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?


1 Answer 1


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 $P_{-2}$.


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