## Tag `09AM`

## 83.12. Comparison with Lichtenbaum-Schlessinger

Let $A \to B$ be a ring map. In [Lichtenbaum-Schlessinger] there is a fairly explicit determination of $\tau_{\geq -2}L_{B/A}$ which is often used in calculations of versal deformation spaces of singularities. The construction follows. Choose a polynomial algebra $P$ over $A$ and a surjection $P \to B$ with kernel $I$. Choose generators $f_t$, $t \in T$ for $I$ which induces a surjection $F = \bigoplus_{t \in T} P \to I$ with $F$ a free $P$-module. Let $Rel \subset F$ be the kernel of $F \to I$, in other words $Rel$ is the set of relations among the $f_t$. Let $TrivRel \subset Rel$ be the submodule of trivial relations, i.e., the submodule of $Rel$ generated by the elements $(\ldots, f_{t'}, 0, \ldots, 0, -f_t, 0, \ldots)$. Consider the complex of $B$-modules \begin{equation} \tag{83.12.0.1} Rel/TrivRel \longrightarrow F \otimes_P B \longrightarrow \Omega_{P/A} \otimes_P B \end{equation} where the last term is placed in degree $0$. The first map is the obvious one and the second map sends the basis element corresponding to $t \in T$ to $\text{d}f_t \otimes 1$.

Definition 83.12.1. Let $A \to B$ be a ring map. Let $M$ be a $(B, B)$-bimodule over $A$. An

$A$-biderivationis an $A$-linear map $\lambda : B \to M$ such that $\lambda(xy) = x\lambda(y) + \lambda(x)y$.For a polynomial algebra the biderivations are easy to describe.

Lemma 83.12.2. Let $P = A[S]$ be a polynomial ring over $A$. Let $M$ be a $(P, P)$-bimodule over $A$. Given $m_s \in M$ for $s \in S$, there exists a unique $A$-biderivation $\lambda : P \to M$ mapping $s$ to $m_s$ for $s \in S$.

Proof.We set $$ \lambda(s_1 \ldots s_t) = \sum s_1 \ldots s_{i - 1} m_{s_i} s_{i + 1} \ldots s_t $$ in $M$. Extending by $A$-linearity we obtain a biderivation. $\square$Here is the comparison statement. The reader may also read about this in [Andre-Homologie, page 206, Proposition 12] or in the paper [Doncel] which extends the complex (83.12.0.1) by one term and the comparison to $\tau_{\geq -3}$.

Lemma 83.12.3. In the situation above denote $L$ the complex (83.12.0.1). There is a canonical map $L_{B/A} \to L$ in $D(A)$ which induces an isomorphism $\tau_{\geq -2}L_{B/A} \to L$ in $D(B)$.

Proof.Let $P_\bullet \to B$ be a resolution of $B$ over $A$ (Remark 83.5.5). We will identify $L_{B/A}$ with $\Omega_{P_\bullet/A} \otimes B$. To construct the map we make some choices.Choose an $A$-algebra map $\psi : P_0 \to P$ compatible with the given maps $P_0 \to B$ and $P \to B$.

Write $P_1 = A[S]$ for some set $S$. For $s \in S$ we may write $$ \psi(d_0(s) - d_1(s)) = \sum p_{s, t} f_t $$ for some $p_{s, t} \in P$. Think of $F = \bigoplus_{t \in T} P$ as a $(P_1, P_1)$-bimodule via the maps $(\psi \circ d_0, \psi \circ d_1)$. By Lemma 83.12.2 we obtain a unique $A$-biderivation $\lambda : P_1 \to F$ mapping $s$ to the vector with coordinates $p_{s, t}$. By construction the composition $$ P_1 \longrightarrow F \longrightarrow P $$ sends $f \in P_1$ to $\psi(d_0(f) - d_1(f))$ because the map $f \mapsto \psi(d_0(f) - d_1(f))$ is an $A$-biderivation agreeing with the composition on generators.

For $g \in P_2$ we claim that $\lambda(d_0(g) - d_1(g) + d_2(g))$ is an element of $Rel$. Namely, by the last remark of the previous paragraph the image of $\lambda(d_0(g) - d_1(g) + d_2(g))$ in $P$ is $$ \psi((d_0 - d_1)(d_0(g) - d_1(g) + d_2(g))) $$ which is zero by Simplicial, Section 14.23).

The choice of $\psi$ determines a map $$ \text{d}\psi \otimes 1 : \Omega_{P_0/A} \otimes B \longrightarrow \Omega_{P/A} \otimes B $$ Composing $\lambda$ with the map $F \to F \otimes B$ gives a usual $A$-derivation as the two $P_1$-module structures on $F \otimes B$ agree. Thus $\lambda$ determines a map $$ \overline{\lambda} : \Omega_{P_1/A} \otimes B \longrightarrow F \otimes B $$ Finally, We obtain a $B$-linear map $$ q : \Omega_{P_2/A} \otimes B \longrightarrow Rel/TrivRel $$ by mapping $\text{d}g$ to the class of $\lambda(d_0(g) - d_1(g) + d_2(g))$ in the quotient.

The diagram $$ \xymatrix{ \Omega_{P_3/A} \otimes B \ar[r] \ar[d] & \Omega_{P_2/A} \otimes B \ar[r] \ar[d]_q & \Omega_{P_1/A} \otimes B \ar[r] \ar[d]_{\overline{\lambda}} & \Omega_{P_0/A} \otimes B \ar[d]_{\text{d}\psi \otimes 1} \\ 0 \ar[r] & Rel/TrivRel \ar[r] & F \otimes B \ar[r] & \Omega_{P/A} \otimes B } $$ commutes (calculation omitted) and we obtain the map of the lemma. By Remark 83.10.4 and Lemma 83.10.3 we see that this map induces isomorphisms $H_1(L_{B/A}) \to H_1(L)$ and $H_0(L_{B/A}) \to H_0(L)$.

It remains to see that our map $L_{B/A} \to L$ induces an isomorphism $H_2(L_{B/A}) \to H_2(L)$. Choose a resolution of $B$ over $A$ with $P_0 = P = A[u_i]$ and then $P_1$ and $P_2$ as in Example 83.5.9. In Remark 83.11.6 we have constructed an exact sequence $$ \wedge^2_B(J_0/J_0^2) \to \text{Tor}_2^{P_0}(B, B) \to H^{-2}(L_{B/A}) \to 0 $$ where $P_0 = P$ and $J_0 = \mathop{\mathrm{Ker}}(P \to B) = I$. Calculating the Tor group using the short exact sequences $0 \to I \to P \to B \to 0$ and $0 \to Rel \to F \to I \to 0$ we find that $\text{Tor}_2^P(B, B) = \mathop{\mathrm{Ker}}(Rel \otimes B \to F \otimes B)$. The image of the map $\wedge^2_B(I/I^2) \to \text{Tor}_2^P(B, B)$ under this identification is exactly the image of $TrivRel \otimes B$. Thus we see that $H_2(L_{B/A}) \cong H_2(L)$.

Finally, we have to check that our map $L_{B/A} \to L$ actually induces this isomorphism. We will use the notation and results discussed in Example 83.5.9 and Remarks 83.11.6 and 83.10.5 without further mention. Pick an element $\xi$ of $\text{Tor}_2^{P_0}(B, B) = \mathop{\mathrm{Ker}}(I \otimes_P I \to I^2)$. Write $\xi = \sum h_{t', t}f_{t'} \otimes f_t$ for some $h_{t', t} \in P$. Tracing through the exact sequences above we find that $\xi$ corresponds to the image in $Rel \otimes B$ of the element $r \in Rel \subset F = \bigoplus_{t \in T} P$ with $t$th coordinate $r_t = \sum_{t' \in T} h_{t', t}f_{t'}$. On the other hand, $\xi$ corresponds to the element of $H_2(L_{B/A}) = H_2(\Omega)$ which is the image via $\text{d} : H_2(\mathcal{J}/\mathcal{J}^2) \to H_2(\Omega)$ of the boundary of $\xi$ under the $2$-extension $$ 0 \to \text{Tor}_2^\mathcal{O}(\underline{B}, \underline{B}) \to \mathcal{J} \otimes_\mathcal{O} \mathcal{J} \to \mathcal{J} \to \mathcal{J}/\mathcal{J}^2 \to 0 $$ We compute the successive transgressions of our element. First we have $$ \xi = (d_0 - d_1)(- \sum s_0(h_{t', t} f_{t'}) \otimes x_t) $$ and next we have $$ \sum s_0(h_{t', t} f_{t'}) x_t = d_0(v_r) - d_1(v_r) + d_2(v_r) $$ by our choice of the variables $v$ in Example 83.5.9. We may choose our map $\lambda$ above such that $\lambda(u_i) = 0$ and $\lambda(x_t) = - e_t$ where $e_t \in F$ denotes the basis vector corresponding to $t \in T$. Hence the construction of our map $q$ above sends $\text{d}v_r$ to $$ \lambda(\sum s_0(h_{t', t} f_{t'}) x_t) = \sum\nolimits_t \left(\sum\nolimits_{t'} h_{t', t}f_{t'}\right) e_t $$ matching the image of $\xi$ in $Rel \otimes B$ (the two minus signs we found above cancel out). This agreement finishes the proof. $\square$

Remark 83.12.4 (Functoriality of the Lichtenbaum-Schlessinger complex). Consider a commutative square $$ \xymatrix{ A' \ar[r] & B' \\ A \ar[u] \ar[r] & B \ar[u] } $$ of ring maps. Choose a factorization $$ \xymatrix{ A' \ar[r] & P' \ar[r] & B' \\ A \ar[u] \ar[r] & P \ar[u] \ar[r] & B \ar[u] } $$ with $P$ a polynomial algebra over $A$ and $P'$ a polynomial algebra over $A'$. Choose generators $f_t$, $t \in T$ for $\mathop{\mathrm{Ker}}(P \to B)$. For $t \in T$ denote $f'_t$ the image of $f_t$ in $P'$. Choose $f'_s \in P'$ such that the elements $f'_t$ for $t \in T' = T \amalg S$ generate the kernel of $P' \to B'$. Set $F = \bigoplus_{t \in T} P$ and $F' = \bigoplus_{t' \in T'} P'$. Let $Rel = \mathop{\mathrm{Ker}}(F \to P)$ and $Rel' = \mathop{\mathrm{Ker}}(F' \to P')$ where the maps are given by multiplication by $f_t$, resp. $f'_t$ on the coordinates. Finally, set $TrivRel$, resp. $TrivRel'$ equal to the submodule of $Rel$, resp. $TrivRel$ generated by the elements $(\ldots, f_{t'}, 0, \ldots, 0, -f_t, 0, \ldots)$ for $t, t' \in T$, resp. $T'$. Having made these choices we obtain a canonical commutative diagram $$ \xymatrix{ L' : & Rel'/TrivRel' \ar[r] & F' \otimes_{P'} B' \ar[r] & \Omega_{P'/A'} \otimes_{P'} B' \\ L : \ar[u] & Rel/TrivRel \ar[r] \ar[u] & F \otimes_P B \ar[r] \ar[u] & \Omega_{P/A} \otimes_P B \ar[u] } $$ Moreover, tracing through the choices made in the proof of Lemma 83.12.3 the reader sees that one obtains a commutative diagram $$ \xymatrix{ L_{B'/A'} \ar[r] & L' \\ L_{B/A} \ar[r] \ar[u] & L \ar[u] } $$

The code snippet corresponding to this tag is a part of the file `cotangent.tex` and is located in lines 2295–2566 (see updates for more information).

```
\section{Comparison with Lichtenbaum-Schlessinger}
\label{section-compare-higher}
\noindent
Let $A \to B$ be a ring map. In \cite{Lichtenbaum-Schlessinger}
there is a fairly explicit determination of $\tau_{\geq -2}L_{B/A}$
which is often used in calculations of versal deformation spaces of
singularities. The construction follows.
Choose a polynomial algebra $P$ over $A$
and a surjection $P \to B$ with kernel $I$. Choose generators
$f_t$, $t \in T$ for $I$ which induces a surjection
$F = \bigoplus_{t \in T} P \to I$ with $F$ a free $P$-module.
Let $Rel \subset F$ be the kernel of $F \to I$, in other words
$Rel$ is the set of relations among the $f_t$. Let $TrivRel \subset Rel$
be the submodule of trivial relations, i.e., the submodule of $Rel$
generated by the elements $(\ldots, f_{t'}, 0, \ldots, 0, -f_t, 0, \ldots)$.
Consider the complex of $B$-modules
\begin{equation}
\label{equation-lichtenbaum-schlessinger}
Rel/TrivRel \longrightarrow
F \otimes_P B \longrightarrow
\Omega_{P/A} \otimes_P B
\end{equation}
where the last term is placed in degree $0$. The first map is the obvious
one and the second map sends the basis element corresponding to $t \in T$
to $\text{d}f_t \otimes 1$.
\begin{definition}
\label{definition-biderivation}
Let $A \to B$ be a ring map. Let $M$ be a $(B, B)$-bimodule
over $A$. An {\it $A$-biderivation} is an $A$-linear map $\lambda : B \to M$
such that $\lambda(xy) = x\lambda(y) + \lambda(x)y$.
\end{definition}
\noindent
For a polynomial algebra the biderivations are easy to describe.
\begin{lemma}
\label{lemma-polynomial-ring-unique}
Let $P = A[S]$ be a polynomial ring over $A$. Let $M$ be a $(P, P)$-bimodule
over $A$. Given $m_s \in M$ for $s \in S$, there exists a unique
$A$-biderivation $\lambda : P \to M$ mapping $s$ to $m_s$ for $s \in S$.
\end{lemma}
\begin{proof}
We set
$$
\lambda(s_1 \ldots s_t) =
\sum s_1 \ldots s_{i - 1} m_{s_i} s_{i + 1} \ldots s_t
$$
in $M$. Extending by $A$-linearity we obtain a biderivation.
\end{proof}
\noindent
Here is the comparison statement. The reader may also read about this
in \cite[page 206, Proposition 12]{Andre-Homologie} or in the paper
\cite{Doncel} which extends the complex
(\ref{equation-lichtenbaum-schlessinger}) by one term and the comparison
to $\tau_{\geq -3}$.
\begin{lemma}
\label{lemma-compare-higher}
In the situation above denote $L$ the complex
(\ref{equation-lichtenbaum-schlessinger}).
There is a canonical map $L_{B/A} \to L$ in $D(A)$ which
induces an isomorphism $\tau_{\geq -2}L_{B/A} \to L$ in $D(B)$.
\end{lemma}
\begin{proof}
Let $P_\bullet \to B$ be a resolution of $B$ over $A$
(Remark \ref{remark-resolution}). We will identify $L_{B/A}$ with
$\Omega_{P_\bullet/A} \otimes B$. To construct the map we
make some choices.
\medskip\noindent
Choose an $A$-algebra map $\psi : P_0 \to P$ compatible with the
given maps $P_0 \to B$ and $P \to B$.
\medskip\noindent
Write $P_1 = A[S]$ for some set $S$. For $s \in S$ we may write
$$
\psi(d_0(s) - d_1(s)) = \sum p_{s, t} f_t
$$
for some $p_{s, t} \in P$. Think of $F = \bigoplus_{t \in T} P$
as a $(P_1, P_1)$-bimodule via the maps $(\psi \circ d_0, \psi \circ d_1)$.
By Lemma \ref{lemma-polynomial-ring-unique} we obtain a unique
$A$-biderivation $\lambda : P_1 \to F$ mapping $s$ to the vector with
coordinates $p_{s, t}$. By construction the composition
$$
P_1 \longrightarrow F \longrightarrow P
$$
sends $f \in P_1$ to $\psi(d_0(f) - d_1(f))$ because the map
$f \mapsto \psi(d_0(f) - d_1(f))$ is an $A$-biderivation agreeing with
the composition on generators.
\medskip\noindent
For $g \in P_2$ we claim that $\lambda(d_0(g) - d_1(g) + d_2(g))$
is an element of $Rel$. Namely, by the last remark of the previous
paragraph the image of $\lambda(d_0(g) - d_1(g) + d_2(g))$ in $P$ is
$$
\psi((d_0 - d_1)(d_0(g) - d_1(g) + d_2(g)))
$$
which is zero by Simplicial, Section \ref{simplicial-section-complexes}).
\medskip\noindent
The choice of $\psi$ determines a map
$$
\text{d}\psi \otimes 1 :
\Omega_{P_0/A} \otimes B
\longrightarrow
\Omega_{P/A} \otimes B
$$
Composing $\lambda$ with the map $F \to F \otimes B$ gives a
usual $A$-derivation as the two $P_1$-module structures on
$F \otimes B$ agree. Thus $\lambda$ determines a map
$$
\overline{\lambda} :
\Omega_{P_1/A} \otimes B
\longrightarrow
F \otimes B
$$
Finally, We obtain a $B$-linear map
$$
q :
\Omega_{P_2/A} \otimes B
\longrightarrow
Rel/TrivRel
$$
by mapping $\text{d}g$ to the class of $\lambda(d_0(g) - d_1(g) + d_2(g))$
in the quotient.
\medskip\noindent
The diagram
$$
\xymatrix{
\Omega_{P_3/A} \otimes B \ar[r] \ar[d] &
\Omega_{P_2/A} \otimes B \ar[r] \ar[d]_q &
\Omega_{P_1/A} \otimes B \ar[r] \ar[d]_{\overline{\lambda}} &
\Omega_{P_0/A} \otimes B \ar[d]_{\text{d}\psi \otimes 1} \\
0 \ar[r] &
Rel/TrivRel \ar[r] &
F \otimes B \ar[r] &
\Omega_{P/A} \otimes B
}
$$
commutes (calculation omitted) and we obtain the map of the lemma.
By Remark \ref{remark-explicit-comparison-map} and
Lemma \ref{lemma-relation-with-naive-cotangent-complex} we see that this map
induces isomorphisms $H_1(L_{B/A}) \to H_1(L)$ and $H_0(L_{B/A}) \to H_0(L)$.
\medskip\noindent
It remains to see that our map $L_{B/A} \to L$ induces an isomorphism
$H_2(L_{B/A}) \to H_2(L)$. Choose a resolution of $B$ over $A$ with
$P_0 = P = A[u_i]$ and then $P_1$ and $P_2$ as in
Example \ref{example-resolution-length-two}.
In Remark \ref{remark-elucidate-degree-two} we have constructed an exact
sequence
$$
\wedge^2_B(J_0/J_0^2) \to \text{Tor}_2^{P_0}(B, B) \to H^{-2}(L_{B/A}) \to 0
$$
where $P_0 = P$ and $J_0 = \Ker(P \to B) = I$.
Calculating the Tor group using the short exact sequences
$0 \to I \to P \to B \to 0$ and $0 \to Rel \to F \to I \to 0$
we find that
$\text{Tor}_2^P(B, B) = \Ker(Rel \otimes B \to F \otimes B)$.
The image of the map $\wedge^2_B(I/I^2) \to \text{Tor}_2^P(B, B)$
under this identification is exactly the image of $TrivRel \otimes B$.
Thus we see that $H_2(L_{B/A}) \cong H_2(L)$.
\medskip\noindent
Finally, we have to check that our map $L_{B/A} \to L$ actually induces
this isomorphism. We will use the notation and results discussed in
Example \ref{example-resolution-length-two} and
Remarks \ref{remark-elucidate-degree-two} and \ref{remark-surjection}
without further mention. Pick an element $\xi$ of
$\text{Tor}_2^{P_0}(B, B) = \Ker(I \otimes_P I \to I^2)$.
Write $\xi = \sum h_{t', t}f_{t'} \otimes f_t$ for some
$h_{t', t} \in P$. Tracing through the exact sequences above we
find that $\xi$ corresponds to the image in $Rel \otimes B$
of the element $r \in Rel \subset F = \bigoplus_{t \in T} P$ with
$t$th coordinate $r_t = \sum_{t' \in T} h_{t', t}f_{t'}$.
On the other hand, $\xi$ corresponds to the element of
$H_2(L_{B/A}) = H_2(\Omega)$ which is the image
via $\text{d} : H_2(\mathcal{J}/\mathcal{J}^2) \to H_2(\Omega)$
of the boundary of $\xi$ under the $2$-extension
$$
0 \to
\text{Tor}_2^\mathcal{O}(\underline{B}, \underline{B})
\to
\mathcal{J} \otimes_\mathcal{O} \mathcal{J} \to \mathcal{J}
\to
\mathcal{J}/\mathcal{J}^2 \to 0
$$
We compute the successive transgressions of our element. First we have
$$
\xi = (d_0 - d_1)(- \sum s_0(h_{t', t} f_{t'}) \otimes x_t)
$$
and next we have
$$
\sum s_0(h_{t', t} f_{t'}) x_t = d_0(v_r) - d_1(v_r) + d_2(v_r)
$$
by our choice of the variables $v$ in
Example \ref{example-resolution-length-two}.
We may choose our map $\lambda$ above such that
$\lambda(u_i) = 0$ and $\lambda(x_t) = - e_t$ where $e_t \in F$
denotes the basis vector corresponding to $t \in T$.
Hence the construction of our map $q$ above sends $\text{d}v_r$ to
$$
\lambda(\sum s_0(h_{t', t} f_{t'}) x_t) =
\sum\nolimits_t \left(\sum\nolimits_{t'} h_{t', t}f_{t'}\right) e_t
$$
matching the image of $\xi$ in $Rel \otimes B$ (the two minus signs
we found above cancel out). This agreement finishes the proof.
\end{proof}
\begin{remark}[Functoriality of the Lichtenbaum-Schlessinger complex]
\label{remark-functoriality-lichtenbaum-schlessinger}
Consider a commutative square
$$
\xymatrix{
A' \ar[r] & B' \\
A \ar[u] \ar[r] & B \ar[u]
}
$$
of ring maps. Choose a factorization
$$
\xymatrix{
A' \ar[r] & P' \ar[r] & B' \\
A \ar[u] \ar[r] & P \ar[u] \ar[r] & B \ar[u]
}
$$
with $P$ a polynomial algebra over $A$ and $P'$ a polynomial algebra over $A'$.
Choose generators $f_t$, $t \in T$ for $\Ker(P \to B)$.
For $t \in T$ denote $f'_t$ the image of $f_t$ in $P'$.
Choose $f'_s \in P'$ such that the elements $f'_t$ for
$t \in T' = T \amalg S$ generate the kernel
of $P' \to B'$. Set $F = \bigoplus_{t \in T} P$ and
$F' = \bigoplus_{t' \in T'} P'$. Let $Rel = \Ker(F \to P)$
and $Rel' = \Ker(F' \to P')$ where the maps are given
by multiplication by $f_t$, resp.\ $f'_t$ on the coordinates.
Finally, set $TrivRel$, resp.\ $TrivRel'$ equal to the submodule
of $Rel$, resp.\ $TrivRel$ generated by the elements
$(\ldots, f_{t'}, 0, \ldots, 0, -f_t, 0, \ldots)$
for $t, t' \in T$, resp.\ $T'$. Having made these choices we obtain a
canonical commutative diagram
$$
\xymatrix{
L' : &
Rel'/TrivRel' \ar[r] &
F' \otimes_{P'} B' \ar[r] &
\Omega_{P'/A'} \otimes_{P'} B' \\
L : \ar[u] &
Rel/TrivRel \ar[r] \ar[u] &
F \otimes_P B \ar[r] \ar[u] &
\Omega_{P/A} \otimes_P B \ar[u]
}
$$
Moreover, tracing through the choices made in the proof of
Lemma \ref{lemma-compare-higher}
the reader sees that one obtains a commutative diagram
$$
\xymatrix{
L_{B'/A'} \ar[r] & L' \\
L_{B/A} \ar[r] \ar[u] & L \ar[u]
}
$$
\end{remark}
```

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