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    $\begingroup$ Nice question. I don't know enough scheme theory to be able to start on an answer. One comment, however: reconstruction theorems generally require a bit more than just the (abstract) category "of representations". More precisely, the statement of Tannaka-Krein is that given a category C with nice properties and a functor f:C->Vect with nice properties, there is a unique (and constructible, from the data) group G up to isomorphism so that (C,f) is equivalent to (G-modules,forget). You can also relax the properties a bit and replace "group" by "(quasitriangular) (quasi)Hopf algebra", etc. $\endgroup$ Commented Feb 24, 2010 at 16:29
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    $\begingroup$ I thought I'd point out that yes one needs extra structure such as the tensor product in the triangulated case. For instance consider two smooth projective birational Calabi-Yau 3-folds. They are derived equivalent (by Bridgeland) so one needs some extra data or it won't be possible to recover both schemes. I also thought I'd comment that (in my opinion) there should be some good notion of homological fibre functor which comes up when extending Balmer's theory to the compactly generated setting. Knowing these would determine the spectrum, but they should know other interesting things too! $\endgroup$ Commented Feb 24, 2010 at 20:25
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    $\begingroup$ There is a lemma in Thomason's "Classification of triangulated subcategories" that says a category is tensor-closed if and only if it closed under tensoring by powers of an ample line bundle. Thus, Bondal-Orlov's reconstruction theorem is a consequence of Balmer's result. A proof of Bondal-Orlov's result along these lines is in Rouquier's notes - see paper 38 on his webpage. One caveat: Bondal-Orlov use only the graded structure of the derived category. With this argument, you need the triangles. $\endgroup$ Commented Feb 24, 2010 at 23:36
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    $\begingroup$ If you still have your notes from last semester in the 800 class, Rosenberg gave us his perspective on Tannaka-Krein. $\endgroup$ Commented Feb 25, 2010 at 0:29
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    $\begingroup$ @Shizhuo: In the case of Balmer's approach one will recover the scheme corresponding to the tensor product one picks. If I understand correctly to use Rosenberg's approach one needs an abelian category? In this case one will get the scheme corresponding to the t-structure one picks. Either the choice of t-structure or tensor product fixes the geometry associated to the category. $\endgroup$ Commented Feb 25, 2010 at 20:13