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I have been reading about Bott's iteration theorem on the index of iterates of closed geodesics (see, for example, here p.172 and here -- p.224). Here is a summary of the setup and my question.

Let $M$ be a complete, connected, smooth Riemannian manifold. Define \begin{equation} \Lambda(M)=\{\,c:[0,1]\to M \mid c \text{ is piecewise smooth and } c(0)=c(1)\,\}. \end{equation} The energy of a curve $c$ is given by \begin{equation} E(c)=\frac{1}{2}\int_0^1 \langle \dot{c}(t),\dot{c}(t)\rangle\, dt. \end{equation} The critical points of $E$ are the closed geodesics (and the constant curves).

For a closed geodesic $c$, let \begin{equation} V_\Lambda(c)=\{\,X \text{ along } c \mid X \text{ is piecewise smooth, } \langle X(t),\dot{c}(t)\rangle=0 \text{ for all } t,\ X(0)=X(1)\,\}. \end{equation} The second variation $H$ of $E$ (see formula (1.1) in this paper) has an index (the number of negative directions) and a nullity (the dimension of its kernel on $V_\Lambda(c)$).

Bott's theorem says that there exist nonnegative integer functions $I(z)$ and $N(z)$ on the unit circle $\{z\in\mathbb{C} : |z|=1\}$ such that \begin{equation} \operatorname{ind}(c^q)=\sum_{z^q=1} I(z) \quad \text{and} \quad \operatorname{null}(c^q)=\sum_{z^q=1} N(z), \end{equation} where $c^q$ is the $q$-th iterate of $c$. Also, $I$ is constant at points at which $N(z) = 0$. In particular, if $N(z)=0$ for all roots of unity $z$, then \begin{equation} \operatorname{ind}(c^q)=q\,\operatorname{ind}(c). \end{equation}

Now consider a geodesic $c$ as $z = 0$ on the ellipsoid \begin{equation} x^2+y^2+\frac{z^2}{100}=1. \end{equation} For this geodesic, the second variation is \begin{equation} H(f\nu)=\int_0^{2\pi}\Bigl(100\,(f'(u))^2-f(u)^2\Bigr)\,du, \end{equation} where $\nu$ is a unit normal along $c$ and $f$ is a piecewise smooth function with $f(0) = f(2\pi)$. A constant function $f$ shows that $\operatorname{ind}(c)=1$. Analyzing the second variation, one can prove that $\operatorname{null}(c)=0.$ Similarly, one can check that $\operatorname{null}(c^2)=0$. According to Bott's theorem one would expect \begin{equation} \operatorname{ind}(c^2)=2\,\operatorname{ind}(c)=2. \end{equation} But a direct look at the second variation does not seem to give index 2.

$\textbf{Question:}$ What is the source of this discrepancy? Is there a subtle point in applying Bott's theorem to this example?

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    $\begingroup$ Welcome to math overflow. Have you tried to compute $I(\omega)$ and $N(\omega)$ at roots of unity from the formula in the middle of your question? I have the impression that both are $0$ for roots of unity with sufficiently small exponent (except $I(0)=1$) and see no contradiction. $\endgroup$ Commented Mar 1 at 10:34
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    $\begingroup$ Hi @SebastianGoette, in our case $I(1)=\operatorname{ind}(c)=1$, $N(1)=\operatorname{null}(c)=0$, and $N(-1)=\operatorname{null}(c^2)=0$, so since $I(z)$ is constant where $N(z)=0$, we have $I(-1)=I(1)=1$, yielding $\operatorname{ind}(c^2)=I(1)+I(-1)=2$, although I don't see any nonconstant vector field with negative second variation for $c^2$. $\endgroup$ Commented Mar 1 at 18:46
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    $\begingroup$ It says in Bott's paper (the first one you mention) that $I$ jumps at any point $z\in S^1$ at most by the value of $N(z)$. So it seems, by "constant at $z$", Bott just means "constant in a neighbourhood of $z$". $\endgroup$ Commented Mar 2 at 19:21

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