A new problem

Consider a positively weighted sphere system \(S\) in \(M_n = \#_n S^1 \times S^2\). By Poincaré duality, we can consider \([S] \in H_2(M_n)\) as an element of \(H^1(M_n) = H^1(F_n)\) (where \(F_n\) is the free group of rank \(n\)). Fix some \(\alpha \in H^1(F_n)\). Then we can define the cocycle complex \(X_\alpha\) to be the collection of weighted sphere systems representing \(\alpha\) under this identification.

As for the cycle complex, the question now becomes: is the cocycle complex contractible?

Finiteness properties of groups

Lately, I have been concerned with a number of finiteness properties of groups. Everyone is probably familiar with finite generation and finite presentation, but there are many more. We will define several types, and discuss the relationships between them. The first types of finiteness one may care about is the finiteness of a corresponding Eilenberg-MacLane space.

Definition: A group \(G\) is of type \(F_n\) if it has a \(K(G,1)\) with finite \(n\) skeleton.

Definition: A group \(G\) is of type \(F\) if

When is $latex \mathbb{Z}$ a projective $latex \mathbb{Z}\Gamma$-module?

Let \(\Gamma\) be a group. We claim that \(\mathbb{Z}\) is a projective \(\mathbb{Z}\Gamma\)-module if and only if \(\Gamma\) is trivial.

One direction is obvious: if \(\Gamma\) is trivial, then \(\mathbb{Z}\Gamma \cong \mathbb{Z}\), so \(\mathbb{Z}\) is not just a projective \(\mathbb{Z}\Gamma\)-module, but a free one.

For the other direction, suppose that \(\mathbb{Z}\) is a projective \(\mathbb{Z}\Gamma\)-module. We have a short exact sequence

\(\displaystyle 0 \to K \to \mathbb{Z}\Gamma \overset{\varepsilon}{\to} \mathbb{Z} \to 0,\)

where \(\varepsilon\) is the map defined by \(\varepsilon \left( \sum a_i g_i \right) = \sum a_i\). Since \(\mathbb{Z}\) is projective, this sequence splits, i.e., there is a map \(j: \mathbb{Z} \to \mathbb{Z}\Gamma\) such that \(\varepsilon j = id_{\mathbb{Z}}\). Such a map is determined by its action on \(1 \in \mathbb{Z}\). Say \(j(1) = x = \sum_{g \in G} a_g g \in \mathbb{Z}\Gamma\). Note that \(x\) is nontrivial since \(j \) is a splitting. Since the action of \(G\) on \(\mathbb{Z}\) is trivial, we have

\(\displaystyle \sum_{g \in G} a_g g = x = hx = \sum_{g \in G} a_g hg = \sum_{g \in G} a_{h^{-1}g}g\)

for all \(h \in G\). This implies that \(a_g = a_{h^{-1}g}\). Since the action of \(G\) on \(G\) is transitive, we get that \(a_g = a_{g’}\) for all \(g, g’ \in G\). Since \(x \neq 0\), it follows that \(G\) is finite (since otherwise the sum \(x = \sum_{g \in G} a_g g\) is an infinite sum of nonzero terms). Moreover, we have \(1 = \varepsilon(x) = \sum_{g \in G} a_g = \vert G \vert \cdot a_{g_1}\). Since, \(a_{g_1} \in \mathbb{Z}\), it must be that \(\vert G \vert = 1\), and so \(G\) is trivial.