Derived functors

The Hochschild homology of an algebra $A$ over a field $k$ with coefficients in an $A$-bimodule $M$ can be interpreted as a derived functor

The definition of the derived functor $\Tor_n^{A^e}$ goes as follows. Having an exact sequence of right $A^e$-modules

$$0\to M'\to M\to M''\to 0$$

we tensor it with $A$ over $A^e$ to get a sequence which is exact on the right

$$M'\ox_{A^e} A\to M\ox_{A^e} A\to M''\ox_{A^e} A\to 0,$$

but the map $M'\ox_{A^e} A\to M\ox_{A^e}$ can have a nontrivial kernel, which we define as $\Tor_1^{A^e}(M'', A)$. Next we can define in an analogous way $\Tor_1^{A^e}(M, A)$ and $\Tor_1^{A^e}(M', A)$ which fit into an exact sequence

$$\Tor_1^{A^e}(M', A)\to \Tor_1^{A^e}(M, A)\to \Tor_1^{A^e}(M', A)\to M'\ox_{A^e} A\to M\ox_{A^e} A\to M''\ox_{A^e} A\to 0.$$

General construction uses a resolution of $A$ by free left $A^e$-modules, $C_{\bullet}\epi A\to 0$,

$$\xymatrix{ \hdots \ar[r] & C_2\ar[r] & C_1 \ar[r] & C_0 \ar@{->>}[d] \\ &&& A \ar[d] \\ &&& 0 }$$

Then we define

$$\Tor_n^{A^e}(M, A):= \rH_n(M \ox_{A^e} C_{\bullet}).$$

As a resolution we can take $C_n:= A^e\ox A^{\ox n}$ and obtain the isomorphism $\rH_n(A, M)\isom \Tor_n^{A^e}(M, A)$.

Recall that the simplicial module $C_{\bullet}$ is a functor $\Dl^{op}\to \Mod$, for example $[n]\mapsto M\ox_{A^e} A^n$. The homology of $C_{\bullet}$ with respect to $b=\sum_i (-1)^i d_i$ can be written as a derived functor

$$\rH_n(C_{\bullet})\isom \Tor_n^{\Dl^{op}}(k, C_{\bullet}),$$

where $C_{\bullet}$ is a left module over $\Dl^{op}$, and $k$ is a right module over $\Dl^{op}$, that is a functor $\Dl\to \Mod$, $[n]\mapsto k$. The resolution for $k$ can be given by

$$\xymatrix{ \hdots \ar[r] & k[\Hom_{\Dl}([n], -)] \ar[r] & \h... ...], -)] \ar[r] & k[\Hom_{\Dl}([0], -)]\ar@{->>}[d] \\ &&&& k }$$

In general for a category $\sC$ we have a following correspondence

Category $\sC$	Algebra $A$
Functor $F\:\sC\to \Mod$	Left $A$-module $M$
Functor $G\:\sC^{op}\to \Mod$	Right $A$-module $N$
Tensor product over a category $G\ox_{\sC} F$	Tensor product over algebra $N\ox_A M$

The tensor product over a category is defined as

$$G\ox_{\sC} F:= \bigoplus_{C\in\Ob(\sC)}G(C)\ox F(C)/\sim,$$

where the equivalence relation $\sim$ is given by

$$y\ox f_*(x)\sim f^*(y)\ox x, \quad C\xrightarrow{f} D, \quad x\in F(C),\; y\in G(D),$$

$$F(C)\xrightarrow{f_*} F(D), \quad G(C)\xleftarrow{f^*} G(D).$$

Using cyclic category $\Dl C$ we can present cyclic homology of a cyclic module $C_{\bullet}$ as a derived functor.

We can write $\Tor_0^{\sC}(G, F)$ simply as the tensor product $G\ox_{\sC} F$. To define higher derived functors $\Tor_n^{\sC}(G, F)$ we need a notion of a free module over a category. Let $\sC^{triv}$ be the category with the same objects as $\sC$, but with only the identity morphisms. For a functor $F\:\sC\to \Mod$ there is a corresponding forgetful functor $\mathrm{forget(F)}\:\sC^{triv}\to \Mod$. Suppose we have an adjoint pair

$$\xymatrix{ \Funct(\sC, \Mod) \ar@<1ex>[rr]^{\text{forgetful}... ...\Funct(\sC^{triv}, \Mod) \ar@<1ex>[ll]^{\text{left adjoint}} }$$

Then we say that a functor $F\:\sC\to \Mod$ is free if it is an image of this left adjoint functor to a forgetful functor. For example

$$A-\Mod \to k-\Mod$$

has a left adjoint

$$k^n\mapsto A^n.$$