# completeness of the Tensor product of vector spaces which are complete for filtration topology

The following question actually came up (in a minor role) in research recently and I am having quite a hard time deciding it. Still the straightforward nature of the question made me decide to post it to stackexchange instead of overflow. Any help would be appreciated.

Suppose $$A$$ is a filtered vector space with filtration $$\ldots \supset F^{-l}A\supset F^{1-l}A\supset\ldots F^{-1}A\supset F^0A\supset F^1A\supset\ldots\supset F^kA\supset F^{k+1}A\supset\ldots$$ such that $$\bigcup_kF^kA=A$$ and $$\bigcap_kA=\{0\}$$. Then we define the corresponding metric $$d\colon A\times A\rightarrow \mathbb{R}_{\geq0}$$ by setting $$d(v,w)=0$$ if $$v=w$$ and $$d(v,w)=2^{-|v-w|}$$ is $$v\neq w$$ where for all $$x\neq 0$$ we have $$|x|=\max\{k\mid x\in F^kA\}$$. We now suppose additionally that $$A$$ is complete. Finally consider another vector space $$B$$ with a similar filtration and again such that $$B$$ is complete for the corresponding metric.

Now the (algebraic) tensor product is naturally equipped with the filtration $$F^k(A\otimes B)=\bigoplus_{\substack{i,j\in \mathbb{Z}\\ i+j=k}}F^iA\otimes F^jB$$ and it is striaghtforward to show that this is again a filtration such as the one above. The question is now simple. Is the metric corresponding to this filtration on $$A\otimes B$$ complete? Of course I also want to know why/why not.

I think that it is in fact complete, but I can't quite come up with a proof. The first thing to try would be to show that Cauchy sequences in $$A\otimes B$$ are somehow "built" out of Cauchy sequence in $$A$$ and $$B$$ respectively, but this is not straightforward, since for instance taking any sequences $$a_k\in F^{2k}A$$ and $$b_k\in F^{-k}B$$ yields a Cauchy sequence $$a_k\otimes b_k$$ in $$A\otimes B$$. The sequence $$B_k$$ however does not need to be Cauchy at all! of couse in this specific case the sequence converges to $$0$$. Even just considering Cauchy sequences of pure tensors $$a_k\otimes b_k$$ I seem to be having some trouble.

The tensor product is not in general complete. Namely if we denote the base-field by $$L$$ and we let $$A=L[\![t]\!]$$ and $$B=L[\![s]\!]$$ both the ring of formal power series. We consider the filtrations $$F^kA=t^kA$$ and $$F^kB=s^kB$$ for $$k\geq 0$$, while $$F^kA=A$$ and $$F^kB=B$$ for $$k\leq 0$$. Then these satisfy the requirements. Now the tensor product $$A\otimes B$$ is given by sums $$\sum_{k,l\geq 0}\sum_{i=1}^f a_{k,i}b_{l,i} t^k\otimes s^l.$$ It is filtered by $$F^k(A\otimes B)=\bigoplus_{l=0}^kt^lA\otimes s^{k-l}B$$. Thus we find the Cauchy sequence $$c_n=\sum_{m=0}^n t^m\otimes s^m$$. A limit of this sequence would be given in the form above by the system of equations $$\sum_{i=1}^f a_{k,i}b_{l,i}=\delta_{kl}$$ for all $$k,l\geq 0$$ and where $$\delta_{kl}$$ denotes the Kronecker delta. Now given $$f\in \mathbb{N}$$ we find that no such limit can exist since it would mean the matrices $$A=(a_{k,i})_{\substack {1\leq k\leq f+1\\ 1\leq i\leq f}}$$ and $$B=(b_{l,j})_{\substack {1\leq l\leq f+1\\ 1\leq j\leq f}}$$ determining linear maps $$A, B \colon L^f\rightarrow L^{f+1}$$ would satisfy $$AB^T=I_{f+1}$$, i.e. the identity of $$L^{f+1}$$ would factor through $$L^f$$.