## Curvature of 2d polyhedral spaces [closed]

Consider two flat triangles, choose a vertex in each of them and glue them together by identifying these two distinguished vertices. Equip the resulting space $P$ with the quotient metric. You get a metric bouquet. As such and since flat triangles are non-negatively and non-positively curved the metric bouquet $P$ is non-positively curved (see e.g. Proposition 4.2.9 in [BBI]=[Burago/Burago/Ivanov: "A Course in Metric Geometry"], http://www.math.psu.edu/petrunin/papers/alexandrov/bbi.pdf. On the other hand, $P$ is a 2-dimensional polyhedral space and the sum of the angles around the connecting vertex is clearly not greater than $2\pi$. Hence Theorem 4.2.14 of [BBI] asserts that $P$ is of non-negative curvature.

So, is there any mistake in this argumentation or is $P$ both non-negatively and non-positively curved? If $P$ was both non-negatively and non-positively curved, however, triangles would have to have the same angles as their Euclidean comparison triangles. But this I cannot see. Instead I can see a lot of triangles which are thinner then their comparison traingles ...

Now, if I glue together in the same way as above 90 flat isoscele triangles (at one common vertex), then the angles at the gluing point add up to $30\pi$. Hence, according to the cited theorem the resulting polyhedral space is not non-negatively curved but, as a metric bouquet, it is non-positively curved. Right?

So does curvature here really depend on the number of triangles I glue together? Intuitively that doesn't make much sense to me ...

What is a nice visual example of a 2-dimensional polyhedral space which is non-negatively curved but NOT non-positively curved? Any link to nice illustrations known?

NB: I am aware that these questions are rather trivial (possibly even silly) and not on research level. Simply, I don't know a more suitable place to ask them. So I would be very grateful if someone could shed some light on this!

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The bouquet of two triangles is not nonnegatively curved (in fact, no bouquet is). In the theorem you cite there is an assumption that the space is a manifold, which is not satisfied here. – Sergei Ivanov Dec 4 at 11:43
In general, if you have a 2d polyhedral Euclidean complex, the test for nonpositive curvature is that length of every circuit in every vertex link should be at least $2\pi$. In your case, the only link has no circuits, and, hence you get nonpositive curvature. – Misha Dec 4 at 14:17
@Sergei Ivanov Ahh, OK, I already had the feeling that I was on the wrong track ... that's why I said it's not research level. I am new to the field and am currently working through [BBI] which is a truly great book but I find the stuff about polyhedra quite difficult to grasp. I was aware of the manifold condition, but isn't it possible to embed $P$ homeomorphically into the plane? (OK, I see that this will never be isometric but that's not required, isn't it?) ... or am I completely confused here?? – unknown (google) Dec 4 at 15:27
@Misha: Is the link at the gluing point a (metric) graph which has two disjoint edges whose lengths equal the angles at the gluing point or am I missing the point? – unknown (google) Dec 4 at 15:28
Unknown: Yes, of course. – Misha Dec 5 at 2:25