# How to compute the index of a vectorfield defined by analytic formula?

An analytic local map (or map germ) $$f: (\mathbb{R}^n, 0) \to (\mathbb{R}^n, 0)$$ can be considered as a vector field with zero at the origin. Assume that the origin is an isolated zero of $$f$$. How can we compute the index of the vector field $$f$$ at the origin?

The index is defined as follows. Let $$S^{n-1}_{\epsilon} \subset \mathbb{R}^n$$ be a sphere around the origin with a small enough radius $$\epsilon$$. That means there is no zero of $$f$$ inside $$S^{n-1}_{\epsilon}$$ except at the origin. Then the index of $$f$$ at the origin is the degree of $$f/|f| : S^{n-1}_{\epsilon} \to S^{n-1}$$.

Some examples and approaches:

• If the Jacobian of $$df_0$$ is nondegenerate, then the index is $$\pm 1$$, i.e. it is equal to $$\text{sign} (\det (df_0))$$. In this case the origin is a nondegenerate zero. ( http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Popa.pdf )

• The index is the multiplicity of the zero at the origin. That is, if we take a small stable perturbation $$\tilde{f}$$ of $$f$$, then $$\tilde{f}$$ has nondegenerate zeros, and the sum of the indices of $$\tilde{f}$$ is equal to the index of $$f$$ at the origin. (Do you know some reference for this statement?)

• Let $$f: \mathbb{C} \to \mathbb{C}$$, $$f(z)=z^n$$. Then the index at the origin is equal to $$n$$. Let $$f(z)= \bar{z}^n$$, then the index is $$-n$$.

• Let $$f, g: (\mathbb{C}^2, 0) \to (\mathbb{C}, 0)$$ be holomororhic germs. Assume that they are irreducible in the local ring $$\mathcal{O}_2$$. Then the origin is an isolated zero of $$(f,g): (\mathbb{C}^2, 0) \to (\mathbb{C}^2, 0)$$, and its index is equal to the intersection multiplicity of the local curves $$f^{-1}(0)$$ and $$g^{-1}(0)$$. That is, equal to the complex dimension of $$\mathcal{O}_2 /(f,g)$$, where $$(f,g)$$ denotes the generated ideal.

So the question is: can we calculate the index from the power series of $$f$$ in general in the real case, or for a holomorphic map (germ) $$f: \mathbb{C}^n \to \mathbb{C}^n$$?

In the holomorphic case, the last formula generalizes in the obious way: If $$f \in (\mathcal{O}_{\mathbb{C}^n,0})^{\oplus n}$$ has an isolated zero at $$0$$, then the index is given by $$\textrm{dim}_\mathbb{C} \mathcal{O}_{\mathbb{C}^n,0}/(f_1,\ldots,f_n),$$ see for example Part I.5 of "Arnolʹd, V. I., Guseĭn-Zade, S. M., Varchenko, A. N.: Singularities of differentiable maps. Vol. I". Further sources where this is discussed is for example Chapter 2 of "D'Angelo, J. P.: Several complex variables and the geometry of real hypersurfaces", and Chapter 5 of "Griffiths, P., Harris, J.: Principles of algebraic geometry".
In the real analytic case, or in the real smooth case with a finiteness assumption on $$f \in (C^\infty_{\mathbb{R}^n,0})^{\oplus n}$$, namely that $$\textrm{dim}_\mathbb{R} (C^\infty_{\mathbb{R}^n,0}/(f_1,\ldots,f_n)) < \infty$$, then the index is given by the somewhat more elaborate Eisenbud–Levine–Khimshiashvili signature formula, see for example the original article "Eisenbud, D.; Levine, H. I. An algebraic formula for the degree of a $$C^\infty$$ map germ. Ann. of Math. 106 (1977), no. 1, 19–44". A discussion of this result can be found in "Eisenbud, D. An algebraic approach to the topological degree of a smooth map. Bull. Amer. Math. Soc. 84 (1978), no. 5, 751-764".