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.

EDIT:

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...

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.

EDIT:

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...

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.

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.

EDIT:

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...