# A smooth twisted tensor product of dg algebras?

I want to consider a Z/2Z dg algebra. As an algebra, it is generated over $\mathbb{Q}$ by two elements where x is even and e is odd with the relations $xe=ex$ and $e^2=1$(this makes it in particular non-commutative) The differentials are defined by $de=x^n, dx=0$. I would like to know if this dg algebra is homologically smooth. Any approach is of course welcome, but let me toss out a line of inquiry which lead to yet more questions. First off, I think this is what should be called a twisted tensor product of homologically smooth dg algebras as defined for example in Hess' paper "Twisted Tensor products of DGA's and Adams Hilton Models" (if this isn't that then I don't know what is). It's pretty clear that the tensor product of homologically smooth dg algebras is homologically smooth. It seems quite reasonable that the twisted tensor products should under suitable hypotheses also be homologically smooth.

The way I hoped to investigate this is using a theorem due to Toen-Vaquie for Z-graded dga's (which I presume doesn't work in full generality for Z/2Z graded dga's) that compact homologically smooth Z graded dga's are the same thing as dga's of finite type, namely dga's of the form T(V,d) where V is a finite dimensional graded vector space filtered in such a way so that '$d:V_n---> T(V_{n-1})$'. Now Hess' paper explains how given free models (T(V),d), (T(W),d) for A and B, one can construct a free model for a twisted tensor product of A and B(again the argument given there might not work for Z/2Z graded dga's). The hope is that this construction, under some reasonable hypothesis, would give something of finite type.

I got stuck on this line of inquiry right at the start. It seems tricky to even figure out how to write $Q[e]/e^2-1$, e odd, as an algebra of finite type (maybe the fact that this is Z/2Z graded is really a problem). Namely, one logically begins by adjoining an odd variable for the class e, then one needs a odd class b, such that $db= e^2-1$. Next one needs c to kill eb+be. Then there is something of a hidden class which needs to be killed b^2+ec+ce. One can hope that this is everything but... how to know?

To summarize if anyone can shed light on any of the following four questions it could be great. 1) Is the above Z/2Z dg algebra homologically smooth?

2) Is there a way to write $Q[e]/e^2-1$ as an algebra of finite type?

3) Is there a general theorem with some hypotheses about homological smoothness of twisted tensor products of dg algebras(in the Z graded case or the Z/2Z graded case).

4) I would also be interested in any of the above when there is instead of $e^2=1$, a higher A(infinity) operation $e^{\otimes m} \mapsto 1$.

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How does the square of an odd element equal anything other than zero? Are you working with characteristic 2 coefficients? –  S. Carnahan Jul 26 '10 at 18:58
Hi Scott, no I am not but I don't assume that the dga's are commutative just a graded algebra with a differential d such that d(ab)= ad(b)+(-1)^|a|b*d(a) –  Daniel Pomerleano Jul 26 '10 at 19:18
What does "homologically smooth" mean? –  Tom Goodwillie Jul 30 '10 at 16:29
I expected d(ab)=d(a)*b+(-1)^|a|ad(b) rather than the rule you wrote. Do you really want that rule? It implies, for example, that d(ab)=d(ba) when a and b are both even. But if you use my rule then the Z/2 graded associative algebra generated by even x and odd e subject only to the relation e^2=1 has no differential satisfying de=x^n because you get 0=d(1)=d(e^2)=d(e)*e-ed(e)=x^ne-e*x^n. –  Tom Goodwillie Jul 30 '10 at 17:49
Hi Tom, here smooth means that A is perfect as an $A\otimesA^op$ module or that A is in the full triangulated subcategory generated by $A\otimesA^op$ and it's finite direct summands. You are totally right about the rule...I meant the standard rule for graded derivation, but goofed when I wrote it. But, I also want the relation ex=xe... so I think everything is fine. I thought my notation was clear but I'll clean that up... –  Daniel Pomerleano Jul 30 '10 at 18:13