Is the tensorproduct of a triangulated category with a ring again triangulated? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:28:38Z http://mathoverflow.net/feeds/question/26715 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26715/is-the-tensorproduct-of-a-triangulated-category-with-a-ring-again-triangulated Is the tensorproduct of a triangulated category with a ring again triangulated? Manuel Koehler 2010-06-01T13:38:50Z 2010-06-02T04:30:45Z <p>$\underline{Background}$ : Suppose $\tau$ is a preadditive category and $R$ a ring. Then one may form a new preadditive category $\tau \otimes R$ in the following way: </p> <p>$\tau \otimes R$ has the same objects as $\tau$ and for objects $A$ and $B$, set $\tau \otimes R(A,B) := \tau(A,B)\otimes_{\mathbb{Z}} R$. Composition is given by $\tau(A,B)\otimes_{\mathbb{Z}} R\otimes_{\mathbb{Z}}\tau(B,C)\otimes_{\mathbb{Z}} R \cong \tau(A,B)\otimes_{\mathbb{Z}}\tau(B,C)\otimes_{\mathbb{Z}}R\otimes_{\mathbb{Z}}R \rightarrow \tau(A,C) \otimes_{\mathbb{Z}}R$, where the last arrow is given by the composition in $\tau$ and multiplication in $R$.</p> <p>$\tau \otimes R$ is additve if $\tau$ is additive.</p> <p>$\underline{Question}$ : Suppose $\tau$ is triangulated, is there a (canonical) structure of a triangulated category for $\tau \otimes R$?</p> <p>Actually, I only need the case $R := \mathbb{Z}[\frac{1}{n}, \theta] \subset \mathbb{C}$, where $\theta$ is a primitive root of unity of order $n$ for some $n \in \mathbb{N}$, but I would also be interested in a general statement/counterexample to the general case.</p> <p>Thank you for reading.</p> http://mathoverflow.net/questions/26715/is-the-tensorproduct-of-a-triangulated-category-with-a-ring-again-triangulated/26774#26774 Answer by Greg Stevenson for Is the tensorproduct of a triangulated category with a ring again triangulated? Greg Stevenson 2010-06-02T02:03:44Z 2010-06-02T02:03:44Z <p>I would imagine it is false in general that given a triangulated category $T$ the category $T\otimes R$ is also triangulated.</p> <p>The following is a concrete counterexample. Consider $D^b(\mathbb{Z})$ and let $T = D^b(\mathbb{Z})\otimes \mathbb{Z}[x]$. In order for $T$ to be triangulated the morphism $\mathbb{Z} \stackrel{x}{\to} \mathbb{Z}$ would need to be able to be completed to a distinguished triangle $$\mathbb{Z}\stackrel{x}{\to}\mathbb{Z}\to C_x \to \Sigma\mathbb{Z}$$ in $T$. Suppose that such a triangle existed and consider the exact sequence obtained by application of $Hom_T(\mathbb{Z},-)$ $$0 \to Hom_T(\mathbb{Z}, \Sigma^{-1}C_x) \to \mathbb{Z}[x] \stackrel{x}{\to} \mathbb{Z}[x] \to Hom_T(\mathbb{Z}, C_x) \to 0$$ As multiplication by $x$ has no kernel we have $$Hom_T(\mathbb{Z}, \Sigma^{-1}C_x) = 0$$ and the object $C_x$ would have to satisfy $$\mathbb{Z} \cong Hom_T(\mathbb{Z},C_x) = Hom_{D^b(\mathbb{Z})}(\mathbb{Z}, C_x)\otimes_\mathbb{Z} \mathbb{Z}[x]$$ But decomposing $C_x$ as a sum of suspensions of free groups and torsion groups this is clearly not possible.</p> <p>One situation in which something like what you ask for works is the following: let $(T,\square,\mathbf{1})$ be a tensor triangulated category (that is we have a symmetric monoidal structure on $T$ which is exact in each variable). Given a multiplicative subset of even elements $S$ of $End^*(\mathbf{1})$ one can localize the hom-sets of $T$ at $S$. This can even be realised as a Verdier quotient by a certain subcategory. This is shown in Balmer's "Spectra, spectra, spectra - Tensor triangular spectra versus Zariski spectra of endomorphism rings" section 3.</p> http://mathoverflow.net/questions/26715/is-the-tensorproduct-of-a-triangulated-category-with-a-ring-again-triangulated/26786#26786 Answer by David Ben-Zvi for Is the tensorproduct of a triangulated category with a ring again triangulated? David Ben-Zvi 2010-06-02T04:30:45Z 2010-06-02T04:30:45Z <p>It seems unreasonable to expect tensoring the morphisms in a triangulated category by a ring to be a meaningful operation - after all these morphisms are typically cohomology groups of natural complexes, and we know there are universal-coefficient-theorem type corrections to tensoring cohomology with a ring as opposed to the more natural operation of tensoring a complex.</p> <p>On the other hand, if we enrich our triangulated categories everything works fine. For example on the level of (pretriangulated) differential graded categories or $A_\infty$ categories, or if you prefer, stable $\infty$-categories, there's a perfectly nice operation of tensor product by a ring - in fact one can tensor by another such category (generalizing the case you're asking about, which is tensoring with R-mod). Even more, there's a structure of symmetric monoidal $\infty$-category on such categories (developed in Lurie's Derived Algebraic Geometry series (an exposition and references can be found eg <a href="http://arxiv.org/abs/0805.0157" rel="nofollow">here</a>). </p>