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  1. As we know, if $\textrm{Ext}_{A}^{1}(M,N)=0$ with $M,N\in \textrm{mod-}A$, then all short exact sequences with first item as $N$ and last item as $M$ are split.

    My question is, what is the meaning of $\textrm{Ext}_{A}^{n}(M,N)=0, n>1$? Is there similar good result about the corresponding long exact sequence?

  2. Let $$0\rightarrow P_n\rightarrow P_{n-1}\rightarrow\cdots \rightarrow P_0\rightarrow T\rightarrow 0$$ as a projective resolution of $T$ in $\textrm{mod-}A$. How to prove following result.

    If $\textrm{Ext}_{A}^{i}(T,P_j)=0$ for all $i>0, 0\leq j\leq n-1$, then $\textrm{Ext}_{A}^{i}(T,T)=\textrm{Ext}_{A}^{i+n}(T,P_{n})$.

(This result is used in Lemma 3.4 of "D-split sequences and derived equivalences, Wei Hu, Changchang Xi* ")

Hanno
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jin zhang
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1 Answers1

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Ad 1: The set underlying $\text{Ext}^n(X,Y)$ is in canonical bijection with the set of path-components of the extension category ${\mathscr E}^n(X,Y)$ whose objects are the length-$n$ exact sequences starting resp. ending in $Y$ resp. $X$, and whose morphisms are commutative diagrams. That's general and independent of the value of $n$. What is special about $n=1$, though, is that ${\mathscr E}^1(X,Y)$ is actually a groupoid, i.e. any morphism is an isomorphism (by the five lemma), and hence you can conclude from $\pi_0\left[{\mathscr E}^1(X,Y)\right]=\ast$ that whatever property holds for some short exact sequence $0\to Y\to ?\to X\to 0$ holds for all such. This in particular applies to being split. For $n>1$, I don't know of any interesting property transferred by the mere existence of a morphism of exact sequences.

Ad 2: That's an example of dimension shifting: Split the given long exact sequence into short exact ones and apply the long exact $\text{Ext}$-sequence to each of them. This will allow you to successively go from $\text{Ext}^i_A(T,T)$ to $\text{Ext}^{i+n}_A(T,P_n)$. Let me know if you need details, but I think it's a good exercise to figure out the details yourself.

Hanno
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