# Spectral sequence from standard/Verma filtration/flag to compute Lie algebra cohomology of tensor product with respect to $\mathfrak{n}$

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?

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 $$P_{-2}$$.