I'm not sure this question fully qualifies as a research-level math question, but from my (limited) past experience on stackexchanged I feared this question might not get an answer there.


**Setting**: the category $\cal{O}$ associated to a complex semisimple Lie algebra $\frak{g}=\frak{n}\oplus\frak{h}\oplus\frak{n}^-$. The Verma with highest weight $\lambda$ is denoted $M\_\lambda$, and its irreducible quotients are $L\_\lambda$. The projective cover of $L\_\lambda$ is $P\_\lambda$, which is self-dual precisely when $\lambda$ is $\varrho$-antidominant. Let the contragredient dual be denoted $M^\dagger$, and let $\mathbb C\_\lambda$ denote the one-dimensional space with an $\mathfrak h$-action given by $\lambda$.


**Question summary**: for the tensor of a Verma with a finite-dimensional, take the spectral sequence associated to the filtered complex which is the Chevalley cochain complex filtered by the Verma filtration on the tensor we are interested in -- then we should get a spectral sequence $H^{p+q}(\mathfrak{n} : M\_{\lambda+\lambda\_p})\ {\Longrightarrow\_p}\ H^{p+q}(\mathfrak{n}:M\_\lambda\otimes V)$. However I am applying this to examples such as $M\_{-1}\otimes L\_1=P\_{-2}$ and $M\_0\otimes L\_2=M\_2\oplus P\_{-2}$ in $\frak sl\_2$ and getting the wrong answers. I reproduce my work below and would be very grateful if someone could point out where I went wrong.


**Details**: For a finite-dimensional object $V\in\cal O$, the tensor product $M\_\lambda\otimes V$ admits a "standard filtration" or "Verma flag", $M\_\lambda\otimes V=M(0)\supset M(1)\supset\cdots\supset 0$, whose quotients are $M(i)/M(i+1)=M\_{\lambda\_i}$, where $\{\lambda\_i\}\_i$ are the weights of $V$ labeled such that $\lambda\_i\ge\lambda\_j\implies i\le j$.


Let $C^\bullet$ be the Chevalley cochain complex $\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge\bullet},M\_\lambda\otimes V)$, and consider the filtration $F^p C^{p+q}=\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge(p+q)},M(p))$. This is a bounded decreasing filtration of a bounded cohomologically graded complex. Then this guy has a spectral sequence
$$E\_1^{p,q}=H^{p+q}(\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge(p+q)},M(p))/\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge(p+q)},M(p+1)))=H^{p+q}(\mathfrak{n}:M\_{\lambda+\lambda\_p}) \ {\Longrightarrow\_p}\ H^{p+q}(\mathfrak{n}:M\_\lambda\otimes V),$$
where $\deg \text{d}\_r=(r,1-r)$.


I tried to apply this to $M\_{-1}\otimes L\_1=P\_{-2}$. In that case we would get the first page $E\_1$ has (I will denote $H^k(\mathfrak{n}:M)$ as just $H^k(M)$) $H^0(M\_0)$ in the $(0,0)$ position, $H^1(M\_0)$ in the $(0,1)$ position, $H^1(M\_{-2})$ in the $(1,0)$ position, and $H^0(M\_{-2})$ in the $(1,-1)$ position, and nothing anywhere else. But $M\_{-2}=L\_{-2}$ is self-dual, and recall from category $\cal O$ that $M^\dagger$ has a standard filtration if and only if $H^{>0}(\mathfrak{n}:M)=0$ (the obstruction to the contragredient having a standard filtration is measured by higher Lie algebra cohomology out of $\mathfrak n$). Hence $H^1(M\_{-2})=0$. Then $E\_1$ just has three nonzero terms, each lying in its own row. The differential $\text d\_1$ has degree $(1,0)$ and points to the right. So $E\_2=E\_1$. It's also easy to see that the later differentials, of degree $(r,1-r)$, will never connect two nonzero terms. So actually $E\_1=E\_\infty$.


Let us compute the actual terms of this page. $H^0(M\_0)=\mathbb C\_0\oplus\mathbb C\_{-2}$ and $H^0(M\_{-2})=\mathbb C\_{-2}$ since $H^0(M)$ is just the $\mathfrak n$-invariants of $M$, and $H^1(M\_{0})=\mathbb C\_{-2}$ since in general there is a formula $\operatorname{ch}\_M=\sum\_\lambda \chi(\operatorname{Ext}^\bullet(M\_\lambda,M))\operatorname{ch}\_{M\_\lambda}$. This would imply, by the convergence of the spectral sequence, that
$$H^0(\mathfrak n:M\_{-1}\otimes L\_1)=\mathbb C\_0\oplus\mathbb C\_{-2}\oplus\mathbb C\_{-2},\qquad H^1(\mathfrak n:M\_{-1}\otimes L\_1)=\mathbb C\_{-2}.$$


HOWEVER this cannot be correct, since $M\_{-1}\otimes L\_1=P\_{-2}$ is both injective and projective, so in particular the $-2$-weight space of $H^1(\mathfrak n:P\_{-2})$, which is given by $\operatorname{Ext}^1(M\_{-2},P\_{-2})$, must vanish. The $H^0$ is also incorrect, since $\text{ch}\_{P\_{-2}}=\text{ch}\_{M\_0}+\text{ch}\_{M\_{-2}}$ then forces $\dim\operatorname{Ext}^0(M\_{-2},P\_{-2})=\dim\operatorname{Ext}^0(M\_{0},P\_{-2})=1$, i.e. $H^0(\mathfrak n:P\_{-2})=\mathbb C\_0\oplus\mathbb C\_{-2}$. So the right answer should be
$$H^0(\mathfrak n:P\_{-2})=\mathbb C\_0\oplus\mathbb C\_{-2},\qquad H^1(\mathfrak n:P\_{-2})=0.$$


I've tried a similar computation with $M\_0\otimes L\_2=M\_2\oplus P\_{-2}$ with similarly disastrously wrong answers.


What gives?



Spectral sequence from standard/Verma filtration/flag to compute Lie algebra cohomology of tensor product with respect to $\mathfrak{n}$
I think the issue here is that the subquotients in the standard filtration have weights (your $\lambda\_i$'s) which are ordered the other way. To be clear, if the weights of $\nu\_0$, ..., $\nu\_n$ of $L$ are ordered so that $\nu\_i \le \nu\_j$ implies $i\le j$, then one obtains a standard filtration for $M\_\lambda \otimes L$ with subquotients $M(i)/M(i+1) = M\_{\lambda+\nu\_i}$. Explicitly, the filtration comes from applying the exact functor $\mathcal{U}(\mathfrak g) \otimes\_{\mathcal{U}(\mathfrak{b})}-$ to the filtration of $U(\mathfrak{b})$-modules $N(0)=\mathbb{C}\_\lambda \otimes L \supset N(1) \supset ... \supset N(n)=0$, where $N(k)=U(\mathfrak{b})\cdot \{v\_k,...,v\_n\}$. (Had you ordered the weights the other way, the $N(i)$ would not be $\mathfrak b$-submodules!)


In your first example, $\nu\_0=-1$ and $\nu\_2=1$, so $M=M\_{-1}\otimes L\_1$ has a filtration with $M(0)=M$, $M(1)=M\_{0}$ and $M(0)/M(1)=M\_{-2}$ (so $\lambda\_0=-2$, $\lambda\_1=0$). Since it only has two steps, the spectral sequence amounts to a single long exact sequence computation: the one associated to the short exact sequence of complexes $0 \rightarrow C^\bullet(\mathfrak n : M(1)) \rightarrow C^\bullet(\mathfrak n : M) \rightarrow C^\bullet(\mathfrak n : M\_{-2}) \rightarrow 0$. It is easy to see that this does agree with your character computations for $P\_{-2}$.


