7
$\begingroup$

I am working in the context of fully extended TQFTs and, at the moment, I am trying to find fully dualizable objects in certain $(\infty, 2)-$categories. In particular I know that for the $(\infty, 2)-$category $dgAlg_2$, (objects: dg-algebras; 1-morphims from $A$ to $B$: $(A, B)-$dg modules: 2-morphisms: interwiners; $\dots$) I shoulg get that fully dualizable objects are $smooth$ and $compact$ dg-algebras (a dg-algebra $A$ is smooth if $\sum_iH^i(A) <\infty$ and it is compact if it is smooth as an $A^e-$module). I have no clue how to prove this, so any hint or suggestion for some references would be really appreciated.

Thanks in advance,

Andrea

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Suppose $A$ is smooth and proper. Then $A$ is dualizable as a module over $k$ and over $A^e=A\otimes A^{op}$. To prove full dualizability you need to exhibit adjoints to the evaluation map which is $A$ considered as an $(A^e, k)$-bimodule. Both $A^* = Hom(A, k)$ and $A^! = Hom_{A^e}(A, A^e)$ give you candidates for the adjoints. $\endgroup$
    – Pavel Safronov
    Mar 7, 2016 at 15:52
  • $\begingroup$ @Andrea What you call smooth is usually called proper, while smooth means compact (in the categorical sense) as an $A^e$-module. $\endgroup$
    – Marc Hoyois
    Mar 7, 2016 at 15:54

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.