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I have a problem in understanding the proof of a theorem by Happel-Reiten-Smalø. The original reference is this article http://arxiv.org/abs/0911.4473 .

I write down the text of the theorem and a lemma which is needed in the proof.

Lemma 3.2.4 Let $k$ be a field and $\mathcal A$ a $k$-linear abelian category that is Ext-finite. Let $T\in\mathcal A$ be a tilting object. Then the functor $-\otimes_{\Lambda}T$ induces an isomorphism $$ \text{Ext}_{\Lambda}^n(M,N)\overset{\sim}{\longrightarrow}\text{Ext}_{\mathcal A}^n(M\otimes_{\Lambda}T,N\otimes_{\Lambda}T) $$ for all $M$, $N$ in $\mathcal Y(T)$ and $n\geq 0$. [$\mathcal Y(T) = \{M\in\text{mod }\Lambda\mid \text{Tor}_1^{\Lambda}(M,T) = 0\}$]

Theorem 3.2.5 (Happel-Reiten-Smalø). Let $\mathcal A$ be a $k$-linear abelian category that is Ext-finite. Let $T$ be a tilting object in $\mathcal A$ and $\Lambda=\text{End}_{\mathcal A}(T)$. Then the functor $$ -\otimes_{\Lambda}^{\mathbf{L}} T\colon\mathbf{D}^b(\text{mod } \Lambda) \to \mathbf{D}^b(\mathcal A) $$ is an equivalence of triangulated categories and its right adjoint $\mathbf{R}\text{Hom}_{\mathcal A}(T,-)$ is a quasi-inverse.

Proof. Set $F_T = -\otimes_{\Lambda}^{\mathbf L}T$ [the left derived functor of $\otimes_{\Lambda}$]. We identify objects in $\text{mod }\Lambda$ and $\mathcal A$ with complexes that are concentrated in degree $0$. For instance, $F_T M = M\otimes_{\Lambda} T$ for each $M$ in $\mathcal Y(T)$. We need to show that for each pair of complexes $X$, $Y$ in $\text{mod }\Lambda$, the induced map $$ \phi_{X,Y}\colon\text{Hom}_{\mathbf D^b(\text{mod }\Lambda)}(X,Y)\to\text{Hom}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY) $$ is bijective. Set $$ \ell(X) = \operatorname{card}\{n\in\mathbb Z\mid X_n\neq 0\} $$ and note that each bounded complex $X\neq 0$ fits into an exact triangle $X'\to X\to X''\to X'[1]$ such that $\ell(X') = \ell(X)-1$ and $\ell(X'') = 1$.

Using the five lemma and induction on $\ell(X)+\ell(Y)$, one shows that $\phi_{X,Y}$ is bijective. The case $\ell(X) = \ell(Y) = 1$ follows from lemma 3.2.4. To be precise, one uses that each $\Lambda$-module $M$ fits into an exact sequence $0\to M'\to P\to M \to 0$ with $M'$ and $P$ in $\mathcal Y(T)$, which yields an exact triangle $M'\to P\to M\to M'[1]$. [...]

Next they show that the given functor is dense, but I am stuck on this passage. I can't see how I could merge the given tools (induction, five lemma, every $\Lambda$-module fits into such an exact sequence) in order to get to the assertion.

I think this theorem is not difficult to find in the literature, for example in Happel-Reiten-Smalø, "Tilting in abelian categories and quasitilted algebras", theorem 4.6. Anyway, the proof always requires torsion pairs, and I have no knowledge about them, so I would like to understand a proof which avoids them.

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  • $\begingroup$ Let $A \rightarrow B \rightarrow C \rightarrow$ be a triangle. The five lemma tells you that if two out of $\phi_{A[n],Y}$, $\phi_{B[n],Y}$, $\phi_{C[n],Y}$ are bijective for all $n$, then so is the third. Similarly in the other variable. We know that $\phi_{X[n],Y}$ is bijective for all $n$ when $X, Y \in \mathcal Y(T)$. The proof above shows one way to chose triangles to get the result for arbitrary $X,Y$. $\endgroup$ – Dag Oskar Madsen Jun 11 '15 at 22:44
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    $\begingroup$ It's not what you asked but if you want to learn classical results of torsion pairs, I would recommend you the following reference on torsion pairs which is quite elementary: the book of Assem, Simson and Skowronski, Elements of the Representation Theory of Associative Algebras: Volume 1 Chapter VI.1 and after $\endgroup$ – Bulois Michael Jun 15 '15 at 7:48
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Here are some more details.

Lemma 3.2.4 written in the notation of Theorem 3.2.5 becomes the statement that $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M,N$ in $\mathcal Y(T)$ and $n \geq 0$. It is also an isomorphism when $n<0$ since then both sides are $0$.

For any $N$ in $\operatorname{mod} \Lambda$ there is a triangle $$N'\to P\to N\to N'[1]$$ with $N',P$ in $\mathcal Y(T)$, and for all $n \in \mathbb Z$ we get a commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(M,N'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,P[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TP[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN'[n+1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TP[n+1]) \end{CD}

Since all the other downward arrows are isomorphisms, it follows from the five lemma that the middle downward arrow is an isomorphism. So $$\phi_{M,N[n]}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(M,N[n]) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TM,F_TN[n])$$ is an isomorphism for all $M$ in $\mathcal Y(T)$, $N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$. A similar argument in the other variable gives that $\phi_{M,N[n]}$ is an isomorphism for all $M,N$ in $\operatorname{mod} \Lambda$ and $n \in \mathbb Z$.

This is the basis for the induction. The inductive step uses triangles of the form $$Y' \to Y \to Y'' \to Y'[1]$$ with $\ell(Y') = \ell(Y)-1$ and $\ell(Y'') = 1$. We apply the five lemma to the commutative diagram $\require{AMScd}$ \begin{CD} \scriptsize {\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b({\rm{mod}}\Lambda)}(X,Y'[n+1])\\ @VV{\wr}V @VV{\wr}V @VVV @VV{\wr}V @VV{\wr}V\\ \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n-1]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY[n]) @>>> \scriptsize {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY''[n]) @>>> \scriptsize{\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY'[n+1]) \end{CD}

and a similar version for the other variable to conclude that $$\phi_{X,Y}\colon {\rm{Hom}}_{\mathbf D^b(\operatorname{mod} \Lambda)}(X,Y) \to {\rm{Hom}}_{\mathbf D^b(\mathcal A)}(F_TX,F_TY)$$ is an isomorphism for all $X,Y$ in ${\mathbf D^b(\operatorname{mod} \Lambda)}$.

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  • $\begingroup$ How do I get the commutative diagrams to fit on the screen? $\endgroup$ – Dag Oskar Madsen Jul 19 '15 at 10:22

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