By Bruno Iochum
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4 Lemma. p, then so are E* and *E. Proof. 4]. 3] we thus obtain an isomorphism Since the last one is a free on generators, the result follows. The left linear case is analogous. 5 Proposition. 26] is an isomorphism. Proof. Consider the special case of * ( E * ) . 4]. 2] twice gives us the following diagram: We leave it to the reader to verify that this diagram is actually a commutative diagram. Now let be an such that where the are free on a single homogeneous generator. 8]. Applying the commutative diagram to the finite family proves that is an isomorphism because for free on a single homogeneous generator the are isomorphisms.
In this section some technical proofs are given to show sufficient conditions for these identifications to be isomorphisms. It turns out that this is the case if the are finitely generated and projective. The condition finitely generated and projective for modules is equivalent to the condition finite dimensional for vector spaces. At the end of this section a summary of the more interesting identifications can be found. 1 Definitions. ) and such that The and are not supposed to be unique. An E is called finitely generated, or of finite type, if there exists a finite set of generators G.
However, this is hardly necessary. If E is an homogeneous and we have the relation This shows that a (left) generating set of homogeneous elements is also generating for the right module structure. And if a set of homogeneous elements is (left) independent, it is also independent for the right module structure. Hence for homogeneous sets, there is no difference between the notions of generating and independence for the left or right module structures. Moreover, by splitting into homogeneous components, any generating set can be made homogeneous.
Cônes autopolaires et algèbres de Jordan by Bruno Iochum