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I am working on a graded vector space $V = \bigoplus_{i\in \mathbb{N}}V_i$ (which is a parabolic Verma module in the sense of [1], but let's ignore such specifics) with a positive definite inner product $(\cdot, \cdot):V \times V \to \mathbb{C}$ and a nilpotent operator $\mathcal{Q}: V_i \to V_{i+1}$ with adjoint $\mathcal{Q}^{\dagger} := \mathcal{S}:V_i \to V_{i-1}$. Each $V_i$ is finite dimensional. I want to compute the cohomology classes of $\mathcal{Q}$ for all $V_i$, that is, I want to find all distinct equivalence classes $[v] \in \ker \mathcal{Q} / \text{im}\, \mathcal{Q}$. Using the Hodge decomposition, the Hodge theorem (corollary 1.1) then states there is a unique mapping between cohomology classes and harmonic elements of the Laplacian $\Delta = \{\mathcal{Q}, \mathcal{S} \}: V_i \to V_i$. Thus, by finding harmonic elements at each level $V_i$, the problem is solved.

Now I want to refine the above setup by working on the quotient space $V_i/W_i$ where $W_i \subset V_i$ is a linear subspace of the corresponding level $V_i$ such that for $w \in W_i$ in general then $\mathcal{Q}(w) \in W_{i+1}$. Since we have an inner product, we know that $V_i/W_i \cong W_i^{\perp}$ although not canonically so. My question is then the following.

Is there a way of constructing a map $\phi_i : V_i/W_i \to W_i^{\perp} \subset V_i$, such that one can modify the definition of $\mathcal{Q} \to \mathcal{Q}_{\perp}$ so $\mathcal{Q}_{\perp}: V_i/W_i \to V_{i+1}/W_{i+1}$ and its adjoint $\mathcal{S}_{\perp}:V_i/W_i \to V_{i-1}/W_{i-1}$ respect the embedding $\phi_i$ so we can subsequently define the modified Laplacian $\Delta_{\perp} = \{ \mathcal{Q}_{\perp}, \mathcal{S}_{\perp}\}$ and utilize Hodge's theorem to find harmonic elements on $W_i^{\perp}$ and thereby compute the cohomology classes on $W_i^{\perp}\cong V_i/W_i$?

I believe I would need to redefine $\mathcal{Q} \to \mathcal{Q}_{\perp}$ using the maps $\phi_i$ such that the action of $\mathcal{Q}_{\perp}$ by construction always "subtracts off" the right amount from a vector $w \in V_{i+1}$ to consistently map it to $W_{i+1}^{\perp}$. In essence, if $\pi: V_i \to V_i/W_i$ is the canonical projection map defined by $\pi(v) = [v]$, then once the map $\phi_i$ is constructed, I think I can define $\mathcal{Q}_{\perp} := \phi_{i+1}\circ \pi \circ \mathcal{Q} \circ \phi_i$. However, I am uncertain about this and do not know if there is a general method for constructing the $\phi_i$ and therefore choosing $W_i^{\perp}$ since it isn't canonical.

An explicit, simple, example illustrating the the computational process in principle would be very helpful.

This question has been migrated from Math Stack Exchange.

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  • $\begingroup$ Are the inclusions $W_i \subset V_i$ somehow compatible with $\mathcal{Q}$? If not, there's probably no unique way to define $\mathcal{Q}_\perp$ and you could do what you like. You should probably add the relevant information about the relation between $W_i$ and $\mathcal{Q}$ to your question. $\endgroup$ Commented Mar 25, 2021 at 16:00
  • $\begingroup$ @IgorKhavkine I'm not exactly sure what you mean by compatible, but in general if $w \in W_i$ then $\mathcal{Q}(w) \in W_{i+1}$. I will put this in the question. My main difficulty lies in actually constructing $\phi_i$, if that is the correct approach. I'm not sure how to do this in principle. $\endgroup$
    – EdRich
    Commented Mar 25, 2021 at 16:37

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If $W \subset V$ is a complemented subspace (you seem to be working in the finite dimensional context, so it always is) and $W'\subset V$ is a complement, this means that you get $W' \cong V/W$ and a direct sum decomposition $V \cong W \oplus W'$. According to your updated question, $W$ is an invariant subspace $\mathcal{Q}W \subset W$. So, with respect to the direct sum decomposition, the operator $\mathcal{Q}$ takes the block matrix form $$ \mathcal{Q} = \begin{bmatrix} * & * \\ 0 & \mathcal{Q}_\perp \end{bmatrix} . $$ The $\mathcal{Q}_\perp$ block is the operator that you want. If you change the choice of the complement $W'$ to $W''$, $Q_\perp$ will change by conjucation with the isomorphism between $W'$ and $W''$, which would obviously not change any of its intrinsic properties.

You seem to be favoring $W' \cong W^\perp$ with respect to your inner product. So all you need is to express $Q$ with respect to the direct sum decomposition $V \cong W \oplus W^\perp$ and extract the $\mathcal{Q}_\perp$ block.

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