nLab tangent category

This page is about the construction of “the tangent category of a category” by abelianization. For categories equipped with an abstract “tangent bundle” construction on their objects, see tangent bundle category.

Idea

The notion of tangent category of a category $C$ is an approximation to the notion of tangent (∞,1)-category in ordinary category theory.

For the moment see there for further motivation.

The tangent category $T_C$ of $C$ is effectively the fiberwise abelianization of the codomain fibration $cod : [I,C] \to C$:

we may think of it as obtained from the codomain fibration by replacing each overcategory fiber $[I,C]_A = C/A$ by the corresponding category of abelian group objects and restricting the morphisms such as to respect the abelian group object structure.

Definition

Let $C$ be a category with pullbacks. Then the tangent category $T_C$ of $C$ is the category whose

• objects are pairs $(A,\mathcal{A})$ with $A \in Ob(C)$ and with $\mathcal{A}$ a Beck module over $A$, i.e. an abelian group object in the overcategory $C/A$;

  notice that for $\mathcal{B} \to B$ an object in the overcategory that is equipped with the structure of an abelian group object – notably with a product $prod_{\mathcal{B}} : \mathcal{B} \times_B \mathcal{B} \to \mathcal{B}$ – and for $f \colon A \to B$ any morphism in $C$, the pullback $f^* \mathcal{B} \coloneqq A \times_B \mathcal{B}$ in $C$ is naturally equipped with the structure of an abelian group object internal to the slice category $C/A$;

• morphisms $(f, \mathcal{f}) : (A,\mathcal{A}) \to (B, \mathcal{B})$ are commuting squares

  $$\array{ \mathcal{A} &\stackrel{\mathcal{f}}{\to}& \mathcal{B} \\ \downarrow && \downarrow \\ A &\stackrel{f}{\to}& B }$$

  such that the induced morphism $\mathcal{A} \to f^*\mathcal{B}$ is a morphism of abelian group objects in $C/A$;

• composition of morphisms is given in the evident way by $(f_2, \mathcal{f}_2) \circ (f_1, \mathcal{f}_1) = (f_2 \circ f_1, \mathcal{f}_2 \circ \mathcal{f}_1)$.

Properties

General

There is an evident functor $p : T_C \to C$, the underlying codomain fibration:

$$p \;\;: \;\; \left( \array{ \mathcal{A} &\stackrel{\mathcal{f}}{\to}& \mathcal{B} \\ \downarrow && \downarrow \\ A &\stackrel{f}{\to}& B } \right) \;\; \mapsto \;\; (A \stackrel{f}{\to} B) \,.$$

The fiber over $Id_A$ of this functor is the category of abelian group objects in the overcategory $C/A$:

$$(T_C)_A \simeq Ab(C/A) \,.$$

There is also another functor $q : T_C \to C$, inherited from the domain cofibration

$$q \;\;: \;\; \left( \array{ \mathcal{A} &\stackrel{\mathcal{f}}{\to}& \mathcal{B} \\ \downarrow && \downarrow \\ A &\stackrel{f}{\to}& B } \right) \;\; \mapsto \;\; (\mathcal{A} \stackrel{\mathcal{f}}{\to} \mathcal{B}) \,,$$

where we first forget the abelian group object structure and then project onto the domains.

(we should be claiming that this functor has a left adjoint which is a section and computes the Kähler differentials of objects in $C$).

Of a site

We discuss morphisms of sites from a site to its tangent category.

check

Let $C$ be a category with finite limits and let $T_C \to C$ be its tangent category.

There is then the 0-section $i : C \to T_C$ which sends $A$ to the terminal object $Id : A\to A$ in the overcategory, equipped, necessarily, with the trivial group structure. This exhibits $C$ as a retract of $T_C$

$$(C \stackrel{i}{\to} T_C \stackrel{p}{\to} C) = Id_C \,.$$

Assume now that $C^{op}$ has pullbacks and is equipped with a coverage, hence with the structure of a site.

Equip $(T_C)^{op}$ with the coverage where $\{f_i : U_i \to U\}$ is a cover in $(T_C)^{op}$ precisely if its image $\{p(f_i) : p(U_i) \to p(U)\}$ is a cover in $C^{op}$.

Then the 0-section $C^{op} \to (T_C)^{op}$ preserves covers.

We claim it also preserves limits: i.e. that $i : C \to T_C$ preserves colimits:

let $F : K \to C$ be a diagram and $\lim_\to F$ its colimit in $C$. Then let $Q$ be any cocone under $i \circ F$ in $T_C$. By applying $p$ to that cocone we find that there is a unique morphism of cocones $\lim_\to F \to p(Q)$ in $C$. But any morphism of the form $A \to p(B)$ for $A \in C$ and $B \in T_C$ has a unique lift to a morphism $i(A) \to B$ in $T_C$ (because the trivial ablian group is initial, so that the morphism in $T_C$ is fixed by its underlying morphism in $C$).

So for any coverage on $C^{op}$ and the above induced coverage on $(T_C)^{op}$, the 0-section $i : C^{op} \to (T_C)^{op}$ is a morphism of sites.

Accordingly, we obtain a geometric morphism of sheaf toposes

$$Sh((T_C)^{op}) \stackrel{\leftarrow}{\underset{}{\to}} Sh(C^{op}) \,.$$

Examples

Modules as tangents to rings

Proposition

For $C =$ CRing we have $T_C \simeq$ Mod.

(More generally for $C =$ Ring then $T_C$ is the category of bimodules, see at Beck module – Over associative algebras).

Proof

Consider the functor

$$F: Mod \to T_{CRing}$$

that sends an $R$-module $N$ to the square-0 extension ring $R \oplus N \stackrel{p_1}{\to} R$, regarded as an abelian group object in $CRing/R$.

The action on morphisms is given as follows: if $(R_1,N_1)$ and $(R_2,N_2)$ are two objects in Mod, then a morphism between them is a pair $(f : R_1 \to R_2, f_*:N_1 \to f^* N_2)$ consisting of a ring homorphism and a morphism of $R_1$ modules from $N_1$ to $R_1 \otimes_f N_2$; the corresponding morphism of rings $R_1\oplus N_1\to R_2\oplus N_2$ is $(r_1,n_1)\mapsto (f(r_1),f_*(n_1))$. The induced morphism of rings $R_1\oplus N_1\to R_1\times_{R_2}(R_2\oplus N_2)\cong R_1\oplus f^*N_2$ is explicitly given by $(r_1,n_1)\mapsto (r_1,f_*(n_1))$ and is easily checked to be a morphism of abelian group objects over $R_1$.

Moreover, by the natural isomorphism $R_1\times_{R_2}(R_2\oplus N_2)\cong R_1\oplus f^*N_2$ in $Ab(CRing/R)$, showing that $F:Mod\to T_{CRing}$ is an equivalence is reduced to showing that $F$ is a fibrewise equivalence, i.e., that that for any fixed ring $R$,

$$F_R: Mod_R \to Ab(CRing/R)$$

is an equivalence of categories. This is shown at module.

Proposition

The domain projection $Mod \to CRing$ has a left adjoint, namely the functor assigning to each commutative ring $A$ the pair $(A, \Omega_A)$, where $\Omega_A$ is the $A$-module of Kähler differentials.

Proof

Let $A$ and $B$ be commutative rings, let $M$ be a $B$-module, and consider $B \oplus M$ as a ring as in the previous proof. Then, to give a ring homomorphism $f : A \to B \oplus M$ is the same as giving a ring homomorphism $f_0 : A \to B$ and an additive homomorphism $f_1 : A \to M$ such that

$$f_1 (x y) = f_0 (x) f_1 (y) + f_0 (y) f_1 (x)$$

for all $x$ and $y$ in $A$. But by the universal property of $\Omega_A$, this is the same as giving a morphism $(A, \Omega_A) \to (B, M)$ in $Mod$.

Modules over smooth algebras

Let $SmoothAlg$ (or $C^\infty Ring$) be the category of smooth algebras. Notice that there is a canonical forgetful functor

$$U : SmoothAlg \to Ring$$

to the underlying ordinary rings.

Theorem

There is an equivalence of categories

$$T_{SmoothAlg} \stackrel{\simeq}{\to} SmoothAlg \times_{Ring} T_{Ring} \,,$$

where on the right we have the strict pullback (i.e. taken in the 1-category Cat).

We give the proof below. First some remarks and corollaries.

Remark

We may regard an object in $T_{SmoothAlg}$ as a module over a smooth algebra. The above says in particular that modules over smooth algebras are just modules over the underlying ordinary rings. However, the category structure on $T_{SmoothAlg}$ does reflect that modules over smooth algebras have a different nature than just bare modules, notably in that the left adjoint to the projection $T_{SmoothAlg} \to SmoothAlg$ produces the correct $C^\infty$-derivations and $C^\infty$-Kähler differentials (see there) as opposed to the purely algebraic ones.

Corollary

For any category $S$ we have that

$$(T_{SmoothAlg})^S \simeq SmoothAlg^S \times_{Ring^S} (T_{Ring})^S \,.$$

So in particular for $S = \Delta$ the simplex category we have that simplicial modules over simplicial smooth algebras are as objects just ordinary simplicial modules over the underlying simplicial rings.

For proving the above theorem the main step is the following lemma.

Lemma

For a fixed smooth algebra $R$, the forgetful functor

$$U : Ab(SmoothAlg/R) \to Ab(Ring/U(R))$$

is an equivalence of categories.

Proof

The statement was suggested at some point by Thomas Nikolaus in discussion with Urs Schreiber, who then asked Herman Stel to prove it. A writeup is in (Stel).

We discuss in detail that the functor is injective on objects, in that for an any abelian group object in $SmoothAlg/R$ its smooth algebra structure on the underlying ring structure is the unique such smooth algebra that makes it an abelian group object over $R$. Whith this it is then easy to see that $U$ is in fact an isomorphism of categories.

The crucial property underlying this statement is that the Lawvere theory $T =$ CartSp over wich smooth algebras are $T$-algebras is in fact a Fermat theory in that Hadamard's lemma holds for smooth functions in particular on Cartesian spaces.

This implies that for every $k \in \mathbb{N}$ and every smooth function $f : \mathbb{R}^k \to \mathbb{R}$ there are smooth functions $\{h_{i,j} \in C^\infty(\mathbb{R}^k \times \mathbb{R}^k, \mathbb{R})\}_{i,j = 1}^n$ such that the function

$$f \circ + : \mathbb{R}^k \times \mathbb{R}^k \to \mathbb{R}$$

has an expansion given for all $p, w \in \mathbb{R}^k$ by

$$f(p+w) = f(p) + \sum_{l = 1}^k w_l \cdot \frac{\partial f}{\partial x_l}(p) + \sum_{i,j} w_i \cdot w_j h_{i,j}(p,w) \,.$$

We now use that any smooth algebra $A$ regarded as a product-preserving functor $A : CartSp \to Set$ reflects these relations in that for all $r, e \in A(k) = U(A)^k$ we have that

$$A(f)(r+e) = A(f)(r) + \sum_{l = 1}^k w_l \cdot A\left(\frac{\partial f}{\partial x_l}\right)(r) + \sum_{i,j} e_i \cdot e_j A(h_{i,j})(r,e) \,.$$

Now if $R \in SmoothAlg$ and $A$ is an object in $Ab(SmoothAlgebra/R)$ then in particular its underlying ring will be an object in $Ab(Ring/U(R))$. By the above theorem this means that the underlying ring is a square-zero extension $U(R) \oplus N$ by some $N \in U(R) Mod$.

So it follows every element of $A(1)$ is of the form $(r, \epsilon)$ with $\epsilon \in N$ and we can always write it as

$$(r,0) + (0, \epsilon) \,.$$

Moreover, since $A$ is by assumption a group object over $R$, it follows that for all $f \in C^\infty(\mathbb{R}^k , \mathbb{R})$ and for all $r \in R(1)$ we have

$$A(f)(r) = R(f)(r) \,.$$

So we only need to know how $A$ acts on mixed terms. The point now is that the above Hadamard-quotient formula reduces the action of any smooth function to just operations of this form $A(f)(r)$ and to ordinary multiplication and addition, so it actually fixes $A(f)$ from the restriction of $A(f)$ to elements of the form $(r,0)$ and the module structure on $N$:

$$A(f)(r +\epsilon) = A(f)(r) + \sum_{l = 1}^k w_l \cdot A\left(\frac{\partial f}{\partial x_l}\right)(r) + 0$$

since $\epsilon_i \cdot \epsilon_j = 0$ in the underlying square-0 extension of $A$ and hence also in $A$.

In summary this shows that the forgetful functor $U$ is injective on objects. The above formula also directly implies, conversely, that the functor is surjective on objects, hence an isomorphism on objects, and moreover that it is a full and faithful functor.

Finally we come to the proof of the full theorem above

Proof

The above lemma shows that $T_{SmoothAlg} \simeq SmoothAlg \times_{Ring} T_{Ring}$ is a bijection on objects.

Since the pullbacks that are involved in the definition of the tangent category $T_{SmoothAlg}$ are preserved by the right adjoint forgetful functor $U : SmoothAlg \to Ring$ (a special case of the general facts about Relative free T-algebra adjunctions), checking bijection on hom-sets

$$T_{SmoothAlg}(A\to R_1,B \to R_2) \to SmoothAlg \times_{Ring} T_{Ring} ((U(A), R_1), (U(B), R_2))$$

amounts to checking for each $f : R_1 \to R_2$ bijections of hom-sets of abelian group objects

$$Ab(SmoothAlg/R_1)(A, f^* B) \to Ab(Ring/U(R_1))(U(A), f^* U(B)) \,.$$

That this is a bijection is the statement of the above lemma.

Related concepts

• tangent (∞,1)-category.

References

The original observation that $T_{Ring} \simeq Mod$ is due to

• Daniel Quillen, …

A discusson of $T_{SmoothAlg}$ is in

• Herman Stel, ∞-Stacks and their Function Algebras