I have a feeling that this may be a very easy question for some people on MO, but it isn't for me.

Take a finite pointed set $X$, with $*$ the base-point. Build a cosimplicial set which in degree $n \ge 1$ is $X^n$ (the cartesian product of $n$ copies of $X$), and in degree $0$ is $\{ * \}$; the cofaces are:

$d^0(x_1, \ldots, x_n) = (*, x_1, \ldots, x_n)$

$d^i(x_1, \ldots, x_n) = (x_1, \ldots, x_i, x_i, x_{i+1}, \ldots, x_n)$

$d^{n+1}(x_1, \ldots, x_n) = (x_1, \ldots, x_n, *)$

Now apply the functor "free $k$-module on", where $k$ is your favorite ring. You get a cosimplicial $k$-module $A^*$, so you can build the associated cochain complex where the differential is the alternating sum of the cofaces. Note that $A^n = (A^1)^{\otimes n}$.

What is the cohomology of this complex?

Ideally someone will say something like "this is the cobar construction, it computes the cohomology of the loop space on the discrete space $X$, so the cohomology is $0$ in positive degrees", or something close. And it would be awesome. (The buzzword "cobar construction" seems to show up a lot among the papers I've skimmed through online.)

Thank you so much for your help!


  • $\begingroup$ You have some clashes between $k$s and $A$s, unless $A$ is the free $k$-module on $(X,\ast)$. In that case, is $\ast$ a generator of this module or not? And when you say cohomology of loop space of $X$, what coefficients? $k$ or $A$? And although it doesn't matter in this instance, you need to specify free or based loop space (based, I suppose, since $X$ is pointed). $\endgroup$
    – David Roberts
    Commented May 26, 2011 at 6:09
  • $\begingroup$ @David Roberts: $k$ is OP's ground ring. $A^1$ is the free $k$ module with basis $X$ --- note that $\ast \in X$ in OP's notation. $A = A^\bullet = \bigoplus (A^1)^{\otimes \bullet}$ is the tensor algebra of $A$ as a graded $k$-module, and it is given some "cobar"-like differential. $\endgroup$ Commented May 26, 2011 at 6:13
  • $\begingroup$ Well, $\otimes$ is maybe a bad thing to write, since $k$ might not be commutative. The $n$th piece of OP's complex is the free $k$ module with basis $X^{\times n}$. $\endgroup$ Commented May 26, 2011 at 6:15
  • $\begingroup$ @Theo: yep, all of that is correct. And I meant for $k$ to be commutative, in fact I only care about the case when $k$ is a field, really. @David: I don't want to be specific about the answer which I expect, I guess it may have loop spaces, but I'm not sure... $\endgroup$
    – Pierre
    Commented May 26, 2011 at 6:37

2 Answers 2


If I've understood well your construction, the complex $\hom_k(A,k)$ is the Hochschild complex of the cochain $k$-algebra $C^\star(X,k)$ of the (discrete) space $X$ with coefficients in the $C^\star(X,k)$-module $k$. The $C^\star(X,k)$-module structure on $k$ is given via the augmentation $C^\star(X,k)\rightarrow k$ induced by the inclusion of the base point in $X$. Therefore

$$H_\star\hom_k(A,k)\cong HH_\star(C^\star(X,k),k).$$

Since $X$ is discrete then $C^\star(X,k)=k\times\stackrel{n}\cdots\times k=k^n$ concentrated in degree $0$, where $n=|X|$ is the number of points. Now the Künneth formula shows that

$$H_0\hom_k(A,k)\cong HH_0(C^\star(X,k),k)\cong k,$$

$$H_d\hom_k(A,k)\cong HH_d(C^\star(X,k),k)=0,\quad d\neq 0,$$


$$H^0A\cong k,$$

$$H^dA=0,\quad d\neq 0.$$


No matter what cosimplicial set $X^\bullet$ is used, the complex will always have $H^n=0$ for $n>0$. This follows from the fact that if elements $x,y\in X^n$ are not in the image of any of the face maps $d^j:X^{n-1}\to X^n$ then the equation $d^ix=d^jy$ implies $x=y$ and $i=j$. (This suggests that there is no interesting homotopy theory of cosimplicial sets!)

A basis for $H^0$ is the set af all $x\in X^0$ such that $d^0x=d^1x$.

  • $\begingroup$ I like this answer very much, too, but I'm not sure if I can fill in the details. I can see how the "fact" you mention implies that the cohomology is 0, but how do you prove that fact based on the cosimplicial identities alone?? Do you assume the presence of codegeneracies? $\endgroup$
    – Pierre
    Commented May 26, 2011 at 19:08
  • $\begingroup$ Yes, I was assuming codegeneracies because that's normally what "cosimplicial" means. I think you have degeneracies, by omitting one x in the list. $\endgroup$ Commented May 26, 2011 at 19:59
  • $\begingroup$ I see! It's all making sense. Was this a well-known lemma? Btw I'm going to select Fernando's answer, based on the fact that... he's got less reputation points! But your answer is very instructive, too! $\endgroup$
    – Pierre
    Commented May 26, 2011 at 21:13
  • $\begingroup$ I don't know how well known. I noticed it for myself some years ago, but it must be well known in some sense. $\endgroup$ Commented May 26, 2011 at 21:46

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