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The connectivity $\eta(X)$ of a simplicial complex $X$ is defined as the $$1+\min_j\{j \mid \tilde{H}_j(X)\neq 0\}.$$ If no such $j$ exists, then $\eta(X):=\infty$. (See here for this definition, which is also related to homological connectivity.)

I hear that if $M$ is a matroid, then $$\eta(M)\ge rank(M).$$

I am wondering if anyone knows some reference of this theorem. (I heard that a matroid looks like a wedge sum of spheres. Not sure if it helps.)

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  • $\begingroup$ When $X$ is a simplex, $\eta(X)$ is defined to be infinity. So the bound still holds. $\endgroup$
    – Connor
    Commented Aug 7, 2022 at 14:03

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Converting my comments to an answer.

It is known from work of Björner in the late 80s/early 90s that a matroid (independence) complex is shellable: see e.g. his "Homology and shellability of matroids and geometric lattices" available online at https://www.cefns.nau.edu/~falk/classes/690/extras/bjorner.pdf.

Hence, such a complex has the homotopy type of a wedge of spheres. More precisely, since a the complex of matroid $M$ is a pure $(r(M)-1)$-dimensional complex (where $r(M)$ is the rank), it has the homotopy type of a wedge of $\widetilde{\chi}(M)$ $(r(M)-1)$-spheres. Here $\widetilde{\chi}(M)$ is the reduced Euler characteristic of the complex, which is equal to a Tutte polynomial evaluation $\widetilde{\chi}(M)=T_M(0,1)$.

So if this number $T_M(0,1)$ is nonzero, then we will have an exact equality $\eta(M) = r(M)$. Otherwise, the matroid complex of $M$ is contractible so $\eta(M)=\infty$ (this happens e.g. for a simplex, but shouldn't happen for most matroids...). Either way your bound holds.

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