Cobordism monopole Floer homology

Question

From the famous book: Monopole and three manifold, Kronheimer and Mrowka(https://www.maths.ed.ac.uk/~v1ranick/papers/kronmrowka.pdf). It is known that: Let $Y$ be a closed oriented $3$ manifold, choosing a spinc structure $\mathfrak s$ and metric $g$ and a generic perturbation $p$, one can construct the monopole Floer homology groups: $$\check{HM}_*(Y,\mathfrak s, g,p),~\hat{HM}_*(Y,\mathfrak s, g,p),~\overline{HM}_*(Y,\mathfrak s, g,p).$$ The groups are graded over a set $\mathbb J_s$ admitting a $\mathbb Z$ action.( details are given in Section 20-22). We define the negative completions(Definition 23.1.3 of the book) by $$\check{HM}_\bullet(Y,\mathfrak s, g,p),~\hat{HM}_\bullet(Y,\mathfrak s, g,p),~\overline{HM}_\bullet(Y,\mathfrak s, g,p).$$ If we want to consider all spinc structures at the same time, we need to consider the completed monopole Floer homology $$\check{HM}_\bullet(M,F;\mathbb F)=\bigoplus_\mathfrak s\check{HM}_\bullet(M,F,\mathfrak;\mathbb F).$$

To show that these homology groups are independent of the metric and the perturbation, the authors gave a property: a cobordism between 3-manifolds gives rise to homomorphisms between their monopole Floer homologies(see Section 23-26). They construct a homomorphism from $\check{HM}_\bullet(Y,g_1,p_1)$ to $\check{HM}_\bullet(Y',g_2,p_2)$, where there is a cobordism from $Y$ to $Y'$.

Q I do not understand the two points below:

• Why the authors use the negative completion, where we need it?

• If we just want to show that the monopole Floer homology $\check{HM}_*(Y,\mathfrak s)$ is independent of the metric and perturbation, can we just using the trivial cobordism $[0,1]\times Y$ to show a homomorphism $\check{HM}_*(Y,\mathfrak s,g_1,p_1) \to \check{HM}_*(Y,\mathfrak s, g_2,p_2)$? The homomorphism is given by counting the number of solutions of the zero-dim moduli space $M([a_1],W^*,[b_2])$, where $W^*=(-\infty,0]\times Y\cup I\times Y\cup[1,\infty)\times Y$, and $[a_1]$ and $[b_2]$ are the critical points of $(Y,\mathfrak s,g_1,p_1)$ and $(Y,\mathfrak s,g_2,p_2)$ respectively. I think the arguments of Section23-25 also work before taking the negative completion .

PS Let $G_*$ be an abelian group graded by the set $\mathbb J$ equipped with a $\mathbb Z$-action. Let $O_a(a\in A)$ be the set of free $\mathbb Z$-orbits in $\mathbb J$ and fix an element $j_a\in O_a$ for each $a$. Consider the subgroups $$G_*[n]=\bigoplus_a\bigoplus_{m\geq n} G_{j_a-m},$$ which form a decreasing filtration of $G_*$. We define the negative completion of $G_*$ as the topological group $G_\bullet\supset G_*$ obtained by completing with respect to this filtration.

Answer 1

The first bullet is definitely explained in the book! Surely around where it was introduced, it has to do with summing over all spin-c structures. We need to pass to the completion because the 4-manifold can have infinitely many spin-c structures that would need to be used.

The second bullet, yes. In general we should not expect results concerning completions of a graded group to also hold for the uncompleted group. But here we consider the trivial cobordism, and spin-c structures on $[0,1]\times Y$ are the same as spin-c structure on $Y$, so no completion is needed in this situation.