specific long exact sequence of derived functors in category $\mathcal O$

Question

I am currently reading Gaitsgorys notes on Category $\mathcal O$ and geometric representation theory (https://people.math.harvard.edu/~gaitsgde/267y/catO.pdf) and I'm having some trouble with the proof of Lemma 4.11 . The setup for this Lemma is as follows. Let $ \mathcal M$ be a Module in $\mathcal O$ let $\lambda$ be a maximal weight of $\mathcal M \ \mathcal M (\lambda)$ be it's $\lambda$ weight space and let $M_{\lambda}$ be the Verma module of weight $\lambda$. Then there is a morphism of $\mathfrak{g}$ modules $ M_{\lambda} \otimes_k \mathcal M (\lambda) \to \mathcal M$ which is an isomorphism on the $\lambda$ weight space as we have $\mathfrak{n}^+ v = 0$ for $v \in \mathcal M(\lambda)$ by maximality. Denote by $N_1$ and $N_2$ the kernel and cokernel of this map respectively. The Lemma now asserts that $Ext^{1}_{\mathcal O}(N_2,M_{\mu}^{\vee})=0 $ for any $ \mu$ where $M_{\mu}^{\vee}$ is the dual verma module. Now in the proof of this there are two long exact sequences of derived functors neither of which I can quite figure out.

For both of these long exact sequences I am not sure about the meaning of $Hom_{\mathcal O}((M_{\lambda} \otimes_k \mathcal M (\lambda) \to \mathcal M),M_{\mu}^{\vee}) $. I've tried to associate a module to this morphism but this doesn't seem to be straightforward.

I've also tried to get these sequences or at least similar ones as long exact sequences of derived functors associated to short exact sequences but I'm stuck here as well especially for the second long exact sequence as there it seems one index is shifted by $-2$. The only exact sequence I can see is

but this isn't even a short exact sequence and so I don't really know what to do. Do I need to invoke some spectral sequence? Any help would be greatly appreciated!

Answer 1

Philosophical answer: $(M_{\lambda} \otimes_k \mathcal M (\lambda) \to \mathcal M)$ is a mapping cone/cylinder/fibre (I’m never sure what the terminology is)?

(The discussion below is mostly philosophical, technical details are slightly tedious and explained afterwards).

Basically, all the discussion below is an invitation to thinking about derived categories (in this case, the category of bounded above complexes, where we quotient by maps homotopic to zero and and quasi-isomorphisms are formally invertible).

The important thing is that (under reasonable assumptions which are met here):

1. derived functors already factor through the derived category
2. the boundary maps of cohomology long exact sequences already exist in the derived category.

In the discussion below, I will make no difference between an object and a projective resolution thereof, precisely because they’re (almost) the same thing in the derived category (but in the technical part I do not use derived categories because I don’t understand them too well formally).

It turns out that the role of $(M_{\lambda} \otimes_k \mathcal M (\lambda) \to \mathcal M)$ is that of a certain very nice complex of projectives built from (projective resolutions of) $M_1=M_{\lambda} \otimes_k \mathcal M(\lambda)$ and $M_2=\mathcal{M}$: in fact, it’s in the middle of an “exact sequence” with $\mathcal{M}$ on the right and a shifted version (accounting for the shift in homology) of $M_{\lambda} \otimes_k \mathcal M(\lambda)$ (and the boundary map is precisely the map $f: M_{\lambda} \otimes_k \mathcal M (\lambda) \to \mathcal M$).

The point of $N_1$ (resp. $N_2$) is that they sit in exact sequences with the domain (or codomain for $N_2$) of $f$ and its image $I$. Thus, from the homology long exact sequences, we have degree-shifting maps $N_2 \to I$ and $I \to N_1$.

The remarkable thing is that the cone of the composition $N_2 \to N_1$ (remember, this represents the composition of two boundary maps in the cohomology) injects into that of the original map, and the quotient is homotopic to zero, so the map is an isomorphism in cohomology and the same group is thus in the two exact sequences.

Roughly speaking, in the cohomology of $(M_{\lambda} \otimes_k \mathcal M (\lambda) \to \mathcal M)$, the image of this map appears both on the right and on the left, and thus “compensates itself”.

Technical version:

Let $\mathcal{A}$ be an abelian category with enough projectives, and $U$ be a contravariant left-exact additive functor from $\mathcal{A}$ to another category.

By “complex”, I mean “bounded above complex of projectives of $\mathcal{A}$.

If $C$ is a complex, I’m usually calling $\delta_C$ its differential and $C[n]$ is the complex $C[n]_i=C_{i+n}$ with differential $(-1)^n\delta_C$.

Lemma 1: Let $f: C \rightarrow D$ be a map of complexes. Then there is a complex $E$ verifying the following two properties:

1. we have an exact sequence $0 \rightarrow D \rightarrow E \rightarrow C[-1] \rightarrow 0$ split in every degree.
2. the boundary map $H^i(U(D)) \rightarrow H^{i+1}(U(C[-1]))=H^i(U(C))$ is exactly $F(f)$. We denote $E=[C \rightarrow D]$.

Proof: Define $E=D \oplus C[-1]$, with the following differential: $\delta_E(d,c)=(-f(c)+\delta_D d, -\delta_C c)$. Everything is now basically tautological.

Lemma 2: assume that we have an exact sequence of complexes $0 \to A \to B \to C \to 0$ with $C$ homotopic to zero. Then $A \to B$ is a homotopy equivalence.

Proof: We have maps $h: C_n \rightarrow C_{n+1}$ such that $h\delta_C+\delta_Ch=\mathrm{id}$. We can then lift them to maps $h’: B_n\rightarrow B_{n+1}$. Then $h’$ canonically maps $A_n$ into $A_{n+1}$, and therefore it’s easy to see that $id-\delta_B h’-h’\delta_B$ is a map of complexes $B\rightarrow A$ such that composing it (pre- or post-) with the map $A \to B$, we get an endomorphism tautologically homotopic to the identity.

Lemma 3: Let $F,G,H$ be three complexes, $C=F \oplus H$, $D = H \oplus G$ with differentials respectively given by $\begin{pmatrix}\delta_F & u_C\\0 &\delta_H\end{pmatrix}$ and $\begin{pmatrix}\delta_H & u_D \\ 0&\delta_G\end{pmatrix}$. Then

1. $u_C: H \rightarrow F[-1]$, $u_D: G \rightarrow H[-1]$, $\pi: C \rightarrow H$ and $\iota: H \to D$ are maps of complexes. 2) we have a natural homotopy equivalence $[G[1] \overset{-u_Cu_D}{\rightarrow} F[-1]] \to [C \overset{\iota\circ \pi}{\rightarrow} D]$.

Proof: For 1), $\pi$ and $\iota$ are formal, for $u_C$ and $u_D$ you just square the matrix differentials and write down what it means for these squares to be zero. We note $f=\iota\pi$.

For 2), let’s write $E’=F[-1]\oplus G$ with differential $\begin{pmatrix}-\delta_F&-u_C \circ u_D\\0&\delta_G\end{pmatrix}$ which is (naturally isomorphic to) $[G[1] \overset{-u_Cu_D}{\rightarrow} F[-1]]$.

Let $E=F[-1]\oplus H[-1] \oplus H \oplus G$, with differential $\begin{pmatrix}-\delta_F&-u_C&0&0\\0&-\delta_H&0&0\\0&-1&\delta_H&u_D\\0&0&0&\delta_G\end{pmatrix}$, it’s naturally isomorphic to $[C \overset{\iota\circ \pi}{\rightarrow} D]$.

Now, there’s a differential-preserving injective map $(f,g) \in E’ \longmapsto (f,u_D(g),0,g) \in E$. Note that its cokernel is the complex $H’=H[-1]\oplus H$ with differential $\begin{pmatrix}-\delta_H &0\\-1\delta_H\end{pmatrix}$ (through the map $(u,v,w,x) \longmapsto (v-u_D x,w)$ and that the sequence is split in each degree.

But you can compute that $\begin{pmatrix}0&-1\\0&0\end{pmatrix}$ is a homotopy of $H’$ between the identity map and zero. In other words, $0 \rightarrow E’ \rightarrow E \rightarrow H’ \rightarrow 0$ is exact and split in each degree with $H’$ homotopic to zero, which implies that $E’ \rightarrow E$ is a homotopy equivalence.

Corollary: let $f: X \rightarrow Y$ be a morphism of $\mathcal{A}$, $K$ its kernel and $L$ its cokernel. Then there is a family $(M^i)_{i \in \mathbb{Z}}$ such that there are two long exact sequences: $R^iU(Y) \to R^iU(X) \to M^i \to R^{i+1}U(Y)$ (the first map being $R^iU(f)$), and $R^{i-2}U(K) \to R^{i}U(L) \to M^i \to R^{i-1}U(K)$.

Proof: we have exact sequences $0 \to K \to X \to I \to 0$ and $0 \to I \to Y \to L \to 0$ with $I$ as the image of $f$. Let $F$ (resp. $G,H$) be a projective resolution of $K$ (resp. $L,I$). By the horseshoe lemma, we can put differentials on $C=F \oplus H$ and $D=H \oplus G$ as in the statement of Lemma 3 making $C,D$ projective resolutions of $X,Y$ respectively.

Let $E=[C \to D]$, $E’=[G[1] \to F[-1]]$ as in Lemma 3, then we have exact sequences split in every degree $0 \to F[-1] \to E’ \to G \to 0$ and $0 \to D \to E \to C[-1] \to 0$, yielding long cohomology sequences after applying the functor $U$. But $U(E’)$ and $U(E)$ are homotopy equivalent so have the same cohomology. QED.

Answer 2

Here's a way to see this using more general theory (i.e. easy observations about the t-structure).

There is nothing special to this particular map, so let's work for simplicity in a slightly more general setting: Let $f: X \to Y$ be a morphism in an abelian category $\mathfrak{A}$ with kernel $\mathcal N_1$ and cokernel $\mathcal N_2$. Furthermore, I identify objects of $\mathfrak A$ and complexes in degree $0$ (via the degree $0$ embedding).

1. As noted above, the object you were confused about should be the mapping cone. Let's call this $C(f) := X[1] \oplus Y$.
2. By the definition of the derived category $ X \to Y \to C(f) \to $ is an exact triangle. Now, by the theory of cohomology functors, you get the first long exact sequence of Lemma 4.11.

Let's see the second long exact sequence that your question is actually about. Again by the theory of cohomology functors, it suffices to prove the following:

Claim: $\mathcal N_1 [1] \to C(f) \to \mathcal N_2 \to $ is an exact triangle.

As $X$ and $Y$ live as complexes only in degree $0$, $C(f)$ shouldn't be so complicated. Indeed, it is just the complex $\dots \to 0 \overset{d_{-3}}{\to} 0 \overset{d_{-2}}{\to} X \overset{d_{-1}=f}{\to} Y \overset{d_0}{\to} 0 \overset{d_1}{\to} 0 \to \dots $ Observe that the only non-zero cohomologies are $H^{-1}(C(f)) = \mathcal N_1$ and $H^0(C(f))= \mathcal N_2$. We want to sort of seperate these and by using the natural t-structure of the derived category.

Explicitly, this means that the trucation functors give a short exact sequence of complexes: $ 0 \to \tau^{\leq -1} C(f) \to C(f) \to \tau^{\geq 0} C(f) \to 0 $.

I won't draw the full diagram of complexes but note that $\tau^{\leq -1} C(f) = \mathcal N_1[1]$. Similarly, $\tau^{\geq 0} C(f)$ looks like $\dots \to 0 \overset{d_{-2}}{\to} im(f) \overset{d_{-1}}{\to} Y \overset{d_0}{\to} 0 \to \dots $ and it is easy to see that the natural map to $\mathcal N_2$ is a quasi-isomorphism, hence an isomorphism in the derived category. At the same time a short exact sequence of complexes gives rise to an exact triangle in the derived category (e.g. see Claim 3 here or Lemma B.4.5. in the book mentioned below; the point is that for a short exact sequences of complexes $0 \to A \overset{\varphi}{\to} B \to C \to 0$ the last term $C$ is quasi-isomorphic to $C(\varphi)$). Hence, we get an exact triangle $\mathcal N_1 [1] \to C(f) \to \mathcal N_2 \to $, as desired.

For a good overview of the results and definitions used about the derived category, I recommend reading Appendix B of "D-Modules, Perverse Sheaves, and Representation Theory" by Hotta, Takeuchi, and Tanisaki (in particular, Def. B.2.4., Def. B.3.8. and Lemma B.4.5.).