Can a Busemman space be CAT(1)?
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$\begingroup$ Yes, every CAT(0) space is Busemann and CAT(1), so the Euclidean plane for example. $\endgroup$– Lee MosherCommented Jun 20, 2013 at 18:18
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$\begingroup$ My space is not CAT(0). $\endgroup$– ChrisCommented Jun 20, 2013 at 18:27
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$\begingroup$ Did you mean to ask a different question, then? $\endgroup$– Lee MosherCommented Jun 20, 2013 at 18:30
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$\begingroup$ Thank you, yes, maybe, I have this space which is for sure Busemann and NOT CAT(0) but it seems to be CAT(1) which feels a bit weird to me. $\endgroup$– ChrisCommented Jun 20, 2013 at 18:32
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3$\begingroup$ @Chris: So there is a reasonable question here: Is it known whether a CAT(1) Busemann space must also be CAT(0)? But it sounds also as if you have extra information about this question in the form of a potential counterexample, in which case I suggest you take a look at the link "how to ask" up above, particularly under the heading "Provide background and motivation". $\endgroup$– Lee MosherCommented Jun 20, 2013 at 22:11
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1 Answer
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Answering a sensible question appeared in comments: If a geodesic space is Busemann and CAT(1), then it must be CAT(0).
Indeed, CAT(1) implies that the space has well-defined metric angles between geodesic segments. And if a space is Busemann and has well-defined angles, then it is CAT(0).
Indeed, let $X$ be the space in question and $\alpha,\beta:[0,1]\to X$ constant-speed minimizing geodesics with $\alpha(0)=\beta(0)=p$. Then, by the Busemann definition, the function $t\mapsto d(\alpha(t),\beta(t))/t$ is non-decreasing. Hence the comparison angle at $p$ of the triangle $p\alpha(t)\beta(t)$ is non-decreasing. Hence its limit at $t=0$ (which is by definition the angle between $\alpha$ and $\beta$) is no greater than the comparison angle at $p$ of the triangle $p\alpha(1)\beta(1)$. This means that Toponogov's hinge comparison holds. It remains to recall that existence of angles plus hinge comparison is one of the standard definitions of CAT(0).
answered Jun 20, 2013 at 23:25
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$\begingroup$ Thank you very much, I was not familiar with the Toponogov's hinge comparison, I will look into that. In fact, this topic of metric geometry is all new to me, I am working with operator spaces and I found a space of finite dimensional operator spaces which is geodesic and it followed naturally the question about cat(k). If I do not ask too much, could you please answer another question? We have established that this Busemann space is not cat(k). The geodesics vary continously with the end points. Can we say more about this space? Would there be another direction I could go into? $\endgroup$– ChrisCommented Jun 21, 2013 at 2:59
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1$\begingroup$ @Chris : if you have another question, you should ask it separately, but only after thinking it out well and making it sufficiently precise. Please read the FAQ carefully to know how to use this site properly. $\endgroup$ Commented Jun 21, 2013 at 7:56
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$\begingroup$ @Benoît: Thank you, sorry about that. $\endgroup$– ChrisCommented Jun 21, 2013 at 11:48