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Let $Q$ be a quiver without oriented cycles. A result of Schofield says that, for dimension vectors $\alpha$ and $\beta$ of $Q$, $\beta\hookrightarrow\alpha$ iff $\operatorname{ext}(\beta, \alpha-\beta)=0$.

[By $\beta\hookrightarrow\alpha$, I mean that a general representation of dimension vector $\beta$ is a sub representation of a general representation of dimension vector $\alpha$ and by $\operatorname{ext}$, I mean the generic $\operatorname{Ext^1}$]

Now, consider quiver $Q'$ with relations, but without oriented cycles. So, I have a path algebra $\Lambda=\mathbb{K}Q'/I$, where $I$ is an admissible ideal and $\mathbb{K}$ is algebraically closed field.

My question is:

Let $B$ be a brick ($\operatorname{End}_{\Lambda}(B)\cong \mathbb{K})$ which is a general representation of dimension vector $\alpha$ in some irreducible component of $rep_{\alpha}(Q')$. Assume there's an irreducible component of $rep_{\beta}(Q')$, say $\mathcal{C}$, such that a general representation of $\mathcal{C}$ is a non-zero proper sub representation of $B$. Is it true that there exists an irreducible component of $rep_{\alpha-\beta}(Q')$, say $\mathcal{Z}$, such that the generic $\operatorname{Ext^1}$, $\operatorname{ext}(\mathcal{C},\mathcal{Z})=0$?

Edit: I have edited the question based on David's comment.

Edit: The answer of @wcb is a perfectly good counter-example of the question that I had asked before editing my question. I apologize for not posing my question properly.

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    $\begingroup$ For quivers with relations, the notions of "general representation" and "general $\text{Ext}^1$" are unclear, because the moduli spaces of representations often have many components and you have to specify which component you are considering the general element of. $\endgroup$
    – David E Speyer
    Commented Aug 8, 2022 at 15:48
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    $\begingroup$ Crossposted on MSE. $\endgroup$
    – Jeremy Rickard
    Commented Aug 8, 2022 at 20:43

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The question doesn't seem to be a generalization of Schofield's result (which doesn't involve fixing a brick $B$), and seems to be false already for quivers without relations.

Take $Q$ to be the quiver with vertices 1,2,3 and arrows $a:1\to 2$, $b:2\to 3$ and $c:1\to 3$. Let $B$ be the brick with dimension vector $\alpha=(1,1,1)$ with $B(b)=0$ and $B(a),B(c)\neq 0$. There is a unique representation $X$ of dimension $\beta = (0,1,0)$, and it is a subrepresentation of $B$. If $Y$ has dimension $\alpha-\beta = (1,0,1)$ then $\mathrm{Ext}^1(X,Y) \neq 0$, since there is a non-split short exact sequence where the middle term $E$ has dimension $\alpha$ with $E(a)=0$, $E(b)\neq 0$, $E(c) = Y(c)$.

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  • $\begingroup$ wouldn't this also contradict Schofield's result? Also, I meant generic Ext, $\operatorname{ext}(\mathcal{C},\mathcal{Z})$. $\endgroup$
    – It'sMe
    Commented Aug 11, 2022 at 20:09
  • $\begingroup$ It doesn't contradict Schofield's result because $\beta\hookrightarrow\alpha$ means not what you say, but that the general representation of dimension $\alpha$ has a subrepresentation of dimension $\beta$. But the general representation $U$ of dimension $\alpha$ has $U(b)\neq 0$, so doesn't have $X$ as a subrepresentation. In my example $\mathrm{Ext}^1(X,Y)\neq 0$ for any $X$ of dimension $\beta$ (there is only one $X$) and any $Y$ of dimension $\alpha-\beta$, so the general Ext is also non-zero. $\endgroup$
    – wcb
    Commented Aug 11, 2022 at 21:29
  • $\begingroup$ thanks for your example and explanations. I realize that I've been careless in posing my question. What I had in mind was that this brick $B$ is a general representation of an irreducible component of $rep_\alpha(Q')$ $\endgroup$
    – It'sMe
    Commented Aug 11, 2022 at 21:36

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