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This is probably a well-known issue, but I could not find a clear discussion in the literature, and I think others could find it useful.

Consider a real-analytic function germ $f:(\mathbb R^2,0) \rightarrow \mathbb R$, it is represented by a convergent power series $\sum_{i,j}a_{ij}x^iy^j \in \mathbb R\{x,y\}$. Suppose $f$ has a singularity at $0$, i.e. $df(0)=0$. Define its Jacobian ideal as the ideal generated by its partial derivatives: $I_{df} = \mathbb R\{x,y\}\langle f_x,f_y\rangle$, and its real local algebra as: $$Q_f := \frac{\mathbb R\{x,y\}}{I_{df}}.$$

Let $\mu_{\mathbb R} := \dim_{\mathbb R} Q_f$. The singularity is called algebraically isolated if $\mu_{\mathbb R}< \infty$. It is called isolated if, in some small neighborhood of $0$, $df(x,y)=0$ only at $(x,y)=0$ (i.e., the usual meaning).

Now take the complexification of $f$ by simply considering the series as a complex one: $\sum_{i,j}a_{ij}z^iw^j$. It is convergent in some polydisc and defines an holomorphic germ $f^{\mathbb C}:(\mathbb C^2,0) \rightarrow \mathbb C$. Take its (complex) Jacobian ideal, that is: $I_{df^{\mathbb C}} = \mathbb C\{x,y\}\langle f^{\mathbb C}_z,f^{\mathbb C}_w\rangle$, and the complex local algebra $$Q^{\mathbb C}_f := \frac{\mathbb C\{x,y\}}{I_{df^{\mathbb C}}}.$$

The Milnor number of the complexified singularity is defined by: $\mu_{\mathbb C} := \dim_{\mathbb C} Q_f^{\mathbb C}.$ A complex singularity is called isolated if $\mu_{\mathbb C} < \infty$. (In the holomorphic case it can be proven that a singularity is isolated if and only if, in some small neighborhood of $0$, $df(z,w)=0$ only at $(z,w)=0$, so we don't need to specify ''algebraically'' here.)

Questions:

  1. What is the relationship between the real-analytic singularity and its complexification? In particular: between the two local algebras, and between $\mu_{\mathbb R}$ and $\mu_{\mathbb C}$? Is there a class of functions where things work better?
  2. Is true (or false) that a real-analytic singularity is algebraically isolated if and only if its complexification is isolated?
  3. In the real setting: is ''algebraically isolated'' stronger then simply ''isolated''?
  4. Any reference!

Thank you.

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    $\begingroup$ It seems to me that $Q_f^{\mathbb{C}} \cong Q_f \otimes \mathbb{C}$, so $\mu_{\mathbb{R}} = \mu_{\mathbb{C}}$. Am I missing something? $\endgroup$ – David E Speyer Feb 26 '16 at 13:51
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As David Speyer has already suggested, we have $Q_f^\mathbb{C}\simeq Q_f\otimes\mathbb{C}$, so that answers the first question (and the second question).

For the third question, consider $f = (x^2+y^2)^2$. The singularity at $(x,y)=0$ is clearly isolated, but it is not algebraically isolated, since $\mu_\mathbb{R} = \infty$. Of course, over the complex numbers, $(z^2+w^2)^2$ has a non-isolated singularity at $(z,w)= (0,0)$; the two lines of singularities $z = \pm i\,w$ intersect at $(0,0)$.

The various confusions will be cleared up by noting that the ring of germs of analytic functions in two variables is a UFD, and, the singularity will be algebraically isolated as long as $f$ does not have any repeated irreducible factors. 'Isolated' is weaker over the reals simply because all the factors of $f$ might have no real zeros other than at the origin.

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  • $\begingroup$ Thanks to both of you. I missed a clear understanding of $Q_f \otimes \mathbb C\simeq Q_f^{\mathbb C}$. Indeed one should verify a few, trivial here but necessary, things: the map $\mathbb R\{x,y\}\otimes \mathbb C \rightarrow \mathbb C\{z,w\}$, $f\otimes \alpha \mapsto \alpha f^C$, 1) respects operations and convergence, 2) has a well defined inverse (since the ''formal real and imaginary part'' of a convergent complex series, put in $\mathbb R\{x,y\}$, converge), 3) commutes with the operation of "taking $i$-th derivative", so maps $I_{df} \otimes\mathbb C$ onto $I_{df^{\mathbb C}}$. $\endgroup$ – peter Feb 26 '16 at 18:04

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