nLab Beck module

Idea

One usually defines cohomology with respect to some coefficient objects:

• For group cohomology of a group $G$, the coefficients come from a (left) $G$-module.
• For Lie algebra cohomology of a Lie algebra $\mathfrak{g}$, the coefficients come from a (left) $\mathfrak{g}$-module.
• For Hochschild cohomology of an associative algebra $A$, the coefficients come from an $A$-bimodule.

Beck modules are a simultaneous generalisation of all three types of module.

Definition

Let $\mathcal{C}$ be a category with pullbacks and let $A$ be an object in $\mathcal{C}$. A Beck module over $A$ is an abelian group object in the slice category $\mathcal{C}_{/ A}$. In particular, if $A$ is the terminal object, this reduces to the notion of an abelian group object in $\mathcal{C}$. We write $\mathbf{Ab}(\mathcal{C}_{/ A})$ for the category of Beck modules over $A$.

(Beck 67, def. 5)

Properties

Proposition

Let $\mathcal{C}$ be an effective regular category (resp. locally presentable category) and let $A$ be an object in $\mathcal{C}$. Then $\mathbf{Ab}(\mathcal{C}_{/ A})$ is an abelian category (resp. locally presentable category).

Proof

If $\mathcal{C}$ is a effective regular category (resp. locally presentable category), then so is $\mathcal{C}_{/ A}$. Thus, the claim reduces to the fact that the category of abelian group objects in an effective regular category (resp. locally presentable category) is an abelian category (resp. locally presentable category).

Proposition

Let $\mathcal{C}$ be an effective regular category with filtered colimits and let $A$ be an object in $\mathcal{C}$. If filtered colimits in $\mathcal{C}$ preserve finite limits, then $\mathbf{Ab}(\mathcal{C}_{/ A})$ (is an abelian category and) satisfies axiom AB5.

Proof

The forgetful functor $\mathcal{C}_{/ A} \to \mathcal{C}$ creates pullbacks and filtered colimits, so filtered colimits in $\mathcal{C}_{/ A}$ also preserve finite limits. The forgetful functor $\mathbf{Ab}(\mathcal{C}_{/ A}) \to \mathcal{C}_{/ A}$ creates limits and filtered colimits, so filtered colimits in $\mathbf{Ab}(\mathcal{C}_{/ A})$ preserve kernels. In view of the earlier proposition, it follows that $\mathbf{Ab}(\mathcal{C}_{/ A})$ satisfies axiom AB5.

Corollary

Let $\mathcal{C}$ be a locally finitely presentable effective regular category and let $A$ be an object in $\mathcal{C}$. Then $\mathbf{Ab}(\mathcal{C}_{/ A})$ is a Grothendieck category.

Proof

Combine the two propositions above.

Derivations

Let $\mathcal{C}$ be a category with pullbacks, let $A$ be an object in $\mathcal{C}$, and let $U : \mathbf{Ab}(\mathcal{C}_{/ A}) \to \mathcal{C}_{/ A}$ be the forgetful functor. Given a Beck module $M$ over $A$, an $M$-valued derivation of $A$ is a morphism $1_A \to U M$ in $\mathcal{C}_{/ A}$, where $1_A$ is the terminal object in $\mathcal{C}_{/ A}$, and we write

$$Der (A, M) = \mathcal{C}_{/ A} (1_A, U M)$$

for the set of $M$-valued derivations of $A$. The Beck module of differentials over $A$ is an object $\Omega_A$ in $\mathbf{Ab}(\mathcal{C}_{/ A})$ representing the functor $Der (A, -) : \mathbf{Ab}(\mathcal{C}_{/ A}) \to \mathbf{Set}$.

The Beck module $\Omega_A$ is not guaranteed to exist in general. When the functor $U : \mathbf{Ab}(\mathcal{C}_{/ A}) \to \mathcal{C}_{/ A}$ has a left adjoint, $\Omega_A$ is simply the value of the left adjoint at $1_A$.

Proposition

Let $\mathcal{C}$ be a locally presentable category and let $A$ be an object in $\mathcal{C}$. Then the forgetful functor $U : \mathbf{Ab}(\mathcal{C}_{/ A}) \to \mathcal{C}_{/ A}$ has a left adjoint.

Proof

The forgetful functor $\mathbf{Ab}(\mathcal{C}_{/ A}) \to \mathcal{C}_{/ A}$ creates limits and $\kappa$-filtered colimits (for some $\kappa$ large enough), so we may apply the accessible adjoint functor theorem.

Examples

Beck modules over associative algebras

Proposition

Let $\mathcal{C}$ be the category of (not necessarily commutative) rings and let $A$ be a ring. Then $\mathbf{Ab}(\mathcal{C}_{/ A})$ is equivalent to the category of $A$-bimodules.

Proof

Let $\epsilon : B \to A$ be ring homomorphism. To give it the structure of a Beck module over $A$, we must give ring homomorphisms $\eta : A \to B$ and $\mu : B \times_A B \to B$ such that $\epsilon \circ \eta = id_A$, $\epsilon (\mu (b_0, b_1)) = \epsilon (b_0) = \epsilon (b_1)$, as well as various other equations. Given elements $b_0, b_1, b_2, b_3$ of $B$ such that $\epsilon (b_0) = \epsilon (b_2)$ and $\epsilon (b_1) = \epsilon (b_3)$, we have the following interchange law:

$$\mu (b_0 + b_1, b_2 + b_3) = \mu (b_0, b_2) + \mu (b_1, b_3)$$

Hence, if $a = \epsilon (b_1) = \epsilon (b_2)$,

$$\mu (b_1, b_2) = \mu (\eta (a) + b_1 - \eta (a), \eta (a) + b_2 - \eta (a)) = \mu (\eta (a), \eta (a)) + \mu (b_1 - \eta(a), b_2 - \eta (a)) = \eta (a) + \mu (b_1 - \eta(a), b_2 - \eta (a))$$

but $\epsilon (b_1 - \eta (a)) = \epsilon (b_2 - \eta (a)) = 0$, so

$$\mu (b_1 - \eta(a), b_2 - \eta(a)) = \mu (0, b_2 - \eta (a)) + \mu (b_1 - \eta (a), 0) = b_2 - \eta (a) + b_1 - \eta (a)$$

and we conclude that

$$\mu (b_1, b_2) = b_1 - \eta (\epsilon (b_1)) + b_2 = b_1 + b_2 - \eta (\epsilon (b_2))$$

and in particular, $\mu$ is entirely determined by $\eta$ and $\epsilon$. We also have the following interchange law,

$$\mu (b_0 b_1, b_2 b_3) = \mu (b_0, b_2) \mu (b_1, b_3)$$

and in particular,

$$\mu (\eta (\epsilon (b_2)) b_1, b_2 \eta (\epsilon (b_1))) = \mu (\eta (\epsilon (b_2)), b_2) \mu (b_1, \eta (\epsilon (b_1))) = b_2 b_1$$

hence,

$$b_2 b_1 = \eta (\epsilon (b_2)) b_1 - \eta (\epsilon (b_2 b_1)) + b_2 \eta (\epsilon (b_1))$$

so if $\epsilon (b_1) = \epsilon (b_2) = 0$, then $b_2 b_1 = 0$.

Let $M = \ker \epsilon$. The above shows that the internal abelian group structure on $B$ restricts to the pre-existing abelian group structure on $\ker \epsilon$. In addition, the homomorphism $\eta : A \to B$ gives $M$ the structure of an $A$-bimodule, and we see that $B$ is naturally isomorphic to the square-0 extension ring $A \oplus M$, with componentwise addition and the multiplication given below,

$$(a_0, m_0) \cdot (a_1, m_1) = (a_0 a_1, a_0 m_1 + m_0 a_1)$$

regarded as a Beck module over $A$ by defining $\epsilon : A \oplus M \to A$, $\eta : A \to A \oplus M$, and $\mu : (A \oplus M) \times_A (A \oplus M) \to A \oplus M$ as follows:

$$\epsilon (a, m) = a$$

$$\eta (a) = (a, 0)$$

$$\mu ((a, m_0), (a, m_1)) = (a, m_0 + m_1)$$

Thus, we have an equivalence between $\mathbf{Ab}(\mathcal{C}_{/ A})$ and the category of $A$-bimodules, as claimed.

Proposition

Let $A$ be a ring. Then the Beck module $\Omega_A$ is isomorphic to the $A$-bimodule of Kähler differentials (relative to $\mathbb{Z}$).

Proof

Let $M$ be an $A$-bimodule, regard $A \oplus M$ as a ring as above, and let $\epsilon : A \oplus M \to A$ be the obvious projection. A ring homomorphism $\phi : A \to A \oplus M$ satisfying $\epsilon \circ \phi = id_A$ is the same thing as an additive homomorphism $\delta : A \to M$ satisfying the following equations,

$$\delta (a_0 a_1) = \delta (a_0) a_1 + a_0 \delta (a_1)$$

i.e. a derivation $A \to M$ (over $\mathbb{Z}$). Thus, the Beck module $\Omega_A$ has the same universal property as the $A$-bimodule of Kähler differentials.

Beck modules over groups

Proposition

Let $\mathcal{C}$ be the category of (not necessarily abelian) groups and let $G$ be a group. Then $\mathbf{Ab}(\mathcal{C}_{/ G})$ is equivalent to the category of left $G$-modules.

Proof

Let $\epsilon : H \to G$ be group homomorphism. To give it the structure of a Beck module over $G$, we must give group homomorphisms $\eta : G \to H$ and $\mu : H \times_G H \to H$ such that $\epsilon \circ \eta = id_A$, $\epsilon (\mu (h_0, h_1)) = \epsilon (h_0) = \epsilon (h_1)$, as well as various other equations. Given elements $h_0, h_1, h_2, h_3$ of $H$ such that $\epsilon (h_0) = \epsilon (h_2)$ and $\epsilon (h_1) = \epsilon (h_3)$, we have the following interchange law:

$$\mu (h_0 h_1, h_2 h_3) = \mu (h_0, h_2) \mu (h_1, h_3)$$

and in particular,

$$\mu (\eta (\epsilon (h_2)) h_1, h_2 \eta (\epsilon (h_1))) = \mu (\eta (\epsilon (h_2)), h_2) \mu (h_1, \eta (\epsilon (h_1))) = h_2 h_1$$

but on the other hand, if $g = \epsilon (h_1) = \epsilon (h_2)$, then

$$\mu (h_1, h_2) = \mu (\eta (g) \eta (g)^{-1} h_1, \eta (g) \eta (g)^{-1} h_2) = \mu (\eta (g), \eta (g)) \mu (\eta (g)^{-1} h_1, \eta (g)^{-1} h_2) = \eta (g) \mu (\eta (g)^{-1} h_1, \eta (g)^{-1} h_2)$$

and writing $e$ for the unit of $G$ and $H$, we have $\epsilon (\eta (g)^{-1} h_1) = \epsilon (\eta (g)^{-1} h_2) = e$, hence

$$\mu (\eta (g)^{-1} h_1, \eta (g)^{-1} h_2) = \mu (e, \eta (g)^{-1} h_2) \mu (\eta (g)^{-1} h_1, e) = \eta (g)^{-1} h_2 \eta (g)^{-1} h_1$$

so we conclude that

$$\mu (h_1, h_2) = h_2 \eta (g)^{-1} h_1$$

and in particular, $\mu$ is entirely determined by $\eta$.

Let $M = \ker \epsilon$. The above shows that the internal abelian group structure on $B$ restricts to the pre-existing group structure on $\ker \epsilon$. (In particular, $M$ is an abelian group!) We make $M$ into a left $G$-module as follows:

$$g \cdot m = \eta (g) m \eta (g)^{-1}$$

We can then construct the semi-direct product $M \rtimes G$, which has the following multplication:

$$(m_0, g_0) \cdot (m_1, g_1) = (m_0 \eta (g_0) m_1 \eta (g_0)^{-1}, g_0 g_1)$$

There is a group homomorphism $M \rtimes G \to H$ defined by $(m, g) \mapsto m \eta (g)$, and it is bijective: surjectivity is clear, and injectivity is a consequence of the fact that $M \cap \operatorname{im} \eta = \{ e \}$. We may regard $M \rtimes G$ as a Beck module over $G$ by defining $\epsilon : M \rtimes G \to G$, $\eta : G \to M \rtimes G$, and $\mu : (M \rtimes G) \times_G (M \rtimes G) \to M \rtimes G$ as follows:

$$\epsilon (m, g) = g$$

$$\eta (g) = (0, g)$$

$$\mu ((m_0, g), (m_1, g)) = (m_1 m_0, g)$$

Thus, we have an equivalence between $\mathbf{Ab}(\mathcal{C}_{/ G})$ and the category of left $G$-modules, as claimed.

Proposition

Let $G$ be a group and let $M$ be a left $G$-module. Under the above identification of Beck modules over $G$ with left $G$-modules, $M$-valued derivations of $G$ are precisely crossed homomorphisms $G \to M$, i.e. maps $\delta : G \to M$ satisfying the following equation:

$$\delta (g_0 g_1) = \delta (g_0) + g_0 \cdot \delta (g_1)$$

Proof

Let $\epsilon : M \rtimes G \to G$ be the evident projection. A group homomorphism $\phi : G \to M \rtimes G$ such that $\epsilon \circ \phi = id_G$ is the same thing as a map $\delta : G \to M$ satisfying the equation below,

$$(\delta (g_0), g_0) \cdot (\delta (g_1), g_1) = (\delta (g_0 g_1), g_0 g_1)$$

which is equivalent to the defining equation for crossed homomorphisms.

The tangent category

One may assemble the individual categories of Beck modules over the objects of $\mathcal{C}$ into a category fibred over $\mathcal{C}$, called the tangent category.

References

The concept is due to

• Jon Beck, Triples, algebras and cohomology, Ph.D. thesis, Columbia University, 1967, Reprints in Theory and Applications of Categories, No. 2 (2003) pp 1-59 (TAC)

and was popularized in

• Daniel G. Quillen, On the (co-)homology of commutative rings, in Proc. Symp. on Categorical Algebra, 65 – 87, American Math. Soc., 1970.

See also

• Michael Barr, Acyclic models, Chapter 6, §1.

An application to knot theory is given in

• Markus Szymik, Alexander-Beck modules detect the unknot, Fund. Math. 246 (2019) 89-108.