TITLE: Isolated degenerate fixed points
QUESTION [6 upvotes]: Here is a rather naïve question: to apply the classical holomorphic Lefschetz theorem, you often work with biholomorphisms of a complex compact manifold whose fixed points are non degenerate (i.e. $1$ is not an eigenvalue of the differential) even if work of O'Brian http://www.jstor.org/stable/1998512 allows to get rid of the degeneracy assumption and compute explicitly the multiplicities at the fixed points, provided they remain isolated.
However, I have encountered no interesting examples where some fixed point are isolated and degenerate. For instance, this condition implies (using Bochner's linearization) that the automorphism is not of finite order.
Ideally I would like to construct such an exemple with the manifold $X$ satisfying the additional assumptions $h^0(X, \, T_X)=h^1(X, \, T_X)=0$. A finite product of $\mathbb{P}^2$ blown up at 4 generic points seems a reasonable candidate, but I'm not sure this is the simplest path to start.

REPLY [3 votes]: I think the surface constructed in Lemma 4 of https://arxiv.org/abs/1609.06391 should do what you want.  No doubt there is a simpler example, so hopefully someone else will chime in.
It's a rational surface $S$, constructed by blowing up $15$ (carefully chosen) points in $\mathbb P^2$.  There is a smooth rational curve $C \subset S$ and an automorphism $\phi : S \to S$ which fixes $C$ and restricts to $C$ as the automorphism $z \mapsto z+1$ in suitable coordinates.  In particular, there is a unique fixed point $p$ on $C$, given by $\infty$ in these coordinates.  This map acts trivially on the tangent space $T_p C \subset T_p S$, which gives the $1$-eigenvector you are looking for.
I think this should have all the properties you want.  One should check there is not some curve of fixed points through $p$, but I am pretty sure of it.  (Note that $C$ is not a curve of fixed points: although it's mapped to itself, the points on $C$ move around, and only $p$ is fixed.)  The automorphism is indeed of positive entropy, hence not finite order, as you note.