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Simplicial modules


\begin{defn}
A \textbf{simplicial module} is a functor
\begin{displaymath}
\Dl^{op}\to \Mod_k,\quad [n]\mapsto M_n
\end{displaymath}\end{defn}
There is a chain complex associated to simplicial module

\begin{displaymath}
M_{\bullet}: \quad\hdots\to M_n\xrightarrow{b_n} M_{n-1}
\xrightarrow{b_{n-1}} M_{n-1}\to \hdots
\end{displaymath}

where $b=b_n=\sum_{i=0}^n (-1)^i d_i$. We have $b^2=0$ as an immediate consequence of $d_id_j=d_{j-1}d_i$, $i<j$, for example:

\begin{displaymath}
\hdots M_2\xrightarrow{d_0-d_1+d_2} M_1\xrightarrow{d_0-d_1} M_0
\end{displaymath}


\begin{displaymath}
(d_0-d_1)(d_0-d_1+d_2)=\underbrace{d_0d_0-d_0d_1}_{0}+
\underbrace{d_0d_2-d_1d_0}_{0}+\underbrace{d_1d_1-d_1d_2}_{0}=0
\end{displaymath}

We define the homology of a simplicial module as

\begin{displaymath}
\rH_n(M_{\bullet}):=\ker(b_n)/ \im(b_{n-1})
\end{displaymath}

It is well defined for presimplicial module, that is using only face maps.
\begin{lem}
The submodule $M'_n$\ of $M_n$\ spanned be the
degeneracy elements gives a subcomplex $M'_{\bullet}$
of $M_{\bullet}$.
\end{lem}

\begin{proof}
This is a consequence of the relations between $s_j$, $d_i$.
\end{proof}
Define the normalized complex $\Bar{M}_{\bullet}$ as a quotient

\begin{displaymath}
0\to M'_{\bullet} \to M_{\bullet} \to \Bar{M}_{\bullet}\to 0
\end{displaymath}


\begin{thm}
The quotient map $M_{\bullet} \to \Bar{M}_{\bullet}$
is an quasi-isomorphism, i.e. it induces an isomorphism
in homology.
\end{thm}

\begin{proof}
From the long exact sequence in homology
\begin{displaymath}
\hdot...
... construct a contracting homotopy.
Then we can proceed by induction.
\end{proof}
Let $A$ be the $k$-algebra and $M$ an $A$-module. There is a simplicial module

\begin{displaymath}
C_{\bullet}(A, M):=M\ox A^{\ox n}
\end{displaymath}


\begin{align*}
d_i(a_0, a_1, \hdots, a_n) &= (a_0, \hdots, a_i a_{i+1}, \hdots, ...
..., a_1, \hdots, a_n) &= (a_0, \hdots, a_j, 1, a_{j+1}, \hdots, a_n )
\end{align*}
Define

\begin{displaymath}
b:=\sum_{i=0}^n (-1)^i d_i
\end{displaymath}

Then $(C_{\bullet}(A, M), b)$ is called the Hochschild chain complex, and its homology $\rH_*(A; M)$ the Hochschild homology of $A$ with coefficients in $M$. If $M=A$, then we denote

\begin{displaymath}
\rH_*(A; A)=: \HH_*(A)
\end{displaymath}

Suppose that $A$ is augmented and let $\Bar{A}$ be its augmentation ideal in $A$, that is $A=\Bar{A}\oplus k1$. Define the reduced Hochschild complex as

\begin{displaymath}
\Bar{C}_n(A, M):= M\ox \Bar{A}^{\ox n}
\end{displaymath}

If $M=A=\bar{A}\oplus k1$, then $C_{\bullet}(A, A)$ has extra degeneracy

\begin{displaymath}
s_{-1}(a_0, \hdots, a_n)=(1, a_0, \hdots, a_n).
\end{displaymath}

We have
\begin{align*}
d_0(1, a_1, \hdots, a_n) &= (a_1, \hdots, a_n) \\
d_n(1, a_1, \hdots, a_n) &= (a_n, \hdots, a_1)
\end{align*}
Define also two maps on $\Bar{A}^{\ox n}$
\begin{align*}
t(a_1, \hdots, a_n) &:= (-1)^n(a_n, a_1, \hdots, a_{n-1})\\
b' &:= \sum_{i=0}^{n-1}(-1)^i d_i, \quad (b= b'+ (-1)^n d_n)
\end{align*}


next up previous contents
Next: Bicomplexes Up: Cyclic category Previous: Generic example of a   Contents
Pawel Witkowski 2006-11-07