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Multiplicity in Noninjective Coordinate Maps

Learn to recognize repeated coverage in a coordinate map and adjust transformed integrals so each target point is counted exactly as intended.

Advanced 10 min read

What You'll Learn

  • Explain why the one-to-one condition in the standard change-of-variables formula prevents repeated counting.
  • Define the multiplicity of a coordinate map as the number of parameter points mapping to a target point.
  • Use absolute Jacobians and multiplicity together in a transformed integral.
  • Partition a parameter domain into injective pieces to apply the substitution formula branch by branch.
  • Identify when multiplicity varies across the image and why a single constant correction then fails.

Why One-to-One Maps Matter

A coordinate map can be locally well behaved and still cover its target more than once. The absolute Jacobian measures local volume scaling, but it does not record how many different parameter points arrive at the same target point. If repeated coverage is not accounted for, a transformed integral can count parts of the target multiple times.

The standard change-of-variables theorem applies directly when the map is one-to-one on the region being used, along with its usual smoothness and nonsingularity conditions. When a map is not one-to-one, the remedy is not to discard the absolute Jacobian. Instead, separate the distinct contributions: use injective pieces, or explicitly count how many times the map covers each target point.

Key idea: The absolute Jacobian corrects for local volume scaling and orientation. It does not correct for global repeated coverage. The number of preimages must also be considered.

Multiplicity of a Coordinate Map

Let \(T:D\to\mathbb{R}^3\) be a coordinate map from a parameter domain \(D\). For a target point \(\mathbf{x}\), a preimage is a parameter point \(\mathbf{q}\in D\) such that \(T(\mathbf{q})=\mathbf{x}\). The multiplicity counts how many distinct preimages there are.

Definition: The multiplicity of \(T\) at \(\mathbf{x}\) is \(N(T,D,\mathbf{x})=\#\{\mathbf{q}\in D:T(\mathbf{q})=\mathbf{x}\}\), where \(\#\) means the number of distinct points in the set. A point with no preimages has multiplicity zero. A map is one-to-one on \(D\) exactly when every point in its image has multiplicity one.

In applying integral formulas, it is usually enough for the multiplicity to be finite almost everywhere; behavior on boundaries or other sets of volume zero does not change a triple integral. The relevant count is of distinct preimages, not a count weighted by the Jacobian.

Suppose \(T\) is continuously differentiable and, apart from boundary or singular sets of volume zero, its domain can be split into finitely many regions on each of which \(T\) is one-to-one with nonzero Jacobian determinant. Write \(J_T(\mathbf{q})=\det DT(\mathbf{q})\). Then for an integrable target function \(\phi\), the change-of-variables formula with multiplicity is

$$ \int_D \phi(T(\mathbf{q}))\,|J_T(\mathbf{q})|\,d\mathbf{q} = \int_{T(D)} \phi(\mathbf{x})\,N(T,D,\mathbf{x})\,d\mathbf{x}. $$

The formula says that a target point is counted once for each of its parameter preimages. When \(N=1\) almost everywhere on the image, it reduces to the usual one-to-one formula. If the goal is to integrate over the image just once, then the multiplicity on the right must not be left in the answer.

Worked Example: A Symmetric Squaring Map

Worked Example: A Symmetric Squaring Map

Consider \(T(u,v,w)=(u^2,v,w)\) on \(D=[-1,1]\times[0,1]\times[0,1]\), and transform the integral of \(1\). The derivative matrix and its determinant are

$$ DT(u,v,w)= \begin{pmatrix} 2u&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad |J_T(u,v,w)|=|2u|. $$

The image is the unit cube \([0,1]^3\). For \(0<x<1\), there are two preimages, since \(u=\sqrt{x}\) and \(u=-\sqrt{x}\), while \(v\) and \(w\) are unchanged. Thus the multiplicity is two in the interior of the image. The boundary exceptions do not affect the integral.

The integral over the parameter domain, with the absolute Jacobian, is

$$ \begin{aligned} \int_0^1\int_0^1\int_{-1}^1 |2u|\,du\,dv\,dw &=\left(\int_{-1}^0 -2u\,du+\int_0^1 2u\,du\right)(1)(1)\\ &=\left(1+1\right)(1)(1)\\ &=2. \end{aligned} $$

The multiplicity formula gives the same result on the target:

$$ \int_{[0,1]^3} 1\cdot 2\,d\mathbf{x} =2\cdot 1=2. $$

But the volume of the image cube is \(1\), not \(2\). To integrate over the cube once, partition the parameter domain into \(u\geq 0\) and \(u\leq 0\). Each half maps onto the cube, and each half contributes \(1\). Keeping both halves therefore counts every interior target point twice.

Why the Multiplicity Formula Holds

The justification comes from applying the usual change-of-variables theorem separately to injective pieces. Suppose, except on sets that do not affect the integrals, the domain is partitioned into regions \(D_1,\ldots,D_k\), and \(T\) is one-to-one with nonzero Jacobian on each region. The image of each region is \(T(D_j)\). On that branch, the ordinary substitution theorem gives

$$ \int_{D_j}\phi(T(\mathbf{q}))\,|J_T(\mathbf{q})|\,d\mathbf{q} = \int_{T(D_j)}\phi(\mathbf{x})\,d\mathbf{x}. $$

Add these identities over all branches. At a target point \(\mathbf{x}\), the sum of the image integrals includes one copy of \(\phi(\mathbf{x})\) for each branch whose image contains \(\mathbf{x}\). Because each branch is one-to-one, that branch contributes one preimage, not several. The number of contributing branches is therefore exactly \(N(T,D,\mathbf{x})\), apart from any boundary exceptions. Consequently,

$$ \begin{aligned} \int_D\phi(T(\mathbf{q}))\,|J_T(\mathbf{q})|\,d\mathbf{q} &=\sum_{j=1}^k\int_{D_j}\phi(T(\mathbf{q}))\,|J_T(\mathbf{q})|\,d\mathbf{q}\\ &=\sum_{j=1}^k\int_{T(D_j)}\phi(\mathbf{x})\,d\mathbf{x}\\ &=\int_{T(D)}\phi(\mathbf{x})\,N(T,D,\mathbf{x})\,d\mathbf{x}. \end{aligned} $$

This argument also explains the role of the assumptions. A branch must be suitable for the ordinary change-of-variables theorem, and the exceptional boundaries or singular sets must not contribute volume to the integral. The partition makes repeated coverage visible rather than treating the entire map as if it were one-to-one.

Worked Example: A Polar Map Covering a Cylinder Twice

Worked Example: A Polar Map Covering a Cylinder Twice

Use the cylindrical-coordinate map \(T(r,\theta,z)=(r\cos\theta,r\sin\theta,z)\) on \(D=[0,1]\times[0,4\pi]\times[0,1]\). Its Jacobian determinant is \(r\), so its absolute Jacobian is \(r\) on this domain. The image is the cylinder of radius \(1\) and height \(1\).

For a point inside the cylinder and away from its axis, the angle occurs once in each full turn. Since the parameter interval covers two full turns, the multiplicity is two almost everywhere in the cylinder. Direct integration over the parameter domain gives

$$ \begin{aligned} \int_0^1\int_0^{4\pi}\int_0^1 r\,dr\,d\theta\,dz &=\int_0^1\int_0^{4\pi}\frac{1}{2}\,d\theta\,dz\\ &=\int_0^1 2\pi\,dz\\ &=2\pi. \end{aligned} $$

The target-side calculation confirms this value. The cylinder has volume \(\pi\), and the multiplicity is two almost everywhere, so its multiplicity-weighted volume is \(2\pi\). If the goal is the cylinder's volume rather than the repeated parameter integral, split the angle interval into \([0,2\pi]\) and \([2\pi,4\pi]\); either turn maps onto the cylinder and contributes \(\pi\).

At \(r=0\), all angles map to the axis, and the angular endpoints can also describe the same points. These special sets have no effect on the volume calculation. The repeated coverage of the cylinder's interior, by contrast, occurs on a full three-dimensional region and must be accounted for.

When Multiplicity Varies Across the Image

Sometimes a map covers some parts of its image more often than others. In that case, the multiplicity cannot be handled by dividing the entire integral by one constant. The next example has two preimages over one part of the image and only one over another.

Worked Example: A Map with Variable Multiplicity

Take \(T(u,v,w)=(u^2,v,w)\), now on \(D=[-1,2]\times[0,1]\times[0,1]\). Its image is \([0,4]\times[0,1]\times[0,1]\), and the absolute Jacobian remains \(|2u|\). For \(0<x<1\), both \(u=-\sqrt{x}\) and \(u=\sqrt{x}\) lie in \([-1,2]\), so the multiplicity is two. For \(1<x<4\), only \(u=\sqrt{x}\) lies in the domain, so the multiplicity is one.

Compute the transformed parameter integral by splitting at \(u=0\):

$$ \begin{aligned} \int_0^1\int_0^1\int_{-1}^2 |2u|\,du\,dv\,dw &=\left(\int_{-1}^0 -2u\,du+\int_0^2 2u\,du\right)(1)(1)\\ &=1+4\\ &=5. \end{aligned} $$

On the target, the part with \(0<x<1\) has volume \(1\), and multiplicity two. The part with \(1<x<4\) has volume \(3\), and multiplicity one. Thus

$$ \begin{aligned} \int_{[0,4]\times[0,1]^2}N(T,D,\mathbf{x})\,d\mathbf{x} &=2(1)(1)(1)+1(3)(1)(1)\\ &=2+3\\ &=5. \end{aligned} $$

If the desired integral is the ordinary volume of the image, it is \(4\), not \(5\). Dividing the transformed integral by two would give \(2.5\), which is also wrong: the multiplicity is not two throughout the image. The correct options are to use the multiplicity function piece by piece or to choose a single branch that covers the image once.

A Common Mistake

A common error is to notice that a map has a nonzero absolute Jacobian almost everywhere and then apply the one-to-one formula on its entire domain anyway. For the symmetric squaring map above, this mistaken calculation is

$$ \text{volume of }[0,1]^3 \ \overset{\text{incorrect}}{=}\ \int_{[-1,1]\times[0,1]^2}|2u|\,du\,dv\,dw =2. $$

The left side is actually \(1\). The calculation on the right is valid as a parameter integral, but it equals the image integral with multiplicity two. The correction is either to restrict to one injective half, obtaining \(1\), or to state the multiplicity formula and divide by two in this constant-multiplicity example. A nonzero Jacobian does not imply global injectivity: it controls local behavior, while multiplicity records global repeated coverage.

Choosing Between Branches and Multiplicity

For a target integral that should count each point once, partitioning is often the clearest approach. On each piece, apply the standard substitution formula and add the results. This avoids a multiplicity factor in the final target integral because each selected branch covers only the portion assigned to it.

Alternatively, if the entire parameter domain is used, the multiplicity-weighted formula describes exactly what the transformed integral computes. When \(N\) is known and finite, one can also recover an unweighted target integral by including a reciprocal multiplicity on the parameter side:

$$ \int_{T(D)}\phi(\mathbf{x})\,d\mathbf{x} = \int_D \frac{\phi(T(\mathbf{q}))\,|J_T(\mathbf{q})|} {N(T,D,T(\mathbf{q}))}\,d\mathbf{q}, $$

under the same branch and integrability conditions, with exceptional sets omitted as appropriate. Indeed, each target point contributes once for each preimage in the parameter integral, and each of those contributions is divided by the same multiplicity. When multiplicity varies, this correction varies as well.

Takeaway: Before applying a coordinate map to an integral, distinguish local volume scaling from global coverage. The absolute Jacobian handles the former; injective branches or an explicit multiplicity count handle the latter.

Check Your Understanding

For each question, identify whether the issue is local volume scaling, repeated coverage, or both.

  1. What does the absolute Jacobian account for, and what information does it leave out?
  2. For \(T(u,v,w)=(u^2,v,w)\) on \([-1,1]\times[0,1]^2\), what is the multiplicity at an interior image point?
  3. Why can the parameter integral over the angle range from \(0\) to \(4\pi\) be twice the volume of its cylindrical image?
  4. In the variable-multiplicity example, why is dividing the full transformed integral by two incorrect?
  5. Describe how partitioning the parameter domain allows the ordinary change-of-variables theorem to be applied.

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