# Wednesday, February 15

:::{.definition title="Gerstenhaber bracket"}
For $f\in \Hom_k(A\tensorpowerk m, A)$ and $g\in \Hom_k(A\tensorpowerk n, A)$, set 
\[
[f, g] \da f\circ g - (-1)^{m-1} g \circ f
\]
where 
\[
(f \circ g )(a_1 \otimes \cdots \otimes a_{m+n-1}) \da \\ 
\sum_{i=1}^m (-1)^{(n-1)(m-1)} f(a_1 \otimes \cdots \otimes a_{i-1} \otimes g\qty{a_i \otimes \cdots \otimes a_{n+i-1}} \otimes a_{n+i} \otimes \cdots \otimes a_{m+n-1} )
.\]
:::

:::{.lemma title="?"}
Let $f, g$ as above and $h\in \Hom_k(A\tensorpowerk p, A)$. 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 $\abs f = m-1, \abs g = n-1, \abs h = p-1$ and $f g \da f\circ g$.

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

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

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

:::{.remark}
Properties 1 and 2 make \( \bigoplus _{i\geq 0} \Hom_k( {A\tensorpowerk i, A} ) \) into a graded Lie algebra, and property 3 makes it into a DGLA with graded derivation $\delta$: 
for $f$ as above,
\[
\delta(f) \da (-1)^{\abs f}d^*(f), \quad \delta([f,g]) = [\delta(f), g] + (-1)^{\abs f}[f, \delta(g)]
.\]
Thus $\HoH^*(A)$ is a graded Lie algebra.
:::

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

1. 
\[
(-1)^{(\abs f + 1)(\abs g + 1)}f \cupprod g - g\cup f = d^*(g) \circ  f + (-1)^{\abs f + 1} d^*(g \circ f) + (-1)^{\abs f} g \circ d^*(f)
.\]

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

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

:::{.theorem title="?"}
Let $A\in \Assoc\kalg$ for $k\in \CRing$.
Then the cup product on $\HoH^*(A)$ is graded commutative, so \( a \cupprod b = (-1)^{\abs a \abs b }b \cup a \) for $a\in \HoH^m(A), b\in \HoH^n(A)$ and $\abs{a}\da m, \abs b\da n$.
:::

:::{.proof title="?"}
Let $a,b$ be images of cocycles $f,g$ in $\Hom_k(A\tensorpowerk m, A)$ and $\Hom_k(A\tensorpowerk n, A)$ respectively.
By part 1 of the lemma, 
\[
(-1)^{\abs a \abs b} f \circ g - g \circ f = d^*(g) \circ f - (-1)^{\abs a} d^* (g \circ f) + (-1)^{\abs a - 1}g \circ d^*(f)
.\]
Since $f,g$ are cocycles, $d^*(f) = d^*(g) = 0$, so 
\[
(-1)^{\abs a \abs b} f \cupprod g = g \cupprod f + (-1)^{\abs a} d^*(g \circ f) 
.\]
The error term vanishes in homology yielding 
\[
(-1)^{\abs a \abs b} a \cupprod b = b \cupprod a \quad \in \HoH^*(A)
.\]
:::

:::{.lemma title="?"}
Let $a\in \HoH^m(A)$ and $b\in \HoH^n(A)$ and $g \in \HoH^p(A)$.
Then 
\[
[g, a \cupprod b] = [ g, a]\cupprod b + (-1)^{\abs a \cdot (\abs g - 1)} a \cupprod [g, b]
.\]
:::

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

:::{.definition title="Gerstenhaber algebras"}
A **Gerstenhaber algebra** or **$G\dash$algebra** $(H, \cupprod, [])$ is a free $\ZZ\dash$graded $k\dash$module $H$ where $(H, \cupprod)$ 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="?"}
$\HoH^*(A)$ is a Gerstenhaber algebra.
:::