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I see the particular of Klingenberg's lemma:

If $M$ is compact and the sectional curvature $$ K\le \delta $$ then, the injective radius $\text{inj}(M)$ satisfy $$ \text{inj}(M) \ge \min\{\frac{\pi}{\sqrt \delta}, \frac{l(M)}{2}\} $$ where $l(M)$ is the length of the shortest simple closed geodesic in $M$.

Since $M$ is compact, if $M$ is smooth, I feel (just feel, not sure) there is a closed geodesic $\gamma$ in $M$ such that $$ l(\gamma) =l(M). $$ Then, for any $p\in \gamma$, I have $$ \exp_p:T_pM\rightarrow M $$ is injective on $B_0(\frac{l(\gamma)}{2})\subset T_pM$. $B_0(\frac{l(\gamma)}{2})$ is the open ball center at origin and radius is $\frac{l(\gamma)}{2}$. Therefore, I have $$ \text{inj}(M) \le \frac{l(M)}{2} \tag{1} $$ Obviously, it is incompatible with Klingenberg's lemma. Where is my mistake?

Enhao Lan
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  • Your inequality is reversed. The injectivity radius is at most half the length of the shortest closed geodesic, not at least. – Deane Mar 30 '22 at 13:09
  • @Deane Do you mean (1) is reversed ? Sorry for my poor English. I think ‘The injectivity radius is at most half the length of the shortest closed geodesic’ equal to $inj(M) \le \frac{l(M)}{2}$. – Enhao Lan May 24 '22 at 08:52
  • I believe the correct inequality is $$ (M) \le \min{\frac{\pi}{\sqrt \delta}, \frac{l(M)}{2}} $$ – Deane May 24 '22 at 15:54

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