Is it true that the fundamental group of a compact finite dimensional Alexandrov space with curvature bounded below is finitely generated?
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1$\begingroup$ pp.9 of the article mural.maynoothuniversity.ie/10079/1/DW-Fundamental-1998.pdf says Yes, and that a result of Gromov proves the group is generated by at most $(2.5)^{\dim(X)}$ generators. $\endgroup$– JHMCommented Feb 9, 2021 at 20:42
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Under these hypotheses the systole of $X$ is clearly bounded from below, and the usual comparison arguments would give an upper bound on the number of points in an $\epsilon$-net in $X$. Then every loop can be discretized and the finite-generation follows.
answered May 9, 2018 at 10:08
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$\begingroup$ Thank you very much. May I ask in addition whether the stronger statement is true: the minimal number of generators of the fundamental group is bounded above by a constant depending only on dimension, diameter, and a lower bound on curvature? $\endgroup$– asvCommented May 10, 2018 at 5:47
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1$\begingroup$ I am not sure. You might want to try to run this by Anton Petrunin. $\endgroup$ Commented May 10, 2018 at 8:57
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$\begingroup$ It is possible that this would follow from a 2017 publication by Sabourau where he provides a specific lower bound for the systole, modulo suitable hypotheses on $X$. @orbits $\endgroup$ Commented May 10, 2018 at 11:32