The Stacks project

92.12 A spectral sequence of Quillen

In this section we discuss a spectral sequence relating derived tensor product to the cotangent complex.

Lemma 92.12.1. Notation and assumptions as in Cohomology on Sites, Example 21.39.1. Assume $\mathcal{C}$ has a cosimplicial object as in Cohomology on Sites, Lemma 21.39.7. Let $\mathcal{F}$ be a flat $\underline{B}$-module such that $H_0(\mathcal{C}, \mathcal{F}) = 0$. Then $H_ l(\mathcal{C}, \text{Sym}_{\underline{B}}^ k(\mathcal{F})) = 0$ for $l < k$.

Proof. We drop the subscript ${}_{\underline{B}}$ from tensor products, wedge powers, and symmetric powers. We will prove the lemma by induction on $k$. The cases $k = 0, 1$ follow from the assumptions. If $k > 1$ consider the exact complex

\[ \ldots \to \wedge ^2\mathcal{F} \otimes \text{Sym}^{k - 2}\mathcal{F} \to \mathcal{F} \otimes \text{Sym}^{k - 1}\mathcal{F} \to \text{Sym}^ k\mathcal{F} \to 0 \]

with differentials as in the Koszul complex. If we think of this as a resolution of $\text{Sym}^ k\mathcal{F}$, then this gives a first quadrant spectral sequence

\[ E_1^{p, q} = H_ p(\mathcal{C}, \wedge ^{q + 1}\mathcal{F} \otimes \text{Sym}^{k - q - 1}\mathcal{F}) \Rightarrow H_{p + q}(\mathcal{C}, \text{Sym}^ k(\mathcal{F})) \]

By Cohomology on Sites, Lemma 21.39.10 we have

\[ L\pi _!(\wedge ^{q + 1}\mathcal{F} \otimes \text{Sym}^{k - q - 1}\mathcal{F}) = L\pi _!(\wedge ^{q + 1}\mathcal{F}) \otimes _ B^\mathbf {L} L\pi _!(\text{Sym}^{k - q - 1}\mathcal{F})) \]

It follows (from the construction of derived tensor products) that the induction hypothesis combined with the vanishing of $H_0(\mathcal{C}, \wedge ^{q + 1}(\mathcal{F})) = 0$ will prove what we want. This is true because $\wedge ^{q + 1}(\mathcal{F})$ is a quotient of $\mathcal{F}^{\otimes q + 1}$ and $H_0(\mathcal{C}, \mathcal{F}^{\otimes q + 1})$ is a quotient of $H_0(\mathcal{C}, \mathcal{F})^{\otimes q + 1}$ which is zero. $\square$

Remark 92.12.2. In the situation of Lemma 92.12.1 one can show that $H_ k(\mathcal{C}, \text{Sym}^ k(\mathcal{F})) = \wedge ^ k_ B(H_1(\mathcal{C}, \mathcal{F}))$. Namely, it can be deduced from the proof that $H_ k(\mathcal{C}, \text{Sym}^ k(\mathcal{F}))$ is the $S_ k$-coinvariants of

\[ H^{-k}(L\pi _!(\mathcal{F}) \otimes _ B^\mathbf {L} L\pi _!(\mathcal{F}) \otimes _ B^\mathbf {L} \ldots \otimes _ B^\mathbf {L} L\pi _!(\mathcal{F})) = H_1(\mathcal{C}, \mathcal{F})^{\otimes k} \]

Thus our claim is that this action is given by the usual action of $S_ k$ on the tensor product multiplied by the sign character. To prove this one has to work through the sign conventions in the definition of the total complex associated to a multi-complex. We omit the verification.

Lemma 92.12.3. Let $A$ be a ring. Let $P = A[E]$ be a polynomial ring. Set $I = (e; e \in E) \subset P$. The maps $\text{Tor}_ i^ P(A, I^{n + 1}) \to \text{Tor}_ i^ P(A, I^ n)$ are zero for all $i$ and $n$.

Proof. Denote $x_ e \in P$ the variable corresponding to $e \in E$. A free resolution of $A$ over $P$ is given by the Koszul complex $K_\bullet $ on the $x_ e$. Here $K_ i$ has basis given by wedges $e_1 \wedge \ldots \wedge e_ i$, $e_1, \ldots , e_ i \in E$ and $d(e) = x_ e$. Thus $K_\bullet \otimes _ P I^ n = I^ nK_\bullet $ computes $\text{Tor}_ i^ P(A, I^ n)$. Observe that everything is graded with $\deg (x_ e) = 1$, $\deg (e) = 1$, and $\deg (a) = 0$ for $a \in A$. Suppose $\xi \in I^{n + 1}K_ i$ is a cocycle homogeneous of degree $m$. Note that $m \geq i + 1 + n$. Then $\xi = \text{d}\eta $ for some $\eta \in K_{i + 1}$ as $K_\bullet $ is exact in degrees $ > 0$. (The case $i = 0$ is left to the reader.) Now $\deg (\eta ) = m \geq i + 1 + n$. Hence writing $\eta $ in terms of the basis we see the coordinates are in $I^ n$. Thus $\xi $ maps to zero in the homology of $I^ nK_\bullet $ as desired. $\square$

Theorem 92.12.4 (Quillen spectral sequence). Let $A \to B$ be a surjective ring map. Consider the sheaf $\Omega = \Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}$ of $\underline{B}$-modules on $\mathcal{C}_{B/A}$, see Section 92.4. Then there is a spectral sequence with $E_1$-page

\[ E_1^{p, q} = H_{- p - q}(\mathcal{C}_{B/A}, \text{Sym}^ p_{\underline{B}}(\Omega )) \Rightarrow \text{Tor}^ A_{- p - q}(B, B) \]

with $d_ r$ of bidegree $(r, -r + 1)$. Moreover, $H_ i(\mathcal{C}_{B/A}, \text{Sym}^ k_{\underline{B}}(\Omega )) = 0$ for $i < k$.

Proof. Let $I \subset A$ be the kernel of $A \to B$. Let $\mathcal{J} \subset \mathcal{O}$ be the kernel of $\mathcal{O} \to \underline{B}$. Then $I\mathcal{O} \subset \mathcal{J}$. Set $\mathcal{K} = \mathcal{J}/I\mathcal{O}$ and $\overline{\mathcal{O}} = \mathcal{O}/I\mathcal{O}$.

For every object $U = (P \to B)$ of $\mathcal{C}_{B/A}$ we can choose an isomorphism $P = A[E]$ such that the map $P \to B$ maps each $e \in E$ to zero. Then $J = \mathcal{J}(U) \subset P = \mathcal{O}(U)$ is equal to $J = IP + (e; e \in E)$. Moreover $\overline{\mathcal{O}}(U) = B[E]$ and $K = \mathcal{K}(U) = (e; e \in E)$ is the ideal generated by the variables in the polynomial ring $B[E]$. In particular it is clear that

\[ K/K^2 \xrightarrow {\text{d}} \Omega _{P/A} \otimes _ P B \]

is a bijection. In other words, $\Omega = \mathcal{K}/\mathcal{K}^2$ and $\text{Sym}_ B^ k(\Omega ) = \mathcal{K}^ k/\mathcal{K}^{k + 1}$. Note that $\pi _!(\Omega ) = \Omega _{B/A} = 0$ (Lemma 92.4.5) as $A \to B$ is surjective (Algebra, Lemma 10.131.4). By Lemma 92.12.1 we conclude that

\[ H_ i(\mathcal{C}_{B/A}, \mathcal{K}^ k/\mathcal{K}^{k + 1}) = H_ i(\mathcal{C}_{B/A}, \text{Sym}^ k_{\underline{B}}(\Omega )) = 0 \]

for $i < k$. This proves the final statement of the theorem.

The approach to the theorem is to note that

\[ B \otimes _ A^\mathbf {L} B = L\pi _!(\mathcal{O}) \otimes _ A^\mathbf {L} B = L\pi _!(\mathcal{O} \otimes _{\underline{A}}^\mathbf {L} \underline{B}) = L\pi _!(\overline{\mathcal{O}}) \]

The first equality by Lemma 92.5.7, the second equality by Cohomology on Sites, Lemma 21.39.6, and the third equality as $\mathcal{O}$ is flat over $\underline{A}$. The sheaf $\overline{\mathcal{O}}$ has a filtration

\[ \ldots \subset \mathcal{K}^3 \subset \mathcal{K}^2 \subset \mathcal{K} \subset \overline{\mathcal{O}} \]

This induces a filtration $F$ on a complex $C$ representing $L\pi _!(\overline{\mathcal{O}})$ with $F^ pC$ representing $L\pi _!(\mathcal{K}^ p)$ (construction of $C$ and $F$ omitted). Consider the spectral sequence of Homology, Section 12.24 associated to $(C, F)$. It has $E_1$-page

\[ E_1^{p, q} = H_{- p - q}(\mathcal{C}_{B/A}, \mathcal{K}^ p/\mathcal{K}^{p + 1}) \quad \Rightarrow \quad H_{- p - q}(\mathcal{C}_{B/A}, \overline{\mathcal{O}}) = \text{Tor}_{- p - q}^ A(B, B) \]

and differentials $E_ r^{p, q} \to E_ r^{p + r, q - r + 1}$. To show convergence we will show that for every $k$ there exists a $c$ such that $H_ i(\mathcal{C}_{B/A}, \mathcal{K}^ n) = 0$ for $i < k$ and $n > c$1.

Given $k \geq 0$ set $c = k^2$. We claim that

\[ H_ i(\mathcal{C}_{B/A}, \mathcal{K}^{n + c}) \to H_ i(\mathcal{C}_{B/A}, \mathcal{K}^ n) \]

is zero for $i < k$ and all $n \geq 0$. Note that $\mathcal{K}^ n/\mathcal{K}^{n + c}$ has a finite filtration whose successive quotients $\mathcal{K}^ m/\mathcal{K}^{m + 1}$, $n \leq m < n + c$ have $H_ i(\mathcal{C}_{B/A}, \mathcal{K}^ m/\mathcal{K}^{m + 1}) = 0$ for $i < n$ (see above). Hence the claim implies $H_ i(\mathcal{C}_{B/A}, \mathcal{K}^{n + c}) = 0$ for $i < k$ and all $n \geq k$ which is what we need to show.

Proof of the claim. Recall that for any $\mathcal{O}$-module $\mathcal{F}$ the map $\mathcal{F} \to \mathcal{F} \otimes _\mathcal {O}^\mathbf {L} B$ induces an isomorphism on applying $L\pi _!$, see Lemma 92.5.6. Consider the map

\[ \mathcal{K}^{n + k} \otimes _\mathcal {O}^\mathbf {L} B \longrightarrow \mathcal{K}^ n \otimes _\mathcal {O}^\mathbf {L} B \]

We claim that this map induces the zero map on cohomology sheaves in degrees $0, -1, \ldots , - k + 1$. If this second claim holds, then the $k$-fold composition

\[ \mathcal{K}^{n + c} \otimes _\mathcal {O}^\mathbf {L} B \longrightarrow \mathcal{K}^ n \otimes _\mathcal {O}^\mathbf {L} B \]

factors through $\tau _{\leq -k}\mathcal{K}^ n \otimes _\mathcal {O}^\mathbf {L} B$ hence induces zero on $H_ i(\mathcal{C}_{B/A}, -) = L_ i\pi _!( - )$ for $i < k$, see Derived Categories, Lemma 13.12.5. By the remark above this means the same thing is true for $H_ i(\mathcal{C}_{B/A}, \mathcal{K}^{n + c}) \to H_ i(\mathcal{C}_{B/A}, \mathcal{K}^ n)$ which proves the (first) claim.

Proof of the second claim. The statement is local, hence we may work over an object $U = (P \to B)$ as above. We have to show the maps

\[ \text{Tor}_ i^ P(B, K^{n + k}) \to \text{Tor}_ i^ P(B, K^ n) \]

are zero for $i < k$. There is a spectral sequence

\[ \text{Tor}_ a^ P(P/IP, \text{Tor}_ b^{P/IP}(B, K^ n)) \Rightarrow \text{Tor}_{a + b}^ P(B, K^ n), \]

see More on Algebra, Example 15.62.2. Thus it suffices to prove the maps

\[ \text{Tor}_ i^{P/IP}(B, K^{n + 1}) \to \text{Tor}_ i^{P/IP}(B, K^ n) \]

are zero for all $i$. This is Lemma 92.12.3. $\square$

Remark 92.12.5. In the situation of Theorem 92.12.4 let $I = \mathop{\mathrm{Ker}}(A \to B)$. Then $H^{-1}(L_{B/A}) = H_1(\mathcal{C}_{B/A}, \Omega ) = I/I^2$, see Lemma 92.11.2. Hence $H_ k(\mathcal{C}_{B/A}, \text{Sym}^ k(\Omega )) = \wedge ^ k_ B(I/I^2)$ by Remark 92.12.2. Thus the $E_1$-page looks like

\[ \begin{matrix} B \\ 0 \\ 0 & I/I^2 \\ 0 & H^{-2}(L_{B/A}) \\ 0 & H^{-3}(L_{B/A}) & \wedge ^2(I/I^2) \\ 0 & H^{-4}(L_{B/A}) & H_3(\mathcal{C}_{B/A}, \text{Sym}^2(\Omega )) \\ 0 & H^{-5}(L_{B/A}) & H_4(\mathcal{C}_{B/A}, \text{Sym}^2(\Omega )) & \wedge ^3(I/I^2) \end{matrix} \]

with horizontal differential. Thus we obtain edge maps $\text{Tor}_ i^ A(B, B) \to H^{-i}(L_{B/A})$, $i > 0$ and $\wedge ^ i_ B(I/I^2) \to \text{Tor}_ i^ A(B, B)$. Finally, we have $\text{Tor}_1^ A(B, B) = I/I^2$ and there is a five term exact sequence

\[ \text{Tor}_3^ A(B, B) \to H^{-3}(L_{B/A}) \to \wedge ^2_ B(I/I^2) \to \text{Tor}_2^ A(B, B) \to H^{-2}(L_{B/A}) \to 0 \]

of low degree terms.

Remark 92.12.6. Let $A \to B$ be a ring map. Let $P_\bullet $ be a resolution of $B$ over $A$ (Remark 92.5.5). Set $J_ n = \mathop{\mathrm{Ker}}(P_ n \to B)$. Note that

\[ \text{Tor}_2^{P_ n}(B, B) = \text{Tor}_1^{P_ n}(J_ n, B) = \mathop{\mathrm{Ker}}(J_ n \otimes _{P_ n} J_ n \to J_ n^2). \]

Hence $H_2(L_{B/A})$ is canonically equal to

\[ \mathop{\mathrm{Coker}}(\text{Tor}_2^{P_1}(B, B) \to \text{Tor}_2^{P_0}(B, B)) \]

by Remark 92.11.5. To make this more explicit we choose $P_2$, $P_1$, $P_0$ as in Example 92.5.9. We claim that

\[ \text{Tor}_2^{P_1}(B, B) = \wedge ^2(\bigoplus \nolimits _{t \in T} B)\ \oplus \ \bigoplus \nolimits _{t \in T} J_0\ \oplus \ \text{Tor}_2^{P_0}(B, B) \]

Namely, the basis elements $x_ t \wedge x_{t'}$ of the first summand corresponds to the element $x_ t \otimes x_{t'} - x_{t'} \otimes x_ t$ of $J_1 \otimes _{P_1} J_1$. For $f \in J_0$ the element $x_ t \otimes f$ of the second summand corresponds to the element $x_ t \otimes s_0(f) - s_0(f) \otimes x_ t$ of $J_1 \otimes _{P_1} J_1$. Finally, the map $\text{Tor}_2^{P_0}(B, B) \to \text{Tor}_2^{P_1}(B, B)$ is given by $s_0$. The map $d_0 - d_1 : \text{Tor}_2^{P_1}(B, B) \to \text{Tor}_2^{P_0}(B, B)$ is zero on the last summand, maps $x_ t \otimes f$ to $f \otimes f_ t - f_ t \otimes f$, and maps $x_ t \wedge x_{t'}$ to $f_ t \otimes f_{t'} - f_{t'} \otimes f_ t$. All in all we conclude that there is an exact sequence

\[ \wedge ^2_ B(J_0/J_0^2) \to \text{Tor}_2^{P_0}(B, B) \to H^{-2}(L_{B/A}) \to 0 \]

In this way we obtain a direct proof of a consequence of Quillen's spectral sequence discussed in Remark 92.12.5.

[1] A posteriori the “correct” vanishing $H_ i(\mathcal{C}_{B/A}, \mathcal{K}^ n) = 0$ for $i < n$ can be concluded.

Comments (0)


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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08RC. Beware of the difference between the letter 'O' and the digit '0'.