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Hi! This is my first question here, so please excuse me if it is too elementary.

I was wondering if the notion of a simultaneous decomposition into eigenspaces could be generalized in a special way I describe below. Let $V$ be a vector space over an algebraically closed field $k$ and let $T \subset \mbox{End}(V)$ be a finite dimensional subspace consisting of pairwise commuting and diagonizable endomorphisms. Than we have a decomposition

$\begin{align*} V = \bigoplus_{\lambda \in T^{\*}} V_{\lambda}, \end{align*}$

where $V_{\lambda} = \lbrace v\in V \hspace{0.3em}\lvert \hspace{0.3em} xv = \lambda v \mbox{ for all } x\in T \rbrace$ and $T^{\*}$ is the dual space of $T$.

I was wondering now if a very similar thing in another context might be possible as well. Some notations first. Let $V$ be as above and let $f \in \mbox{End}(V)$. Then set $\mbox{Hau}(f,\lambda) = \bigcup_{n\ge 0} \mbox{ker}(f-\lambda\cdot\mbox{id})^n$. It is known that $V = \bigoplus_{\lambda\in k} \mbox{Hau}(f,\lambda)$ if and only if $f$ is locally finite.

Now let $S\subset \mbox{End}(V)$ be an abelian, finitely generated subalgebra such that each $x\in S$ is locally finite. By $S^{\times}$ I denote the set of algebra homomorphisms $S\to k$ (that map $1$ to $1$) and for $\chi\in S^{\times}$ I denote

$\begin{align*} \mbox{Hau}_s(S,\chi) = \bigcap_s\mbox{Hau}(s,\chi(s)), \end{align*}$

where $s$ runs over all $s\in S$. My question now is the following: is it true that

$\begin{align*} V = \bigoplus_{\chi\in S^{\times}}\mbox{Hau}_s(S,\chi) ? \end{align*}$

I have serious difficulties proving it. My attempts so far have been that $S$ must be isomorphic to $k[x_1, \dots, x_l]/I$ for some $l$, and I tied induction over $l$. The above equality seemed basic enough for me to be found in any text on linear algebra - I thought. But I did not find it. I would be very very glad for any pointers to literature or anything else. Or is the statement false in this way?

Thank you very much.


To give you my motivation for such a question: In the representation theory in the context of category $\mathcal{O}$, $\mathcal{O}$ can be decomposed into blocks, parameterised by algebra homomorphisms $\chi: \mbox{Z}(\mathfrak{g})\to k$, where each $M\in \mathcal{O}_{\chi}$ satisfies

$\begin{align*} \forall z\in \mbox{Z}(\mathfrak{g}) \forall v\in M: (z-\chi(z))^n v = 0 \mbox{ for some } n>0 \mbox{ depending on } z. \end{align*}$

Since $M$ is an $\mbox{U}(\mathfrak{g})$-module, we get an algebra homomorphism $\mbox{Z}(\mathfrak{g})\to \mbox{End}(M)$. $\mbox{Z}(\mathfrak{g})$ is known to be isomorphic to a polynomial algebra in finitely many variables. The image of $\mbox{Z}(\mathfrak{g})$ under this morphism would play the role of $S$ in the above paragraph, and if I had the statement I want to prove, it would explain why each $\mbox{U}(\mathfrak{g})$-module decomposes into a direct sum, where each summand belongs to a block...

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2 Answers

up vote 2 down vote accepted

I'm not sure if this precise formulation is standard linear algebra, but it is true. The important point is that $S$ acts locally finitely on $V$ (be careful, this only works because $S$ is commutative): if $v$ is a random vector, and $x_i$'s generators of $S$, then there's some minimal $m_i$ such that $x_i^{m_i}v=a_{m_i-1}x_i^{m_i-1}v+\cdots$, and the space $S\cdot v$ is spanned by monomials in the $x_i$'s where the power of $x_i$ is less than $m_i$ (just check that any linear combination of these times an $x_i$ can written in this form, using the relation above).

Thus, one can decompose any vector $v\in V$ by simply considering $S\cdot v$ and decomposing this using the finite dimensional result. This shows that $V$ is the sum of these subspaces, and their intersections are trivial essentially by definition.

For category $\mathcal{O}$ this is really unnecessary though; you can consider the action of the center on the endomorphism space (which is finite dimensional) of your module, and the projections of the identity to the different generalized eigenspaces will be idempotents projecting to the desired block decomposition.

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First of all: thank you very much! That helps me a lot! But do you mean that my module itself is finite dimensional? One typical example would be a verma module, and they are in generally not finite dimensional. Or do you mean that the endomorphism spaces are finite dimensional? If true, why are they finite dimensional? (The merely linear endomorphisms of a verma would also form an infinite dimensional space). I'm not at all trying to argue with you, just trying to learn. Thank you! –  Sh4pe Dec 12 '11 at 16:20
I meant the endomorphism space. This is finite dimensional since any $\mathfrak{g}$-module map between objects in category $\mathcal{O}$ is determined by what it does on finitely many weight spaces (those which appear as highest weights in composition factors) and the space of vector space maps between those weight spaces is finite dimensional. For example, any endomorphism of a Verma module sends the highest weight vector to a highest weight vector, and thus is scalar multiplication. –  Ben Webster Dec 12 '11 at 19:41
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Ben has addressed the general question here, which as he suggests is not "standard" linear algebra. I guess this set-up might be relevant in categories of modules with fewer finiteness restrictions than category $\mathcal{O}$. Anyway it's probably helpful to add explicit references for the motivating example, since that much is standard by now in representation theory of Lie algebras. The first chapter of my 2008 AMS text on the BGG category discusses central characters. Theorem 1.11 shows explicitly that the Hom space is finite dimensional for any two modules in the category. There is also a discussion of block decomposition in the category (1.13), though I prefer to define "block" more narrowly than is done in the question when the weights involved are non-integral (4.9).

By the way, in the first part of the question one unhelpful phrase should be omitted: "and $T^*$ is the dual space of $T$."

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With this unhelpful phrase I was trying to set up the notation I would like to use, since I've seen other notations for dual spaces as well. I agree that this is somewhat lengthy and too verbose. Apart from that, do you think that such a phrase is harmful? Thank you very much for you suggestions! I appreciate that. –  Sh4pe Dec 14 '11 at 10:00
I was just pointing out that the dual space of T plays no further role in the question here. A more concise formulation of questions is always welcome, as long as essential detail and notation is included. –  Jim Humphreys Dec 14 '11 at 23:25
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