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Any differentiable manifold can be triangulated (and it can be done piecewise-linearly), but I get the impression that the converse is false. For insight, is there a good example of a topological space that can be triangulated but is not a differentiable manifold? If we know that a space can be triangulated, is there some set of additional assumptions under which we can say that it's a manifold?

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    Here's a simplicial complex that is not a manifold: $\bowtie$. – 2017-01-15
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    @RobArthan: I think that could have been an awesome one-character comment, although the SE software would probably not have allowed it :-) I would still be interested in understanding if there are additional assumption that would make it work. – 2017-01-15
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    @BenCrowell What do you mean by a manifold? A topological manifold? A PL manifold?..... – 2017-01-15
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    @MoisheCohen: I would happy to know any set of additional assumptions that would guarantee it to be any specific type of manifold. – 2017-01-15
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    @BenCrowell If links are topological spheres, the complex is a topological manifold. – 2017-01-15
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    @MoisheCohen: Oh, I see -- that's what a PL manifold is, right? If you feel like making that comment into an answer, I would be happy to accept it. – 2017-01-15

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For instance, if links in the complex are triangulated spheres, then the complex is a PL manifold. However, even if links are not spheres, the complex might still be homeomorphic to a topological manifold (the double suspension of a 3-manifold which is a homology 3-sphere is a classical example).