Existence of arithmetic degrees for generic orbits and dynamical Lang–Siegel problem

Let

where Γ f is the graph of 𝑓. Let U ⊂ X be the largest open subset such that

We call 𝑈 the largest open subset over which 𝑓 is finite.

Let U ⊂ X be as in Definition B.1. If V , W ⊂ X are open subsets such that V ∩ I f = ∅ and f | V : V → W is finite, then W ⊂ U .

Proof

Consider the following diagram:

Since f | V ∘ α and 𝛽 are proper, p 1 − 1 ⁢ ( V ) = p 2 − 1 ⁢ ( W ) . As V ∩ I f = ∅ , 𝛼 is isomorphism and hence 𝛽 is finite and p 2 − 1 ⁢ ( W ) ∩ p 1 − 1 ⁢ ( I f ) = ∅ . Thus W ⊂ U . ∎

We define a sequence of open subsets U n ⊂ X , n = 0 , 1 , 2 , … , inductively as follows. Let U ⊂ X be as in Definition B.1. We set U 0 = X . Let n ≥ 0 and suppose we have constructed U n so that f n is finite over U n . Then define

The following diagram summarizes the situation:

Let W 0 , … , W n ⊂ X be open subsets such that

are finite for i = 1 , … , n , i.e.

Then we have W 0 ⊂ U n .

Proof

We prove by induction on 𝑛. If n = 0 , then U 0 = X and the statement is trivial. Let n ≥ 0 and suppose the statement holds for 𝑛. Then we have the following diagram:

By Lemma B.2, 𝑓 is finite over W n , that is, W n ⊂ U . Also, since f n : f − n ⁢ ( U n ) → U n and W n → W 0 are finite, we have W n = f − n ⁢ ( W 0 ) . Thus

We have

We set

We equip the subspace topology from 𝑋. With this topology, U ∞ is a Zariski topological space (i.e. Noetherian topological space such that every irreducible closed subset has unique generic point). We sometimes write U ∞ = X f back .

A sequence κ n : f − n ⁢ ( U n ) → [ 1 , ∞ ) , n = 0 , 1 , 2 , … , of upper semicontinuous functions is called submultiplicative cocycle if

1. min ⁡ { κ n ⁢ ( x ) ∣ x ∈ f − n ⁢ ( U n ) } = 1 for all n ≥ 0 ;

2. for all m , n ≥ 0 and x ∈ f − m − n ⁢ ( U m + n ) , we have κ m + n ⁢ ( x ) ≤ κ n ⁢ ( x ) ⁢ κ m ⁢ ( f n ⁢ ( x ) ) . Here note that the right-hand side is well-defined since x ∈ f − m − n ⁢ ( U m + n ) ⊂ f − n ⁢ ( U n ) and f n ⁢ ( x ) ∈ f − m ⁢ ( U m + n ) ⊂ f − m ⁢ ( U m ) .

For x ∈ U ∞ , the limit

exists. Moreover, for any δ ∈ R > 1 , the set { x ∈ U ∞ ∣ κ − ⁢ ( x ) ≥ δ } is a proper closed subset of U ∞ . In particular, κ − : U ∞ → R is upper semicontinuous.

For all x ∈ U ∞ and l ≥ 1 , we have

Proof

Let M = max 0 ≤ s ≤ l ⁡ max z ∈ f − s ⁢ ( U s ) ⁡ κ s ⁢ ( z ) . For any m ≥ 0 , s ∈ { 0 , … , l } , take y ∈ f − m − s ⁢ ( x ) such that κ − m − s ⁢ ( x ) = κ m + s ⁢ ( y ) . Then

Thus

for all n ≥ 0 and s ∈ { 0 , … , l } . Thus we are done. ∎

Let y ∈ U ∞ . We say 𝑦 is 𝑓-periodic if y ∈ f − l ⁢ ( y ) for some l ∈ Z ≥ 1 .

Let y ∈ U ∞ be an 𝑓-periodic point. Then

1. y ∈ f − m ⁢ ( U m ) for all m ≥ 0 ;

2. f m ⁢ ( y ) is also 𝑓-periodic for all m ≥ 0 .

Proof

Suppose y ∈ f − l ⁢ ( y ) with l ≥ 1 . Then y ∈ f − n ⁢ l ⁢ ( y ) ⊂ f − n ⁢ l ⁢ ( U ∞ ) for all n ≥ 0 . Then, for any n ≥ 0 and s ∈ { 0 , … , l − 1 } , we have

This proves (1).

Next, let m ≥ 0 . Take n ≥ 0 such that n ⁢ l ≥ m . Since y ∈ f − n ⁢ l ⁢ ( y ) , we have

Thus

Moreover, since y ∈ f − n ⁢ l − m ⁢ ( U n ⁢ l + m ) , f n ⁢ l ⁢ ( f m ⁢ ( y ) ) = f m ⁢ ( y ) . This proves (2). ∎

Let y ∈ U ∞ be an 𝑓-periodic point. Then κ n ⁢ ( y ) is well-defined for all n ≥ 0 by Lemma B.11, and we have the following.

1. κ + ( y ) : = lim n → ∞ κ n ( y ) 1 / n exists.

2. For all m ≥ 0 , we have κ + ⁢ ( f m ⁢ ( y ) ) = κ + ⁢ ( y ) .

3. lim inf n → ∞ κ − n ⁢ ( y ) 1 / n ≥ κ + ⁢ ( y ) .

Proof

Let l ≥ 1 be such that y ∈ f − l ⁢ ( y ) . Then we have

Thus, by Fekete’s lemma, lim n → ∞ κ n ⁢ l ⁢ ( y ) 1 / ( n ⁢ l ) exists. For m ≥ 0 and s ∈ { 0 , … , l } , we have

where M = max 0 ≤ s ≤ l ⁡ max z ∈ f − s ⁢ ( U s ) ⁡ κ s ⁢ ( z ) . Thus M − 1 ⁢ κ ( n + 1 ) ⁢ l ⁢ ( y ) ≤ κ n ⁢ l + s ⁢ ( y ) ≤ M ⁢ κ n ⁢ l ⁢ ( y ) . Thus lim n → ∞ κ n ⁢ ( y ) 1 / n exists.

Similarly, we have

Thus we get κ + ⁢ ( f ⁢ ( y ) ) = κ + ⁢ ( y ) . Thus, inductively, we get κ + ⁢ ( f m ⁢ ( y ) ) = κ + ⁢ ( y ) for all m ≥ 0 .

Finally, since κ − n ⁢ l ⁢ ( y ) = max z ∈ f − n ⁢ l ⁢ ( y ) ⁡ κ n ⁢ l ⁢ ( z ) ≥ κ n ⁢ l ⁢ ( y ) , we have

By Lemma B.9, we are done. ∎

Let 𝑇 be a topological space. Let { O λ } λ ∈ Λ be a family of open subsets of 𝑇. Let Y ⊂ T be an irreducible closed subset with generic point η ∈ Y . Then

if and only if η ∈ ⋂ λ ∈ Λ O λ .

Proof

This is obvious. ∎

Let dim X = k , q ∈ { 0 , … , k − 1 } , and let Ω ⊂ U ∞ be a non-empty open subset such that f − 1 ⁢ ( Ω ) ⊂ Ω . Assume there exist n 0 ∈ Z ≥ 1 and δ ∈ R > 1 such that

Then there are n 1 ∈ Z ≥ 1 and a proper closed subset Σ ⊊ U ∞ such that

1. every generic point η ∈ Σ is 𝑓-periodic, and f m ⁢ ( η ) ∈ Σ for m ≥ 0 ;

2. either Σ = ∅ , or Σ ̄ is of pure dimension 𝑞;

3. lim inf n → ∞ κ − n ⁢ ( x ) 1 / n ≥ δ for all x ∈ Σ ;

4. dim { x ∈ Ω ∖ Σ ∣ κ − n 1 ⁢ ( x ) ≥ δ n 1 } ̄ ≤ q − 1 . (When q = 0 , this means the set in the left-hand side is empty.)

Proof

Let Σ ̃ be the union of the irreducible components 𝑍 of { x ∈ Ω ∣ κ − n 0 ⁢ ( x ) ≥ δ n 0 } such that dim Z ̄ = q .

We define closed subsets Σ , Σ ′ , Σ ′′ ⊂ U ∞ as follows. Let

Note that the generic point of U ∞ is 𝑓-periodic and κ + = 1 . Thus Σ ⊊ U ∞ . It is easy to see that Σ satisfies properties (1)–(2). Moreover, for any x ∈ Σ , take 𝑦 as in the definition of Σ so that x ∈ f i ⁢ ( y ) ̄ . Then, by the upper semicontinuity of κ − n ’s on U ∞ and Lemma B.12, we have

Thus (3) also holds. Furthermore, we have f − 1 ⁢ ( Ω ∖ Σ ) ⊂ Ω ∖ Σ . Indeed, let x ∈ f − 1 ⁢ ( Ω ∖ Σ ) . Since f − 1 ⁢ ( Ω ) ⊂ Ω , we only need to show that x ∉ Σ . Suppose x ∈ Σ . Take a generic point y ∈ Σ such that x ∈ { y } ̄ . Since x , y ∈ f − 1 ⁢ ( U 1 ) , we have

Since f ⁢ ( y ) ∈ Σ , we get f ⁢ ( x ) ∈ Σ and this is contradiction.

Next, let

By Lemma B.12, for any generic point z ∈ Σ ′ , we have κ + ⁢ ( z ) < δ . Thus there are 1 ≤ δ 1 < δ and m ≥ 1 such that κ m ⁢ n 0 ⁢ ( z ) ≤ δ 1 m ⁢ n 0 for all generic points z ∈ Σ ′ .

Finally, let

We set 𝑙 to be the number of generic points of Σ ′′ .

Now we set

Then take N ≥ 1 so that

Set n 1 = N ⁢ m ⁢ n 0 . We claim that (4) holds for this n 1 . To this end, it is enough to show that, for any η 0 ∈ Ω ∖ Σ with dim { η 0 } ̄ = : q ′ ≥ q , we have κ − n 1 ⁢ ( η 0 ) < δ n 1 .

For arbitrary η ′ ∈ f − N ⁢ m ⁢ n 0 ⁢ ( η 0 ) , we set

Since f − 1 ⁢ ( Ω ∖ Σ ) ⊂ Ω ∖ Σ , we have η − i ∈ Ω ∖ Σ for i = 0 , … , N ⁢ m . Also, since the morphisms appearing above are all finite, we have dim { η − i } ̄ = dim { η 0 } ̄ = q ′ ≥ q . Thus we have

Once we have proven this, we get

and we are done.