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I'm interested in an example of a simple $\mathcal{g}$-module $M$ over some locally simple Lie algebra say $\mathcal{g}\simeq gl(\infty)$ such that $M$ is not isomorphic to a direct limit of simple finite dimension modules.

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Why not just take a Verma module with a generic highest weight (over, say, $sl_n$)? If the highest weight is generic, this should be a simple, infinite dimensional module, so it has no finite dimensional submodules. (Also –  Peter Samuelson Jun 20 '13 at 16:37
    
I want modules not over $sl_n$ but over $sl_{\infty}$. I didn't thought much about Vermas over $sl_{\infty}$ but I doubt they are integrable. –  Alex Jun 21 '13 at 12:04
    
@Alex: if you want integrable modules, you should specify it in the question. –  Yves Cornulier Sep 19 '13 at 9:52
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Taking the finitary simple Lie algebras $L$, like $\mathfrak{sl}(\infty)$, $\mathfrak{o}(\infty)$, then these locally simple Lie algebras are known to have no non-trivial finite-dimensional module at all, see the work of Penkov. In this sense there are many examples. However, if you mean locally simple, i.e., integrable modules of these Lie algebras, then this means that they are the direct limit of simple $L_i$-modules $M_i$. So we need to find non-integrable modules.
Given an integrable module $M$, the dual module $M^{\ast}$ is integrable if and only if for any $i>0$ $Hom_{L_i}(N,M) = 0$ only for finitely many non-isomorphic simple $L_i$-modules N. This should give non-integrable modules as dual modules, i.e., examples of modules not being a direct limit of finite-dimensional $L_i$-modules.

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I know about the dual module criterion from Penkov's work with Serganova. However the example I was looking for was a simple integrable $\mathfrak{g}$-module $M$ such that it cannot be exhausted by finite dimensional simple modules. But now that I think about it better I guess you can always find a sequence of simple modules at each level and a system of injective morphisms such that $M$ is the direct limit of that sequence with respect to that system. –  Alex Jun 21 '13 at 8:35
    
A simple integrable module which cannot be exhausted by finite dimensional modules need to be uncountable-dimensional, right ? –  Dietrich Burde Jun 21 '13 at 8:55
    
I was under the impression that a simple integrable module has to be countable dimensional. Or can you give me an example of uncountable dimensional simple integrable module? –  Alex Jun 21 '13 at 12:05
    
As far as I can tell, your first two sentences answer the question. If $\mathfrak{g}$ has no nontrivial finite dimensional modules, then any infinite dimensional simple integrable module cannot be a direct sum of finite dimensional simple modules. Did I miss something? –  S. Carnahan Sep 19 '13 at 11:26
    
@S. Carnahan: the question is not about "direct sum", but "direct limit". Still the question is intriguing, because there is no nontrivial way to make a direct limit of simple modules, unless it means simple modules over simple subalgebras of the Lie algebra or so. –  Yves Cornulier Sep 19 '13 at 15:16
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