Fundamental groups of positively curved manifolds with symmetry

Abstract

We obtain a classification for the fundamental groups of closed $n$-manifolds of positive sectional curvature which admit an isometric $T^k$-action with $k \ge \frac{n}{6}+1 (n \ne 11, 15)$.

Appendix: Proof of Theorem A for \(n<23\)

For the sake of completeness and the convenience for readers, we will also present a simplified proof of the cases $n=17, 19, 21$.

It turns out that our proof of Theorem A for $n=17$ requires a strengthened version of (1.6.2) in [3] (see Case 3).

 

Theorem A1

[3] Let $M$ be a closed $n$-manifold of positive sectional curvature, and let $N_j\hookrightarrow M$ be a closed totally geodesic submanifold of dimension $\nu _{j}(j=1, 2)$. Then the following natural homomorphisms,

$$ \pi _i(N_1, N_1\cap N_2)\rightarrow \pi _i(M,N_2),\qquad \qquad \pi _i(N_2, N_1\cap N_2)\rightarrow \pi _i(M,N_1), $$

are isomorphisms for $i\le \nu _{1}+\nu _{2}-n$ and are surjections for $i=\nu _{1}+\nu _{2}-n+1$.

 

Proof of Theorem A for n = 17

By Corollary 1.9, we may assume that any isotropy subgroup has fixed point set of dimension $\le$13.

By Lemma 1.14, we can assume two independent subgroups with torus rank at most 1, $H_1, H_2\subset T^4$, such that $T^4/H_i$ acts effectively on $N_i$ up to a finite subgroup, where $N_i$ denotes an $H_i$-fixed point component with $\dim (N_i)\ge 9$. By (1.6.1), $\pi _1(N_i)\cong \pi _1(M)$ and thus the rest of the proof is to show the desired property for $\pi _1(N_i)$.

If $\dim (N_1)=9$ (or $\dim (N_2)=9)$, then $\pi _1(N_1)$ is cyclic (Theorem 1.5). Hence, in the rest of the proof, we may assume that $\dim (N_i)\ge 11$.

• Case 1. Assume that $H_1$ is finite (or $H_2$ is finite). If $\dim (N_1)\le 13$ (or $\dim (N_2)\le 13)$, then $(N_1,T^4/H)$ satisfies Theorem 1.5 and $\pi _1(N_1)$ is cyclic. In the rest of the proof, we will assume that $H_i$ is a circle subgroup of $T^4$. Let $N_0=N_1\cap N_2$.

• Case 2. Assume that $\dim (N_1)=\dim (N_2)=11$. Then $\dim (N_0)=5$ or $7$. In the former case, $\pi _1(N_0)$ is cyclic (Theorem 1.5), and thus $\pi _1(N_1)$ is cyclic (1.6.1). In the latter case, we consider the isotropy representation of $T^2$ on the normal space of $N_0$ in $M$. It is easy to see that there is a circle subgroup of $T^2$ with fixed point component $F$ of dimension $9$ which contains $N_0$. Then $\pi _1(F)$ is cyclic (Corollary 1.9), and $\pi _1(F)\cong \pi _1(M)$ (1.6.1).

• Case 3. Assume that $\dim (N_1)=11$ and $\dim (N_2)=13$. Then $\dim (N_0)=7$ or $9$. In the latter case, $\pi _1(N_1)$ is cyclic (Corollary 1.9). In the former case, by Theorem A1 we may conclude that $N_1\cap N_2\hookrightarrow N_2$ is $7$-connected because $\pi _i(N_2,N_1\cap N_2)\cong \pi _i(M,N_1)=0$ for $i\le 7$ (see (1.6.1)). Moreover, by Lemma 1.10 one concludes that the universal cover of $N_2$ is a homology sphere, so the universal cover of $M$ is a homology sphere (1.6.1). Then by Lemma 1.8, $\pi _1(N_2)$ is cyclic.

• Case 4. Assume that $\dim (N_1)=\dim (N_2)=13$. Then $\dim (N_0)=9$ or $11$. We may only consider the former case. Because $N_0\hookrightarrow N_1$ is $9$-connected, $\tilde{N}_1$ is a homology sphere (Lemma 1.10). By Lemma 1.8, $\pi _1(N_1)$ is cyclic.

$\square$

 

Proof of Theorem A for n = 19

Without loss of generality, we may assume any isotropy subgroup has fixed point set of dimension $\le 15$ (Corollary 1.9).

By Lemma 1.14, $T^5$ has two independent subgroups with torus rank at most 1, $H_1$ and $H_2$, whose fixed point sets $N_1$ and $N_2$ have dimensions $\ge 11$ on which $T^5/H_i$-actions are effective up to a finite group. Hence, the proof reduces to show the desired property for $\pi _1(N_i)$ (1.6.1). Note that by Theorem 1.5 we may assume that $\dim (N_1)=\dim (N_2)=15$. If $H_i$ is finite, then $5>\frac{\dim (N_1)+1}{4}$, and thus $\pi _1(N_1)$ is cyclic (Theorem 1.5). So we may assume that $H_i$ are circles.

If the $T^5$-fixed point set is not empty, then by the last part of (1.12.1), $\pi _1(M)$ is cyclic. So we can assume that the $T^5$-action has no fixed point. Let $N_0=N_1\cap N_2$. Note that $N_0\hookrightarrow N_1$ is $11$-connected (1.6.2), and by Lemma 1.13 (applied to $(N_1,N_0))$, there is $T^4\subset T^5$ with a connected fixed point set $F$ of dimension $3$. We then complete the proof by applying Lemma 3.2. $\square$

 

Proof of Theorem A for n = 21

Let $H_1$ and $H_2$ be subgroups of $T^5$ of rank at most 1 with fixed point sets $N_1$ and $N_2$ of dimension $\ge 11$ (Lemma 1.14) and such that $T^5/H_i$ acts effectively on $N_i$. Then $\pi _1(M)\cong \pi _1(N_i)$ (1.6.1). Without loss of generality, we may assume that $\dim (N_1)\le 17$ (Corollary 1.9).

If $H_1$ is finite, then $\pi _1(N_1)$ is cyclic (Theorem 1.5). So we may assume both $H_1$ and $H_2$ are circle subgroups. Since the case $n=17$ has been proved, we may assume that $\dim (N_i)\le 15$. If $\dim (N_1)\le 13$, then $\pi _1(N_1)$ is cyclic (Theorem 1.5). Hence, we may assume $\dim (N_i)=15, i=1, 2$. Let $N_0=N_1\cap N_2$. Clearly, we may assume $\dim (N_0)=9, 11$ (Corollary 1.9) and $N_0$ is fixed by a torus $T^2$ generated by $H_i$. If $\dim (N_0)=9$, then $\pi _1(N_0)$ is cyclic (Theorem 1.5) and so is $\pi _1(N_1)$. If $\dim (N_0)=11$, then it is easy to see, by analyzing the isotropy representation of $T^2$ on the normal space of $N_0$, that there is $T^1\subset T^2$ with a fixed point set of dimension 13. By Corollary 1.9, we then conclude that $\pi _1(N_0)$ is cyclic. $\square$

Note that Lemma 2.1 is not required in the above proofs.