Definition 92.13.1. Let $A \to B$ be a ring map. Let $M$ be a $(B, B)$-bimodule over $A$. An $A$-biderivation is an $A$-linear map $\lambda : B \to M$ such that $\lambda (xy) = x\lambda (y) + \lambda (x)y$.
92.13 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
92.13.0.1
\begin{equation} \label{cotangent-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 $\text{d}f_ t \otimes 1$.
For a polynomial algebra the biderivations are easy to describe.
Lemma 92.13.2. 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 [page 206, Proposition 12, Andre-Homologie] or in the paper [Doncel] which extends the complex (92.13.0.1) by one term and the comparison to $\tau _{\geq -3}$.
Lemma 92.13.3. In the situation above denote $L$ the complex (92.13.0.1). There is a canonical map $L_{B/A} \to L$ in $D(B)$ 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 92.5.5). 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 92.13.2 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 92.11.4 and Lemma 92.11.3 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 92.5.9. In Remark 92.12.6 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 92.5.9 and Remarks 92.12.6 and 92.11.5 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 92.5.9. 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 92.13.4 (Functoriality of the Lichtenbaum-Schlessinger complex). Consider a commutative square of ring maps. Choose a factorization 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 Moreover, tracing through the choices made in the proof of Lemma 92.13.3 the reader sees that one obtains a commutative diagram
\[ \xymatrix{ A' \ar[r] & B' \\ A \ar[u] \ar[r] & B \ar[u] } \]
\[ \xymatrix{ A' \ar[r] & P' \ar[r] & B' \\ A \ar[u] \ar[r] & P \ar[u] \ar[r] & B \ar[u] } \]
\[ \xymatrix{ L' : & Rel'/TrivRel' \ar[r] & F' \otimes _{P'} B' \ar[r] & \Omega _{P'/A'} \otimes _{P'} B' \\ L : \ar[u] & Rel/TrivRel \ar[r] \ar[u] & F \otimes _ P B \ar[r] \ar[u] & \Omega _{P/A} \otimes _ P B \ar[u] } \]
\[ \xymatrix{ L_{B'/A'} \ar[r] & L' \\ L_{B/A} \ar[r] \ar[u] & L \ar[u] } \]
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)