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Question: If $X$ is a simplicial complex that's simply connected and $2$-dimensional, does there always exist a contractible subcomplex $Y$ satisfying $X^{(1)} \subseteq Y$?

The statement is true "down a dimension": If $X$ is connected and $1$-dimensional, then there exists a contractible subcomplex $Y$ satisfying $X^{(0)} \subseteq Y$, namely you can use a spanning tree. So, one could think of the desired $Y$ in the original problem as a $2$-dimensional analog of a spanning tree, with contractibility being the key desired property.

It seems there is a higher dimensional analog of "spanning tree" in the literature, at least for finite complexes, e.g., Definition 3.1 of this. Note that Proposition 3.7 of that paper implies that (finite) simply connected $2$-complexes indeed have these sorts of $2$-dimensional "spanning trees". But these are not necessarily contractible, they're more of a homological analog of trees.

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  • $\begingroup$ You would need your complex to be homotopy equivalenct to a wedge of 2-spheres since contracting Y would give a sphere for each triangle not in Y. I don't know if that is sufficient $\endgroup$
    – Benjamin Steinberg
    Commented Jul 6, 2022 at 16:20
  • $\begingroup$ @BenjaminSteinberg That's true, but it turns out every simply connected $2$-complex is homotopy equivalent to a wedge of 2-spheres! (According to math.stackexchange.com/questions/747597/… this is "well known", but I guess I don't actually know a proof.) $\endgroup$
    – Matt Zaremsky
    Commented Jul 6, 2022 at 16:33
  • $\begingroup$ Ok. I didn't know that. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 6, 2022 at 16:45
  • $\begingroup$ Perhaps the proof of this fact gives this contractible spanning complex? $\endgroup$
    – Benjamin Steinberg
    Commented Jul 6, 2022 at 16:56
  • $\begingroup$ Can you just order the 2-cells and mark which don't increase the second Betti number when attached, and then the subcomplex is given by attaching these cells to the 1-skeleton? $\endgroup$
    – Connor Malin
    Commented Jul 6, 2022 at 18:16

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I think the answer is no.

In Hatcher's Algebraic Topology there is an example of an acyclic 2-dimensional complex with one 0-cell, two 1-cells and two 2-cells (Example 2.38). You start with a wedge of two circles, labeled $a$ and $b$, and attach two 2-dimensional cells by the words $a^5b^{-3}$ and $b^2(ab)^{-2}$. One can easily check that it is acyclic. The fundamental group has the presentation $<a, b\mid a^5b^{-3}, b^2(ab)^{-2}>$, and one can show that it is not trivial. Now form a new complex $X$ by attaching one more $2$-dimensional cell, along $b$. The fundamental group of $X$ is easily seen to be trivial. But no proper subcomplex of $X$ containing the $1$-skeleton is simply-connected, let alone contractible.

To make this example into a simplicial complex, choose a triangulation of $X$. This can be done because all the attaching maps are nice piece-wise linear maps. I believe it is still true that no simplicial subcomplex of $X$ that contains the $1$-skeleton is contractible. Removing a two-cell from the triangulation of $X$ is like punching a hole in a two cell in the original CW structure on $X$. We already know that punching a single hole in some of the cells does not make a contractible complex, and punching additional holes only makes the fundamental group bigger.

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  • $\begingroup$ Aha, excellent! I think this works. Is it obvious that $X$ itself is not contractible? That's the only thing I don't see right away. $\endgroup$
    – Matt Zaremsky
    Commented Jul 6, 2022 at 20:34
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    $\begingroup$ $X$ has Euler characteristic 2, so not contractible. $\endgroup$
    – Gregory Arone
    Commented Jul 6, 2022 at 20:36
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    $\begingroup$ Perfect! And actually, I think this same sort of idea, of taking a presentation for the trivial group that becomes non-trivial upon removing any one defining relator, would work for an even simpler example, like the presentation 2-complex of $\langle a\mid a^2,a^3\rangle$. $\endgroup$
    – Matt Zaremsky
    Commented Jul 6, 2022 at 20:42
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    $\begingroup$ Duh. Of course. Well spotted. $\endgroup$
    – Gregory Arone
    Commented Jul 7, 2022 at 5:53

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