Why a Nonlinear Inverse Needs a Branch
A coordinate substitution is useful only when the relationship between input and output coordinates is understood. For a nonlinear map, that relationship may not determine a unique input. The simplest example is \(x=u^2\): if \(x>0\), then both \(u=\sqrt{x}\) and \(u=-\sqrt{x}\) produce the same output.
This is not merely a question of choosing a convenient algebraic formula. A single inverse formula must correspond to a domain on which the original map is one-to-one. That domain restriction is called a branch selection. It turns a many-to-one map into a coordinate patch that has a well-defined inverse.
One-to-one and nonsingular are separate requirements. Being one-to-one ensures that each point in the image comes from only one point in the chosen domain. A nonzero Jacobian determinant ensures local invertibility: near each point, the map does not collapse three-dimensional volume in a way that prevents a differentiable local inverse. For a suitable restriction, both conditions hold throughout the patch.
Restricting a Square Map
For \(F(u,v,w)=(u^2,v,w)\), the coordinates \(v\) and \(w\) do not create ambiguity. The ambiguity comes entirely from \(u^2\). Restricting to \(u>0\) selects the positive square root; restricting to \(u<0\) selects the negative square root. These restrictions yield distinct patches with the same image.
Worked Example: Selecting the Positive Square-Root Branch
Consider \(F(u,v,w)=(u^2,v,w)\) on the domain \(U=\{(u,v,w):u>0\}\). First determine the image and inverse.
Because \(u>0\), we have \(x=u^2>0\). The other two coordinates are unchanged, so the image is \(V=\{(x,y,z):x>0\}\). Solving for the inputs gives
Thus the inverse on \(V\) is \(F^{-1}(x,y,z)=(\sqrt{x},y,z)\). To check it, substitute the inverse into \(F\):
In the other direction, if \(u>0\), then \(\sqrt{u^2}=u\), so \(F^{-1}(F(u,v,w))=(u,v,w)\). The Jacobian matrix and determinant are
The restriction therefore gives both a unique inverse and a nonzero Jacobian determinant throughout the domain.
Using \(u<0\) instead also gives a valid patch, with inverse \(u=-\sqrt{x}\). The determinant is then \(2u<0\), but it is still nonzero. The sign of the determinant does not select the inverse branch; the sign restriction on the domain does.
Branch Selection for a Shifted Map
Not every square appears as an output coordinate by itself. A variable may also enter another coordinate, so the inverse must be solved consistently across the system. The essential task remains the same: choose the sign of the square root to match the restricted domain, then recover the other inputs.
Worked Example: A Square Map with a Coupled Coordinate
Let \(F(u,v,w)=(u^2,v+u,w)\), restricted to \(U=\{(u,v,w):u<0\}\). The first output satisfies \(x=u^2>0\), and the domain restriction selects \(u=-\sqrt{x}\). From \(y=v+u\), solve for \(v\):
The image is the half-space \(V=\{(x,y,z):x>0\}\), since any point in \(V\) gives a unique input by these formulas. Check the inverse by substituting it into \(F\):
For the reverse composition, take \(u<0\). Then \(\sqrt{u^2}=|u|=-u\), so the inverse returns \(-\sqrt{u^2}=u\), while its second component returns \((v+u)+\sqrt{u^2}=v+u-u=v\). It also returns \(w\). Thus it recovers the original input.
The Jacobian matrix is
The determinant is nonzero everywhere on \(U\), as required. The coupling in the second coordinate changes the inverse formula for \(v\), but it does not introduce another branch.
More Than Two Branches
A single square creates two possible signs away from zero. If several independent variables are squared, each can contribute its own sign choice. For instance, the map \((u,v,w)\mapsto(u^2,v^2,w)\) has up to four preimages for an output with \(x>0\) and \(y>0\): the choices \(u=\pm\sqrt{x}\) and \(v=\pm\sqrt{y}\).
A natural strategy is to restrict \(u\) and \(v\) to chosen sign intervals, such as \(u>0\) and \(v<0\). This selects one of those four possibilities. The resulting image has \(x>0\) and \(y>0\), and the inverse is \(u=\sqrt{x}\), \(v=-\sqrt{y}\), \(w=z\). The coordinate axes require care: at \(u=0\) or \(v=0\), the corresponding square has derivative zero, so the Jacobian determinant vanishes. To obtain an open nonsingular patch, exclude those boundaries.
A Nonlinear Map with a Branch Cut
In some maps, the ambiguity is not described by independent sign choices for separate coordinates. A useful example is the complex-square map written in real coordinates. It also illustrates how a restriction of the input domain can exclude a curve from the image.
Worked Example: The Real Form of a Complex Square
Consider the three-dimensional map
Writing the first two output coordinates as \(x=u^2-v^2\) and \(y=2uv\), compute
Therefore \(\sqrt{x^2+y^2}=u^2+v^2\). Adding this equation to \(x=u^2-v^2\) gives \(x+\sqrt{x^2+y^2}=2u^2\). Since \(u>0\), the branch is
These formulas require \(x+\sqrt{x^2+y^2}>0\). This fails exactly when \(y=0\) and \(x\leq 0\). The image is therefore the set \(V=\{(x,y,z): (x,y)\notin\{(x,0):x\leq0\}\}\): the plane with the nonpositive \(x\)-axis removed, extended through all \(z\).
To verify the inverse, let \(r=\sqrt{x^2+y^2}\) and \(u=\sqrt{(x+r)/2}>0\), with \(v=y/(2u)\). Since \(v^2=y^2/(4u^2)=(r-x)/2\), it follows that \(u^2-v^2=x\). Also \(2uv=y\), and \(w=z\). Thus applying \(F\) to the proposed inverse gives \((x,y,z)\). Conversely, for an input with \(u>0\), the identities above recover \(u\) as the positive square root and then recover \(v=y/(2u)\) and \(w=z\).
The Jacobian matrix and determinant are
Since \(u>0\), this determinant is strictly positive. As a numerical check, take \((u,v,w)=(3/5,4/5,2)\). Its image is \((-7/25,24/25,2)\). Here \(r=1\), so the inverse gives \(u=\sqrt{((-7/25)+1)/2}=3/5\) and \(v=(24/25)/(6/5)=4/5\), while \(w=2\). The original input is recovered.
Why the Restriction Makes the Inverse Work
The key point can be proved directly for the complex-square example. Suppose two inputs in the chosen domain have the same output in their first two coordinates. In complex notation, let \(\zeta_1=u_1+iv_1\) and \(\zeta_2=u_2+iv_2\), where \(u_1,u_2>0\). Equality of outputs says \(\zeta_1^2=\zeta_2^2\). Factoring the difference gives
Thus either \(\zeta_1=\zeta_2\) or \(\zeta_1=-\zeta_2\). The second possibility would imply \(u_1=-u_2\), which is impossible because both real parts are positive. Hence \(\zeta_1=\zeta_2\), so \(u_1=u_2\) and \(v_1=v_2\). Equality of the third output coordinate also gives \(w_1=w_2\). This proves that the restriction is one-to-one.
The same reasoning explains why the unrestricted map is not one-to-one: \(\zeta\) and \(-\zeta\) have the same square. Restricting to \(u>0\) allows at most one of these two inputs. The Jacobian calculation separately confirms that the selected domain avoids the singular point \(u=v=0\).
Determine which input coordinates or transformations create multiple preimages, such as a square or a repeated angular value.
Restrict the inputs, for example to \(u>0\) or to a chosen sign combination, so only one inverse choice remains.
Use the restriction to identify the outputs that occur, then solve for the inputs using the selected branch.
Substitute the inverse into the map, check the reverse composition, and verify that the Jacobian determinant is nonzero throughout the chosen open domain.
Common Mistake: Mixing Branches
A frequent error is to write an inverse formula without specifying the domain on which it is valid. For \(x=u^2\), the equation \(u=\sqrt{x}\) is not an inverse on a domain that includes negative \(u\). For example, if \(u=-2\), then \(x=4\), but the proposed formula returns \(u=\sqrt{4}=2\), not \(-2\). The formula has selected the positive branch, so its domain must be restricted accordingly.
The correction is to match the inverse formula and domain: use \(u=\sqrt{x}\) when \(u>0\), or \(u=-\sqrt{x}\) when \(u<0\). In either case, the boundary \(u=0\) cannot be included in a nonsingular patch for this map, because \(dx/du=2u=0\) there. A branch choice is therefore not just a sign convention; it must also be compatible with one-to-one behavior and nonvanishing Jacobian.
Check Your Understanding
For each question, identify the branch restriction and check what it implies for the image or Jacobian.
- For \(F(u,v,w)=(u^2,v,w)\), what is the inverse on the patch \(u<0\), and what is the sign of its Jacobian determinant?
- For \(F(u,v,w)=(u^2,v^2,w)\), how many sign branches are available when both squared output coordinates are positive?
- For \(F(u,v,w)=(u^2,v+u,w)\), derive the inverse when the domain restriction is \(u>0\).
- For the map \(F(u,v,w)=(u^2-v^2,2uv,w)\), why must the nonpositive \(x\)-axis be omitted from the first two output coordinates on the patch \(u>0\)?
- Why does a domain restriction that makes a map one-to-one not, by itself, guarantee that it is a valid coordinate patch?