Finiteness of $\ell^2$-Betti numbers

Question

I'm reading the paper "Improved algebraic fibering" by Sam Fisher (https://arxiv.org/pdf/2112.00397.pdf) and in the proof of lemma 6.4 it claims the followng:

$(\mathcal{D}_{\mathbb{F}K}\ast\mathbb{Z})ax\subset Z_n(\mathcal{D}_{\mathbb{F}G}\otimes_{\mathbb{F}G} P_\ast)$, where $P_\ast$ is a free resolution of $\mathbb{F}$ and $(\mathcal{D}_{\mathbb{F}K}\ast\mathbb{Z})ax$ is an infinite dimensional $\mathcal{D}_{\mathbb{F}K}$-subspace of $Z_n(\mathcal{D}_{\mathbb{F}K}\otimes_{\mathbb{F}K} P_\ast)$. Hence, $b_p^{\mathcal{D}_{\mathbb{F}K}}(K)<\infty$ implies that there exists $b\in\mathcal{D}_{\mathbb{F}K}\ast\mathbb{Z}$ such that $bax=\partial y$ for some $y\in\mathcal{D}_{\mathbb{F}K}\otimes_{\mathbb{F}K}P_{n+1}$.

Here $b_p^{\mathcal{D}_{\mathbb{F}K}}(K)=\dim_{\mathcal{D}_{\mathbb{F}K}}\text{Tor}_p^{{\mathbb{F}K}}(\mathbb{F},\mathcal{D}_{\mathbb{F}K})$ is a generalized version of $\ell^2$-betti numbers (The usual $\ell^2$-Betti numbers are obtained by setting $\mathbb{F}=\mathbb{Q}$ which implies $\mathcal{D}_{\mathbb{F}K}=\mathcal{D}(\mathbb{Q})$, which is the Linnell ring)

Maybe that is an easy question, or a property of $\text{Tor}$ that I am forgetting, but I don't understand why knowing that the betti number is finite give us that information.

Thanks for your help.

Answer 1

(This is more of a comment/sketch than a full answer but it seemed clearer to write things here.)

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To remove some clutter let me write $A$ for ${\mathcal D}_{{\mathbb F}K}$. Then paraphrasing the relevant part of the proof of Lemma 6.4, it says

$\dots (A\ast{\mathbb Z})\cdot ax$ is an infinite-dimensional $A$-subspace of $Z_p(A\otimes_{{\mathbb F}K} F_\bullet$). Since $b_p^A(K)<\infty$, there is a non-zero $b\in A\ast {\mathbb Z}$ such that $bax$ is a boundary (in the appropriate level of this chain complex)

So without checking all the definitions, it seems to me that the intended argument must be someting like the following: we have an infinite-dimensional subspace of the space of $p$-cycles, but we know that the homology in degree $p$ is finite-dimensional, hence there must be some part of this subspace which lies in the kernel of the map $p$-cycles $\to$ $p$th homology group.