# Monday, April 25 ## Triangulated categories :::{.definition title="Triangulated categories"} A **triangulated category** is an additive category $\cat C\in\Add\Cat$ with an additive autoequivalence $T: \cat{C}\to \cat{C}$ and a set of distinguished triangles $X\to Y\to Z\to TX$ satisfying - **TR1**: - $X \mapsvia{\id} X \to 0 \to TX$ is distinguished, - Any triangle isomorphic to a distinguished triangle is again distinguished, - For every $X \mapsvia{u} Y$ there is a distinguished triangle $X \mapsvia{u} Y \to Z\to X[1]$. Idea: $Z\approx Y/X$. - **TR2**: - For every $X\to Y\to Z\to X[1]$, there is a triangle $Y\to Z\to X[1] \mapsvia{Tu} Y[1]$. - **TR3**: - Given 3 triangles \[ X&\to Y\to Z'\to Y&\to Z\to X'\to X&\to Z\to Y'\to \] there is a triangle $Z'\to Y'\to X'$ making the relevant octahedral diagram commute. \begin{tikzcd} & {Y'} \\ {X'} && Z &&&& {} \\ & Y &&&& {} \\ {Z'} && X \arrow[from=4-3, to=3-2] \arrow[from=3-2, to=4-1] \arrow[from=3-2, to=2-3] \arrow[from=2-3, to=2-1] \arrow[from=4-3, to=2-3] \arrow[from=2-1, to=4-1] \arrow[from=2-1, to=3-2] \arrow[from=4-1, to=4-3] \arrow[dashed, from=2-3, to=1-2] \arrow[dashed, from=1-2, to=2-1] \arrow[dashed, from=4-1, to=1-2] \arrow[dashed, from=1-2, to=4-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMiwzLCJYIl0sWzEsMiwiWSJdLFswLDMsIlonIl0sWzIsMSwiWiJdLFswLDEsIlgnIl0sWzUsMl0sWzYsMV0sWzEsMCwiWSciXSxbMCwxXSxbMSwyXSxbMSwzXSxbMyw0XSxbMCwzXSxbNCwyXSxbNCwxXSxbMiwwXSxbMyw3LCIiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNyw0LCIiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMiw3LCIiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNywwLCIiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) This can equivalently be expressed as a braid lemma: \begin{tikzcd} X && Z && {X'} && {Z'[1]} \\ \\ & Y && {Y'} && {Y[1]} \\ \\ && {Z'} && {X[1]} \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=1-1, to=3-2] \arrow[from=3-2, to=1-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=1-3, to=3-4] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=3-4, to=1-5] \arrow[from=1-5, to=3-6] \arrow[from=3-6, to=1-7] \arrow[from=3-2, to=5-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=5-3, to=3-4] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=3-4, to=5-5] \arrow[from=5-3, to=5-5] \arrow[from=5-5, to=3-6] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Equivalently, a 3x3 lemma holds: \begin{tikzcd} X & Y & Z & {X[1]} \\ {X'} & {Y'} & {Z'} & {Y'[1]} \\ {X''} & {Y''} & \textcolor{rgb,255:red,214;green,92;blue,92}{Z''} & {Z''[1]} \\ {X[1]} & {Y[1]} & {Z[1]} & {X[2]} \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=3-1, to=3-2] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=3-2, to=3-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=3-3, to=3-4] \arrow[from=4-1, to=4-2] \arrow[from=4-2, to=4-3] \arrow[from=4-3, to=4-4] \arrow[from=1-1, to=2-1] \arrow[from=2-1, to=3-1] \arrow[from=3-1, to=4-1] \arrow[from=1-2, to=2-2] \arrow[from=2-2, to=3-2] \arrow[from=3-2, to=4-2] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=1-3, to=2-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=2-3, to=3-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, dashed, from=3-3, to=4-3] \arrow[from=1-4, to=2-4] \arrow[from=2-4, to=3-4] \arrow[from=3-4, to=4-4] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) ::: :::{.theorem title="?"} For $\cat{A}\in\Ab\Cat, \derivedcat{\cat A}\in \triang\Cat$. ::: :::{.definition title="?"} For $f\in \Ch\cat{A}(X, Y)$, there is a *cone* complex $\Cone(f) = TX \oplus Y$ with differential $d_{\Cone(f) } = \matt{d_{X[1]}}{0}{f[1]}{d_Y}$ and a *cylinder* complex $\Cyl(f)$: ![](figures/2022-04-25_11-03-11.png) Note that $d_{\Cone(f)} \tv{x_{i+1}, y_i} = \tv{-d_X x_{i+1}, f(x_{i+1}) + d_Y(y_i)}$, and one can check $d^2=0$. ::: :::{.remark} Any distinguished triangle $X \mapsvia{f} Y \to Z\to X[1]$ in $\derivedcat{\cat A}$ is isomorphic to a triangle of the form $X\to \Cyl(f) \to \Cone(f)\to X[1]$. For $\Ch\cat{A}$, define $T^nA \da A[n]$, so $(T^nA)_k = A[n]_k = A_{n+k}$, and $\bd_{TA} \da (-1)^n \bd_A$. :::