Eisenbud–Levine–Khimshiashvili signature formula

Consider the n-dimensional space Rⁿ. Assume that Rⁿ has some fixed coordinate system, and write x for a point in Rⁿ, where x = (x₁, …, x_n).

Let X be a vector field on Rⁿ. For 1 ≤ k ≤ n there exist functions ƒ_k : Rⁿ → R such that one may express X as

${\displaystyle X=f_{1}({\mathbf {x} })\,{\frac {\partial }{\partial x_{1}}}+\cdots +f_{n}({\mathbf {x} })\,{\frac {\partial }{\partial x_{n}}}.}$

To say that X is an analytic vector field means that each of the functions ƒ_k : Rⁿ → R is an analytic function. One says that X is singular at a point p in Rⁿ (or that p is a singular point of X) if X(p) = 0, i.e. X vanishes at p. In terms of the functions ƒ_k : Rⁿ → R it means that ƒ_k(p) = 0 for all 1 ≤ k ≤ n. A singular point p of X is called isolated (or that p is an isolated singularity of X) if X(p) = 0 and there exists an open neighbourhood U ⊆ Rⁿ, containing p, such that X(q) ≠ 0 for all q in U, different from p. An isolated singularity of X is called algebraically isolated if, when considered over the complex domain, it remains isolated.[3][4]

Since the Poincaré–Hopf index at a point is a purely local invariant (cf. Poincaré–Hopf theorem), one may restrict one's study to that of germs. Assume that each of the ƒ_k from above are function germs, i.e. ƒ_k : (Rⁿ,0) → (R,0). In turn, one may call X a vector field germ.

Let A_n,0 denote the ring of analytic function germs (Rⁿ,0) → (R,0). Assume that X is a vector field germ of the form

${\displaystyle X=f_{1}({\mathbf {x} })\,{\frac {\partial }{\partial x_{1}}}+\cdots +f_{n}({\mathbf {x} })\,{\frac {\partial }{\partial x_{n}}}}$

with an algebraically isolated singularity at 0. Where, as mentioned above, each of the ƒ_k are function germs (Rⁿ,0) → (R,0). Denote by I_X the ideal generated by the ƒ_k, i.e. I_X = (ƒ₁, …, ƒ_n). Then one considers the local algebra, B_X, given by the quotient

${\displaystyle B_{X}:=A_{n,0}/I_{X}\,.}$

The Eisenbud–Levine–Khimshiashvili signature formula states that the index of the vector field X at 0 is given by the signature of a certain non-degenerate bilinear form (to be defined below) on the local algebra B_X.[2][4][5]

The dimension of ${\displaystyle B_{X}}$ is finite if and only if the complexification of X has an isolated singularity at 0 in Cⁿ; i.e. X has an algebraically isolated singularity at 0 in Rⁿ.[2] In this case, B_X will be a finite-dimensional, real algebra.

Consider the case n = 2 of a vector field on the plane. Consider the case where X is given by

${\displaystyle X:=(x^{3}-3xy^{2})\,{\frac {\partial }{\partial x}}+(3x^{2}y-y^{3})\,{\frac {\partial }{\partial y}}.}$

Clearly X has an algebraically isolated singularity at 0 since X = 0 if and only if x = y = 0. The ideal I_X is given by (x³ − 3xy², 3x²y − y³), and

${\displaystyle B_{X}=A_{2,0}/(x^{3}-3xy^{2},3x^{2}y-y^{3})\cong \mathbb {R} \langle 1,x,y,x^{2},xy,y^{2},xy^{2},y^{3},y^{4}\rangle .}$

The first step for finding the non-degenerate, bilinear form β is to calculate the multiplication table of B_X; reducing each entry modulo I_X. Whence

Direct calculation shows that J_F = 9(x⁴ + 2x²y² + y⁴), and so [J_F] = 24y⁴. Next one assigns values for ${\displaystyle \scriptstyle \ell }$. One may take

${\displaystyle \ell (1)=\ell (x)=\ell (y)=\ell (x^{2})=\ell (xy)=\ell (y^{2})=\ell (xy^{2})=\ell (y^{3})=0,\ {\text{and}}\ \ell (y^{4})=3.}$

This choice was made so that ${\displaystyle \scriptstyle \ell \left(\left[J_{F}\right]\right)\,>\,0}$ as was required by the hypothesis, and to make the calculations involve integers, as opposed to fractions. Applying this to the multiplication table gives the matrix representation of the bilinear form β with respect to the given basis:

$${\displaystyle \left[{\begin{array}{ccccccccc}0&0&0&0&0&0&0&0&3\\0&0&0&0&0&0&1&0&0\\0&0&0&0&0&0&0&3&0\\0&0&0&3&0&1&0&0&0\\0&0&0&0&1&0&0&0&0\\0&0&0&1&0&3&0&0&0\\0&1&0&0&0&0&0&0&0\\0&0&3&0&0&0&0&0&0\\3&0&0&0&0&0&0&0&0\\\end{array}}\right]}$$

The eigenvalues of this matrix are −3, −3, −1, 1, 1, 2, 3, 3 and 4 There are 3 negative eigenvalues (#N = 3), and six positive eigenvalues (#P = 6); meaning that the signature of β is #P − #N = 6 − 3 = +3. It follows that X has Poincaré–Hopf index +3 at the origin.

With this particular choice of X it is possible to verify the Poincaré–Hopf index is +3 by a direct application of the definition of Poincaré–Hopf index.[6] This is very rarely the case, and was the reason for the choice of example. If one takes polar coordinates on the plane, i.e. x = r cos(θ) and y = r sin(θ) then x³ − 3xy² = r³cos(3θ) and 3x²y − y³ = r³sin(3θ). Restrict X to a circle, centre 0, radius 0 < ε ≪ 1, denoted by C_0,ε; and consider the map G : C_0,ε → C_0,1 given by

${\displaystyle G\colon X\longmapsto {\frac {X}{||X||}}.}$

The Poincaré–Hopf index of X is, by definition, the topological degree of the map G.[6] Restricting X to the circle C_0,ε, for arbitrarily small ε, gives

${\displaystyle G(\theta )=(\cos(3\theta ),\sin(3\theta )),\,}$

meaning that as θ makes one rotation about the circle C_0,ε in an anti-clockwise direction; the image G(θ) makes three complete, anti-clockwise rotations about the unit circle C_0,1. Meaning that the topological degree of G is +3 and that the Poincaré–Hopf index of X at 0 is +3.[6]