Subject: question on bar construction Date: Thu, 28 Feb 2002 14:31:48 -0800 From: John H Palmieri To: Don Davis Hi Don, Here's a question for the list: I'm confused about grading and the bar construction. Let (A,d_A) be an augmented differential graded algebra, with d of degree -1. Let I be the augmentation ideal. Then the bar construction BA on A is the tensor coalgebra on sI, where sI means the suspension of I, with a differential d. (This is the part where I'm confused: the grading on the differential.) d is the sum of two pieces, d = d_0 + d_1, where d_0 is induced by the differential on A: d_0 [ sa_1 | sa_2 | ... | sa_n ] = Sum (+/-) [ sa_1 | ... | s d_A(a_i) | ... | sa_n ] (sx in sI means the suspension of x in I.) So d_0 preserves tensor-length and decreases "internal degree" by 1. d_1 is induced by the product on A: d_1 [ sa_1 | sa_2 | ... | sa_n ] = Sum (+/-) [ sa_1 | ... | s (a_{i-1} a_i) | ... | sa_n ] So d_1 decreases tensor-length by 1 and decreases internal degree by 1 (because of there are n suspensions s in the source and n-1 suspensions in the target). I would feel more comfortable if d_0 had bidegree (0,-1) and d_1 had bidegree (-1,0), but instead, d_1 has bidegree (-1,-1). This seems odd to me. Can anyone reassure me? What is the role of the suspension s (which seems to be absent in the original papers of Eilenberg-MacLane and Adams)? -- J. H. Palmieri Dept of Mathematics, Box 354350 mailto:palmieri@math.washington.edu University of Washington http://www.math.washington.edu/~palmieri/ Seattle, WA 98195-4350