Let $\mathcal{X}$ be a finite acyclic quiver, and $v_1$ be a source vertex of $\mathcal{Q}$. Let $\mathcal{X}$ be a representation in $\mathrm{Re}(\mathcal{Q},R)$, where $R$ is a commutative noetherian ring. Let $\mathcal{X}'\in \mathrm{Rep}(\mathcal{Q},R)$ be such that $\mathcal{X}'_{v_1}=0$, and $\mathcal{X}'_v=\mathcal{X}_v$ if $v\neq v_1$, and for any arrow $a:v \to w$, $\mathcal{X}'_a=\mathcal{X}_a$ if $v \neq v_1$. Can I show that if $\mathrm{Ext}^{>0}_{\mathrm{rep}}(\mathcal{X},R\mathcal{Q})=0$, then $\mathrm{Ext}^{>0}_{\mathrm{rep}}(\mathcal{X}',R\mathcal{Q})=0$?
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$\begingroup$ This is true when $R$ is a field, by applying $\operatorname{Hom}_{\text{rep}}(-,R\mathcal{Q})$ to the short exact sequence $0\to\mathcal{X}'\to\mathcal{X}\to C\to0$, which exists since $v_1$ is a source, and using that $RQ$ has global dimension $1$ in this case. In the general case, it is true (by the same argument) if and only if $\operatorname{Ext}^i_{\text{rep}}(C,R\mathcal{Q})=0$ for all $i\geq2$. I suspect this is false, but I don't have a counterexample immediately. $\endgroup$– Matthew PresslandCommented Jul 27, 2017 at 14:56
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$\begingroup$ Thank you so much. So if we add the condition that $R$ is a semisimple ring we can prove the claim. Is it true? $\endgroup$– Homa81Commented Jul 28, 2017 at 7:55
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$\begingroup$ Sure: in that case you are essentially working with a bunch of copies of $k\mathcal{Q}$ (although possibly for different fields $k$). (I don't actually have a counterexample to the general statement, but if it is true it has to be for a quite different reason, since it is certainly possible for the global dimension of $R\mathcal{Q}$ to be greater than $1$ in general.) $\endgroup$– Matthew PresslandCommented Jul 28, 2017 at 9:14
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This is true when $R$ is a field (or semisimple, so a product of fields), for the following reason.
Since $v_1$ is a source, there is a monomorphism $\mathcal{X}'\to\mathcal{X}$, which fits into a short exact sequence $$0\to\mathcal{X}'\to\mathcal{X}\to C\to 0.$$
Applying the functor $\operatorname{Hom}(-,R\mathcal{Q})$ to this sequence and looking at the long exact sequence, one sees
$$\cdots\to\operatorname{Ext}^i_{\text{rep}}(\mathcal{X},R\mathcal{Q})\to\operatorname{Ext}^i_{\text{rep}}(\mathcal{X}',R\mathcal{Q})\to\operatorname{Ext}^{i+1}_{\text{rep}}(C,R\mathcal{Q})\to\operatorname{Ext}^{i+1}_{\text{rep}}(\mathcal{X},R\mathcal{Q})\to\cdots$$
for each $i>0$. By the assumption on $\mathcal{X}$, it follows that $\operatorname{Ext}^i_{\text{rep}}(\mathcal{X}',R\mathcal{Q})\cong\operatorname{Ext}^{i+1}_{\text{rep}}(C,R\mathcal{Q})$. If $R$ is semisimple, then $\operatorname{Ext}^i_{\text{rep}}(-,-)=0$ for $i\geq2$, and the desired result follows.
In the general case, the result holds if and only if $\operatorname{Ext}^{i+1}_{\text{rep}}(C,R\mathcal{Q})=0$ for all $i\geq1$. I can't think of a reason why this should be true, but I don't have an explicit counterexample at the moment.
answered Jul 28, 2017 at 9:19