Generated group acts without fixed points

Question

A friend of mine gave me a problem and while trying to solve it I came up with a question/conjecture:

Let $M$ be a complete oriented Riemannian manifold and let $S\subset ISO_{+}(M)$ be a collection of orientation preserving isomorphisms from $M$ to $M,$ such that no $\phi \in S$ has fixed point. Does it follow that the group generated by $S$ acts without fixed points on $M$?

If this is not true in general, what are some „natural\easy“ assumptions for it to be true? Simply connectedness? Non-positive sectional curvature? Do restrictions on the dimension help?

Thanks in advance.

EDIT: As was pointed out by Jason de Vito the general form was clearly wrong. So I‘d like to restate my question now with the restriction that the sectional curvature is bounded above by a non positive constant, i.e. $K_M\leq \delta\leq 0.$

Answer 1

It's certainly not true in general. Consider, for example, the $S^1$ action on $S^3$ given by $z(w_1, w_2) = (zw_1 ,z^2 w_2)$. Here, I am thinking of $S^1$ as the unit complex numbers, and I am thinking of $S^3$ as $S^3 = \{(w_1, w_2)\in \mathbb{C}^2: |w_1|^2 + |w_2|^2 = 1\}.$

Take $S = \{i\}$. Then the element $i\in S^1$ doesn't fix any points: if $iw_1 = w_1$, then $w_1 = 0$. Likewise, if $-1 w_2 = w_2$, then $w_2 = 0$. But $(w_1,w_2) = (0,0)$ is not in $S^3$.

On the other hand, $i^2 = -1$ does fix points. In fact, $-1(0,w_2) = w_2$ for any $(0,w_2)\in S^3$.

Note that $S^3$ is simply connected. If you also want the group to be simply connected, instead, just replace all uses of the word "complex" above with "quaternionic". This gives an action of $S^3$ on $S^7$ for which $S= \{i\}$ doesn't fix anything, but $i^2 = -1$ does.

I am not aware of any nice conditions which guarantee what you want, other than trivial things like "The whole group acts fixed point free" or something like that. I don't think restricting the dimension will count because you can now consider actions of $S^3$ on $S^7 \times M$ where $M$ is your favorite manifold (and $S^3$ acts trivially on it). One will still have $i$ fixing nothing, but $-1$ fixing even more.