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I am trying to prove that if $\varphi: \mathbb{P}^{n} \to \mathbb{P}^{n}$ is an algebraic morphism of degree $d > 1$ (by which I mean $\varphi^{*}(\mathcal{O}(1)) = \mathcal{O}(d)$, so the topological degree is $d^{n}$), there exists a periodic cycle whose points all miss the critical hypersurface of $\varphi$. For such a periodic cycle, the action of $\varphi$ on the tangent space of every point would have nonzero eigenvalues. Now, by a result of Fakhruddin (see http://arxiv.org/abs/math/0212208), we know that the periodic points are Zariski-dense, but since what I want is a point for period $k$ that misses the critical hypersurface of the iterate $\varphi^{k}$, this isn't enough - I need the entire cycle of my period-$k$ point to miss the critical hypersurface.

I believe I have a proof of this fact, provided I can find a periodic point $x$ such that at least one of the eigenvalues of $\varphi_{*}T_{x}$ is nonzero, and $x$ does not lie on any periodic components of the critical hypersurface. Finding a periodic point not lying on any periodic critical components is easy by the above result of Fakhruddin; it's proving that there exists a point with a nonzero eigenvalue that is hard.

I have tried using the holomorphic index formula in several variables. If all fixed points of $\varphi$ have only zero eigenvalues, then we can apply https://en.wikipedia.org/wiki/Holomorphic_Lefschetz_fixed-point_formula and get a contradiction. But the fixed point with a nonzero eigenvalue may lie on a fixed critical component.

I have also tried a point count. Here is the argument I wanted to use in my paper, except that it is wrong. The Jacobian matrix of $\varphi$, $J_{\varphi}$ is $(n+1)$ by $(n+1)$ and has entries of degree $d-1$. To get only a nilpotent action on $T_{x}$, we need all of $J_{\varphi}(x)$'s eigenvalues but one to be zero. This means all terms in the characteristic polynomial of $J_{\varphi}$ below the $n$th power should be zero. We get $n$ equations, of degrees $(n+1)(d-1), \ldots, 2(d-1)$. Overall, $(d-1)^{n}(n+1)!$ points. If we replace $\varphi$ by $\varphi^{k}$ then we get $(d^{k}-1)^{n}(n+1)!$. The number of period-$k$ cycles grows roughly as $d^{nk}/k$, so we can't quite do a point count.

What I want to say is that if a point $x$ has only zero eigenvalues under $\varphi$, then its preimages under $\varphi^{k-1}$ have only zero eigenvalues under $\varphi^{k}$. This is analogous to how the preimages of a critical point of $\varphi$ are critical under iterates of $\varphi$. We can only have one periodic cycle in each orbit, so we can throw away preimages of already-found zero-eigenvalue cycles. So what we really need to do is take $(d^{k}-1)^{n}(n+1)!$ and subtract every preimage of a zero-eigenvalue point of period up to $k-1$, of which there are $d^{n}(d^{k-1}-1)^{n}(n+1)!$. We do the subtraction and, voila!, we get an expression that grows more slowly than $d^{nk}/k$. We can also use this to deal with periodic critical components - the number of periodic points on those components grows as $d^{(n-1)k}$.

Except, well, it's not true that if $x$ has only zero eigenvalues under $\varphi$, then its preimages under $\varphi^{k-1}$ only have zero eigenvalues under $\varphi^{k}$. The whole question of eigenvalues is weird at non-fixed points - it's not invariant under coordinate-change; what's invariant is the rank of $\varphi_{*}T_{x}$. If $x$ is a fixed point of $\varphi$ with only zero eigenvalues, then $\varphi^{n}_{*}T_{x}$ is the zero matrix, and then $\varphi^{n+k}_{*}T_{\varphi^{-k}(x)}$ is the zero matrix, but then the point count stops working.

Am I missing something? Is there some way to push the point count through anyway, or to use the index formula more precisely to avoid periodic critical hypersurfaces? Or is there something I'm missing?

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    $\begingroup$ I would try the following. 1) Work over a finite field and compare Hrushovski's estimate for the number of periodic points with some estimate for the number of points of an hypersurface. 2) Lift a good periodic point from a finite field to characteristic zero (as in Fakhruddin's argument). $\endgroup$
    – ACL
    Sep 10, 2015 at 5:16
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    $\begingroup$ As @ACL says, the proof of Zariski density actually gives a stronger statement. If $x$ is a periodic point (over a finite field) produced by Hrushovski's theorem, then $\phi(x)$ is a Galois conjugate of $x$. If $V$ is any subvariety defined over the field of definition of $\phi$ and $x \notin V$, it follows that $\phi^n(x) \notin V$. Applying this with $V$ the critical hypersurface and then lifting $x$ gives what you want. $\endgroup$
    – naf
    Sep 10, 2015 at 6:36
  • $\begingroup$ @ulrich Yes, I think this works. The point $x \in \overline{\mathbb{F}_{q}}$ produced by Fakhruddin (and Hrushovski) satisfies $\varphi(x) = x^{q^m}$ for some $m$ and then it follows that a) it is periodic and b) its orbit consists of Galois conjugates over $\mathbb{F}_{q}$, so they all miss the critical locus provided $\varphi$ is defined over $\mathbb{F}_{q}$. $\endgroup$
    – Alon Levy
    Sep 10, 2015 at 15:20
  • $\begingroup$ I had no doubts that it did! $\endgroup$
    – naf
    Sep 11, 2015 at 8:43

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