数理解析研究所講究録
第 1795 巻 2012 年 154-162

154

THE NUMBER OF ARROWS IN THE QUIVER OF
TILTING MODULES OVER A PATH ALGEBRA OF TYPE
$A$

AND

$D$

RYOICHI KASE

Department of Pure and Applied Mathematics Graduate School ofInformation Science and Technology,Osaka University, Toyonaka, Osaka
560-0043, Japan
E-mail:r-kase@cr.math.sci.osaka-u.ac.jp

INTRODUCTION
In this report we use the following notations. Let be a finite dimensional
algebra over an algebraically closed field , and let mod-A be the categoryof finite dimensional right A-modules. For $M\in mod-A$ we denote by $pd_{A}M$
the projective dimension of $M$ , and by add $M$ the full subcategory of directsums of direct summands of $M$ . Let $Q=(Q_{0}, Q_{1})$ be a finite connected(resp. ) be the set of vertices
quiver without loops and cycles, and(resp.arrows) of (we use this notation for an arbitrary quiver). We denoteover , and by rep the category of finite
by $kQ$ the path algebra ofdimensional representations of the quiver which is category equivalent to$mod-kQ$ . For $M\in$ rep , denote by
the vector space of $M$ associatedthe linear map $M_{a}arrow M_{b}$ of $M$ . For a
to a vertex , and denote by
be the quiver obtained from by reversing all arrows
vertex of , letstarting at or ending at . A module $T\in$ mod-A is called a tilting module
provided the following three conditions are satisfied:(a) pd $T<\infty$ ,(b) $Ext^{i}(T, T)=0$ for all $i>0$ ,(c) there exists an exact an sequence$0arrow Aarrow T_{0}arrow T_{1}arrow\cdotsarrow T_{r}arrow 0(T_{i}\in$ add $T)$
$A$

$k$

$Q_{1}$$Q_{0}$

$Q$

$Q$

$k$

$Q$

$Q$

$Q$

$M_{a}$$M_{aarrow b}$

$a$

$a$

$Q$

$a$

$Q$

$\sigma_{a}Q$$a$in mod-A. In the hereditary case the tilting condition above is equivalent
to the following:(a) $Ext^{1}(T, T)=0$ ,(up to isomor(b) the number of indecomposable direct summands ofphism) is equal to the number of simple modules.
Then we determine the number of arrows of the tilting quiverof typeor $D$ . Note that the(c.f. section 1) for any Dynkin quiver
may be embeded into the exchange graph, orunderlying graph ofthe cluster complex, of the corresponding cluster algebra of finite $type:the$tilting modules of $kQ$ correspond to positive clusters (cf.[3] and [12]). The$T$$\vec{\mathcal{K}}(kQ)$$Q$$\tilde{\mathcal{K}}(kQ)$

$A$

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RYOICHI KASE

number of positive clusters when the orientation is alternating is given in[6, prop. 3.9]. However, the number of edges of this subdiagram of positive
clusters is not known in the cluster tilting theory. Note also that if we
consider the similar problem for the exchange graph, it is not interesting,because the number of edges is(the number of vertices), and the numberof vertices is given in [6, prop. 3.8]. The following is known.[6, prop. 3.9].
$\frac{n}{2}\cross$$\#\vec{\mathcal{K}}(kQ)_{0}=\{\begin{array}{ll}\frac{1}{n+1}[Matrix] if Q is a Dynkin quiver of type A_{n},\frac{3n-4}{2n}(^{2(n-1)}n-1) if Q is a Dynkin quiver of type D_{n}.\end{array}$

The main result is as follows.

Theorem 0.1. (1) $:LetQ$ be a Dynkin quiver. Then
dent of the $0$ entation.

$\#\vec{\mathcal{K}}(kQ)_{1}$is indepen-(2) :$\#\vec{\mathcal{K}}(kQ)_{1}=\{\begin{array}{ll}[Matrix] if Q is a Dynkin quiver of type A_{n},(3n-4)(^{2(n-2)}n-3) if Q is a Dynkin quiver of type D_{n}.\end{array}$

1. PRELIMINARIES, and recall its some properIn this section we define a tilting quiverties. First, for a tilting module , we define the right perpendicular category$\vec{\mathcal{K}}(A)$$T$$T^{\perp}=\{X\in mod- A |Ext_{A}^{>0}(T, X)=0\}$.

Lemma 1.1. (cf. [9, lemma 2.1 $(a)]$ ) For two tilting modules $T,$ $T’$ thefollowing conditions are equivalent:(1),$:T\in T’\perp$(2).
$:T^{\perp}\subset T’\perp$

Denote by Tilt $(A)$ the set of isomorphic classes of basic tilting modulesof mod-A.

Definition 1.2. We define a partial order on Tilt $(A)$ by$T\leq T’\Leftrightarrow T^{\perp}\ f\subset T’\perp\Leftrightarrow T\in T’\perp$

for

$T,$

$T’\in Tilt(A)$,.

1.1. A tilting quiver.
is defined as
Definition 1.3. The tilting quiver
follows.
(1),(2) $T’arrow T$ in, for $T,$ $T’\in Tilt(A)$ , if $T’=M\oplus X,$ $T=M\oplus Y$ with
$X,$ $Y\in$ ind
and there is a non-split short exact sequence$\vec{\mathcal{K}}(A)=(\vec{\mathcal{K}}(A)_{0},\vec{\mathcal{K}}(A)_{1})$$\vec{\mathcal{K}}(A)_{0}=Tilt(A)$$\vec{\mathcal{K}}(A)$$A$$0arrow Xarrow\overline{M}arrow Yarrow 0$

with$\overline{M}\in$

add $M$ .

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THE NUMBER OF ARROWS IN THE QUIVER OF TILTING MODULES

is the Hasse-diagram of (Tilt $(A),$Theorem 1.4. (cf. [8, thm 2.1])( $i.e$ . ifand $T\geq T’’\geq T’$ then $T”=T$ or $T”=T’$ ).
$\vec{\mathcal{K}}(A)$$\leq$$Tarrow T’\in\tilde{\mathcal{K}}(A)_{1}$$)$Proposition 1.5. ( $cf.[8$ , cor 2.2]) If$\vec{\mathcal{K}}(A)=C$$\vec{\mathcal{K}}(A)$.

Example 1.6.

$C$

, then$(A_{3})$$0arrow 0arrow 0$$Q$

has a finite component$0arrow 0arrow 0$$0\inftyarrow 0$$\nearrow^{o}\backslash$$OO\downarrow$$\vec{\mathcal{K}}(Q)$$\nwarrow_{O,1}\prime^{O}o$$0$1.2. A local structure of a tilting quiver. Let $Q=(Q_{0}, Q_{1})$ be a quiver
without loops and cycles and $A=kQ$ . For $T\in Tilt(A)$ , let$s(T)$$=$$\#\{T’\in Tilt(A)|Tarrow T’ in \mathcal{K}\vec{(}A)\}$$e(T)$$=$$\#\{T’\in Tilt(A)|T’arrow T in \mathcal{K}\vec{(}A)\}$and define $\delta(T)=s(T)+e(T)$ . Then the following is key proposition for
proof of main result.
Proposition 1.7. ( $cf.[10$ , prop 3.2]) $\delta(T)=n-\#\{a\in Q_{0}|(\underline{\dim}T)_{a}=1\}$ ,where $n=\# Q_{0}$ .

2. A THEOREM OF LADKANIIn this section, we review [11]. Let be a quiver without loops and cycles,
be a source of and $Q’=\sigma_{x}Q$ . Let Tilt $(Q)$ $:=Tilt(kQ)$ and define
$Q$

$x$

$Q$

Tilt

$(Q)^{x}$$:=\{T\in Tilt(Q)|S(x)|T\}$ ,where $S(x)$ is the simple module associated to .
$x$Definition 2.1. Let $(X, \leq x),(Y, \leq Y)$ be two posets and : $Xarrow Y$ anoforder-preserving function. Then we define two partial orders$xuY$ as follows.
$f$$\leq_{+}^{f},$$a\leq_{+}^{f}b\Leftrightarrow\{\begin{array}{ll}a\leq xb if a, b\in X,a\leq Yb if a, b\in Y,f(a)\leq Yb if a\in X and b\in Y.\end{array}$$a\leq xb$$ifa,b\in Xifa,b\in Y,$$a\leq^{\underline{f}}b\Leftrightarrow\{$’$a\leq Yb$$a\leq Yf(b)$if $a\in Y$ and $b\in X$ .

$\leq^{\underline{f}}$157
RYOICHI KASE

Define$j_{*}$: rep$(Q\backslash \{x\})arrow$rep$N_{a}$$(a\neq x)$$\oplus_{xarrow y}N(y)$$(a=x)$$(j_{*}N)_{a}=\{$$(a\neq x)$$N_{a}$

and

$Q$

,

$j_{*}’$: rep$(Q\backslash \{x\})arrow$rep$Q^{l}$

as follows$(j_{*}N)_{aarrow b}=\{\begin{array}{ll}N_{aarrow b} (a\neq x)(j_{*}N)_{x}^{projectim}arrow N_{b} (a=x)\end{array}$,$(j_{*}’N)_{aarrow b}=\{$$(j_{*}’N)_{a}=\{\begin{array}{ll}N_{aarrow b} (b\neq x)(j_{*}’N)_{a}^{injecti\sigman}arrow N_{x} (b=x)\end{array}$$(a=x)$$\oplus_{yarrow x}N(y)$Theorem 2.2.
$\iota_{x}:T\mapsto S(x)\oplus j_{*}T$and$\iota_{\acute{x}}:T’\mapsto S’(x)\oplus j_{*}’T’$

enduce the order-preserving functions
$($

Tilt

$(Q\backslash \{x\}),$$\leq)arrow(Tilt(Q), \leq)$$(Q\backslash \{x\}),$$\leq)arrow(Tilt(Q’), \leq)$

,

and
$($

Tilt

.

More over there is a commutative diagram of the posets

Tilt $(Q)\backslash Tilt(Q)^{x}arrow Tilt(Q’)\sim p_{x}\backslash Tilt(Q’)^{x}$with(Tilt $(Q),$$\leq$) $\simeq(Tilt(Q)\backslash Tilt(Q)^{x}uTilt(Q)^{x}, \leq^{\underline{f}})$ ,and(Tilt $(Q’),$In particular iforientation.

$Q$$\leq$) $\simeq(Tilt(Q’)\backslash Tilt(Q’)^{x}uTilt(Q’)^{x}, \leq_{+}^{f’})$ .

is a Dynkin quiver, then $\# Tilt(Q)$ is independent of an

Remark 2.3. In [11] the partial order on Tilt $(A)$ is defined by$T\geq T’=T^{\perp}\subset T’\perp$(opposite to our definition).

3. MAIN RESULTS
in the caseIn this section we determine the number of arrows of$D$
is a Dynkin quiver of type or . Then, from the facts of section 2, we getfollowing lemma.
$\vec{\mathcal{K}}(Q)$

$A$

$Q$

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THE NUMBER OF ARROWS IN THE QUIVER OF TILTING MODULES

Lemma 3.1. If

is a sink then$x$$\{\alpha\in\vec{\mathcal{K}}(Q)_{1}|s(\alpha)\in Tilt(Q)^{x}, t(\alpha)\in Tilt(Q)\backslash Tilt(Q)^{x}\}Tilt(Q)^{x}\underline{1:1}$

If

$x$

.

is a source then$\{\alpha\in\vec{\mathcal{K}}(Q)_{1}|t(\alpha)\in Tilt(Q)^{x}, s(\alpha)\in Tilt(Q)\backslash Tilt(Q)^{x}\}Tilt(Q)^{x}\underline{1:1}$

Where, for$Tarrow\alpha T’,$$s(\alpha)=T$.

and $t(\alpha)=T’$ .

Corollary 3.2.
$\neq\vec{\mathcal{K}}(Q)_{1}=\#\vec{\mathcal{K}}(Q’)_{1}$In particular, if is a Dynkin quiver thenunderlying graph of .
$Q$.

$\#\vec{\mathcal{K}}(Q)_{1}$depends only on the$Q$

Proof. By Theorem2.2 and lemma3.1,#JC$(Q)_{1}$$\#\tilde{\mathcal{K}}(Q\backslash \{x\})_{1}+\#\vec{\mathcal{K}}(Tilt(Q)\backslash Tilt(Q)^{x})_{1}+\# Tilt(Q)^{x}$$=$$\#\vec{\mathcal{K}}(Q’)_{1}$$=$.
$\square$3.1. case$A$. In this subsection we consider the quiver,$(\vec{A}_{n}=)Q=^{12}0arrow 0arrow\cdotsarrow on$.

By Gabriel‘ theorem, ind $kQ=\{L(i,j)|0\leq i<j\leq n\}$ where$s$$L(i,j)=\{$$k$$(i<a\leq j),$$0$otherwise,$L(i,j)_{aarrow b}=\{\begin{array}{l}1 (i<a, b\leq j),0 otherwise.\end{array}$And$\tau L(i,j)=\{\begin{array}{ll}L(i+1,j+1) (j<n),0 (j=n),\end{array}$where$\tau$

is a Auslander-Reiten translation.

Definition 3.3. A pair of intervals $([i,j], [i’,j’])$ is compatible if$[i,j]\cap[i’,j’]=\emptyset$

or $[i,j]\subset[i’,j’]$ or $[i’,j’]\subset[i,j]$ .

Applying Auslander-Reiten duality,
DExt $(M, N)\cong Hom(N, \tau M)(D=Homk(-, k))$ ,we get the following lemma.
Lemma 3.4. We have$Ext(L(i,j), L(i’,j’))=0=Ext(L(i’,j’), L(i,j))$if and only if $([i,j], [i’,j’])$ is compatible.
Lemma 3.5. For any $T\in Tilt(Q)$ , we get $\delta(T)=n-1$ .

Now it is easy to check the number of arrows into$\frac{1}{2}\sum_{T\in Tilt(Q)}\delta(T)$Corollary 3.6.

$\#\vec{\mathcal{K}}(Q)_{1}=\frac{n-1}{2(n+1)}(\begin{array}{l}2nn\end{array})=(\begin{array}{l}2n-ln-2\end{array})$.

$\vec{\mathcal{K}}(Q)$, because it is equal

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RYOICHI KASE

3.2. case$D$. Through this subsection, we consider the quiver$Q=Q_{n}=0arrowarrow 12$.

$arrow o_{\backslash }n-\nu^{O}n^{+}$$on^{-}$

Then ind $kQ=\{L(a, b)|0\leq a<b\leq n-1\}$$\cup\{L^{\pm}(a, n)|0\leq a\leq n-1\}\cup\{M(a, b)|0\leq a<b\leq n-1\}$

where$L(a, b)_{i}$$=$$\{\begin{array}{l}k if a<i\leq b,0 otherwise,\end{array}$$L(a, b)_{iarrow j}$$=$$\{\begin{array}{l}1 if a<i<b,0 otherwise,\end{array}$$L(a, n)_{i}^{\pm}$$=$$\{\begin{array}{l}k if a<i\leq n-1 or i=n^{\pm},0 otherwise,\end{array}$$L(a, n)_{iarrow j}^{\pm}$$=$$\{\begin{array}{l}1 if a<i<n-1 or i=n-1,j=n^{\pm},0 otherwise,\end{array}$$M(a, b)_{i}$$=$$\{\begin{array}{ll}k if a<i\leq b or i=n^{\pm},k^{2} if b<i\leq n-1,0 otherwise,\end{array}$$M(a, b)_{iarrow j}$$=$$\{\begin{array}{ll}1 if a<i<b,[Matrix] if i=b,(1, 0) if i=n-1,j=n^{+},(0,1) if i=n-1,j=n^{-},[Matrix] if b<i<n-1,0 otherwise.\end{array}$Then$\tau L(a, b)=\{\begin{array}{ll}L(a+1, b+1) if b<n-1,M(0, a+1) if b=n-1,\end{array}$$\tau L^{+}(a, n)=L^{-}(a+1, n)$ ,$\tau L^{-}(a, n)=L^{+}(a+1, n)$ ,160THE NUMBER OF ARROWS IN THE QUIVER OF TILTING MODULES$\tau M(a, b)=\{\begin{array}{ll}M(a+1, b+1) if b<n-1,0 if b=n-1.\end{array}$

Then the $Ext=0$ conditions are as follows.

Lemma 3.7.
(1) $Ext(L(a, b), L(a’, b’))=0=Ext(L(a^{J}, b’), L(a, b))$$\Leftrightarrow([a, b], [a’, b’])$: compatible.

(2) $Ext(L(a, b), L^{\pm}(a’, n))=0=Ext(L^{\pm}(a’, n), L(a, b))$$\Leftrightarrow([a, b], [a’, n])$: compatible.

(3) $Ext(L(a, b), M(a’, b’))=0=Ext(M(a’, b’), L(a, b))$$\Leftrightarrow([a, b], [a’, n]),$$([a, b], [b’, n])$ : compatible.

(4) $Ext(M(a, b), L^{\pm}(a’, n))=0=Ext(L^{\pm}(a^{J}, n), M(a, b))$$\Leftrightarrow a\leq a’\leq b$.

(5) $Ext(L^{\pm}(a, n), L^{\pm}(a’, n))=0=Ext(L^{\pm}(a’, n), L^{\pm}(a, n))$ for all $a,$

$a’$

.

(6) $Ext(L^{+}(a, n), L^{-}(a’, n))=0=Ext(L^{-}(a’, n), L^{+}(a, n))$$\Leftrightarrow a=a^{J}$.

(7) $Ext(M(a, b), M(a’, b’))=0=Ext(M(a’, b’), M(a, b))$$\Leftrightarrow[a, b]\subset[a’, b’]$or $[a’, b’]\subset[a, b]$ .

Lemma 3.8. Let $T\in Tilt(Q)$ then $\delta(T)\geq n-1$ , and $\delta(T)=n-1$ if andonly if $L^{\pm}(0, n)|T$ and other indecomposable direct summands of havethe form $L(a, b)(0\leq a<b\leq n-1)$ . In particular,$T$.

$\#\{T\in Tilt(kQ)|\delta(T)=n-1\}=\frac{1}{n}(\begin{array}{ll}2(n -1)n -1\end{array})= \frac{1}{n-1}(\begin{array}{ll}2(n -1)n -2\end{array})$

Now we define subsets of Tilt $(Q)$ by$\mathcal{T}0$$\mathcal{T}_{1}$$\mathcal{T}_{2}$$:=$$\{T\in Tilt(Q)|\delta(T)=n+1\}$ ,$:=$$\{T\in Tilt(Q)|\delta(T)=n\}$ ,$:=$$\{T\in Tilt(Q)|\delta(T)=n-1\}$ .

The above lemma shows that$\neq\tau_{2}=\frac{1}{n}(^{2(n-1)}n-1)=\frac{1}{n-1}(\begin{array}{ll}2(n -1)n -2\end{array})$

Let us define the following subsets of$\mathcal{T}_{1}$:$\mathcal{A}_{i}:=\{T\in \mathcal{T}_{1}|(\underline{\dim}T)_{i}=1\}$

Then we get the following correspondences.

.

.

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RYOICHI KASE

Lemma 3.9.
(1):

(2) :

$rightarrow^{1:1}$$A_{\eta}\pm$$\underline{1:1}$$\mathcal{A}_{i}$Tilt.
Tilt $(\vec{A}_{i-1})\cross Tilt(Q_{n-1})(i\neq n^{\pm})$ .
$(\vec{A}_{n})\backslash Tilt(\vec{A}_{n-1})$

Remark 3.10. In (1) we identify $\{T’\in Tilt(\vec{A}_{n})|L(0, n-1)|T’\}$ with.
Tilt$(\vec{A}_{n-1})$

By lemma 3.9, we can calculate$\#\mathcal{T}_{1}$and$\#\mathcal{T}_{0}$.

Corollary 3.11. we have$\#\mathcal{T}_{1}=3(\begin{array}{ll}2(n -1)n -2\end{array}),$$\neq\tau_{0}=\frac{3(n-1)}{n+1}(\begin{array}{ll}2(n -1)n -2\end{array})$.

Theorem 3.12.
$\#\vec{\mathcal{K}}(Q)_{1}=(3n-1)(^{2(n-1)}n-2)$

Proof. In fact,$\#\vec{\mathcal{K}}(Q)_{1}$.

is equal to$\frac{1}{2}\{\frac{n-1}{n-1}(^{2(n-1)}n-2)+3n(\begin{array}{ll}2(n -1)n-2 \end{array})+3(n-1) (\begin{array}{ll}2(n -1)n-2 \end{array})\}$$=(3n-1)(\begin{array}{ll}2(n -1)n -2\end{array})$.

$\square$Example 3.13.

$(D_{4})$(5)$0\}\}$(6)

6
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$D_{n}$