Let $(\mathcal T, \otimes)$ be a tensor tringulated (tt-)category. Balmer defined a functor from the category of tt-categories to the category of locally ringed spaces, called the Balmer spectra or tt-spectra. He showed that the functor is not essentially injective if we set the target to be the category of topological spaces, but I am wondering if there is any known counter-example in our setting. In particular, when the Bamler spectra $X$ of $(\mathcal T, \otimes)$ is a qcqs scheme (or a smooth proper variety over a field if that makes a difference), is it true that we have $(\mathcal T, \otimes) \simeq (Perf X, \otimes_{\mathcal O_X}^{\mathbb L})$ as tt-categories? Thank you in advance.


1 Answer 1


No this is not true in general. In Tensor Triangulated Categories in Algebraic Geometry, Prop 4.0.9, Sosna shows that if X is a connected noetherian scheme then it is possible to put a tt-structure $\boxtimes$ on the derived category of $X\amalg X$ such that $Spc(D^{b}(X\amalg X),\boxtimes)\cong X$, yet $D^{b}(X\amalg X)\not\simeq D^{b}(X)$ so there cannot be an equivalence as tt-categories between $(D^{b}(X),\otimes_{X}^{\mathbb{L}})$ and $D^{b}(X\amalg X,\boxtimes)$. This structure is just like a square zero extension for the $\otimes_{X}^{\mathbb{L}}$ tt-structure on $D^{b}(X)$.

You need some more control on either the variety or the sort of tt-structure you want to put on the triangulated category. For example disregard spaces like $X\amalg X$ or fix some conditions on the unit.

  • $\begingroup$ Thank you very much for your reference and suggestions. Is it straightforward to see that $Spc D^b(X \bigsqcup X, \boxtimes)$ is isomorphic to $X$ as ringed spaces? $\endgroup$
    – P. Usada
    May 18, 2023 at 2:14
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    $\begingroup$ @P.Usada I think so. You have to keep in mind that the unit of this tt-structure is given by (Ox,0) and that as described, the spectrum is given by subcategories of the form Q+Db(X) with Q in the spec of Db(X), and so the open subsets of this space are too just of the form U+Db(X). Then you can calculate by hand the structure sheaf and check it must coincide with that of X. $\endgroup$
    – AT0
    May 18, 2023 at 7:05

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