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Parroting the link below (I imagine this is a standard definition), a topological space $T$ is $k$-connected ($k ≥ 0$) if, for every $0 ≤ r ≤ k$, every continuous map $f : S^r → T$ extends to a continuous map $f : B^{r+1} → T$. ($B^r$ is the $r$-dimensional unit ball.) A simplicial complex is $k$-connected if its geometric realization is.

http://web.cs.elte.hu/~lovasz/kurzusok/topol13.pdf#page=2

Parroting the link below, two simplices $\sigma$ and $\tau$ are $k$-connected if there is a sequence of simplices $\sigma$; $\sigma_1$; $\sigma_2$; $\dots$; $\sigma_n$; $ \tau$; such that any two consecutive simplices have at least $k+1$ vertices in common. A simplicial complex is $k$-connected if any two simplices of dimension greater than or equal to $k$ are $k$-connected.

https://math.la.asu.edu/~helene/papers/atheory_final.pdf#page=2

Are the above two definitions equivalent? If not, what is a more combinatorial way of describing what the first definition means?

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They appear not to be equivalent. Take two tetrahedra (full complexes on four vertices) and join at a point. This gives us a simplicial complex on $7$ vertices whose geometric realisation is contractible.

By your first definition, this simplicial complex is $2$-connected. By your second, it appears not to be. It's not even $1$-connected.

A more combinatorial way of describing the first definition might be to say something like:

Two simplices $\sigma$ and $\tau$ are $k$-connected if there is a sequence of simplices $\sigma,...,\tau$ such that any two consecutive simplices are joined by a simplex.

However, I wouldn't go just altering your definitions without very strong motivation - whilst I know little about the topic of the second paper, there is sure to be good reasoning behind their definition.

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  • $\begingroup$ Thank you for your answer. $\endgroup$ – Tri Jul 23 '19 at 17:28
  • $\begingroup$ Feel free to mark your question with a tick as answered if you're satisfied with this :) $\endgroup$ – Matt Jul 23 '19 at 19:12

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