Vector fields and the Poincaré–Hopf theorem
Measure the local index of an isolated zero and relate the global sum to Euler characteristic under the correct boundary condition.
On this page
The question for today
What you’ll understand
- Define the index of an isolated vector-field zero using a small surrounding sphere.
- Calculate nondegenerate indices from a tangent-space determinant.
- Apply Poincaré–Hopf with compactness and an outward boundary condition stated.
A vector field can vanish at a few points while its directions rotate around those zeros in different ways. The index measures that local turning through a degree. Poincaré–Hopf relates the sum of these local integers to the topology of a compact manifold. Boundary behavior is decisive: on a disk a nonvanishing constant field exists, but it does not point outward all around the boundary. The theorem's outward condition prevents that example from contradicting the disk's nonzero Euler characteristic.
Normalize the field around an isolated zero
Let a smooth vector field X have an isolated zero p on an n-manifold with n≥1. Choose coordinates around p and a sufficiently small sphere enclosing no other zero. Divide the coordinate field by its length on that sphere to obtain a map to the unit sphere. Its degree is the index. The field must be nonzero on the surrounding sphere so that normalization is defined. Shrinking the sphere without crossing another zero gives a homotopy and preserves the integer.
The index is independent of the coordinate chart, because a coordinate change affects the source sphere and vector components compatibly. At a nondegenerate zero the derivative of the coordinate field is invertible. The nonlinear field can then be compared locally with that linearization without introducing a zero on a sufficiently small sphere, giving index equal to the determinant sign. This derivative acts on tangent coordinates of the manifold; using an unrelated ambient Jacobian can give the wrong dimension and determinant.
The global index sum needs compactness and boundary control
For a compact smooth oriented manifold and a smooth vector field with isolated zeros, Poincaré–Hopf states that the sum of the indices is its Euler characteristic, provided the field is strictly outward pointing everywhere on any boundary. This outward condition includes being nonzero there, so all zeros occur in the interior. In the closed case there is no boundary condition to check. Compactness and isolated zeros then make the set of zeros finite, allowing an ordinary finite sum.
The Euler characteristic is an alternating sum of homology ranks, so it does not depend on the chosen vector field. Local changes can create or remove collections of zeros whose total index cancels, but cannot change the permitted global sum. A nowhere-zero field on a closed manifold with nonzero Euler characteristic is therefore impossible. The same conclusion for a manifold with boundary requires the outward hypothesis; without it, boundary direction contributes information absent from the stated formula.
A saddle and a radial field can have different local signs
In the plane, a radial field has identity derivative at its zero and index +1. A linear saddle with one expanding and one contracting axis has negative determinant and index −1. A planar rotation also has positive determinant and index +1, even though its arrows are tangent to surrounding circles rather than radial. Thus index is not a label for outward flow, attraction, or stability. It measures the degree of the direction map around the isolated zero.
This distinction matters when applying the theorem on a region. A rotating field on a disk has an isolated central zero but is tangent to the boundary, so the outward version is not justified merely because its index happens to match the Euler characteristic. Check tangency, outward component, and zero isolation separately. On a curved manifold, first verify that a proposed ambient formula is tangent, then locate zeros and linearize in the actual tangent space. A good drawing suggests these checks but cannot replace them.
Work it through
A tangent field with two zeros on the sphere
On the unit sphere S²⊂R³, define X(p)=e₃−z p, where z is the third coordinate. Show that X is tangent, find its zeros, and compute the index sum.
Because p has unit length, , so the field is tangent. The equation e₃=zp forces the north and south poles p=±e₃, and both are zeros.
For a tangent variation v at either pole, its third component is zero. Differentiating gives DX(v)=−zv on the tangent plane. At the north pole this is −I₂; at the south it is +I₂. Both determinants are +1.
Each zero is nondegenerate and contributes +1. The sum is 2, agreeing with χ(S²)=2. Compactness and the absence of boundary put this field in the closed Poincaré–Hopf setting.
The field has two zeros with total index 2. Tangent-plane linearization, rather than an ambient three-dimensional determinant, supplies the correct local signs.
The living notebook
Explore the model
Interactive lab · fields
Separate local index from an outward boundary condition
Pause and Reset; set Source strength q=1,Rotation Ω=0 and predict outward arrows on every centered circle. For a fresh comparison Reset and set q=0,Ω=1: predict tangent rotational arrows instead. Both nonzero fields away from the origin have a central direction index +1, but only the radial choice is strictly outward on a disk boundary. The softened vector-field instrument illustrates these models, not arbitrary fields on a curved manifold or a proof of the global theorem.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA local index and the boundary direction are different checks; matching one integer does not establish the theorem's hypotheses.
Pause & reason
Check your understanding
Your turn
Try it yourself
On the closed unit disk in R², take X(x,y)=(x,y). Verify the boundary condition, compute the isolated-zero index, and compare with the field Y=(1,0).
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The outward unit normal on the unit circle is (x,y). For X use the derivative at the origin; for Y inspect its normal component around the circle.
Compare with the solution
For X, the boundary dot product with the outward normal is x²+y²=1>0. Its only zero is the origin, where DX=I₂ and index is +1, matching χ(disk)=1. The constant field Y has no zeros, but its boundary normal component is x, positive on one side and negative on the other. It fails the outward hypothesis and is outside this theorem's stated boundary setting.
Keep the idea
Take it with you
- The index is a local degree around an isolated zero.
- Nondegenerate indices are determinant signs in tangent coordinates.
- The boundary version of Poincaré–Hopf requires a strictly outward field on the whole boundary.
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