Generators and dgas

Let D be a triangulated category. Let E be an object of D. We say that E is a generator if for every nonzero object K we have that Hom(E, K[n]) is nonzero for some n. As discussed earlier on this blog, if X is a quasi-compact and quasi-separated scheme, then D_{QCoh}(X) has a perfect generator E and in this case we get D_{Qcoh}(X) = D(A) where A is the dga of endomorphisms of E. In a formula

A = RHom(E, E)

(To get an actual dga you find a K-injective complex of O_X-modules representing E and you take its endomorphisms in the differential graded category of O_X-modules.) Moreover, it then follows that D^b_{Coh}(X) = D_{compact}(A).

Let X be a smooth projective curve over a field k. Let E be a vector bundle on X. Then it turns out that E is a generator for D_{QCoh}(X) if and only if for every nonzero vector bundle F on X either H^0(X, F ⊗ E) or H^1(X, F ⊗ E) is nonzero. (We omit the proof of this statement.) Take for example E = L ⊕ N for some invertible sheaves L and N. Then it follows from Riemann-Roch that E is a generator if L and N have different degrees.

Let’s specialize even further. Assume the genus g of X is not 1. Take L = O_X and N general of degree g – 1. Then H^0(N) = H^1(N) = 0 and H^0(N^*) = 0. Thus the differential graded algebra A has cohomology H^*(A) given by

H^0(A) = k x k

(put these two factors k along the diagonal in a 2×2 matrix) and

H^1(A) = H^1(O_X) ⊕ H^1(N^*) ⊕ H^1(O_X)

(put these in an upper triangular 2×2-matrix) and all other cohomologies are zero. It follows from this that the algebra structure on H^*(A) depends only on the integer g and not on the isomorphism class of X (or N for that matter). However, since we know that the derived category of X determines the isomorphism class of X we see that there are moduli hinding in the dga A! In particular this shows (in a very roundabout way) that there is a positive dimensional family of isomorphism classes of dgas A with cohomology algebra as indicated above.

I have no doubt there are simpler examples of this phenomenon. Please leave a comment or email me if you have one or a reference. Thanks!