\partial G^{d}[\mathbf{x},\mathbf{y}]}{\partial y^{b}}\frac{\partial y^{b}[\mathbf{x}]}{\partial x^{j}}$. $\frac{\partial y^{i}[\mathbf{G}]}{\partial G^{d}}\frac{\partial G^{d}[\mathbf{x},\mathbf{y}]}{\partial x^{j}}=-\frac{\partial y^{i}[\mathbf{x}]}{\partial x^{j}}$ So the expression $-\frac{\partial y^{i}[\mathbf{G}]}{\partial x^{j}}=\frac{\partial y^{i}[\mathbf{x}]}{\partial x^{j}}$ is, in fact true. – Steven Thomas Hatton Jun 5 '17 at 15:33
• @StevenHatton These are derivatives of quite different functions, see the added section. – Lutz Lehmann Jun 15 '17 at 8:17
• I will need to wait until the coffee kicks in before I can think through your example. I suspect you are saying the same thing I intended by writing the same function name with a different parameter. I guess this is what happens when I write mathematics as if I were writing computer code or physics. $\mathbf{y}[\mathbf{G}]$ is not the same function as $\mathbf{y}[\mathbf{x}]$. I blame the mathematician, Stroustrup. This experience is helping me appreciate what I used to consider a needless proliferation of symbols in math books. – Steven Thomas Hatton Jun 15 '17 at 14:22
• I do believe I intended the same thing as your much clearer exposition communicates. I posted an "answer" of my own. I'm hoping that, at least the graphic will prove insightful. – Steven Thomas Hatton Jun 16 '17 at 19:18

I am aware my notation leaves something to be desired. I will work on that. This is not as good of an answer as the one already provided. I just want to share my observations regarding the problem, when reduced to 3-dimensional context.

In the graph, the blue sheet is the surface $\{x,y,G[x,y]\}$. The solid blue curve depicts $G[x_{0},y]=z$. The solid red curve depicts $G[x,y_{0}]=z$. I'm hoping the rest will be self-evident.

$G:\mathbb{R}^{2}\rightarrow\mathbb{R}$

$y[x]:\mathbb{R}\rightarrow\mathbb{R}$

$G[x,y[x]]=0$

$G[x_{0},y_{0}]=0$

$\frac{\partial G}{\partial y}[x_{0},y_{0}]\ne0$ implies there is some function $y[G[x_{0},y]]=y$.

In terms of the graph, this means, given the height of a point on the blue line, the value of $y$ can be determined. The correspondence between the value of $G$ and $y$ is aproximately $\Delta y\approx\frac{dy[G]}{dG}[x_{0},y_{0}]\Delta G$. Now, if $y$ is held fixed, and $x$ is varied$\Delta G\approx\frac{\partial G}{\partial x}[x_{0},y_{0}]\Delta x$.

Substituting $\Delta G$ in the first relationship gives $\Delta y\approx\frac{dy[G]}{dG}[x_{0},y_{0}]\frac{\partial G}{\partial x}[x_{0},y_{0}]\Delta x$, or $\frac{\Delta y}{\Delta x}\approx\frac{dy[G]}{dG}\frac{\partial G}{\partial x}$. The line through $\{x_{0},y_{0}\}$ with slope $\frac{\Delta y}{\Delta x}=\frac{dy[G]}{dG}\frac{\partial G}{\partial x}$ is depicted in green.

What this means geometrically is that if the point of evaluation of $G[x,y]$ is moved in the positive $x$ direction by $\Delta x$, it must be moved in the negative $y$ direction by $\Delta y\approx-(\frac{dy[G]}{dG}[x_{0},y_{0}]\frac{\partial G}{\partial x}[x_{0},y_{0}]\Delta x)$ in order that $G[x,y]=0$.

Notice that this motion is represented in the smaller rectangle by the diagonal opposite that of the green line. The direction of that motion is parallel to the yellow line tangent to the solution curve of $G[x,y[x]]=0$.