Cotangent complexes
DAG Winter School Flumserberg - Nov 2013
Georg Oberdieck, ETH Zürich
We work over a field of char 0. Let B −→ A be a map of dgas and let M be an A-module.
Definition 1. A B-derivation of A into M is a B-linear map
φ : A −→ M
such that φ(ab) = φ(a)b + aφ(b) for all a, b ∈ A. In particular, this implies φ(c) = 0 for all c ∈ B.
Let DerB (A, M ) be the set of such derivations. This defines a functor
DerB (A, ·) : A − Mod −→ Sets.
Definition 2 (Def/Prop). DerB (A, ·) is corepresentable: There is a A-module ΩA/B (unique
up to unique isomorphism), called the module of Kähler differentials, and a universal derivation
δ : A −→ ΩA/B such that the natural morphism
HomA−Mod (ΩA/B , M ) −→ DerB (A, M )
given by precomposing with δ is an isomorphism for all A-modules M .
Sketch of Proof, following Manetti Prop 2.3. Define the free A-module generated by the formal
symbols δx for x ∈ A homogeneous,
M
F =
Aδx
x∈A
x hom.
and let I be the submodule generated by expressions
δ(ab) − δ(a)b − aδ(b)
δ(a + b) − δ(a) − δ(b)
for all a, b ∈ A
for all c ∈ B
δ(c)
F has a natural grading when we set the degree of δx to be the degree of x. To define the
differential, consider
δ
/ Fk
Ak
A
d
k+1
δ
/F
?
k+1
We define the differential by making the diagram commutative, i.e. we set d(δx) := δ(dx) and
extend by the Leibniz rule
d(aδx) = (da)δx + (−1)|a| adδx
Check that the differential d preserves the submodule I. Hence ΩA/B = F/I is a A-dg module
and δ : x 7→ δx defines a derivation A −→ ΩA/B . The assertion of the proposition follows then
from not too hard arguments.
We like to lift the above construction to the homotopy category, that is, given an element in
the homotopy category of dg-algebras, we like to speak of its module of differentials or
expressed less sophisticated, we like to take derivatives of objects in the homotopy category.
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But Ω·/B does not preserve weak equivalences of A: it is easy to come up with an example of
·x
B-algebras A1 ∼
= A2 such that ΩA1 /B ΩA2 /B (Let B = k[x], A1 = k, A2 = k[x] −→ k[x]).
There are several ways to define a derived version of Ω, for now we give an ad-hoc definition
and later justify its properties.
For deriving the Kaehler differentials of a morphism B −→ A, we will work in the category
dgaB /A of B-dg algebras over A. Objects are B-dgas X together with a mapping X −→ A
and morphisms are morphisms of B-dgas respecting the map to A. The natural model
structure on dgaB induces a model structure on dgaB /A, where in particular the semifree
resolutions P together with the natural map to A are cofibrant replacements.
Definition 3. Let P −→ A be a semifree resolution of A in dgaB . We define the cotangent
complex of A over B by
LA/B := ΩP/B ⊗P A
∈ Ho(A − Mod)
Note that we dont need to derive the tensor product, as ΩP/B is a free P -module (see example
1 below).
Lemma 4. The above definition is independent of the semifree resolution P −→ A and LA/B
is well-defined in the derived category.
Proof. If P −→ A ←− P 0 are two semifree resolutions, then P and P 0 are weak equivalent
cofibrant objects, hence they are homotopically equivalent. The statement follows then from
Proposition 5.1. in Manetti.
Example 5. (1) A semifree over B: Recall that A semifree means
L the underlying graded
ring |A| is isomorphic to the free polynomial algebra B[V ] = n≥0 V ⊗n /Sn of a graded
vector space V (taking in the graded-commutative sense). Following the construction in
the proposition it is easy to see that the underlying graded algebra of ΩA/B is given by
B[V ] ⊗ V . We will conveniantly write the elements in the second factor as δv for v ∈ V .
The universal derivation will then be given by δ : v ∈ V ⊂ A 7→ δv. As in the proposition,
the differential on B[V ] ⊗ V can then be defined by commutativity with δ and the Leibniz
rule. To memorize this fact, we may
L write the symmetric product in the natural analogy
with the exponential as B[V ] = n≥0 V ⊗n /Sn =: exp(V ) and the Kaehler differential
functor as differential δ. Then the above just says, δ exp(V ) = exp(V ) otimesδV .
(2) A = B/I complete intersection: Let B = k[x1 , . . . , xn ] be the polynomial ring in n
variables and I an ideal generated by a regular sequence f1 , . . . , fk (Recall: f1 , . . . , fk is
regular if fi is a non-zero divisor in B/(f1 , . . . , fi−1 ) and (f1 , . . . , fk ) is proper). There is
a natural map d : B ⊕k −→ B sending the i-th generator yi of B ⊕k to fi ∈ B. The Koszul
resolution of A over B is then given by
^•
P = k[B ⊕k [1]] ∼
(B ⊕k ) −→ A
=
B
where the differential d is given by the Leibniz rule, e.g.
d(y1 y2 ) = d(y1 )y2 − y1 d(y2 ) = f1 y2 − f2 y1 . Since f1 , . . . , fk is regular, this is a semifree
resolution (see e.g. Fulton Appendix A.5). We calculate
LA/B = ΩP/B ⊗ A = (P ⊗ hδy1 , . . . , δyk i) ⊗ A = Aδy1 ⊕ · · · ⊕ Aδyk = A⊕k [1]
where naturally yi is in degree −1.
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For the cotangent complex of A over the ground field k, we need to take a resolution of A
that is semifree with respect to k. As we have chosen B to be a free k-algebra, the Koszul
resolution P = B[y1 , . . . , yk ] = k[x1 , . . . , xn , y1 , . . . , yk ] suffices. Hence
h
i
d
LA/k = hδx1 , . . . , δyk i ⊗ A = Aδy1 ⊕ . . . Aδyk −→ Aδx1 ⊕ · · · ⊕ Aδxn
as A lives in degree 0. Relativ to B the differential was zero, on the other hand here we
∂fi
have d(δyi ) = δdyi = δfi = ∂x
δxj . Let x : A −→ k be a point of A. Tensoring the above
j
sequence and dualizing, we find the tangent complex at A by
h ∂
∂
∂
∂ i
dFx
h
TA,x = h
,...,
i −→
...
i
∂x1
∂xn
∂y1
∂yk
P ∂F
∂
with dFx ( ∂x
) = j ∂x1j (x) ∂y∂ j the usual Jacobian of the function F = (f1 , . . . , fk ).
i
H 0 (TA,x ) can then be identified with the usual tangent space to Spec A while H 1 (TA,x ),
the cokernel of dFx , is the excess space. We see the derived geometry captures naturally
the excess spaces of non-transverse intersections.
Note: The better way to do this calculation is to use the exact triangle associated to the
maps Spec A −→ Spec B −→ Spec k. This shows that for any smooth B and complete
d
intersection A, LA/k ∼
= [0 −→ I/I 2 −→ ΩB ⊗ A −→ 0]. When B −→ A is only a closed
embedding into a smooth B and not necessarily, the same argument proves this
expression only for the cutoff τ≥−1 LA/k .
(3) C = A1 ⊗LB A2 derived intersection: Let A1 , A2 be dg algebras over B and lets consider
the derived intersection A1 ⊗LB A2 . This fits into the cofiber square
A1 ⊗LB A2 o
O
AO 2
A1 o
B
(1)
By the general theory of cotangent complexes, we get a homotopy cofiber square
o
LC
O
LA2 ⊗ C
O
LA1 ⊗ C o
LB ⊗ C
For A1 = k a point, we obtain the exact triangle LB −→ LA2 −→ LC . More generally, C
is also the homotopic colimit of the diagram
CO o
Bo
A1 ×O A2
∆
B×B
/B
and by using that we get a homotopy cofiber square relative to B and a little exercise, we
find that we have an exact triangle
LB ⊗ C −→ (LA1 ⊕ LA2 ) ⊗ C −→ LC .
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In the case where A1 , A2 , B are classical affine schemes, with B smooth and B −→ A1 , A2
closed immersion, let us again restrict to a point x of C, dualize and take the long exact
sequence in cohomology. One finds
0 −→ H 0 (TC,x ) −→ H 0 (TA1 ,x ) ⊕ H 0 (TA2 ,x ) −→ H 0 (TB,x )
−→ H 1 (TC,x ) −→ H 1 (TA1 ,x ) ⊕ H 1 (TA2 ,x ) −→ 0
which is
0 −→ H 0 (TC,x ) −→ TA1 ,x ⊕ TA2 ,x −→ TB,x −→ H 1 (TC,x ) −→ EA1 /B ⊕ EA2 /B −→ 0
where EAi /B are the excess spaces of Ai to B. We see that H 1 (TC,x ) = Ext1 (LC,x , k) is a
measure for both the non-transversality of the intersection as well as the
non-transversality of the defining equation for A1 and A2 .
Remark 6. (a) The classical intersection B/(IA1 + IA2 ) = A1 ⊗B A2 might be very
singular. In fact, its cotangent complex is usually supported in all negative degrees.
(b) If B is only a complete intersection instead of smooth, the derived intersection might
have a contangent complex supported in [−2, 0]. This is impossible for classical
schemes. An easy by hand computable example is B = k[x, y]/(xy) and
A1 = k[x, y]/y, A2 = k[x, y]/x. The derived intersection point is C = k[t], t in degree
−2. The cotangent complex is then k[2], non-zero in degree −2! What happens
when A1 = A2 = k[x, y]/x ?
(c) Actually, we can use the methods of (iii) to compute (ii). Just consider the zero fiber
of the map F : An −→ Ak .
After we have seen how to calculate the cotangent complex in some cases, let us try to
understand what kind of information the cotangent complex encodes. Let B −→ A and M as
above. Let A ⊕ M be the trivial square zero extension of A by M , i.e. we set m1 · m2 = 0 for
all m1 , m2 ∈ M .
Lemma 7. There is a 1-to-1 correspondence between derivations δ : A −→ M and morphisms
φ ∈ HomdgaB /A (A, A ⊕ M ).
Generalizing this idea, set
DerB (A, M ) := MapHo(dgaB /A) (A, A ⊕ M )
∈ Ho(sSet)
Proposition 8. DerB (A, −) : Ho(A − Mod) −→ Ho(sSet) is corepresentable by the A-module
LA/B (as defined above), i.e.
DerB (A, M ) = MapHo(A−Mod) (LA/B , M )
for all A-modules M
Proof. Let P −→ A be a semifree resolution. Then by quasi-isomorphism and since the
homotopy categories of dgaB /A and dgaB /P are the same, we have
MapHo(dgaB /A) (A, A ⊕ M ) = MapHo(dgaB /A) (P, P ⊕ M ) = MapHo(dgaB /P ) (P, P ⊕ M )
As P is cofibrant, we can now use HOM, the underived mapping space, to reduce to the
following claim: For any A we have
HOMdgaB /A (A, A ⊕ M ) = HOMA−Mod (ΩA/B , M )
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Proof of claim: Note that HOMdgaB / (A, A ⊕ M ) is defined by the following fiber square
/ HOMdga (A, A)
OB
HOMdgaB (A, A ⊕ M )
O
id
HOMdgaB /A (A, A ⊕ M )
/∗
When restricting to level n, the map id send the point to the constant homotopy at the
identity. We see
HOMdgaB /A (A, A ⊕ M )n = HomdgaB /A (A, Ω∗ (∆n ) ⊗ (A ⊕ M )) (constant on first factor)
= HomdgaB /A (A, A ⊕ (Ω∗ (∆n ) ⊗ M )
= HomdgaB /A (A, A ⊕ (C ∗ (∆n ) ⊗ M )
= HomA−Mod (ΩA/B , C ∗ (∆n ) ⊗ M )
= HOMA−Mod (ΩA/B , M )n
where Ω∗ (∆n ) is the polynomial de Rham complex on the n-simplex and C ∗ are the cochains
(see the Toen notes).
This has the following two easy consequences. For convenience, we will supress the base ring B
from the notation.
Lemma 9. Let x : A −→ R be a point (R some ring) and N a R-module. Then
ExtiR−Mod (LA ⊗ R, N ) = HomHo(x/Aff) (Spec(R ⊕ N [i]), Spec A)
(Here HomHo(x/Aff) are the morphisms respecting the map x)
For R = k and x : A −→ k a k-valued point and N = k, we have
Exti (LA,x , k) = H i (TA,x ) = Homx/ (Spec(k ⊕ k[i]), Spec A)
This shows that the higher cohomology groups of the tangent complex parametrize i-th
infinitesimal tangent vectors.
Proof. This is completely formal:
R HomHo(R−Mod) (LA ⊗ R, N ) = R HomHo(R−Mod) (LA , A N ) = DerB (A, A N )
= MapHo(dga /A) (A, A ⊕ A N ) = MapHo(dga /R) (A, R ⊕ N )
Replacing N with N [i] and taking H 0 we find
ExtiR−Mod (LA ⊗ R, N ) = H 0 (R Hom(LA ⊗ R, N [i]) = HomHodga /R (A, R ⊕ N [i])
as desired.
The second consequence is that the cotangent complex controls the obstructions of extending a
map by a square zero extension. That is, given a square zero extension
p : A0 −→ A
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with kernel M and a map x : Spec A −→ Y (Y affine lets say), we can ask, when does there
exist a lift x0 : Spec A0 −→ Y ? In diagrams
Spec A0 o
Spec A
p
∃x0 ?
#
Y
|
x
The answer is
Proposition 10. There is a class α(x, p) ∈ HomHo(A−Mod) (LY ⊗ A, M [1]) that vanishes iff x
lifts to a map x0 . Moreover the space of lifts is given by MapHo(A−Mod) (LY ⊗ A, M ).
This is not hard to prove (HAG II, 1.4.2.5), but we will just consider a concrete example.
Consider again the derived intersection in example 3 and recall the tangent complex
0 −→ H 0 (TC,x ) −→ H 0 (TA1 ,x ) ⊕ H 0 (TA2 ,x ) −→ H 0 (TB,x )
−→ H 1 (TC,x ) −→ H 1 (TA1 ,x ) ⊕ H 1 (TA2 ,x ) −→ 0
Lets denote Z = Spec C. For a point x : C −→ k and the trivial extension k ⊕ k, the
obstruction problem is trival: There always exist the zero tangent vector. Lets consider
v : Spec(k[]/2 ) −→ Z a given tangent vector, and ask, when can we extend it to a map
v 0 : Spec k[]/3 −→ Z? By the proposition there is a natural obstruction class
α(v) ∈ Ext1 (LC,x , k) = H 1 (TC,X ) that vanishes iff such a lift exist. The above sequence then
can be given the following interpretation: If the image of α(v) in H 1 (TA1 ,x ) ⊕ H 1 (TA2 ,x )
vanishes, then there exist a lift of v into A1 and A2 . In this case, the obstruction α(v) comes
from a tangent vector of B at x, measuring the inifinitesimal difference between some choice of
lifts of v to A1 and A2 resp. If we can write this tangent vector as a linear combination of a
vector in A1 and a vector in A2 , we can correct the two lifts to obtain a lift of v to C.
The above also applies to lifts of the form k[]/n → k[]/n−1 . The obstructions will always lie
in Ext1 (LZ,x , k). On the other hand, if we consider extensions of k[]/2 by k[, t]/(2 , t) with t
in degree −1, we will obtain obstructions in Ext2 and the space of lifts is Ext1 . More generally,
we can say that Exti captures obstructions for lifts of i − 1-th derived infinitesimal extension,
while Exti−1 is the space of such lifts. In essence, we see that taking derivatives in the derived
sense provides a good framework for infinitesimal derived geometry.
Acknowledgements: I like to thank Matthieu Anel for a lot of help in the preparation of the
talk and the organizers, especially Claudia Scheimbauer, for their successful work.
References:
(1) HAG II: For the functorial properties of cotangent complexes, see Prop. 1.2.1.6
(2) Manetti, The cotangent complex in characteristic 0,
www1.mat.uniroma1.it/people/manetti/DT2011/marco2.pdf
(3) Vezzosi, A note on the cotangent complex in DAG, arXiv:1008.0601
(4) Toen, Introduction to derived algebraic geoemetry (notes for GAeL 2012)
(5) Fulton, Intersection Theory.
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