Differential Tensor Algebras and their Module Categories

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Bocses ditalgebras and modules. Triangular ditalgebras. Exact structures in AMod.

Quotient ditalgebras. Frames and Roiter ditalgebras.

What is a Tensor? Lesson 20: Algebraic Structures II - Modules to Algebras

Horntensor relations and dual basis. Critical ditalgebras. Reduction functors. Modules over nonWild ditalgebras. Tameness and wildness. It can be shown that the Weyl algebra is a left and right Noetherian ring.

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Moreover, it is simple , that is to say, its only two-sided ideal are the zero ideal and the whole ring. These properties make the study of D -modules manageable. Notably, standard notions from commutative algebra such as Hilbert polynomial , multiplicity and length of modules carry over to D -modules. The associated graded ring is seen to be isomorphic to the polynomial ring in 2 n indeterminates. In particular it is commutative.


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The Hilbert polynomial is defined to be the numerical polynomial that agrees with the function. It is bounded by the Bernstein inequality. As mentioned above, modules over the Weyl algebra correspond to D -modules on affine space. The Bernstein filtration not being available on D X for general varieties X , the definition is generalized to arbitrary affine smooth varieties X by means of order filtration on D X , defined by the order of differential operators.

The characteristic variety is defined to be the subvariety of the cotangent bundle cut out by the radical of the annihilator of gr M , where again M is equipped with a suitable filtration with respect to the order filtration on D X. As usual, the affine construction then glues to arbitrary varieties.

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The Bernstein inequality continues to hold for any smooth variety X. While the upper bound is an immediate consequence of the above interpretation of gr D X in terms of the cotangent bundle, the lower bound is more subtle. Holonomic modules have a tendency to behave like finite-dimensional vector spaces.

For example, their length is finite. Holonomic modules can also be characterized by a homological condition: M is holonomic if and only if D M is concentrated seen as an object in the derived category of D -modules in degree 0.

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  • This fact is a first glimpse of Verdier duality and the Riemann—Hilbert correspondence. It is proven by extending the homological study of regular rings especially what is related to global homological dimension to the filtered ring D X. Another characterization of holonomic modules is via symplectic geometry.

    By using our site, you acknowledge that you have read and understand our Cookie Policy , Privacy Policy , and our Terms of Service. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. With regards to my background, I learned differential and Riemannian geometry from O'Neill and Lee's series.

    Your Answer

    I'm working on my algebra background which is admittedly a bit weak and trying to think algebraically about some of the constructions I'm familiar with in differential geometry. Now it doesn't make sense physically as far as I know to add tensors of different ranks or variances, but it does make sense to take their tensor product. The algebraic approach is simple: whether or not it makes sense to add tensors of different ranks, that's just what you do. If you can't simplify a sum of two terms, you just leave it as a sum of two terms.

    Braided Tensor Categories and Extensions of Vertex Operator Algebras

    This is a graded algebra over the same ring as the module, where the homogeneous components can be taken to be the tensor powers. The construction of the tensor algebra of a module is functorial, meaning any homomorphism of modules induces a graded homomorphism of tensor algebras. The construction of the homomorphism is simple; simply map the module elements, which are the homogeneous elements of degree 1, via the homomorphism into the other module, then extend to the whole algebra using the product.

    I basically had the same question as you.