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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Hochschild Homology
  }
    \\ {\normalsize Lectures by Tekkin. University of Georgia, Spring
2023} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in Hochschild Homology taught by Tekkin at the
University of Georgia in Spring 2023. As such, any errors or
inaccuracies are almost certainly my own. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2023-02-15 }
\end{flushleft}


\newpage

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\tableofcontents
\newpage

\hypertarget{wednesday-january-18}{%
\section{Wednesday, January 18}\label{wednesday-january-18}}

\begin{remark}

Some review:

\begin{itemize}
\item
  \(M\in {}_{k} \mathsf{Alg}\iff M\in {}_{k}{\mathsf{Mod}}\cap\mathsf{Ring}\)
  and \(\exists m: M\otimes_k M\to M\) a multiplication map.
\item
  \(M\in {\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}\iff \exists [{-},{-}]: M\otimes_k M\to M\)
  satisfying the usual identities.

  \begin{itemize}
  \tightlist
  \item
    E.g.
    \({ \operatorname{End} }_k(V) \in{\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}\)
    when \(V\in {}_{k}{\mathsf{Mod}}\).
  \end{itemize}
\item
  \(\mathop{\mathrm{Der}}_k(M)\) is \emph{not} closed under composition,
  but is a Lie algebra under
  \([\delta_1 \delta_2] \coloneqq\delta_1 \circ \delta_2 - \delta_2 \circ \delta_1\).

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

  \begin{itemize}
  \tightlist
  \item
    1-cocycles are derivations
  \item
    2-cocyles are
    \(\delta: M{ {}^{ \scriptscriptstyle\otimes_{k}^{2} } } \to M\) such
    that \(\delta(ab,c) - \delta(a, bc) = a\delta(b,c) - \delta(a,b)c\).
  \item
    \(n{\hbox{-}}\)cocycles will be
    \(\delta: M{ {}^{ \scriptscriptstyle\otimes_{k}^{n} } } \to M\)
    satisfying
    \begin{align*}
     \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} . 
    .\end{align*}
  \end{itemize}
\item
  Define \(Z^n(M)\) to be \(n{\hbox{-}}\)cycles -- this is not a Lie
  algebra for \(n\geq 2\) unless the bracket is trivial.
\item
  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.
\item
  Define boundaries \(B_n(M)\) and
  \({\operatorname{HH}}^n(M) \coloneqq Z^n(M)/B^n(M)\).

  \begin{itemize}
  \tightlist
  \item
    \({\operatorname{HH}}^1(M) = Z(M)\) is the center.
  \item
    \({\operatorname{HH}}^2(M) = \mathop{\mathrm{Der}}(M)\) when \(M\)
    is commutative.
  \end{itemize}
\item
  Recall the definitions of chain complexes and their morphisms.
\item
  Recall the different formulations of projectives \(P\) in
  \({}_{R}{\mathsf{Mod}}\):

  \begin{itemize}
  \tightlist
  \item
    \(\exists F \in {}_{R}{\mathsf{Mod}}^{\mathrm{free}}\) with
    \(F \cong P \oplus T\) for some \(T\in {}_{R}{\mathsf{Mod}}\) (not
    necessarily free).
  \item
    Every \(B\twoheadrightarrow P\) and \(B'\to P\) lifts to
    \(B'\to B\).
  \item
    Every SES \(A\hookrightarrow B\twoheadrightarrow P\) splits.
  \end{itemize}
\item
  Some useful resolutions:

  \begin{itemize}
  \tightlist
  \item
    \({\mathbf{Z}}\xrightarrow{\cdot n} {\mathbf{Z}} \xrightarrow[]{{\varepsilon}} { \mathrel{\mkern-16mu}\rightarrow }\, {\mathbf{Z}}/n{\mathbf{Z}}\)
    for \(R = {\mathbf{Z}}\) where \({\varepsilon}\) is the quotient and
    \(\ker {\varepsilon}= n{\mathbf{Z}}\).
  \item
    \(k[x] \xhookrightarrow{\cdot x} k[x] \xrightarrow[]{{\varepsilon}(x) = 0} { \mathrel{\mkern-16mu}\rightarrow }\, k \in {}_{R}{\mathsf{Mod}}\)
    for \(R = k[x]\), where the kernels are all
    \(\left\langle{x}\right\rangle\).
  \item
    \(\cdots \to k[x] \xrightarrow{\cdot x} k[x] \xrightarrow{\cdot x} k[x] \xrightarrow[]{{\varepsilon}(x) = 0} { \mathrel{\mkern-16mu}\rightarrow }\, k\)
    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.
  \end{itemize}
\end{itemize}

\end{remark}

\hypertarget{monday-january-23}{%
\section{Monday, January 23}\label{monday-january-23}}

\begin{remark}

Recall

\begin{itemize}
\tightlist
\item
  \(({-})\otimes_R B\) is right-exact for any
  \(B\in {}_{R}{\mathsf{Mod}}\) and \(\mathop{\mathrm{Hom}}_R({-}, B)\)
  is left-exact.
\item
  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.
\item
  \(\operatorname{Tor}_0^R(A, B) = A\otimes_R B\). Here
  \(B[n] \coloneqq\left\{{b\in B{~\mathrel{\Big\vert}~}nb=0}\right\}\).
\item
  \(\operatorname{Ext}_R^*(A, B) = H_*(\mathop{\mathrm{Hom}}_R(P_A, B))\).
\end{itemize}

\end{remark}

\begin{example}[?]

\begin{align*}
\operatorname{Tor}_*^R(C_n, B) \cong B/nB\cdot t^0 + B[n]\cdot t^1
\end{align*}
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,
\begin{align*}
 \operatorname{Ext}^*_{\mathbf{Z}}(C_m, B) = B[m]\cdot t^0 + B/mB \cdot t^1
.\end{align*}

\end{example}

\hypertarget{wednesday-february-01}{%
\section{Wednesday, February 01}\label{wednesday-february-01}}

\begin{exercise}[?]

Show
\({\operatorname{HH}}^* k[x] = k[x]{ {}^{ \scriptscriptstyle\oplus^{2} } }\)
and find \({\operatorname{HH}}_* k[x]\). Use the complex
\begin{align*}
k[x]{ {}^{ \scriptscriptstyle\oplus^{2} }  } \hookrightarrow k[x]{ {}^{ \scriptscriptstyle\oplus^{2} }  } \twoheadrightarrow k[x]
.\end{align*}

\end{exercise}

\begin{example}[?]

Let \(A = k[x]/\left\langle{x^n}\right\rangle\) and consider
\begin{align*}
\cdots \xrightarrow{v} A^e \xrightarrow{u} A^e \xrightarrow{v} A^e \xrightarrow{u} A^e \xrightarrow{\pi} A \to 0
\end{align*}
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
\begin{align*}
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
.\end{align*}
Apply \(\mathop{\mathrm{Hom}}_{A^e}({-}, A)\) to get
\begin{align*}
0 \to A \xrightarrow{u^*} A \xrightarrow{v^*} A \xrightarrow{u^*} \cdots
,\end{align*}
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 \(f_a(1\otimes 1) = a\):
\begin{align*}
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
\end{align*}
and similarly
\begin{align*}
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
.\end{align*}
This yields
\begin{align*}
0 \to A \xrightarrow{0} A \xrightarrow{nx^{n-1}\cdot} A \xrightarrow{0} A \to \cdots
.\end{align*}

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

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

\end{example}

\begin{exercise}[?]

How can you interpret \({\operatorname{HH}}(A; M)\) in low degrees?

\end{exercise}

\hypertarget{wednesday-february-15}{%
\section{Wednesday, February 15}\label{wednesday-february-15}}

\begin{definition}[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
\begin{align*}
[f, g] \coloneqq f\circ g - (-1)^{m-1} g \circ f
\end{align*}
where
\begin{align*}
(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} )
.\end{align*}

\end{definition}

\begin{lemma}[?]

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

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Graded anticommutativity: \([f,g] = (-1)^{(m-1)(n-1)}[g, f]\)
\item
  Graded Jacobi identity:
  \begin{align*}
  (-1)^{(m-1)(p-1)}[f, [g,h]] + (-1)^{(n-1)(m-1)} [g, [h,f]] + (-1)^{(p-1)(n-1)} [h, [f,g]]
  .\end{align*}
\item
  Graded derivation:
  \(d^*([f,g]) = (-1)^{n-1}[d^*(f), g] + [f, d^*(g)]\).
\end{enumerate}

\end{lemma}

\begin{proof}[?]

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

\textbf{Part 1}:
\begin{align*}
[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]
.\end{align*}

\textbf{Part 2}:
\begin{align*}
&(-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
,\end{align*}
and everything cancels.

\end{proof}

\begin{exercise}[?]

Check part 3, this shows why the bracket is generally difficult to
compute.

\end{exercise}

\begin{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,
\begin{align*}
\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)]
.\end{align*}
Thus \({\operatorname{HH}}^*(A)\) is a graded Lie algebra.

\end{remark}

\begin{lemma}[?]

Let \(f, g\) as above, then

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item

  \begin{align*}
  (-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)
  .\end{align*}
\item
  \([f, \pi] = -d^*(f)\) where \(\pi\) is multiplication.
\end{enumerate}

\end{lemma}

\begin{proof}[?]

Follows from a direct calculation.

\end{proof}

\begin{theorem}[?]

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
\({\left\lvert {a} \right\rvert}\coloneqq m, {\left\lvert {b} \right\rvert}\coloneqq n\).

\end{theorem}

\begin{proof}[?]

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,
\begin{align*}
(-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)
.\end{align*}
Since \(f,g\) are cocycles, \(d^*(f) = d^*(g) = 0\), so
\begin{align*}
(-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) 
.\end{align*}
The error term vanishes in homology yielding
\begin{align*}
(-1)^{{\left\lvert {a} \right\rvert} {\left\lvert {b} \right\rvert}} a \smile b = b \smile a \quad \in {\operatorname{HH}}^*(A)
.\end{align*}

\end{proof}

\begin{lemma}[?]

Let \(a\in {\operatorname{HH}}^m(A)\) and
\(b\in {\operatorname{HH}}^n(A)\) and
\(g \in {\operatorname{HH}}^p(A)\). Then
\begin{align*}
[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]
.\end{align*}

\end{lemma}

\begin{proof}[?]

See \emph{The Cohomology Structure of an Associative Algebra},
Gerstenhaber 1963.

\end{proof}

\begin{definition}[Gerstenhaber algebras]

A \textbf{Gerstenhaber algebra} or \textbf{\(G{\hbox{-}}\)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.

\end{definition}

\begin{theorem}[?]

\({\operatorname{HH}}^*(A)\) is a Gerstenhaber algebra.

\end{theorem}

\addsec{ToDos}
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\end{document}
