# Wednesday July 15th

Let $R$ be a unital (not necessarily commutative) ring, then we can consider two categories:

- $\rmod$: the category of left $R\dash$modules
- $\modr$: the category of right $R\dash$modules

Let $X\in \modr$ a consider the functor
\[
X\tensor_R \wait: \rmod &\to \ZZ\dash\text{mod} \\
Y &\mapsto X\tensor_R Y
.\]
Given a short exact sequence, we obtain a right-exact sequence, i.e. this functor is right-exact 

:::{.warning}
The induced map $X\tensor_R A \to X\tensor_R B$ need not be injective.
\begin{tikzcd}
0 \ar[r] \ar[d, "{\ZZ/(2) \tensor_\ZZ \wait }"] & \ZZ \ar[r, "\cdot 2"] & \ZZ \ar[r] & \ZZ/(2) \ar[r] & 0 \\
?? \ar[r] & \ZZ/(2) \ar[r, "{\cdot 2^*}"] & \ZZ/(2) \ar[r] & \ZZ/(2) \tensor_\ZZ \ZZ/(2) \ar[r] & 0
\end{tikzcd}

:::


Then $f_*$ is not injective, since $\qty{\id \tensor_\ZZ (\cdot 2)} (T\tensor 1)) = T\tensor 2 = 2T\tensor 1 = 0$.

Recall that this has derived functors $\tor_i^R(X, \wait)$ which vanish if either component is projective and yields a LES.

:::{.definition title="Flat Modules"}
An $R\dash$module $X$ is *flat* if $X\tensor_R \wait$ is exact.
:::


Recall that free $\implies$ projective $\implies$ flat.

Construction of $\tor_i^R(X, Y)$:

1. $\rmod$ has enough projectives, so take a projective resolution of $Y$:
\[ 
P_* \da \qty{ \cdots \to P_2 \to P_1 \to P_0 \to Y \to 0 }
\] 
  with each $P_j$ projective.
2. Apply $X\tensor_R \wait$ to $P_*$: 
\[
X\tensor_R \wait P_\wait: \cdots \to X \tensor_R P_2 \to X\tensor_R P_1 \to X\tensor_R P_0 \to 0
.\] 
  Note that this is a chain complex, but not exact in general.

3. Define 
\[
\tor_i^R(X, Y) \definedas H_i(X\tensor_R P_\wait)
,\]
i.e. take kernels mod images.

Note that choices were made, is this independent (up to natural isomorphism) of the chosen projective resolution $P_\wait$?

:::{.lemma title="?"}
Let $f: Y\to Y' \in \rmod$, and let $P_\wait, P_\wait'$ be respective projective resolutions.
Then there exists a map of chain complexes $\tilde f: P_\wait \to P_\wait'$ such that

\begin{tikzcd}
\cdots \ar[r]& P_2 \ar[r]\ar[d, "\tilde f_2"] & P_1 \ar[r]\ar[d, "\tilde f_1"] & P_0 \ar[r]\ar[d, "\tilde f_0"] & Y \ar[d, "f"] \\
\cdots \ar[r] & P_2 \ar[r] & P_1 \ar[r] & P_0 \ar[r] & Y' \\
\end{tikzcd}
where all of the squares commute, and $\tilde f$ is unique up to chain homotopy.

:::

So take $f = \id_Y: Y\to Y$, by the lemma there is a map of chain complexes $\tilde \id: P_\wait \to P_\wait'$, and applying $X\tensor_R \wait$ yields a map of chain complexes $X\tensor_R P_\wait \to X\tensor_R P_\wait'$.
Since homotopic maps yield the same map in homology, we get maps $H_i(X\tensor_R P_\wait) \to H_i(X\tensor_R P_\wait')$, and it can be checked that these maps are isomorphisms for every $i$ by using the lemma again but reversing all of the maps.

:::{.remark}
For $A\in \text{mod}\dash\ZZ$, $A$ is torsionfree $\iff \tor_1^\ZZ(A, B) = 0$ for every other $B\in \ZZ\dash\text{mod}$.
As a special case, for $k$ a field and $R$ a $k\dash$algebra, $R\tensor_k R\op$ is a $k\dash$algebra when equipped with the product $(a\tensor b)\cdot(a'\tensor b') = (aa') \tensor (bb')$.
This is called the **enveloping algebra** of $R$ 

> Note: has nothing to do with universal enveloping algebras in Lie theory.

There is an isomorphism of categories
\[
R\tensor_R R\op\dash\text{mod} &\mapsvia{\cong} R\dash\text{bimod} \\
(r_1\tensor r_2) \cdot m &\mapsfrom
r_1 \cdot m \cdot r_2 \in RMR 
.\]

The point of this is that we may not be able to make sense of projective resolutions on the RHS, but we can think of them as usual modules over the enveloping algebra instead.
:::

:::{.warning}
Some care must be taken with the monoidal structure.
The monoidal structur in the LHS is $\tensor_{R\tensor_R R\op}$, whereas it is $\tensor_R$ in the RHS.
:::

:::{.definition title="Hochschild Homology"}
Define the $i$th **Hochschild homology** of $R$ as the functor
\[
HH_i(R, \wait) \definedas \tor_i^{R\tensor_R R\op}(R, \wait) : R\dash\text{Bimod} &\to k\dash\text{mod}
.\]
:::


:::{.remark}
Note that this lands in $k\dash$modules instead of $\ZZ\dash$modules.
Moreover, everything works in the graded setting and yields a functor $R\dash\text{grBimod} \to k\dash\text{grMod}$.
:::


How does this relate to the HOMFLY-PT polynomial?
Recall that we had a trace 
\[ 
\mch(S_n) \mapsvia{\tr}  \ZZ[ v, v\inv]
,\] 
and we think of $HH$ as the categorification of the trace on the Hecke algebra.
It has trace-like behavior, namely
\[
HH_\wait(R, M\tensor_R N) \cong HH_\wait(R, N\tensor_R M) \quad \in k\dash\text{Mod}
,\]
which is similar to $\tr(ab) = \tr(ba)$.

:::{.example}
Consider $HH_\wait(\CC[t], \CC[t])$ with $\abs t = 2$.

1. Write a free resolution of $\CC[t]$ as a $\CC[t] \tensor_\CC \CC[t]\dash$module:
\begin{tikzcd}
0 \ar[r] & \CC[t] \tensor_\CC \CC[t] \ar[rr, "m_t \tensor 1 - 1\tensor m_t"] && \CC[t] \tensor_C \CC[t] \ar[r] & \CC[t] \ar[r]      & 0 \\
         & f\tensor g \ar[rr]                                                && tf\tensor g - f\tensor tg      &                    &   \\
         &                                                                   && f\tensor g \ar[r]              & f\cdot g           &
\end{tikzcd}


2. Apply $\CC[t] \tensor_{\CC[t] \tensor_\CC \CC[t]} \wait$.
\begin{tikzcd}
0 \ar[r] & \CC[t] \tensor_{\CC[t] \tensor_\CC \CC[t]} \qty{\CC[t] \tensor_\CC \CC[t] }(2) \ar[r] & \CC[t] \tensor_{\CC[t] \tensor_\CC \CC[t]} \qty{\CC[t] \tensor_\CC \CC[t] } \ar[r] & 0 \\
 & f\tensor g \tensor h \ar[r] & f\tensor tg \tensor h - f\tensor g\tensor th 
\end{tikzcd}

Note that the image is equal to
\[
f \tensor (tg\tensor h) \cdot (1\tensor 1) - f\tensor (g\tensor th)\cdot (1\tensor 1) = 
tgfh \tensor 1\tensor 1 - gfth \tensor 1 \tensor 1 = 0
,\]
so this is in fact the zero map.

We thus have
\begin{tikzcd}
0 \ar[r] & \CC[t](2) \ar[r, "0"] & \CC[t] \ar[r] & 0
\end{tikzcd}

3. Read off 
\[  
HH_0(\CC[t], \CC[t]) &\cong \CC[t] \\
HH_1(\CC[t], \CC[t]) &\cong \CC[t](2)
.\]

:::