when is the geodesic flow an isometry? let $g^t: SM \to SM$ be the geodesic flow of a Riemannian manifold $(M, g)$. I am reading Bess's book "Manifolds all of whose geodesics are closed", the authors said: The flow is an isometry if and only if the sectional curvatures are constant $=1$.
But I cannot imaging that the flow on the Euclidean space is not an isometry.
Did I miss something trivial?
 A: As mentioned in the comments, we have to work in the double tangent bundle $TSM$, i.e. the tangent bundle of the unit tangent bundle $SM$.
Let's review the setup in the general case and then look at what happens in the Euclidean case.
The double tangent bundle splits naturally into a horizontal and a vertical component via the Riemannian metric.
Elements in $TSM$ then take the form $X = (X_H,X_V)$.
(In case you have not seen this, the splitting works as follows:
Denote by $\nabla$ the Levi-Civita connection of the Riemannian metric.
If $c(t) = (x(t),u(t))$ is any path in $SM$ with $\dot{c}(t) = X$, then $X_H = \dot{x}(0)$ and $X_V = \nabla_tu(0)$.)
The natural Riemannian metric on $SM$ is the Sasaki metric $G_S$ defined by
$$G_S(X,Y) = g(X_H,Y_H) + g(X_V,Y_V).$$
If nothing else is mentioned, $SM$ is usually equipped with this metric.
So, saying that the geodesic flow is an isometry of $SM$ means that it preserves $G_S$.
In Euclidean space, we have $SM = \mathbb{R}^n \times S^{n-1}$ and the geodesic flow is simply $g^t(x,u) = (x+tu,u)$.
Its differential acts on a vector $X \in TSM$ by $$(dg^t)_{(x,u)}(X) = (X_H+tX_V,X_V).$$
From this, you see that the Sasaki metric is not preserved.
Indeed,
$$G_S((dg^t)_{(x,u)}(X),(dg^t)_{(x,u)}(Y)) = G_S(X,Y) + tg(X_V,Y_H) + tg(X_H,Y_V) + t^2g(X_V,Y_V).$$
In the general case, the differential of the geodesic flow is given in terms of Jacobi fields, and the Jacobi equation is the link to the curvature of the manifold.
