The tensor product extends to graded maps. Let $f: A \to B$ and $g:X \to Y$ be two graded maps, then their tensor product $f \tensor g: A \tensor B \to X \tensor Y$ is defined as:
The graded modules together with graded maps of degree $0$ form the category $\grMod{\k}$ of graded modules. From now on we will simply refer to maps instead of graded maps. Together with the tensor product and the ground ring, $(\grMod{\k}, \tensor, \k)$ is a monoidal category. This now dictates the definition of a graded algebra.
The sign is due to \emph{Koszuls sign convention}: whenever two elements next to each other are swapped (in this case $g$ and $a$) a minus sign appears if both elements are of odd degree. More formally we can define a swap map
$$\tau : A \tensor B \to B \tensor A : a \tensor b \mapsto(-1)^{\deg{a}\deg{b}} b \tensor a. $$
The graded modules together with graded maps of degree $0$ form the category $\grMod{\k}$ of graded modules. From now on we will simply refer to maps instead of graded maps. Together with the tensor product and the ground ring, $(\grMod{\k}, \tensor, \k)$ is a symmetric monoidal category (with the symmetry given by $\tau$). This now dictates the definition of a graded algebra.
\begin{definition}
A \emph{graded algebra} consists of a graded module $A$ together with two maps of degree $0$:
@ -48,8 +53,6 @@ Again these objects and maps form a category, denoted as $\grAlg{\k}$. We will d
$$ xy =(-1)^{\deg{x}\deg{y}} yx. $$
\end{definition}
\todo{Add a remark about the signs somewhere}
\subsection{Differential graded algebra}
@ -59,15 +62,22 @@ Again these objects and maps form a category, denoted as $\grAlg{\k}$. We will d
A differential graded module $(M, d)$ with $M_i =0$ for all $i < 0$ is a \emph{chain complex}. A differential graded module $(M, d)$ with $M_i =0$ for all $i > 0$ is a \emph{cochain complex}. It will be convenient to define $M^i = M_{-i}$ in the latter case, so that $M =\bigoplus_{n \in\N} M^i$ and $d$ is a map of \emph{upper degree}$+1$.
The tensor product of two differential graded modules is again a differential graded module if we define the differential as follows. \todo{Define this}
\begin{definition}
Let $(M, d_M)$ and $(N, d_N)$ be two differential graded modules, their tensor product $M \tensor N$ is a differential graded module with the differential given by:
Finally we come to the definition of a differential graded algebra. This will be a graded algebra with a differential. Of course we want this to be compatible with the algebra structure, or stated differently: we want $\mu$ and $\eta$ to be chain maps.
\begin{definition}
A \emph{differential graded algebra (DGA)} is a graded algebra $A$ together with an differential $d$ such that in addition:
A \emph{differential graded algebra (DGA)} is a graded algebra $A$ together with an differential $d$ such that in addition the \emph{Leibniz rule} holds:
$$ d(xy)= d(x) y +(-1)^{\deg{x}} x d(y)\quad\text{ for all } x, y \in A. $$
\end{definition}
\todo{Define the notion of derivation?}
It is not hard to see that this definition precisely defines the monoidal objects in the category of differential graded modules. The category of DGAs will be denoted by $\DGA_\k$, the category of commutative DGAs (CDGAs) will be denoted by $\CDGA_\k$. If no confusion can arise, the ground ring $\k$ will be suppressed in this notation.
Let $M$ be a DGA, just as before $M$ is called a \emph{chain algebras} if $M_i =0$ for $i < 0$. Similarly if $M^i =0$ for all $i < 0$, then $M$ is a \emph{cochain algebra}.
@ -94,6 +104,13 @@ For differential graded algebras we can consider the (co)homology by forgetting
is a graded algebra.
\end{lemma}
\begin{proof}
\todo{}
\todo{Maybe just state this?}
\end{proof}
\TODO{Discuss:
\titem The Künneth theorem (especially in the case of fields)
\titem The tensor algebra $T : Ch^\ast(\Q)\to\DGA_\Q$ and free cdga $\Lambda : Ch^\ast(\Q)\to\CDGA_\Q$
\titem Coalgebras and Hopfalgebras?
\titem Define reduced/connected differential graded things
\titem If a map has LLP/RLP wrt fib/cof, it is a cof/fib
\titem Fibs are preserved under pullbacks/limits
\titem Cofibrantly generated mod. cats.
\titem Small object argument
}
\todo{Define homotopy category}
\todo{Cofibrantly generated mod cats?}
\todo{Small obj. argument?}
\subsection{Quillen pairs}
In order to relate model categories and their associated homotopy categories we need a notion of maps between them.
In order to relate model categories and their associated homotopy categories we need a notion of maps between them. We want the maps such that they induce maps on the homotopy categories.
In this section we will define a model structure on CDGAs over a field $\k$\todo{Can $\k$ be a c. ring here?}, where the weak equivalences are quasi isomorphisms and fibrations are surjective maps. The cofibrations are defined to be the maps with a left lifting property with respect to trivial fibrations.
\TODO{First discuss the model structure on (co)chain complexes. Then discuss that we want the adjunction $(\Lambda, U)$ to be a Quillen pair. Then state that (co)chain complexes are cofib. generated, so we can cofib. generate CDGAs.}
In this section we will define a model structure on CDGAs over a field $\k$ of characteristic zero\todo{Can $\k$ be a c. ring here?}, where the weak equivalences are quasi isomorphisms and fibrations are surjective maps. The cofibrations are defined to be the maps with a left lifting property with respect to trivial fibrations.
\begin{proposition}
There is a model structure on $\CDGA_\k$ where $f: A \to B$ is
@ -51,7 +53,7 @@ Next we will prove the factorization property [MC5]. We will do this by Quillen'
The maps $i_n$ are trivial cofibrations and the maps $j_n$ are cofibrations.
\end{lemma}
\begin{proof}
Since $H(T(n))=\k$ we see that indeed $H(i_n)$ is an isomorphism. For the lifting property of $i_n$ and $j_n$ simply use surjectivity of the fibrations. \todo{give a bit more detail}
Since $H(T(n))=\k$\todo{Note that this only hold when characteristic = 0}we see that indeed $H(i_n)$ is an isomorphism. For the lifting property of $i_n$ and $j_n$ simply use surjectivity of the fibrations. \todo{give a bit more detail}
Joshua Moerman
11 years ago