Notation 1.1.4. Let $a<b\in {\mathbb {R}}\cup \{\pm \infty \}$ . By the notation $|a,b|$ , we mean one of the following interval subsets of ${\mathbb {R}}$ : $(a,b), (a,b], [a,b),[a,b]$ . For example, $M_{|a,b|}$ refers to one of $M_{(a,b)}$ , $M_{(a,b]}$ , $M_{[a,b)}$ , $M_{[a,b]}$ where intervals cannot be closed at $+\infty $ and $-\infty $ .

Notation 2.2.6. Let V and $W=\bigoplus W_j$ be objects in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . Let $X_{V,W}=\bigoplus W_{j_k}$ be a choice of summands of W giving the left minimal $\mathop {\text {add}}(W)$ -approximation of V and let $Y_{V,W}$ be the other summands of W. Dually, let ${_{V,W}}X$ denote summands of V giving the right minimal $\mathop {\text {add}}(V)$ -approximation of W and let ${_{V,W}}Y$ denote the other summands of V.

Proof. Setup and Base Case. We assume V is indecomposable and prove the lemma by induction on the number of summands of W. Our base case is 1; that is W is also indecomposable. Note that, for any object, X in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ , we have $T_r(X[1])=(T_r(X))[1]$ . The base case then follows from Theorem 1.2.6 by rotating distinguished triangles.

Induction Setup and Trivial Case. For induction assume statements (1) and (2) hold when V is indecomposable and W has n or fewer indecomposable summands. Suppose $W\cong \bigoplus _{i=1}^n W_i$ where each $W_i$ is indecomposable. For each summand $W_i$ , let $f_i:V\to W_i$ be such that $f=[f_1\, \cdots \, f_n]^t$ . Let $W_{n+1}$ be an indecomposable and $f_{n+1}:V\to W_{n+1}$ a morphism. If $f_{n+1}=0$ , the result follows using the induction hypothesis.

$\textbf {Hom}{\boldsymbol {(}\mathbf{W}\boldsymbol {,}\mathbf{W}_{\mathbf{n}\boldsymbol {+}\mathbf{1}}\boldsymbol {)}\boldsymbol {\neq } \mathbf{0}} \textbf { or } \textbf {Hom}{\boldsymbol {(}\mathbf{W}_{\mathbf{n}\boldsymbol {+}\mathbf{1}}\boldsymbol {,}\mathbf{W}\boldsymbol {)}\boldsymbol {\neq } \mathbf{0}}$ . Assume $f_{n+1}\neq 0$ . By exchanging the roles of a summand $W_i$ of W and $W_{n+1}$ , we may assume, without loss of generality, that $\operatorname {\mathrm {Hom}}(W_{n+1},W)\neq 0$ . Then there is at least one summand $W_j$ of W such that $\operatorname {\mathrm {Hom}}(W_{n+1},W_j) \neq 0$ . Then $f_j$ factor through $f_{n+1}$ and we may apply Lemma 2.1.2 to $f\oplus f_{n+1}$ and $T_r(f\oplus f_{n+1})$ with $W_j$ playing the role of $W_2$ in Lemma 2.1.2 and the statement follows by induction on n.

$\textbf {Hom}{\boldsymbol {(}\mathbf{W}\boldsymbol {,}\mathbf{W}_{\mathbf{n}\boldsymbol {+}\mathbf{1}}\boldsymbol {)}\boldsymbol {=}\mathbf{0}\boldsymbol {=} \textbf {Hom}\boldsymbol {(}\mathbf{W}_{\mathbf{n}\boldsymbol {+}\mathbf{1}}\boldsymbol {,}\mathbf{W}\boldsymbol {)}}$ . Suppose $f_{n+1}\neq 0$ , $\operatorname {\mathrm {Hom}}(W,W_{n+1})=0$ , and $\operatorname {\mathrm {Hom}}(W_{n+1},W)=0$ . Let $X_{V,W}$ be as in Notation 2.2.6. If $X_{V,W}$ has fewer than n indecomposable summands then we are done since $X_{V,W}\oplus W_{n+1}$ has n or fewer indecomposable summands and we may then apply the induction hypothesis.

Now suppose $X_{V,W}=W$ . Note $\operatorname {\mathrm {Hom}}(W,W_{n+1})=0$ but $f_i\neq 0$ for all $1\leq i\leq n+1$ . Thus the $W_i$ , for all $1\leq i \leq n+1$ , are totally ordered by y-coordinate and position (2 is greater than 1 and 4 which are greater than 3). Reindex the $W_i$ ’s such that $W_i < W_{i+1}$ in the total order and replace W with $\bigoplus _{i=1}^n W_i$ in the new index. We have the following distinguished triangles:

By our base case $U'$ has at most two indecomposable summands and by the rest of Theorem 1.2.6, the slopes of the line segments from $W_{n+1}$ to each summand of $U'$ is $\pm (1,1)$ . By [Reference Rock19, Lem. 5.2.9], $W_n$ can map to at most one of the indecomposable summands of $U'$ and that is the only possible summand that can map to $W_n[1]$ .

Thus W can map to at most one indecomposable summand of $U'$ . Then by induction E is determined by geometry and $f\circ h'[-1]$ . By the octahedral axiom, we obtain the distinguished triangles:

Note that if we apply the base case to the first distinguished triangle, this proves statement 2. We repeat the same argument for $T_r$ applied to each morphism and indecomposable. Thus, E is determined by geometry and $[ f\, f_{n+1}]^t$ . Furthermore, the cone has the desired number of indecomposable summands. The case where W is indecomposable is similar.

Proof. The proof is by induction on the number of indecomposable summands of V, using Lemma 2.2.7 as the base case. Assume the lemma holds for all positive integers $\leq n$ and let $V_i$ for $1\leq i\leq n+1$ be indecomposables and let $V=\bigoplus _{i=1}^n V_i$ . Let $f:V\to W$ and $f_{n+1}:V_{n+1}\to W$ be nontrivial morphisms. Consider the following distinguished triangles:

Applying the octahedral axiom, we obtain the following distinguished triangles:

By the induction hypothesis, E is determined by geometry and $g_{n+1}\circ f$ .

Proof. Suppose that the slope of the line segment from $\mathcal G V$ to $\mathcal G W$ in the AR-space of $\mathcal D^b (A_{{\mathbb {R}},S'})$ is greater than $(1,1)$ or less than $-(1,1)$ . Then $\operatorname {\mathrm {Hom}}_{\mathcal D^b(A_{{\mathbb {R}},S'})}(\mathcal G V,\mathcal G W)=0$ by [Reference Rock19, Lem. 5.2.9]. This contradicts that $\mathcal G$ is an equivalence of categories.

Now suppose the slope is less than $(1,1)$ and greater than $-(1,1)$ . Then, by [Reference Rock19, Lem. 5.2.9], $\operatorname {\mathrm {Hom}}_{\mathcal D^b(A_{{\mathbb {R}},S'})}(\mathcal G W,\mathcal G V[1])\neq 0$ . By Theorem 1.2.6, we have a distinguished triangle $\mathcal G V \to U\oplus U' \to \mathcal G W\to \mathcal G V [1]$ and any morphism $f:\mathcal G V\to \mathcal G W$ factors through either U or $U'$ but not both. This contradicts the fact that any morphism from V to W in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ that factors through some X and $X'$ factors through both either as $V\to X\to X'\to W$ or $V\to X'\to X\to W$ .

Proof. By using the translation T in Definition 2.3.2 for any indecomposable V in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ , $\mathcal F V$ is well-defined and indeed $\mathcal F$ induces a bijection on indecomposable objects. Furthermore, since ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ and $\mathcal D^b(A_{{\mathbb {R}},S'})$ are both Krull–Schmidt, we have a bijection on all objects. By definition of $\mathcal F(f)$ given above, $\mathcal F$ gives an isomorphism

$$ \begin{align*} \mathcal F:\operatorname{\mathrm{Hom}}_{{\mathcal{D}^b(A_{{\mathbb{R}},S})}}(V,W)\cong \operatorname{\mathrm{Hom}}_{\mathcal D^b(A_{{\mathbb{R}},S'}}(\mathcal F(V),\mathcal F(W)) \end{align*} $$

for all $V,W$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . Therefore, $\mathcal F$ is an additive equivalence of categories.

We have already pointed out that $\mathcal F(V[1])=(\mathcal FV)[1]$ on objects since $\iota $ and $T_r$ commute with shift on objects.

Proof. First, suppose W is indecomposable.

By Theorem 1.2.6, $U\xrightarrow f V\xrightarrow g W$ is a rectangle or almost complete rectangle as shown on the right. Each hom set, $\operatorname {\mathrm {Hom}}(U,V_1)$ , $\operatorname {\mathrm {Hom}}(U,V_2)$ , and so on, is equal to k. So morphisms are given by scalars $a,b,c,d,e\in k$ . These scalars must satisfy $ad-bc=0$ . By an automorphism of the distinguished triangle we may assume $(a,b,c,d)=(1,1,1,-1)$ .

Take the morphism $\mathcal F W\to \mathcal F U[1]$ and complete it to a distinguished triangle in $\mathcal D^b(A_{{\mathbb {R}},S'})$ : $\mathcal FU\to X\to \mathcal FW\to \mathcal FU[1]$ . By Theorem 1.2.6, $\mathcal FU\to X\to \mathcal FW$ is a rectangle or almost complete rectangle. This forces $X\cong \mathcal FV$ . By the same argument as above, we may adjust the scalars corresponding to $a,b,c,d$ in the figure above and we obtain a distinguished triangle

$$ \begin{align*} \mathcal FU\xrightarrow{\mathcal F(f)} \mathcal FV\xrightarrow{\mathcal F(g)} \mathcal FW\xrightarrow \varphi \mathcal FU[1]. \end{align*} $$

Since $\operatorname {\mathrm {Hom}}(\mathcal FW,\mathcal FU[1])\cong k$ , $\varphi $ must be a scalar multiple of $\mathcal F(h)$ , say $\varphi =c_V\mathcal F(h)$ .

We claim that the scalar $c_V$ is independent of V. This is easiest to see in the case when U is in (micro) position 1. In that case, there is a universal choice of V, call it $\widetilde V$ , making $U\to \widetilde V\to \widetilde W$ and $\mathcal FU\to \mathcal F\widetilde V\to \mathcal F\widetilde W$ into almost split sequences. Any other distinguished triangle $U\to V\to W\to U[1]$ or $\mathcal FU\to \mathcal FV\to W' \to \mathcal FU[1]$ is a pushout of these universal sequences as shown on the left side of the following diagram. Since W and $W'$ are determined by geometry, we must have $W'\cong \mathcal FW$ as shown on the right side.

Then $h'$ must be equal to $c_{\widetilde V}\mathcal Fh$ . This is clear when W is indecomposable. Otherwise we take a component $W_i$ and complete $h_i:W_i\to U[1]$ to a distinguished triangle $U\to V_i\to W_i\to U[1]$ and we see that the ith component of $h'$ must be $c_{V_i}\mathcal Fh_i$ which is equal to $c_{\widetilde V}\mathcal Fh_i$ by the easy case. So, the lemma holds with $\Psi _U$ being multiplication by $c_{\widetilde V}$ .

When U is not in position 1, we take a “virtual almost split triangle” which is a family of distinguished triangles $U\to V_\varepsilon \to W_\varepsilon \to U[1]$ for $\varepsilon>0$ with $W_\varepsilon $ converging to U as $\varepsilon \to 0$ . Then any distinguished triangle $U\to V\to W\to U[1]$ is a pushout of such a triangle for sufficiently small $\varepsilon $ and we can repeat the above argument. To show that $\Psi _U$ is a natural isomorphism, we need to show that $c_V$ is independent of U. To do this, take $U,U'$ indecomposable, complete $U\to U'$ to a distinguished triangle, then rotate the triangle $\mathcal FU\to \mathcal FU'\to \mathcal FW\to \mathcal FU[1]$ to see that the scalar for U is the same as that for $U'$ .

Proof. If (c) holds then by Proposition [Reference Igusa, Rock and Todorov14, Prop. 3.2.1] the representation categories are equivalent as abelian categories and so the derived categories are also be equivalent as triangulated categories. If (a) or (b) holds then the categories are already equivalent as additive categories, by Lemma 2.3.3. It remain to show that $\mathcal F$ is a equivalence of triangulated categories, that is, we need to extend Lemma 2.3.4 to the case when U is not indecomposable. So, let $U\stackrel {f}{\to } V\stackrel {g}{\to } W \stackrel {h}{\to } U[1] $ be distinguished in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ with $U=\bigoplus U_i$ . Then [Reference Neeman18, Def. 2.1.1] we need to show that

(*) $$ \begin{align} \mathcal FU\xrightarrow{\mathcal F(f)} \mathcal FV\xrightarrow{\mathcal F(g)} \mathcal FW\xrightarrow{\Psi\circ\mathcal F(h)} \mathcal FU[1] \end{align} $$

is distinguished in $\mathcal D^b(A_{{\mathbb {R}},S'})$ where $\Psi :\mathcal FU[1]\cong \mathcal FU[1]$ is multiplication by a fixed constant c. By Lemma 2.3.4, the analogous statement holds for $U_i$ : completing the ith component $h_i:W\to U_i[1]$ to a distinguished triangle gives $U_i\stackrel {f_i}{\to } V_i\stackrel {g_i}{\to } W \stackrel {h_i}{\to } U_i[1] $ . Lemma 2.3.4 gives the distinguished triangle $\mathcal FU_i\stackrel {\mathcal F f_i}{\to } \mathcal FV_i\stackrel {\mathcal Fg_i}{\to } W \stackrel {\Psi \circ \mathcal Fh_i}{\to } \mathcal FU_i[1] $ . This gives a morphism of distinguished triangles:

However, $U\stackrel {f}{\to } V\stackrel {g}{\to } W$ is the pull back of $U\to \bigoplus V_i\to \bigoplus W$ along the diagonal $\Delta :W\to \bigoplus W$ . Since $\mathcal F$ is an equivalence of categories, $V'=\mathcal FV$ is the pull back in the diagram above and we conclude that $(\ast )$ is a distinguished triangle. So, $(\mathcal F,\Psi )$ is an equivalence of triangulated categories.

Since every continuous quiver of type A with finitely many sinks and sources is derived equivalent we show that classes (b) and (c) are disjoint from (a) using the straight descending orientation. Afterwards, we show (b) and (c) are also disjoint.

Let $A_{{\mathbb {R}},S}$ be a continuous quiver of type A with infinitely many sinks and sources. Let $A_{{\mathbb {R}}}$ be the straight descending continuous quiver of type A. Let $\mathcal P$ be the set of indecomposable projectives of $\text {rep}_k(A_{{\mathbb {R}}})$ included as indecomposables in degree 0 in $\mathcal D^b(A_{{\mathbb {R}}})$ .

For contradiction, suppose there is an equivalence of categories $\mathcal G:\mathcal D^b(A_{{\mathbb {R}}})\to {\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . By Lemma 2.3.1, we know that for any pair $P,P'\in \mathcal P$ , the slope of any line segment from $\mathcal G P$ to $\mathcal G P'$ (switching roles if necessary) is $\pm (1,1)$ . Since $\mathcal G$ is an equivalence we know that if there exists V in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ such that $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\mathcal G P,V)\cong \operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(V,\mathcal G P')\cong k$ for $P,P'\in \mathcal P$ then $V \cong \mathcal G P''$ for some $P''\in \mathcal P$ .

Note that $\mathcal P$ induces an almost-complete line segment in the AR-space of $\mathcal D^b (A_{{\mathbb {R}}})$ . So, there does not exist an indecomposable object V in $\mathcal D^b(A_{{\mathbb {R}}})$ such that (i) V is not isomorphic to an element of $\mathcal P$ and (ii) $\operatorname {\mathrm {Hom}}_{\mathcal D^b(A_{{\mathbb {R}}})}(V,P)\cong k$ for all $P\in \mathcal P$ . We also know that if $\operatorname {\mathrm {Hom}}_{\mathcal D^b(A_{{\mathbb {R}}})}(P_{+\infty },U)\cong \operatorname {\mathrm {Hom}}_{\mathcal D^b(A_{{\mathbb {R}}})}(P_{+\infty },U') \cong k$ then $\operatorname {\mathrm {Hom}}_{\mathcal D^b(A_{{\mathbb {R}}})}(U,U')\cong k$ or $\operatorname {\mathrm {Hom}}_{\mathcal D^b(A_{{\mathbb {R}}})}(U',U)\cong k$ . Since $\mathcal G$ is an equivalence of categories these properties are preserved. This forces the image $\mathcal G\mathcal P$ to induce an almost complete line segment whose existing endpoint $\mathcal G P_{+\infty }$ is on the top or bottom boundary of the AR-space of ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . We describe how we arrive at a contradiction and include pictures after this paragraph. Since $A_{{\mathbb {R}},S}$ has infinitely many sinks and sources we must split the set $\mathcal G \mathcal P$ into (at least) two pieces since there is a missing graph of a $\lambda $ function from the image of $\mathop {\boldsymbol {\Gamma }}^b$ . The set $\mathop {\boldsymbol {\Gamma }}^b \mathcal G\mathcal P$ is homeomorphic to the disjoint union of (at least) two intervals. The set $(\mathop {\boldsymbol {\Gamma }}')^b \mathcal P$ is homeomorphic to one interval. We cannot reorder the elements in the almost-complete line segment and we cannot allow a strict inclusion of $\mathcal P$ into the almost-complete line segment in the AR-space of ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . This finally leads us to the contradiction. In Figure 7, we depict two of the issues with mapping the projectives by observing their image under $\mathop {\boldsymbol {\Gamma }}^b$ .

Figure 7 A visual depiction of the obstruction to a derived equivalence $\mathcal G:\mathcal D^b(A_{{\mathbb {R}}})\to {\mathcal {D}^b(A_{{\mathbb {R}},S})}$ for $S={\mathbb {N}}$ and $S={\mathbb {Z}}$ in the proof of Theorem 2.3.5.

Now, we show that (b) and (c) are also disjoint. Suppose $A_{{\mathbb {R}},S}$ has half bounded sinks and sources and $A_{{\mathbb {R}},S'}$ has unbounded sinks and sources on both sides. The image in ${\mathbb {R}}^2$ under $\mathop {\boldsymbol {\Gamma }}^b$ of an almost complete line segment in the AR-space of ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ has at most two path components. However, one may construct an almost complete line segment in the AR-space of $\mathcal D^b(A_{{\mathbb {R}},S'})$ whose image in ${\mathbb {R}}^2$ under $(\mathop {\boldsymbol {\Gamma }}')^b$ has three path components. Knowing this we apply a similar argument as before and see that an equivalence of categories cannot exist between ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ and $\mathcal D^b(A_{{\mathbb {R}},S'})$ .

Proof. Suppose $\operatorname {\mathrm {Hom}}_{{\mathcal {C}(A_{{\mathbb {R}},S})}}(V,W)\neq 0$ . We know

$$ \begin{align*}\operatorname{\mathrm{Hom}}_{{\mathcal{C}(A_{{\mathbb{R}},S})}}(V,W)= \bigoplus_{n\in{\mathbb{Z}}}\operatorname{\mathrm{Hom}}_{{\mathcal{D}^b(A_{{\mathbb{R}},S})}}(V,W[n])\end{align*} $$

and so there exists at least one n such that $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(V,W[n])\cong k$ . By [Reference Rock19, Prop. 4.4.2 and Lem. 5.2.9], if $V\not \cong W$ we know that $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(V,W[m])=0$ for $m>n$ and $m<n$ . If $V\cong W$ we note that, since $\operatorname {\mathrm {Ext}}^n_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(V,V)=0$ for $n\neq 0$ ,

$$ \begin{align*} \operatorname{\mathrm{Hom}}_{{\mathcal{C}(A_{{\mathbb{R}},S})}}(V,W)=\bigoplus_{n\in {\mathbb{Z}}}\operatorname{\mathrm{Hom}}_{{\mathcal{D}^b(A_{{\mathbb{R}},S})}}(V,V[n])=\operatorname{\mathrm{Hom}}_{{\mathcal{D}^b(A_{{\mathbb{R}},S})}}(V,V)\cong k. \end{align*} $$

Proof. When V and W are in $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ or ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ this follows from [Reference Rock19, Sec. 4.2 and Lem. 5.2.9]. Now suppose V and W are in ${\mathcal {C}(A_{{\mathbb {R}},S})}$ and $f\neq 0$ . Since $\operatorname {\mathrm {Hom}}_{{\mathcal {C}(A_{{\mathbb {R}},S})}}(V,W)=\bigoplus _{n\in {\mathbb {Z}}} \operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(V,W[n])$ some shift of W must be isomorphic to V in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . Then $V\cong W$ in ${\mathcal {C}(A_{{\mathbb {R}},S})}$ . Then f is an isomorphism since $\operatorname {\mathrm {Hom}}(V,V)\cong k$ and $f\neq 0$ .

Proof. If V is degenerate then $\mathop {\boldsymbol {\Gamma }}^b V=(x,\pm \frac \pi 2)$ . So, $\mathop {\boldsymbol {\Gamma }}^b V[1]=(x+\pi ,\mp \frac \pi 2)$ . So, $V[1]$ is degenerate if and only if V is degenerate. So, $\mathcal N$ satisfies the first condition to be a null system. By Proposition 3.2.2, it satisfies the second condition. Hence, $\mathcal N$ is a null system. It follows from Proposition. 3.2.4 that $\mathcal NQ$ admists a left and right calculus of fractions.

Proof. In ${\mathcal {C}(A_{{\mathbb {R}},S})}$ , $W\cong W[n]$ for all $n\in {\mathbb {Z}}$ . Thus, by replacing W with $W[n]$ , and choosing the appropriate n, we can prove (2) for $\mathop {\boldsymbol {\Gamma }}^b V= \mathop {\boldsymbol {\Gamma }}^b W$ .

First suppose $\mathop {\boldsymbol {\Gamma }}^b V =\mathop {\boldsymbol {\Gamma }}^b W$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . Without loss of generality, first suppose $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(V,W)\cong k$ . By Theorem 1.2.6 any nonzero indecomposable summands of U in the distinguished triangle $W\to U\to V[1]\to W[1]$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ are degenerate. Then U is degenerate and so $V\cong W$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ . Furthermore, if V and W are instead in ${\mathcal {C}(A_{{\mathbb {R}},S})}$ then U is still degenerate and so $V\cong W$ in ${\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ .

Now suppose $V\cong W$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ . Then U in the distinguished triangle $W\to U\to V[1]\to W[1]$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ is degenerate. We use Theorem 1.2.6 again and see $\mathop {\boldsymbol {\Gamma }}^b V=\mathop {\boldsymbol {\Gamma }}^b W$ . Now suppose $V\cong W$ in ${\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ ; so U is degenerate in the distinguished triangle $W\to U\to V\to W[1]$ in ${\mathcal {C}(A_{{\mathbb {R}},S})}$ . As an orbit category by almost-shift there are choices of lifts W, U, and $V[1]$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ that yield the triangle in ${\mathcal {C}(A_{{\mathbb {R}},S})}$ . Again we apply Theorem 1.2.6 and see $\mathop {\boldsymbol {\Gamma }}^b V=\mathop {\boldsymbol {\Gamma }}^b W$ .

Proof. Suppose $\operatorname {\mathrm {Hom}}_{\mathcal D}(V_1,W)=0$ . Consider a roof in $\mathcal D[(\mathcal NQ)^{-1}]$ . Since $f\in \mathcal NQ$ , we know (shifting if necessary in ${\mathcal {C}(A_{{\mathbb {R}},S})}$ ) that $\mathop {\boldsymbol {\Gamma }}^b V=\mathop {\boldsymbol {\Gamma }}^b U$ . Then $\mathop {\boldsymbol {\Gamma }}^b V_1=\mathop {\boldsymbol {\Gamma }}^b U=\mathop {\boldsymbol {\Gamma }}^b V$ and so $\operatorname {\mathrm {Hom}}_{\mathcal D}(V_1,U)\cong k\cong \operatorname {\mathrm {Hom}}_{\mathcal D}(V_1,V)$ .

Let $s:V_1\to U$ be a nontrivial morphism. We then have the commutative diagram in $\mathcal D$ on the right. Since $f\circ s\in \mathcal NQ$ , we see that these two roofs are equivalent in $\mathcal D[(\mathcal NQ)^{-1}]$ . Denote the bottom roof by $\underline {f}'$ .

If $g=0$ then $\underline {f}$ was 0 all along. If $g\neq 0$ we have $g\circ s=0$ but $f\circ s\in \mathcal NQ$ and so $\underline {f}'=0$ . Thus, $\underline {f}$ must be 0 and so $\operatorname {\mathrm {Hom}}_{\mathcal D[(\mathcal NQ)^{-1}]}(V,W)=0$ .

Now suppose $\operatorname {\mathrm {Hom}}_{\mathcal D}(V_1,W)\neq 0$ . Choose nonzero morphisms $f:V_1\to V$ and $g:V_1\to W$ . Then, we have the roof . For contradiction, suppose $\underline {f}=0$ .

Then there is a roof

where $g'=0$ and the commuting diagram in $\mathcal D$ shown on the right. However, this means (up to shifting in ${\mathcal {C}(A_{{\mathbb {R}},S})}$ ) $\mathop {\boldsymbol {\Gamma }}^b \tilde {U}=\mathop {\boldsymbol {\Gamma }}^b V_1$ and so $\tilde {U}\cong V_1$ , a contradiction as then the right side of the diagram would not commute.

Therefore, there exists a nonzero $\underline {f}\in \operatorname {\mathrm {Hom}}_{\mathcal D[(\mathcal NQ)^{-1}]}(V,W)$ .

Proof. Let V and W be indecomposable in $\mathcal D[\mathcal N Q^{-1}]$ so that $\operatorname {\mathrm {Hom}}_{\mathcal D[\mathcal N Q^{-1}]}(V,W)\neq 0$ . Let $V_1$ be an indecomposable in $\mathcal D$ such that $\mathop {\boldsymbol {\Gamma }}^b V_1=\mathop {\boldsymbol {\Gamma }}^b V$ and the position of $V_1$ is 1. By Lemmas 3.2.8 and 3.2.9, we know $V\cong _{\mathcal D[(\mathcal NQ)^{-1}]} V_1$ and $\operatorname {\mathrm {Hom}}_{\mathcal D}(V_1,W)\neq 0$ . We show that $\operatorname {\mathrm {Hom}}_{\mathcal D[(\mathcal NQ)^{-1}]}(V,W) \cong \operatorname {\mathrm {Hom}}_{\mathcal D}(V_1,W) $ by defining two maps

Fix a nonzero morphism $f':V_1\to V$ in $\mathcal D$ . As we saw in the proof of Lemma 3.2.9 every nonzero morphism $V\to W$ in $\mathcal D[(\mathcal NQ)^{-1}]$ is equivalent to a roof whose middle term is $V_1$ . For contradiction, suppose $\underline {f}=0$ . Then there is a roof in $\mathcal D[(\mathcal NQ)^{-1}]$ . Since $\operatorname {\mathrm {Hom}}_{\mathcal D}(V_1,V)\cong k$ , there is a unique $s:V_1\to V_1$ in $\mathcal D$ such that $f\circ s= f'$ . Similarly, there is a unique $g':V_1\to W$ in $\mathcal D$ such that $g'=g\circ s$ . So we set $\Phi (\underline {f})=g'$ .

Let $g':V_1\to W$ be a morphism in $\mathcal D$ . Then there is a roof in $\mathcal D[(\mathcal NQ)^{-1}]$ . So we set $\Psi (g')=\underline {f}'$ .

Note that $\Psi \Phi (\underline {f})=\underline {f}'$ but the diagram on the right commutes in $\mathcal D$ . Thus, $\underline {f}'=\underline {f}$ in $\mathcal D[(\mathcal NQ)^{-1}]$ and so $\Psi \Phi (\underline {f})=\underline {f}$ . We see $\Phi \Psi (g') = g'$ since in this case the s from our definition of $\Phi ( \Psi (g) )$ is the identity. Therefore, $\Phi = \Psi ^{-1}$ and $\Psi = \Phi ^{-1}$ .

Finally, it is straightforward to check that $\Phi $ and $\Psi $ preserve addition and send 0 to 0. Thus, we have the desired isomorphism and the proposition holds.

Proof. One quickly verifies that G induces a bijection on isomorphism classes of objects and bijections on Hom spaces. It remains to show that cones in distinguished triangles are taken to cones in distinguished triangles.

Let $V \to U \to W\to V[1]$ be a distinguished triangle in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ such that V, U, and W are all nonzero and distinct and V and W are indecomposable. Then this comes from a triangle $\tilde {V}\to \tilde {U}\to \tilde {W}\to \tilde {V}[1]$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ and so $U=U_1\oplus U_2$ where $U_1$ and $U_2$ are indecomposable and at most one is 0. Furthermore, $\tilde {V}$ and $\tilde {W}$ are indecomposable. Then by Theorem 1.2.6, there is a rectangle or almost complete rectangle in the AR-space of ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ whose sides have slopes $\pm (1,1)$ and whose left and right corner are $\tilde {V}$ and $\tilde {W}$ , respectively. Thus, $\tilde {U}$ has at most two indecomposable summands.

Since U is not 0, $\mathop {\boldsymbol {\Gamma }}^b \tilde {V}, \mathop {\boldsymbol {\Gamma }}^b\tilde {W}$ , and $\mathop {\boldsymbol {\Gamma }}^b \tilde {U}$ form the corners of a rectangle in ${\mathbb {R}}^2$ whose left and right corners are $\mathop {\boldsymbol {\Gamma }}^b \tilde {V}$ and $\mathop {\boldsymbol {\Gamma }}^b\tilde {W}$ , respectively. Since $U\neq 0$ no more than one indecomposable summand of $\tilde {U}$ may be sent to ${\mathbb {R}}\times (-\frac {\pi }{2},\frac {\pi }{2})$ by $\mathop {\boldsymbol {\Gamma }}^b$ .

We give the coordinates of the image of $\mathop {\boldsymbol {\Gamma }}^b$ of the lifts of each indecomposable and compute G on each of our indecomposables below. First, a few notes. If one of $\tilde {U}_1$ or $\tilde {U}_2$ is 0 then there is no $\mathop {\boldsymbol {\Gamma }}^b$ of that indecomposable. If one of $U_1$ or $U_2$ is 0 then G of that indecomposable will be 0. However, we compute $\mathop {\boldsymbol {\Gamma }}^b$ and G for both possibilities in each case for when those situations arise.

$$ \begin{align*} {\mathop{\boldsymbol{\Gamma}}}^b \tilde{V}\,& =(x,y), & GV\, &= M(x-y,x+y), \\ {\mathop{\boldsymbol{\Gamma}}}^b \tilde{U}_1& =(x+\alpha, y-\alpha), &GU_1 &= M(x-y+2\alpha,x+y), \\ {\mathop{\boldsymbol{\Gamma}}}^b \tilde{U}_2& =(x+\beta, y+\beta), &GU_2 &= M(x-y,x+y+2\beta), \\ {\mathop{\boldsymbol{\Gamma}}}^b \tilde{W}&= (x+\alpha+\beta,y-\alpha+\beta), &GW &= M(x-y+2\alpha,x+y+2\beta). \end{align*} $$

By the description in [Reference Igusa and Todorov15, Sec. 2.5] the four indecomposables in the image of G also form a distinguished triangle. In particular, if $\alpha $ or $\beta $ are 0 then the distinguished triangles in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ and $\mathcal D_\pi $ are split/trivial. With the same techniques used to prove Theorems 2.2.9 and 2.3.5, we see that G takes cones in distinguished triangles to cones in distinguished triangles.

Proof. Let $\mathcal C$ be the orbit category of ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ via doubling and almost-shift as in Definitions 3.1.1 and 3.1.2. Since ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ is triangulated equivalent to $\mathcal D_\pi $ , by Theorem 3.3.3, we see there must be a triangulated equivalence $H_2:\mathcal C\to \mathcal C_\pi $ . We define a triangulated equivalence $H_1:{\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}] \to \mathcal C$ . Afterwards we let $H = H_2 \circ H_1$ .

Since $\mathcal C$ is an orbit category we choose our fundamental domain. We choose those indecomposables V that come from an indecomposable $\tilde {V}$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ such that $(\alpha ,\beta )=\mathop {\boldsymbol {\Gamma }}^b V$ satisfy $ -\frac {\pi }{2} < \beta < \frac {\pi }{2}$ and $ \beta \leq \alpha < \pi - \beta. $ This is precisely the image of the 0th degree indecomposables in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ , excluding the injective indecomposables from $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ .

Recall ${\mathcal {C}(A_{{\mathbb {R}},S})}$ has the same objects as ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ but different isomorphism classes and similarly for $\mathcal C$ and ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ , respectively. For each indecomposable V in ${\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ there exists a $\tilde {V}$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ in degree 0 such that after taking the orbit, $\tilde {V}$ is sent to V in the localization of ${\mathcal {C}(A_{{\mathbb {R}},S})}$ . We define $H_1 V$ to be the indecomposable in $\mathcal C$ that comes from an indecomposable $\hat {V}$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ that also comes from $\tilde {V}$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ .

Let V and W be indecomposables in ${\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ . We show $\operatorname {\mathrm {Hom}}_{{\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]}(V,W)\cong k$ if and only if $\operatorname {\mathrm {Hom}}_{\mathcal C}(H_1 V, H_1 W)\cong k$ and similarly for 0 hom spaces. First, there are $\tilde {V}$ and $\tilde {W}$ in degree 0 in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ that are sent to V and W after taking the orbit and localization. Then either $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V},\tilde {W})\cong k$ or $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V},\tilde {W}[1])\cong k$ . Let $\tilde {V}_1$ be an indecomposable in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ such that $\mathop {\boldsymbol {\Gamma }}^b \tilde {V}_1=\mathop {\boldsymbol {\Gamma }}^b \tilde {V}$ and $\tilde {V}_1$ has position 1. If $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V}_1,\tilde {W})$ and $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V}_1,\tilde {W}[1])$ were both 0 then $\operatorname {\mathrm {Hom}}_{{\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]}(V,W)$ would be 0. Since this is not the case, $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V}_1,\tilde {W})\cong k$ or $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V}_1,\tilde {W}[1])\cong k$ . Then $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]}(\hat {V},\hat {W})\cong k$ or $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]}(\hat {V},\hat {W}[1])\cong k$ . In either case $\operatorname {\mathrm {Hom}}_{\mathcal C}(H_1 V,H_1 W)\cong k$ . In the case that $\operatorname {\mathrm {Hom}}_{{\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]}(V,W)=0$ we have $\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V}_1,\tilde {W})=0=\operatorname {\mathrm {Hom}}_{{\mathcal {D}^b(A_{{\mathbb {R}},S})}}(\tilde {V}_1,\tilde {W}[1]$ and so $\operatorname {\mathrm {Hom}}_{\mathcal C}(H_1 V,H_1 W)=0$ as well.

Now, as in Definition 2.3.2 for our triangulated equivalence of derived categories for different continuous quivers of type A, we can choose representatives from each isomorphism class of indecomposables and fix isomorphisms between indecomposables and their respective representatives. With such a construction we set $H_1 (\operatorname {\mathrm {Hom}}_{{\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]}(V,W)) := \operatorname {\mathrm {Hom}}_{\mathcal C}(H_1 V,H_1 W)$ for each pair of representatives V and W in ${\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ . This gives us an equivalence of categories but must still check triangles.

However, each distinguished triangle $U\to V\to W\to U[1]$ in ${\mathcal {C}(A_{{\mathbb {R}},S})}[(\mathcal NQ)^{-1}]$ comes from a triangle $\tilde {U}\to \tilde {V}\to \tilde {W}\to \tilde {U}[1]$ in the doubling of ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . But after taking localization and then the orbit, the image of $\tilde {U}\to \tilde {V}\to \tilde {W}\to \tilde {U}$ is a distinguished triangle in $\mathcal C$ . This is, by definition, precisely the image of $U\to V\to W\to U[1]$ under $H_1$ . Thus, $H_1$ is a triangulated equivalence and so is $H=H_2\circ H_1$ .

Proof. We know $\{Y,X\}$ is not $\mathbf P$ -compatible but $T=(T'\setminus \{Y\})\cup \{X\}$ is a $\mathbf P$ -cluster.

Proof. Let $P_V$ , $P_W$ , $Q_V$ , and $Q_W$ be the projectives in $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ whose image is in the isomorphism classes indicated by the g-vectors. If there is an extension $V\hookrightarrow U\twoheadrightarrow W$ in $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ then there is a nontrivial morphism $W\to V[1]$ in ${\mathcal {D}^b(A_{{\mathbb {R}},S})}$ . By Theorem 1.1.9 there is, up to scaling and isomorphism, a unique extension. This extension exists because there is a morphism $Q_W \to P_V$ that does not factor through $Q_V\oplus P_W$ .

By the proof of [Reference Igusa, Rock and Todorov14, Prop. 3.2.4], we see this means there must be some indecomposable summand of $Q_W$ that maps to at least one indecomposable summand of $P_V$ but does not factor through $Q_V$ or $P_W$ . By Theorem 1.1.7, $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ is hereditary. So, $Q_W$ is a subrepresentation of $P_W$ and $Q_V$ is a subrepresentation of $P_V$ . Thus, $\langle [P_V]-[Q_V]\, ,\, [P_W]-[Q_W] \rangle <0$ .

If we start with incompatible g-vectors then we reverse the argument and see that, up to symmetry, there is a morphism $Q_W\to P_V$ that does not factor through $Q_V\oplus P_W$ . Thus there is an extension $V\hookrightarrow U\twoheadrightarrow W$ in $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ .

Proof. Without loss of generality suppose V is the left corner and W is the right corner of a rectangle or almost complete rectangle in the AR-space of $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ . Then by Theorem 1.2.6 there is a distinguished triangle $V\to U\to W\to V[1]$ . This corresponds to an extension $V\hookrightarrow U\twoheadrightarrow W$ in $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ and so by Proposition 4.2.4, we know $[P_V]-[Q_V]$ and $[P_W]-[Q_W]$ are not $\mathbf E$ -compatible.

Now suppose $[P_V]-[Q_V]$ and $[P_W]-[Q_W]$ are not $\mathbf E$ -compatible. Without loss of generality suppose $\langle [P_W]-[Q_W]\, ,\, [P_V]-[Q_V] \rangle < 0$ . Then in $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ there is an extension $V\hookrightarrow U\twoheadrightarrow W$ . Thus by Theorem 1.2.5, there is a rectangle or almost complete rectangle in the AR-space of $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ whose left corner is V and right corner is W.

Proof. We prove: if $\{X,U_1,U_2\}$ is not $\mathbf E$ -compatible then either $\{X,W\}$ or $\{X,V\}$ is not $\mathbf E$ -compatible. Thus, without loss of generality, we assume $\{X,U_1\}$ is not $\mathbf E$ -compatible. We need only to prove that $\{X,V\}$ or $\{X,W\}$ is not $\mathbf E$ -compatible.

Suppose one of $\{X,V\}$ or $\{X,W\}$ is $\mathbf E$ -compatible. We shall assume $\{X,V\}$ is $\mathbf E$ -compatible as the other assumption is symmetric. By Proposition 4.2.5, since $\{X,U_1\}$ is not $\mathbf E$ -compatible there is a (almost complete) rectangle in the AR-space of $\operatorname {\mathrm {rep}}_k(A_{{\mathbb {R}},S})$ whose left and right corners are X and $U_1$ (possibly not respectively).