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0-stable Modules

By trdunlap2

For Nakajima’s analogue of MV-cycles we need to understand what it takes for a “module” to be 0-stable. Let me review first what that means.

Fix a finite set I={1 .. r} and a directed graph (I,\Omega). Let H=\Omega\sqcup\bar\Omega. For two I-graded vector spaces V and W we define
E(V,W)=\bigoplus_{h\in H} Hom(V_{s(h)},W_{e(h)}
L(V,W)=\bigoplus_{i\in I} Hom(V_i,W_i)
M(V,W)=E(V,V)\oplus L(W,V) \oplus L(V,W).
(where s and e denote the start and end vertices of an edge.)
We have a function
\epsilon:E(V,V)\rightarrow E(V,V)
which multiplies by -1 components corresponding to h\in\bar\Omega and fixes components corresponding to h\in\Omega. We have a composition
L(V,W)\times L(W,X) \rightarrow L(V,X)
which is straightforward. We have a composition
E(V,W)\times E(W,X) \rightarrow L(V,X)
where (CB)_i=\sum_{e(h)=i} C_hB_{\bar h}. And we have a “moment map vanishing at the origin”
\mu:M(V,W)\rightarrow L(V,V)
where \mu(B,a,b)=(\epsilon B)B+ab.

From now on we fix V and W.

A point (B,a,b)\in\mu^{-1}(0) is called a “module”.

A “sub-module” of (B,a,b) is a B-invariant I-graded vector space V'\subset V which either contains Im(a) or is contained in Ker(b).

A module (B,a,b) is said to be “0-stable” if the only sub-modules are 0 and V.

**EDIT: I should say (B,a,b) is zero-stable if the only sub-module containing Im(a) is V and the only submodule contained in Ker(b) is 0.

This entry was posted on April 24, 2009 at 4:24 pm and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “0-stable Modules”

  1. trdunlap2 Says:
    April 24, 2009 at 4:55 pm | Reply

    Tomorrow I give an alternate definition of 0-stable and next week some example calculations.

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