# Cobordism monopole Floer homology

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.

• This is explained in the remark below Theorem 23.1.5. However, if the restriction map $\text{Spin}^c(W) \to \text{Spin}^c(Y) \times \text{Spin}^c(Y')$ has finite fibers, as is the case for $W = Y \times I$, then there is a cobordism map at the level of the uncompleted grading. So you get invariance of the uncompleted groups, but not full functoriality. (The topological energy, which you need to bound to get compactness, is bounded by the expected dimension of the resulting moduli space and a quantity depending on the underlying spin^c structure.) – Mike Miller May 27 '20 at 14:20

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.