# Wednesday, January 18

::: {.remark}
Some review:

-   \( M\in  {}_{k}  \mathsf{Alg}\iff M\in   {}_{k}{\mathsf{Mod}}\cap\mathsf{Ring} \) and \( \exists m: M\otimes_k M\to M \) a multiplication map.

-   \( M\in {\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}\iff \exists [{-},{-}]: M\otimes_k M\to M \) satisfying the usual identities.

    -   E.g. \( { \operatorname{End} }_k(V) \in{\mathsf{Lie}{\hbox{-}} \mathsf{Alg}} \) when \( V\in   {}_{k}{\mathsf{Mod}} \).

-   \( \mathop{\mathrm{Der}}_k(M) \) is *not* closed under composition, but is a Lie algebra under \( [\delta_1 \delta_2] \coloneqq\delta_1 \circ \delta_2 - \delta_2 \circ \delta_1 \).

    -   Counterexample: \( \mathop{\mathrm{Der}}_k(k[x]) = k[x] {\frac{\partial }{\partial x}\,} \cong k[D] \) but \( {\frac{\partial }{\partial x}\,}\circ {\frac{\partial }{\partial x}\,} \) s not a derivation.

-   \( {\operatorname{HH}}(M) \) will make meaningful higher analogs of derivations, \( \delta^n: A{ {}^{ \scriptscriptstyle\otimes_{k}^{n} }  } \to A \).

    -   1-cocycles are derivations
    -   2-cocyles are \( \delta: M{ {}^{ \scriptscriptstyle\otimes_{k}^{2} }  } \to M \) such that \( \delta(ab,c) - \delta(a, bc) = a\delta(b,c) - \delta(a,b)c \).
    -   \( n{\hbox{-}} \)cocycles will be \( \delta: M{ {}^{ \scriptscriptstyle\otimes_{k}^{n} }  } \to M \) satisfying
        \[
         \sum_{i=1}^n(-1)^i \delta\left(a_1, a_2, \cdots, a_i a_{i+1}, \cdots, a_{n+1}\right)=-a_1 \delta\left(a_2, \cdots, a_{n+1}\right)+(-1)^n \delta\left(a_1, \cdots, a_n\right) a_{n+1} . 
        .\]

-   Define \( Z^n(M) \) to be \( n{\hbox{-}} \)cycles -- this is not a Lie algebra for \( n\geq 2 \) unless the bracket is trivial.

-   Gerstenhaber's idea: define a new bracket \( [{-}, {-}]: Z^m(M) \otimes_k Z^n(M) \to Z^{n+m-1}(M) \) with for \( m=n=1 \) is the commutator; this makes \( Z^*(M) \) into a graded Lie algebra.

-   Define boundaries \( B_n(M) \) and \( {\operatorname{HH}}^n(M) \coloneqq Z^n(M)/B^n(M) \).

    -   \( {\operatorname{HH}}^1(M) = Z(M) \) is the center.
    -   \( {\operatorname{HH}}^2(M) = \mathop{\mathrm{Der}}(M) \) when \( M \) is commutative.

-   Recall the definitions of chain complexes and their morphisms.

-   Recall the different formulations of projectives \( P \) in \(   {}_{R}{\mathsf{Mod}} \):

    -   \( \exists F \in   {}_{R}{\mathsf{Mod}}^{\mathrm{free}} \) with \( F \cong P \oplus T \) for some \( T\in   {}_{R}{\mathsf{Mod}} \) (not necessarily free).
    -   Every \( B\twoheadrightarrow P \) and \( B'\to P \) lifts to \( B'\to B \).
    -   Every SES \( A\hookrightarrow B\twoheadrightarrow P \) splits.

-   Some useful resolutions:

    -   \( {\mathbf{Z}}\xrightarrow{\cdot n} {\mathbf{Z}} \xrightarrow[]{{\varepsilon}} { \mathrel{\mkern-16mu}\rightarrow }\,  {\mathbf{Z}}/n{\mathbf{Z}} \) for \( R = {\mathbf{Z}} \) where \( {\varepsilon} \) is the quotient and \( \ker {\varepsilon}= n{\mathbf{Z}} \).
    -   \( k[x] \xhookrightarrow{\cdot x} k[x]  \xrightarrow[]{{\varepsilon}(x) = 0} { \mathrel{\mkern-16mu}\rightarrow }\,  k \in   {}_{R}{\mathsf{Mod}} \) for \( R = k[x] \), where the kernels are all \( \left\langle{x}\right\rangle \).
    -   \( \cdots \to k[x] \xrightarrow{\cdot x} k[x] \xrightarrow{\cdot x} k[x] \xrightarrow[]{{\varepsilon}(x) = 0} { \mathrel{\mkern-16mu}\rightarrow }\,  k \) for \( R= k[x]/\left\langle{x^2}\right\rangle \) where the kernels are all \( \left\langle{x}\right\rangle \). Note that this is an infinite periodic resolution.
:::

# Monday, January 23

::: {.remark}
Recall

-   \( ({-})\otimes_R B \) is right-exact for any \( B\in   {}_{R}{\mathsf{Mod}} \) and \( \mathop{\mathrm{Hom}}_R({-}, B) \) is left-exact.
-   For \( A\in  {\mathsf{Mod}}_{R} \) and \( B\in  {}_{R}{\mathsf{Mod}} \), define \( \operatorname{Tor}_*^R(A, B) \) as \( H_*(P_A \otimes_R B) \) where \( P_A\rightrightarrows A \) is a projective resolution.
-   \( \operatorname{Tor}_0^R(A, B) = A\otimes_R B \). Here \( B[n] \coloneqq\left\{{b\in B{~\mathrel{\Big\vert}~}nb=0}\right\} \).
-   \(  \operatorname{Ext}_R^*(A, B) = H_*(\mathop{\mathrm{Hom}}_R(P_A, B)) \).
:::

::: {.example title="?"}
\[
\operatorname{Tor}_*^R(C_n, B) \cong B/nB\cdot t^0 + B[n]\cdot t^1
\]
for any \( B\in   {}_{{\mathbf{Z}}}{\mathsf{Mod}} \) using \( {\mathbf{Z}}\xhookrightarrow{\cdot n} {\mathbf{Z}}\twoheadrightarrow C_n \) to get \( P_B = (0\to B\to B\to 0) \). Similarly,
\[
 \operatorname{Ext}^*_{\mathbf{Z}}(C_m, B) = B[m]\cdot t^0 + B/mB \cdot t^1
.\]
:::

# Wednesday, February 01

::: {.exercise title="?"}
Show \( {\operatorname{HH}}^* k[x] = k[x]{ {}^{ \scriptscriptstyle\oplus^{2} }  } \) and find \( {\operatorname{HH}}_* k[x] \). Use the complex
\[
k[x]{ {}^{ \scriptscriptstyle\oplus^{2} }  } \hookrightarrow k[x]{ {}^{ \scriptscriptstyle\oplus^{2} }  } \twoheadrightarrow k[x]
.\]
:::

::: {.example title="?"}
Let \( A = k[x]/\left\langle{x^n}\right\rangle \) and consider
\[
\cdots \xrightarrow{v} A^e \xrightarrow{u} A^e \xrightarrow{v} A^e \xrightarrow{u} A^e \xrightarrow{\pi} A \to 0
\]
where \( u = (x\otimes 1 - 1\otimes x)\cdot \) and \( v = \qty{ (x^{n-1}\otimes x^0) + (x^{n-2} \otimes x^1) + (x^{n-3}\otimes x^2) + \cdots + (x^0 \otimes x^{n-1})} \cdot \). Compute \( uv(x^i \otimes x^j) = 0 \) and \( vu= 0, \pi u = 0 \) to verify that this is a complex. Show it is exact using the contracting homotopy \( s_{-1}(1) = 1\otimes 1 \) and
\[
s_{2m}(1\otimes x^j) = - \sum_{\ell =1}^j x^{j-\ell} \otimes x^{\ell-1},\qquad s_{2m-1}(1\otimes x^j) = \delta_{j, n-1}\otimes 1
.\]
Apply \( \mathop{\mathrm{Hom}}_{A^e}({-}, A) \) to get
\[
0 \to A \xrightarrow{u^*} A \xrightarrow{v^*} A \xrightarrow{u^*} \cdots
,\]
using \( \mathop{\mathrm{Hom}}_{A^e}(A^e, A) \cong A \) via \( f\mapsto f(1\otimes 1) \). Show that \( u^*(a) = 0 \) for \( a\in A \) corresponding to \( f_a \) where \( f_a(1\otimes 1) = a \):
\[
u^*(a) = u^*(f_a(1\otimes 1)) = (u^* f_a)(1\otimes 1) = f_a(u(1\otimes 1)) = f_a(x\otimes 1 - 1\otimes x) = x f_a(1\otimes 1) - f(1\otimes 1)x = xa-ax = 0
\]
and similarly
\[
v^*(a) = v^*(f_a(1\otimes 1)) = (v^* f_a)(1\otimes 1) = f_a v(1\otimes 1) = f_a(x^{n-1}\otimes 1 + \cdots + 1\otimes x^{n-1}) = x^{n-1} f_a(1\otimes 1) + \cdots + f_a(1\otimes 1) x^{n-1} = x^{n-1} a + x^{n-2}ax + \cdots + ax^{n-1} = nx^{n-1} a
.\]
This yields
\[
0 \to A \xrightarrow{0} A \xrightarrow{nx^{n-1}\cdot} A \xrightarrow{0} A \to \cdots
.\]

So the homology depends on if \( \operatorname{ch}k \divides n \):

-   If so, \( HH^* A = A + \sum_{n\geq 0} \left\langle{x}\right\rangle t^{2n+1} + \sum_{n\geq 0} A/\left\langle{x^{n-1}}\right\rangle \).
-   If not, check!
:::

::: {.exercise title="?"}
How can you interpret \( {\operatorname{HH}}(A; M) \) in low degrees?
:::

# Wednesday, February 15

::: {.definition title="Gerstenhaber bracket"}
For \( f\in \mathop{\mathrm{Hom}}_k(A{ {}^{ \scriptscriptstyle\otimes_{k}^{m} }  }, A) \) and \( g\in \mathop{\mathrm{Hom}}_k(A{ {}^{ \scriptscriptstyle\otimes_{k}^{n} }  }, A) \), set
\[
[f, g] \coloneqq f\circ g - (-1)^{m-1} g \circ f
\]
where
\[
(f \circ g )(a_1 \otimes \cdots \otimes a_{m+n-1}) \coloneqq\\ 
\sum_{i=1}^m (-1)^{(n-1)(m-1)} f(a_1 \otimes \cdots \otimes a_{i-1} \otimes g\qty{a_i \otimes \cdots \otimes a_{n+i-1}} \otimes a_{n+i} \otimes \cdots \otimes a_{m+n-1} )
.\]
:::

::: {.lemma title="?"}
Let \( f, g \) as above and \( h\in \mathop{\mathrm{Hom}}_k(A{ {}^{ \scriptscriptstyle\otimes_{k}^{p} }  }, A) \). Then

1.  Graded anticommutativity: \( [f,g] = (-1)^{(m-1)(n-1)}[g, f] \)
2.  Graded Jacobi identity:
    \[
    (-1)^{(m-1)(p-1)}[f, [g,h]] + (-1)^{(n-1)(m-1)} [g, [h,f]] + (-1)^{(p-1)(n-1)} [h, [f,g]]
    .\]
3.  Graded derivation: \( d^*([f,g]) = (-1)^{n-1}[d^*(f), g] + [f, d^*(g)] \).
:::

::: {.proof title="?"}
Define \( {\left\lvert {f} \right\rvert} = m-1, {\left\lvert {g} \right\rvert} = n-1, {\left\lvert {h} \right\rvert} = p-1 \) and \( f g \coloneqq f\circ g \).

**Part 1**:
\[
[f, g] = fg - (-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert}} g f 
&= -(-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert}}(g f - (-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert}} f g ) \\
&= -(-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert}} [g f]
.\]

**Part 2**:
\[
&(-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {h} \right\rvert}} [f, g h - (-1)^{{\left\lvert {g} \right\rvert} {\left\lvert {h} \right\rvert}} h g] \\
&+ (-1)^{{\left\lvert {g} \right\rvert} {\left\lvert {f} \right\rvert}} [g, h f - (-1)^{{\left\lvert {h} \right\rvert} {\left\lvert {f} \right\rvert}} f h] \\
&+ (-1)^{{\left\lvert {h} \right\rvert} {\left\lvert {g} \right\rvert}} [h, f g - (-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert}} g f] \\ \\
\, =& 
(-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {h} \right\rvert}} [fgh - (-1)^{{\left\lvert {g} \right\rvert} {\left\lvert {h} \right\rvert}} f h g - (-1)^{{\left\lvert {f} \right\rvert}\cdot({\left\lvert {g} \right\rvert} + {\left\lvert {h} \right\rvert}) } \qty{g h f - (-1)^{{\left\lvert {g} \right\rvert} {\left\lvert {h} \right\rvert}} h g f } \\
&(-1)^{{\left\lvert {g} \right\rvert} {\left\lvert {f} \right\rvert}} [ghf - (-1)^{{\left\lvert {h} \right\rvert} {\left\lvert {f} \right\rvert}} g f h - (-1)^{{\left\lvert {g} \right\rvert}\cdot({\left\lvert {f} \right\rvert} + {\left\lvert {h} \right\rvert}) } \qty{h f g - (-1)^{{\left\lvert {h} \right\rvert} {\left\lvert {f} \right\rvert}} f h g } \\
&(-1)^{{\left\lvert {h} \right\rvert} {\left\lvert {g} \right\rvert}} [hfg - (-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert}} h g f - (-1)^{{\left\lvert {h} \right\rvert}\cdot({\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert})} \qty{f g h - (-1)^{{\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert}} g f h } \\ \\
\, =&
(-1)^{(m-1)(p-1)} f g h - (-1)^{(p-1)(m+n-2)} f h g - (-1)^{(m-1)(n+2p-3)} g h f + (-1)^{mn + np - m - p} h g f \\
&(-1)^{(n-1)(m-1)} g h f - (-1)^{(m-1)(n+p-2)} g f h - (-1)^{(n-1)(p+2m-3)} h f g + (-1)^{mp + np -m - n} h f g \\
&(-1)^{(p-1)(n-1)} h f g - (-1)^{(n-1)(m+p-2)} h f g - (-1)^{(p-1)(m+2n-3)} f g h + (-1)^{mp + mn - p -n} f g h
,\]
and everything cancels.
:::

::: {.exercise title="?"}
Check part 3, this shows why the bracket is generally difficult to compute.
:::

::: {.remark}
Properties 1 and 2 make \( \bigoplus _{i\geq 0} \mathop{\mathrm{Hom}}_k( {A{ {}^{ \scriptscriptstyle\otimes_{k}^{i} }  }, A} ) \) into a graded Lie algebra, and property 3 makes it into a DGLA with graded derivation \( \delta \): for \( f \) as above,
\[
\delta(f) \coloneqq(-1)^{{\left\lvert {f} \right\rvert}}d^*(f), \quad \delta([f,g]) = [\delta(f), g] + (-1)^{{\left\lvert {f} \right\rvert}}[f, \delta(g)]
.\]
Thus \( {\operatorname{HH}}^*(A) \) is a graded Lie algebra.
:::

::: {.lemma title="?"}
Let \( f, g \) as above, then

1.  
    \[
    (-1)^{({\left\lvert {f} \right\rvert} + 1)({\left\lvert {g} \right\rvert} + 1)}f \smile g - g\cup f = d^*(g) \circ  f + (-1)^{{\left\lvert {f} \right\rvert} + 1} d^*(g \circ f) + (-1)^{{\left\lvert {f} \right\rvert}} g \circ d^*(f)
    .\]

2.  \( [f, \pi] = -d^*(f) \) where \( \pi \) is multiplication.
:::

::: {.proof title="?"}
Follows from a direct calculation.
:::

::: {.theorem title="?"}
Let \( A\in \mathsf{Assoc} {}_{k}  \mathsf{Alg} \) for \( k\in \mathsf{CRing} \). Then the cup product on \( {\operatorname{HH}}^*(A) \) is graded commutative, so \( a \smile b = (-1)^{{\left\lvert {a} \right\rvert} {\left\lvert {b} \right\rvert} }b \cup a \) for \( a\in {\operatorname{HH}}^m(A), b\in {\operatorname{HH}}^n(A) \) and \( {\left\lvert {a} \right\rvert}\coloneqq m, {\left\lvert {b} \right\rvert}\coloneqq n \).
:::

::: {.proof title="?"}
Let \( a,b \) be images of cocycles \( f,g \) in \( \mathop{\mathrm{Hom}}_k(A{ {}^{ \scriptscriptstyle\otimes_{k}^{m} }  }, A) \) and \( \mathop{\mathrm{Hom}}_k(A{ {}^{ \scriptscriptstyle\otimes_{k}^{n} }  }, A) \) respectively. By part 1 of the lemma,
\[
(-1)^{{\left\lvert {a} \right\rvert} {\left\lvert {b} \right\rvert}} f \circ g - g \circ f = d^*(g) \circ f - (-1)^{{\left\lvert {a} \right\rvert}} d^* (g \circ f) + (-1)^{{\left\lvert {a} \right\rvert} - 1}g \circ d^*(f)
.\]
Since \( f,g \) are cocycles, \( d^*(f) = d^*(g) = 0 \), so
\[
(-1)^{{\left\lvert {a} \right\rvert} {\left\lvert {b} \right\rvert}} f \smile g = g \smile f + (-1)^{{\left\lvert {a} \right\rvert}} d^*(g \circ f) 
.\]
The error term vanishes in homology yielding
\[
(-1)^{{\left\lvert {a} \right\rvert} {\left\lvert {b} \right\rvert}} a \smile b = b \smile a \quad \in {\operatorname{HH}}^*(A)
.\]
:::

::: {.lemma title="?"}
Let \( a\in {\operatorname{HH}}^m(A) \) and \( b\in {\operatorname{HH}}^n(A) \) and \( g \in {\operatorname{HH}}^p(A) \). Then
\[
[g, a \smile b] = [ g, a]\smile b + (-1)^{{\left\lvert {a} \right\rvert} \cdot ({\left\lvert {g} \right\rvert} - 1)} a \smile[g, b]
.\]
:::

::: {.proof title="?"}
See *The Cohomology Structure of an Associative Algebra*, Gerstenhaber 1963.
:::

::: {.definition title="Gerstenhaber algebras"}
A **Gerstenhaber algebra** or **\( G{\hbox{-}} \)algebra** \( (H, \smile, []) \) is a free \( {\mathbf{Z}}{\hbox{-}} \)graded \( k{\hbox{-}} \)module \( H \) where \( (H, \smile) \) is a commutative associative algebra and \( (H, []) \) is a graded Lie algebra, where the two operations are compatible as in the lemma above.
:::

::: {.theorem title="?"}
\( {\operatorname{HH}}^*(A) \) is a Gerstenhaber algebra.
:::