2 responses to the question about the bar construction.......DMD
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Subject: Re: question abt bar construction
Date: Fri, 1 Mar 2002 17:37:52 +0100 (MET)
From: Johannes.Huebschmann@agat.univ-lille1.fr (Johannes Huebschmann)

Answer to Palmeri's question:

Eilenberg-Mac Lane's description of the bar construction
may be found e. g. in:
On the groups H(\pi,n), I. Annals of Math. 58 (1953) 55-106.
In this paper, the suspension is hidden in what is called
"simplicial dimension". Given A, an element
[a_1|a_2| ... |a_m] of BA has "tensor dimension"
|a_1|+ ... + |a_m| and simplicial dimension m; thus, when
q=|a_1|+ ... + |a_m|, the element [a_1|a_2| ... |a_m] lies in B_{m,q}.
The operator d_0 has bidegree (0,-1) while the operator
d_1 has bidegree (-1,0). Thus the correct bidegree of
sA_{i_1} \otimes sA_{i_2}  \otimes ...\otimes  sA_{i_m} is
(m,i_1+i_2+...+i_m).

HUEBSCHMANN Johannes
Professeur de Mathématiques
USTL, UFR de Mathématiques
UMR 8524 AGAT
F-59 655 Villeneuve d'Ascq Cédex  France

TEL. (33) 3 20 43 41 97
     (33) 3 20 43 42 33 (secretariat)
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e-mail Johannes.Huebschmann@agat.univ-lille1.fr
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Subject: Re: question abt bar construction
Date: Fri, 1 Mar 2002 13:49:11 -0500
From: Tom Goodwillie <tomg@math.brown.edu>

Two ways to look at it, and you just have to stick with one or the other:

First way:

BA is the direct sum of the tensor powers of I.
It is bigraded:
First grading is "tensor-length".
Second grading is "internal grading".
So if a_i is in degree n_i in A then [a_1|...|a_k] is in bidegree
(k,n_1+...n_k).
The differentials have bidegrees (0,-1) and (-1,0).
Total degree means the sum of the two degrees.

Second way:

BA is the direct sum of the tensor powrers of sI.
It is bigraded:
First grading is "tensor-length".
Second grading is "internal grading".
So if a_i is in degree n_i in A then [sa_1|...|sa_k] is in bidegree
(k,k+ n_1+...n_k).
The differentials have bidegrees (0,-1) and (-1,-1).
Total degree means the internal grading.