On the complement of the dense orbit for a quiver of type $A$

A multisegment $M$ consists of a union of intervalls $[i,j]$ with $1\le i\le j\le t$, written as $M=\oplus [i,j{]}^{a_{i,j}}$ (since a multisegments represents an isomorphism class of representations we write ’direct sum’ instead of ’union’). The dimension vector of such a multisegemt is defined as

There are natural bijections between the multisegments of dimensions vector $d$, the isomorphism classes of representations of $A_t$, and the orbits of the $\text{Gl}\left(d\right)$–action on $R(Q,d)$. Moreover, for any dimension vector $d$ there exists a unique multisegment $M\left(d\right)$ corresponding to the dense orbit. This multisegment can be constructed recursively as follows: Define $a_{1,t}$ to be the minimum of the entries $d_i$ in $d$. Then we consider $d^1:=d-a_{1,t}(1,\dots ,1)$ and consider the longest interval $[i,j]$ in $d^1$ with minimal $i$. Then

is nonnegative for some maximal $a_{i,j}$ and we proceed with $d^2$ instead of $d^1$ in the same way. Eventually, we obtain a multisegment $M\left(d\right)$ with at most $t$ different direct summands. A second way to obtain this multisegment is described in in [H1], Section 8 (this is a similar, but not the same, construction as in [BHRR]) as follows: consider the unique diagram with $d_i$ vertices in the $i$th column and connect each vertex in the $i$th column and the $k$th row with the vertex in the $(i+1)$th column and the $k$th row (if it exists). Roughly one connects all neighboured vertices in the same row. The connected components of this diagram are the direct summands and this diagram represents the multisegment $M\left(d\right)$ (see Section 5 for examples).

Definition. A dimension vector $d$ is generic if $M\left(d\right)$ contains precisely $t$ pairwise different segments. A dimension vector is pure if $d_1=d_t$, $d_l\ge d_1=d_t$ for each $1\le l\le t$ and this condition holds recursively for each connected component in the support of $d(\ge a):=(max\{d_1-a,0\},\dots ,max\{d_t-a,0\left\}\right)$. Examples can be found in Section 5, see 5.1 and 5.4.

The category of finite dimensional representations of $A_t$ is a hereditary category and the Euler characteristic $⟨-,-⟩=\dim \mathrm{Hom}(-,-)-\dim \mathrm{Ext}(-,-)$, respectively the Hom- and Ext-spaces are

All this follows from direct calculations using a projective or an injective resolution

Proof. a) and the first claim of b) is a direct consequence from the formula for the extension groups above. The uniqueness in b) follows either directly from the construction, or since $R(Q,d)$ is irreducible (it can contain at most one dense orbit). To prove c) one uses that ${\mathrm{Ext}}^1$ is additive, thus there is at most one non–vanishing extension group. Finally, to prove d) we note that for $d=\underset{–}{dim}N$ we have $\dim \mathrm{End}(N,N)=\dim \mathrm{End}\left(M\right(d),M(d\left)\right)+1$ by a simple computation of the Euler characteristic $⟨M\left(d\right),M\left(d\right)⟩=⟨N,N⟩.$ Thus, the stabilizer of the orbit of $M\left(d\right)$ and $N$ differ by one and then the dimension of the orbits also differ by one. The closure of the orbit of $M\left(d\right)$ obviously contains the orbit of $N$. Now assume $M$ is a multisegmet as in the claim and it is neither generic nor almost generic. Take two direct summands $[a,b]$ and $[c,d]$ of $M$ with $\mathrm{Ext}\left(\right[a,b],[c,d\left]\right)\ne 0$. We define a new multisegment $M^{\prime }$ of the same dimension vector by deleting $[a,b]$ and $[c,d]$ and replacing it by $[c,b]\oplus [a,d]$ (if $c=b+1$ we replace it just by $[a,d]$). Then $M$ is in the closure of the orbit of $M^{\prime }$ and $M$ is almost generic precisely when $M^{\prime }$ is already generic. This in turn is equivalent to the second condition in d), proving one direction of the claim.
Now assume $N$ satisfies $\mathrm{Ext}(N,N)=k$. Then the closure of the orbit of $N$ is of codimension one in the space of all representations of dimensions vector $\underset{–}{dim}N$. Thus it is some irreducible component in the complement of the dense orbit. Now we add the remaining segments to $N$ so that we obtain $M=M^{\prime }\oplus N$. The multisegment $M^{\prime }$ is then also a direct summand of $M\left(d\right)$, since on can get $M\left(d\right)$ from $N\oplus M^{\prime }$ by extending only two segments in $N$.

Assume $M$ contains two indecomposable direct summands $[a,b]$ and $[c,d]$, both occuring with multiplicity at least two and $\mathrm{Ext}\left(\right[a,b],[c,d\left]\right)\ne 0$, then we can again (using extensions) construct an orbit that is not generic and contains $M$ in ist closure. Consequently, such an $M$ is not almost generic. $\square$

The proof also follows directly from Zwara’s result [Z] that the partial order of the Ext-degeneration and the partial order for the geometric degeneration coincide. In the proof above we only used the trivial direction.

To any representation $A$ of $Q$ one can associate the ranks of the compositions of the corresponding matrices. Consider $A=\left(A_{i,i+1}\right)\in R(Q,d)$. Then we define the rank triangle

Moreover, it is convenient to define the extended rank triangle with $r_{i,i}:=d_i$ and to define $r_{i,j}=0$ whenever $i\le 0$ or $j>t$. Obviously, we must have $r_{i,j}\left(A\right)\le r_{i,j}:=\text{min,}\{d_l\mid i\le l\le j\}$ and (using generic matrices) the set

is open and dense in $R(Q,d)$. In fact, the set $O\left(d\right)$ consists of all representations isomorphic to $M\left(d\right)$, since $r_{i,j}\left(M\right(d\left)\right)=r_{i,j}$ by construction.

We fix a dimension vector $d$ and consider any triangle $s=\left(s_{i,j}\right)$ of non-negative integers $s_{i,j}$ satisfying $s_{i,j}\le r_{i,j}$. Then

defines an open (possibly empty) subvariety $X_s^0$ in a closed, non-empty algebraic subvariety $X_s$ (not necessarily irreducible) of $R(Q,d)$. The rank triangles are partially ordered by $s\le u$ iff $u-s$ has only non-negative entries. It turns out that some of the $X_s$ are irreducible (we determine which ones) and the rank conditions are very useful for determining the components in the orbit closures. Moreover, one can reconstruct the multisegment $M$ from the rank condition $s$, where the orbit of $M$ is dense in $X_s$ with $s$ minimal: A direct sum $[i,j{]}^a$ is a direct summand of $M$ (with maximal possible $a$) if and only if $a=r_{i,j}-r_{i+1,j}-r_{i,j+1}+r_{i+1,j+1}$. Consequently, $X_s^0$ is empty, if some $r_{i,j}-r_{i+1,j}-r_{i,j+1}+r_{i+1,j+1}$ is negative. Otherwise $X_s^0$ is dense in $X_s$.

Conversely, given a multisegment $M$ we can easily determine its rank vector $r\left(M\right)=\left(r\right(M{)}_{i,j})$ as follows

In the particular case of a segment $[k,l]$, we obtain just the characteristic function of a triangle as the rank triangle

Proof. Assertion a) is obvious, since $\mathrm{rk}{A}_{(i,j)}\le a$ implies $\mathrm{rk}{A}_{(i,j)}\le b$ for any $b>a$.

To prove b) we decompose $X_s$ in a disjoint union of $\text{Gl}\left(d\right)$–orbits. This is possible, since $X_s$ is $\text{Gl}\left(d\right)$–invariant. Thus we obtain a set of multisegments $M_s$ with

Consequently, $X_s$ is the union of a finite number of orbit closures $\overline{\text{Gl}\left(d\right)M}$ for a finite number of multisegments $M$. We can assume this set is minimal. Thus $X_s$ is irreducible precisely when $X_s=\overline{\text{Gl}\left(d\right)M}$ for some maximal $M$ in $M_s$.

For c), note that $X_s^0$ is nonempty, precisley when there exists a multisegment $M$ with $s=\mathrm{rk}M$. This is also equivalent to $s={\sum }_{(i,j)}{a}_{(i,j)}\mathrm{rk}[i,j]$ is the sum of rank functions of segments. To prove the last characterization we note that for $s={\sum }_{(i,j)}{a}_{(i,j)}\mathrm{rk}[i,j]$ we obtain $s_{i,j}-s_{i+1,j}-s_{i,j+1}+s_{i+1,j+1}=a_{(i,j)}\ge 0$. Conversely, if $s_{i,j}-s_{i+1,j}-s_{i,j+1}+s_{i+1,j+1}\ge 0$ then we define $a_{(i,j)}=s_{i,j}-s_{i+1,j}-s_{i,j+1}+s_{i+1,j+1}$. $\square$

Proof. We consider the projection of a representation of $Q$ to the quiver $Q^{\prime }$ with vertices $i,i+1,\dots ,j-1,j$ and its subvarieties $Y_{i,j}$ in $R(Q,d)$ and $Y_{i,j}^{\prime }$ in $R(Q^{\prime },d^{\prime })$ defined by $\mathrm{rk}{A}_{(i,j)}<r_{i,j}$. Then $R(Q,d)$ is a direct product of $R(Q^{\prime },d^{\prime })$ with some affine space and $Y_{i,j}$ is a product of $Y_{i,j}^{\prime }$ with some affine space. Thus $Y_{i,j}$ is irreducible precisely when $Y_{i,j}^{\prime }$ is irreducible. Consequently, it is sufficient to prove the claim for $Y_{1,t}$ in $R(Q,d)$.

We now assume $(i,j)=(1,t)$ and $d_i>d_1,d_t$ for any $1<i<t$.

Now we consider a multisegment $M$ consisting of $[1,t-1]\oplus [2,t]$ and $M\left(e\right)$ for $e=d-(1,2,2,\dots ,2,2,1)$. A computation of the ranks $r_{i,j}\left(M\right)$ yields $r_{1,t}\left(M\right)=r_{1,t}-1$ and $r_{i,j}\left(M\right)=r_{i,j}$ for all $(i,j)\ne (1,t)$. Thus the equation $s_{1,t}=r_{1,t}-1$ and $s_{i,j}=r_{i,j}$ for $(i,j)\ne (1,t)$ defines an orbit $X_s$ and $X_s$ is the closure of this orbit containing $M$. Consequently it is irreducible, and it coincides with $Y_{1,t}$.

Finally, we need to compute the codimension of the orbit closure $Y_{1,t}$. For this we compute the dimension of the stabilizer of $M\left(d\right)$ and of $M$ constructed above. To make the computation easier, we delete the common direct summands that contribute with the same dimension to the stabilizer and assume without loss of generality $d_1\ge d_t$. Then we need to compute

Back to $M\left(d\right)$, we decompose it into $M\left(d\right)=[2,t-1]\oplus [1,t{]}^a\oplus [1,t-1{]}^b\oplus M^{\prime }$ with maximal $a$ and $b$. Then $M$ is $[2,t]\oplus [1,t{]}^{a-1}\oplus [1,t-1{]}^{b+1}\oplus M^{\prime }$ and

Consequently, the codimension of the orbit of $M$ equals $b+1=d_1-d_t+1$ and this equals the codimension of $Y_{1,t}$. Finally, note that under the reduction from arbitrary $Y_{i,j}$ to $Y_{1,t}$ the codimension does not change. $\square$

Proof. Without loss of generality we may assume $d_i\le d_j$ in the proof.
a) Consider the maps $A_{(i,j)}:V_i⟶V_j$, $A_{(i,l)}:V_i⟶V_l$, and $A_{(l,j)}:V_l⟶V_j$. We consider two cases.

$d_i\ge d_l$:
Assume $\mathrm{rk}{A}_{(i,j)}<d_l$, then $\mathrm{rk}{A}_{(i,l)}<d_l$ or $\mathrm{rk}{A}_{(l,j)}<d_l$ since $A_{(i,j)}:V_i⟶V_l⟶V_j$ factors through $V_l$ with $\dim V_l\le \dim V_i,\dim V_j$.

$d_i<d_l$:
Assume $\mathrm{rk}{A}_{(i,j)}<d_i$, then $\mathrm{rk}{A}_{(i,l)}<d_i$

b) Consider the maps $A_{(i,j)}:V_i⟶V_j$ and $A_{(i,l)}:V_i⟶V_l$. Since $(i,j)\in J\left(d\right)\setminus I\left(d\right)$ there exist some $l>j$ with $d_i\le d_l<d_j$ and $(i,l)\in J\left(d\right)$. Then from $\mathrm{rk}{A}_{(i,j)}<d_i$ follows $\mathrm{rk}{A}_{(i,l)}<d_i$.

c) This case is opposite to case b). $\square$

Proof. The dense orbit is defined by the condition $\mathrm{rk}{A}_{(i,j)}=r_{i,j}$. Thus, the complement satisfies $\mathrm{rk}{A}_{(i,j)}<r_{i,j}$ for at least one pair $(i,j)$ with $r_{i,j}>0$. Since $r_{i,j}=\text{min,}\{d_i\mid 1\le i\le t\}>0$ we finish the proof of the first equality.

To prove the second one we use the proposition above. From Proposition 3.2 a) we obtain ${\bigcup }_{1\le i<j\le t}{Y}_{i,j}\subseteq {\bigcup }_{(i,j)\in J\left(d\right)}{Y}_{i,j}$ and from part b) and c) ${\bigcup }_{(i,j)\in J\left(d\right)}{Y}_{i,j}\subseteq {\bigcup }_{(i,j)\in I\left(d\right)}{Y}_{i,j}$. $\square$

Proof. Thanks to Proposition 3.1, we only need to prove that $Y_{i,j}$ is not irreducible for $(i,j)$ not in $J\left(d\right)$. Take $(i,j)$ not in $J\left(d\right)$, thus there exists an $l$ with $i<l<j$ and $d_l\le \text{max,}\{d_i,d_j\}$ is minimal. Then by Proposition 3.2 a) we have $Y_{i,j}\subseteq Y_{i,l}\cup Y_{l,j}$. Assume first that $d_l\le min\{d_i,d_j\}.$ We claim that $Y_{i,j}=Y_{i,l}\cup Y_{l,j}$ and this is a proper decomposition, none contains the other. To see the equality, consider any element $A$ in $Y_{i,l}\cup Y_{l,j}$. Then $\mathrm{rk}(A{)}_{i,l}<r_{i,l}$ or $\mathrm{rk}(A{)}_{l,j}<r_{l,j}$. From each of the inequalities follows $\mathrm{rk}(A{)}_{i,j}<r_{i,j}$. On the other hand, there exists a representation with $\mathrm{rk}(A{)}_{i,l}=r_{i,l}$ and $\mathrm{rk}(A{)}_{l,j}<r_{l,j}$ and vice versa, proving also the last claim.
In the second case $max{d}_i,d_j\ge d_l>min\{d_i,d_j\}.$ We construct two different subvarieties that contain $Y_{i,j}$ and none contains the other. To simplify the arguments, we assume without loss of generality $d_i\ge d_l>d_j$ and, using the first case, $d_l$ is the minimal entry of $d$ between $d_i$ and $d_j$. The first variety is just the orbit closure $Y_{l,j}$, the second one is defined by $r_{i,l}\left(A\right)<r_{i,l}$ and $r_{i,j}\left(A\right)<r_{i,j}$. Using multisegments (or rank conditions) one can show that we obtain at least two irreducible components in this way ($Y_{l,j}$ is irreducible, the other variety need not to be). Anyway, we obtain at least two irreducible components. $\square$