# What is the cohomology of this complex?

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!

Pierre

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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). –  David Roberts May 26 '11 at 6:09
@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. –  Theo Johnson-Freyd May 26 '11 at 6:13
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 $X^{\times n}$. –  Theo Johnson-Freyd May 26 '11 at 6:15
@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... –  Pierre May 26 '11 at 6:37

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

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

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

Therefore

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

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

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Thanks, this is a great answer! –  Pierre May 26 '11 at 19:06
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 $d^0x=d^1x$.