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\title[Periodic points]{85 years of Nielsen theory: Periodic Points}
\author[Staecker]{P. Christopher Staecker}
\institute[Fairfield U.]{Fairfield University, Fairfield CT}
\date[]{Nielsen Theory and Related Topics 2013}

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\frame{
Fixed point theory is about $f(x)=x$.
\fpause
We want to generalize the ideas to $f^n(x)=x$ for various $n$. \pause These are \emph{periodic points} with period $n$. \pause If this $n$ is minimal, we say $x$ has ``minimal period $n$''.
\fpause
A very simplistic approach:\pause\ A periodic point with period $n$ is a fixed point of $f^n$.
\fpause
So we can use $L(f^n)$ and $N(f^n)$ to count periodic points, and 
\[ N(f^n) \le MF(f^n). \]
\pause
This is all true, but not quite what we want.
}

\frame{
Why isn't this good enough?
\fpause
Consider $S^1 \subset \mathbb{C}$, and $f:S^1\to S^1$ by $f(x) = \bar x$, the complex conjugate.
\fpause
Then $f^2(x) = x$ for all $x$, so $f^2$ is the degree 1 map on $S^1$, so $L(f^2) = |1-1| = 0$ and $N(f^2) = |1-1|=0$.
\fpause
BUT: \pause $f:S^1 \to S^1$ is degree $-1$, so $N(f) = L(f) = 2$ so all maps homotopic to $f$ have at least 2 fixed points\pause, and thus at least 2 periodic points of period 2.
\fpause
So $N(f^2) = 0$ even though all maps homotopic to $f$ have at least 2 periodic points of period 2.
}

\frame{
What happened?
\fpause
The problem is that there's a difference between:
\[ MF(f^2) = \min \{ \#\Fix(g) \mid g \htp f^2 \} \]
and 
\[ \min \{ \#\Fix(g^2) \mid g \htp f \} \]
\fpause
In our example, $MF(f^2) = 0$ but the second quantity is $\ge 2$.
\fpause
In other words, when we look at $f^n(x)=x$, we should only be changing $f$ by homotopy\pause, not $f^n$.
}

\frame{
There are more subtleties: if we know $\ind(f,x)$, what does this tell us about $\ind(f^n,x)$?
\fpause
Not much! \pause In our example $\ind(f,x) = 1$ and $\ind(f^2,x)=0$.
\fpause
The sequence of fixed point indices $(\ind(f,x), \ind(f^2,x), \dots)$ can be fairly unpredictable.
\fpause
Even the sequence of Leftchetz numbers $(L(f), L(f^2), \dots)$ has a complicated structure.
\fpause
This was studied by Dold (1983)
}

\frame{
\begin{thm} (Dold) For each $n$, we have
\[\sum_{k\mid n} \mu(k) L(f^{n/k}) = 0 \mod n \]
where $\mu$ is the M\"obius function.
\end{thm}
\fpause
The above equations are called the \emph{Dold congruences}\pause, and they are also satisfied by the sequence of indices.
\fpause
In fact, Dold proved a converse:
\begin{thm} (Dold)
Let $(i_n)$ be a sequence which satisfies the Dold congruences. Then there is a selfmap of an ENR such that $(i_n)$ is the sequence of fixed point indices.
\end{thm}
\fpause
Lots more work on this followed.
}

\frame{
By the way, the asymptotic behavior of the sequences of Lefschetz and Nielsen numbers are also studied.
\fpause
We can use the sequences to define zeta functions \pause (Dugardein's talk)
\fpause
Jiang also defines the \emph{asymptotic Nielsen number} $N^\infty(f)$, the exponential growth rate of the sequence of Nielsen numbers.
\fpause
Jiang showed $\log N^\infty(f)$ is a lower bound for the topological entropy of $f$. 
}

\frame{
Summary: \pause The behavior of the sequences $(L(f^n)), (N(f^n)), (\ind(f^n,x))$ is complicated and interesting.
\fpause
We'll focus mainly on getting information about \emph{specific} iterations. \pause Not the whole sequences.
}

\frame{
This theory is not 85 years old- \pause\ the basics are by Jiang, in his 1983 book.
\fpause
Jiang's work was based on unpublished papers by Halpern from the 1970s.
\fpause
There are two basic invariants, which J\&M call the \emph{Nielsen-Jiang periodic numbers}:
\pause
\begin{itemize}
\item $NP_n(f)$ counts the number of periodic points with minimal period $n$.\pause
\item $N\Phi_n(f)$ counts the number of all periodic points with period $n$.
\end{itemize}
}

\frame{
We'll discuss the definitions of $NP_n(f)$ and $N\Phi_n(f)$, along with some basic properties and relations between them.
\fpause
For example, for any map we have:
\[ \#\{\text{ points with period $n$ } \}= \sum_{k \mid n} \#\{\text{ points with minimal period $k$}\} \]
\pause
So we could hope that 
\[ N\Phi_n(f) \overset?= \sum_{k\mid n} NP_k(f). \]
\pause
This is true for nice spaces, but not always.
}

\frame{
The definitions for $NP_n$ and $N\Phi_n$ are a bit subtle.
\fpause
As usual we'll use the Reidemeister classes and the fixed point index\pause, but what about minimality of periods?
\fpause
It turns out this can be approached algebraically using the Reidemeister classes.
}

\frame{
For any map there is an inclusion $\Fix(f^k) \subset \Fix(f^n)$ when $k$ divides $n$. 
\fpause
We want something like this for Reidemeister classes.
\fpause
If $x\in \Fix(f^k)$ has Reidemeister class $\alpha\in \Reid(f^k)$, what is the Reidemeister class of $x$ when we view $x\in \Fix(f^n)$?
\fpause
It can't be ``the same'', since $\Reid(f^k)$ and $\Reid(f^n)$ are different sets. 
\fpause
For example on the circle, if $f$ is degree $d$, then $\Reid(f^k) = \Z_{|1-d^k|}$ and $\Reid(f^n) = \Z_{|1-d^n|}$.
\fpause
So what we need is a map $\Reid(f^k) \to \Reid(f^n)$ which respects the periods correctly.
}

\frame{
Here is the map $\iota_{k,n}:\Reid(f^k) \to \Reid(f^n)$, called the \emph{boost from level $k$ to level $n$}. 
\fpause
For $\alpha \in \pi_1$ define:
\[ \iota_{k,n}([\alpha]^k) = [\alpha \, f_\#^k(\alpha) \, f_\#^{2k}(\alpha) \dots f_\#^{n-k}(\alpha)]^n. \]
\pause
The superscript in $[\alpha]^k$ just reminds us that this is the Reidemeister class of $\alpha$ in $\Reid(f^k)$.
}

\frame{
The boost $\iota_{k,n}$ respects periods of points:
\fpause
If $x \in \Fix(f^k)$ has Reidemeister class $[\alpha]^k \in \Reid(f^k)$, then $x \in \Fix(f^n)$ has Reidemeister class $\iota_{k,n}([\alpha]^k) \in \Reid(f^n)$. 
\fpause
The boost also composes nicely. 
\fpause
When $m\mid k \mid n$, we have
\[ \iota_{m,n} = \iota_{k,n} \circ \iota_{m,k} \]
}

\frame{
When $[\alpha]^n$ is in the image of some $\iota_{k,n}$ for $k\mid n$, we say that $[\alpha]^n$ is \emph{reducible}. \pause Otherwise it's \emph{irreducible}.
\fpause
This is the algebraic version of some point having nonminimal period.
\fpause
If we want a Nielsen number for minimal periods, it might be good to define $NP_n(f)$ as 
\[\# \text{ of essential irreducible classes of $\Reid(f^n)$ } \]
\pause
This is not quite good enough.
}

\frame{
The periodic points live in \emph{orbits}: 
\[ x\in \Fix(f^n) \text{ has } \{x, f(x), \dots, f^{n-1}(x)\}. \]
\pause
When $x$ has minimal period $n$, the points of the orbit are all distinct\pause, and they all have minimal period $n$.
\fpause
It turns out there are times when $x\in \Fix(f^n)$ has the same Reidemeister class as $f(x) \in \Fix(f^n)$.
\fpause
Actually it's possible that every point in the orbit of  $x$ has the same Reidemeister class as $x$.
}

\frame{
\[ x\in \Fix(f^n) \text{ has } \{x, f(x), \dots, f^{n-1}(x)\}. \]
Possible to have $n$ distinct periodic points with minimal period $n$, but only one Reidemeister class containing them all.
\fpause
In that case, $\Reid(f^n)$ has only 1 essential irreducible class\pause, but $n$ different \emph{points} with minimal period $n$.
\fpause
So the number of essential irreducible classes is not a lower bound for the number of points with minimal period $n$.
\fpause
We need to be careful about the orbits.
}

\frame{
Luckily, we can approach the orbits algebraically too.
\fpause
For a class $[\alpha]^n \in \Reid(f^n)$, the \emph{Reidemeister orbit} of $\alpha$ is:
\[ \{ [\alpha]^n, [f_\#(\alpha)]^n, \dots, [f^{n-1}_\#(\alpha)]^n \}. \]
\pause
An orbit is \emph{reducible} if it contains a reducible class, and \emph{essential} if it contains an essential class.
}

\frame{
The invariant we're looking for is:
\[ NP_n(f) = (\# \text{ essential irreducible orbits in $\Reid(f^n)$ }) \cdot n \]
\pause
Times $n$ because each orbit indicates $n$ points of minimal period $n$.
\fpause
Once you show all of this is well-defined, it's not hard to show:
\[ NP_n(f) \le \min \{ \# P_n(g) \mid g \htp f \} \]
where $P_n$ is the set of periodic points with minimal period $n$.
}

\frame{
Let's do an example on a circle.
\fpause
On circles (and tori), things are well behaved:
\begin{itemize}
\item When $f$ is degree $d \neq 1$, we have $\Reid(f^n) = \Z_{|1-d^n|}$ and all classes are essential. \pause
\item All Reidemeister orbits at level $n$ have $n$ distinct classes.
\pause 
\begin{align*}
NP_n(f) &= (\# \text{ essential irreducible orbits in $\Reid(f^n)$ }) \cdot n \\
&= \# \text{ of essential irreducible classes of $\Reid(f^n)$ } 
\end{align*}

\end{itemize}
}

\frame{
Let $f:S^1 \to S^1$ be degree $4$.
\fpause
For $NP_1(f)$: \pause All classes at level 1 are irreducible, so 
\begin{align*}
\onslide<3->{NP_1(f) &= \# \text{ of essential irreducible classes of $\Reid(f^1)$ }}  
\onslide<4-> {\\
&= \# \text{ of essential classes of $\Reid(f)$ }}
\onslide<5->{= N(f)} 
\onslide<6->{= |1-4| = 3.}
\end{align*}
}

\frame{
For $NP_2$, we have $\Reid(f^2) = \Z_{|1-4^2|} = \Z_{15}$. 
\fpause
All are essential, which are reducible?\pause\ What's the image of $\iota_{1,2}$?
\pause
\[ \iota_{1,2} ([\alpha]^1)= [\alpha]^2 + [f(\alpha)]^2 = [\alpha] + 4[\alpha] = 5[\alpha]. \]
\pause
So $\iota_{1,2} :\Z_3 \to \Z_{15}$ is multiplication by 5.
\fpause
So in $\Z_{15}$, $\{0,5,10\}$ are reducible, the other 12 are irreducible.
\fpause
So $NP_2(f) = 12$.
}

\frame{
$NP_3$ is similar. \pause $\Reid(f^3) = \Z_{|1-4^3|} = \Z_{63}$.
\pause
\[ \iota_{1,3} = 1 + 4 + 4^2 = 21 \]
\pause
There is no $\iota_{2,3}$.
\fpause
So in $\Z_{63}$ the multiples of 21 are reducible.
\fpause
So we have 3 reducible classes, so $NP_3(f) = 63-3=60$.
}

\frame{
$NP_4$ is a bit more interesting since we have $2$ nontrivial boosts.
\fpause
We have 
\begin{align*}
\onslide<2->{\iota_{1,4} &= 1 + 4 + 4^2 + 4^3 = 85}
\onslide<3->{\\ \iota_{2,4} &= 1 + 4^2 = 17}
\end{align*}
\pause
So in $\Reid(f^4) = \Z_{|1-4^4|} = \Z_{255}$, multiples of 85 and multiples of 17 are reducible.
\fpause
But $85 = 5 \cdot 17$, so really it's just the multiples of 17\pause\ and $255 = 15 \cdot 17$, so we have 15 reducible classes.
\fpause
So $NP_4(f) = 255 - 15 = 240$.
}

\frame{
This type of computation can be done pretty easily for tori\pause, extended to nilmanifolds and some solvmanifolds by Heath \& Keppelmann, late 1990s. 
\fpause
It turns out all these spaces have some very nice properties which make these computations possible. 
\fpause
We've repeatedly used the fact that the Reidemeister orbits at level $n$ contain $n$ distinct classes.
\fpause
It's also true for tori that $\iota_{k,n}$ is injective (when $L(f^n) \neq 0$). \pause Then we often don't need to compute the map $\iota_{k,n}$ exactly.
}

\frame{
What about $N\Phi_n(f)$?
\fpause
This is meant to be a lower bound for 
\[ \min \{ \#\Fix(g^n) \mid g \htp f \} \]
\pause
$N(f^n)$ is inadequate for this.
\fpause
Our earlier example: complex conjugate on $S^1$.
}

\frame{
$f^2$ is the degree 1 map on $S^1$, so $N(f^2) = |1-1|=0$.
\fpause
But $N(f) = 2$ so all maps homotopic to $f$ have at least 2 fixed points\pause, and thus at least 2 periodic points of period 2.
\fpause
So $N(f^2) = 0$ even though all maps homotopic to $f$ have at least 2 periodic points of period 2.
\fpause
The issue here is that the 2 points of period 2 are an inessential class in $\Reid(f^2)$, but they are preceded by an essential class in $\Reid(f)$. 
\fpause
So really we need to count those as being essential.
}

\frame{
For a definition of $N\Phi_n(f)$:
\fpause
Given a Reidemeister orbit at level $n$, we need to consider all possible reductions to see if it is preceded by an essential orbit at a lower level. 
\fpause
Each such preceding essential orbit contributes to $N\Phi_n(f)$. 
\fpause
Specifically, a preceding essential orbit at level $k$ should increase $N\Phi_n(f)$ by $k$.
}

\frame{
We'll need to look at the entire union
\[ \bigcup_{k\mid n} \OR(f^k) \]
\pause
A set of orbits $\mathcal G$ is called a \emph{preceding [$n$-]system} when every essential orbit of the union reduces to something in $\mathcal G$.
\fpause
Every orbit has a \emph{depth}: the lowest level to which it reduces.
\fpause
$\mathcal G$ is a \emph{minimal} preceding system if its depth sum is minimal.
}

\frame{
Then $N\Phi_n(f)$ is defined as:
\[ N\Phi_n(f) = \sum_{O \in \mathcal G} d(O), \]
where $d$ is the depth and $\mathcal G$ is any minimal preceding $n$-system.
\fpause
So any preceding orbit at level $k$ contributes $k$ to the sum, which is what we wanted.
\fpause
Pretty complicated!
}

\frame{
If you're lucky, you'll be able to compute $N\Phi_n(f)$ by other means. 
\fpause
For example, 
\[ N\Phi_n(f) \overset{?}{=} \sum_{k\mid n}NP_k(f) \]
\fpause
Let's talk about this. 
\fpause
We want to relate a preceding $n$-system to the total number of all essential irreducible orbits.
}

\frame{
Any preceding $n$-system automatically contains every essential irreducible orbit.
\fpause
An essential irreducible orbit at level $k$ has depth $k$ since it's irreducible. \pause So:
\[ N\Phi_n(f) \ge \sum_{k\mid n}  (\# \text{ essential irreducible orbits in $\Reid(f^k)$ }) \cdot k \]
\pause
So
\[ N\Phi_n(f) \ge \sum_{k\mid n} NP_k(f). \]
\fpause
So half of our equality is always true. \pause The other direction is not always true.
}

\frame{
Consider the antipodal map on $S^2$. 
\fpause
$\Reid(f^k) = 1$ for every $k$, since $\pi_1$ is trivial.
\fpause
So all classes at all levels reduce to level 1.
\fpause
But the level 1 class is inessential because $f$ is fixed point free.
}

\frame{
All classes reduce except the bottom inessential one. \pause (The even level classes are essential)
\fpause
So there is never any essential irreducible class, so $NP_k(f) = 0$ for all $k$.
\fpause
But $f^2$ is the identity, so $N(f^2) = 1$\pause, so any preceding system will contain the class at level 1.
\fpause
Thus $N\Phi_n(f) = 1$ but $\sum_{k\mid n} NP_k(f) = 0$.
}

\frame{
The issue here is that we had essential classes at level 2 reducing down to inessential classes at level 1.
\fpause
Such a class will be counted in $N\Phi_n(f)$, but not in $NP_n(f)$. 
\fpause
To get the summation formula, we require that this never happens.
}

\frame{
$f$ is \emph{essentially reducible} if the reduction of an essential class is essential.
\fpause
\begin{thm} (Heath \& You, 1992)
If $f$ is essentially reducible, then
\[ N\Phi_n(f) = \sum_{k\mid n} NP_k(f). \]
\end{thm}
\pause
This makes $N\Phi_n$ much easier to compute.
\fpause
The essential reducibility condition holds for all maps on tori and all nil and solvmanifolds.
}

\frame{
So for nice spaces, $N\Phi_n(f)$ can be computed in terms of $NP_k(f)$. 
\fpause
But it seems reasonable that $NP_n(f)$ could be computed in terms of $N\Phi_n(f)$ by inclusion-exclusion.
\fpause
For example, 
\[ \#P_6(f) = \#\Fix(f^6) - \#\Fix(f^3) - \#\Fix(f^2) + \#\Fix(f^1) \]
\fpause
We ``include'' or ``exclude'' based on how exactly the levels divide one another.
}

\frame{
This inclusion-exclusion idea actually works if we assume essential reducibility. 
\fpause
\begin{thm}
If $f$ is essentially reducible, then
\[ NP_n(f) = \sum_{\tau \subset p(n)} (-1)^{\#\tau} N\Phi_{n:\tau} (f). \]
\end{thm}
\pause
This is obtained directly by M\"obius inversion of the previous theorem.
}

\frame{
There are several other identities based on stronger assumptions of reducibility and other things. 
\fpause
Sometimes you can even express $N\Phi_n(f)$ in terms of various $N(f^k)$.
\fpause
In particular this is true for tori. 
}

\frame{
Let's talk about Wecken theorems.
\fpause
Is it really true that $N\Phi_n(f) = \min \{ \#\Fix(g^n) \mid g \htp f \}$?
\fpause
And $NP_n(f) = \min \{ \#P_n(g) \mid g\htp f\}$?
\fpause
Probably we'll need to assume manifolds of dimension $\neq 2$.
}

\frame{
For $N\Phi_n(f)$, the theorem we need is:
\begin{thm}
If $X$ is a manifold of dimension $\neq 2$ and $f$ is a selfmap, then there is some map $g \htp f$ with 
\[ N\Phi_n(f) = \#\Fix(g^n). \]
\end{thm}
\pause
This was stated by Halpern in 1980 assuming that $\dim X \ge 5$. \pause Called the ``Halpern conjecture.''
\fpause
Proved in mid 2000s by Jezierski, for PL-manifolds\pause, first for $\dim X \ge 4$, then 3.
\fpause
Realizing by a smooth map is different. (Jezierski's talk today)
}

\frame{
What about $NP_n(f)$? We want to say:
\fpause
If $X$ is a manifold of dimension $\neq 2$ and $f$ is a selfmap, then there is some map $g \htp f$ with 
\[ NP_n(f) = \#P_n(g). \]
\fpause
This is also true. \pause (I think)
}

\frame{
A stronger Wecken property would be the following:
\fpause
If $X$ is a manifold of dimension $\neq 2$ and $f$ is a selfmap, then there is some map $g \htp f$ with 
\[ N\Phi_n(f) = \#Fix(g^n) \text{ for all $n$ } \]
\fpause
This would be a \emph{``simultaneous Wecken theorem''}. \pause (Similarly for $NP_n$)
\fpause
For $N\Phi_n$ this is an open question.
}

\frame{
For $NP_n$ the simultaneous Wecken theorem does not hold.
\fpause
Recall our example: the antipodal map on $S^2$.
\fpause
Here $NP_n(f)=0$ for all $n$, since the class at level 1 is inessential, and all classes at other levels reduce to it.
\fpause
The simultaneous Wecken theorem would mean that all periodic points could be removed simultaneously.
}

\frame{
In this example $N(f) = 0$ but $N(f^2) = 1$ so all maps homotopic to $f$ have a periodic point of period 2.
\fpause
This point could be of minimal period 2, or it could be a fixed point.
\fpause
In any case, we cannot remove \emph{every} periodic point of $f$ simultaneously.
}

\frame{
That's all for now!
\fpause
Next time, coincidences.
}


\end{document}