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Homogeneous right multiplication reconstructs the graded kernel action

Statement

Let $k$ be a field, $A, B$ graded $k$ -algebras and $F : {GrMod}_{0} (A) \to {GrMod}_{0} (B)$ $k$ -linear (hence additive) and coherently shift-compatible with comparisons $θ$ (Coherently shift-compatible functors and natural transformations). Put $M : = F (A) \in {GrMod}_{0} (B)$ .

For homogeneous $a \in A_{d}$ the map $r_{a} : A {d} ⟶ A, r_{a} (x) : = x a,$ is degree-zero and $A$ -linear; the prescription $m \cdot a : = F (r_{a}) θ_{A, d}^{- 1} (m) (m \in M, a \in A_{d} homogeneous)$ defines a degree-zero map $M {d} \to M$ , that is, a homogeneous right action of $A$ of degree $d$ , and extending $k$ -bilinearly over $A = ⨁_{d} A_{d}$ makes $M$ a graded $(B, A)$ -bimodule.
The action satisfies $m \cdot 1 = m, m \cdot (a + a^{'}) = m \cdot a + m \cdot a^{'}, (m \cdot a) \cdot b = m \cdot (a b)$ for homogeneous $a \in A_{d}$ , $b \in A_{e}$ , commutes with the left $B$ -action ( $b (m \cdot a) = (b m) \cdot a$ ) and with the central $k$ -scalars, and is homogeneous: $M_{g} A_{d} \subseteq M_{g + d}$ .
The construction uses only the shift comparisons and the functoriality, additivity and $k$ -linearity of $F$ : degree-zero endomorphisms of $A$ alone would recover only $A_{0}$ , the shifts $A {d}$ are what make the homogeneous action accessible. No choice is used.

Facts & Assumptions

Given: A field $k$ , graded $k$ -algebras $A, B$ , a $k$ -linear coherently shift-compatible functor $F : {GrMod}_{0} (A) \to {GrMod}_{0} (B)$ with comparisons $θ$ , homogeneous $a, a^{'} \in A_{d}$ , $b \in A_{e}$ , elements $m \in M : = F (A)$ and $t, λ \in k$ .

[L1]

Coherently shift-compatible functors carry natural degree-zero isomorphisms $θ_{X, r} : F (X {r}) \to F (X) {r}$ with $θ_{X, 0} = 1$ and the cocycle $θ_{X, r + s} = (θ_{X, r} {s}) \circ θ_{X {r}, s}$ (Coherently shift-compatible functors and natural transformations).

[L2]

The internal shift satisfies ${0} = i d$ and ${r} {s} = {r + s}$ on the nose, acts as the identity on underlying sets and vectors, so the shift of a morphism is the same underlying map, and it preserves degreewise kernels and cokernels (Internal shifts are autoequivalences and commute with the graded tensor product).

[L3]

A graded $(B, A)$ -bimodule is a $(B, A)$ -bimodule that is graded as a $k$ -module and homogeneous under both actions, the two actions commuting and inducing the same $k$ -action: $η_{B} (t) m = t m = m η_{A} (t)$ ; a map is degree-zero when it is $A$ -linear and preserves degrees (Associative graded algebras, bimodules, and internal shifts).

[L4]

An $(S, R)$ -bimodule is an abelian group that is a left $S$ -module and a right $R$ -module with commuting actions ( $(S, R)$ -bimodules and commuting left and right scalar actions).

[L5]

A left $R$ -module satisfies $r (m + m^{'}) = r m + r m^{'}$ , $(r + r^{'}) m = r m + r^{'} m$ , $(r r^{'}) m = r (r^{'} m)$ and $1_{R} m = m$ , and symmetrically for right modules (Unital left and right modules over a ring; unqualified module means left module).

[L6]

An additive functor satisfies $F (f + g) = F f + F g$ for parallel morphisms (Additive functor).

[L7]

A functor satisfies $F (1_{X}) = 1_{F (X)}$ and $F (g \circ f) = F g \circ F f$ (Covariant functor, identity functor, composite functor, and contravariant functor).

[L8]

A functor between $k$ -linear categories is $k$ -linear when each induced map of hom-spaces is $k$ -linear, so $F (λ f) = λ F (f)$ for parallel $f$ and $λ \in k$ (k-linear categories and k-linear functors).

[L9]

In a vector space the scalar action is additive in the vector and scalar and satisfies $(λ μ) m = λ (μ m)$ , $1 m = m$ (Vector space over a field).

[L10]

A field has a commutative multiplication and distinguished $0 \neq 1$ (Field).

[L11]

Every field is a commutative ring with the same operations and units (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring).

Proof

technique · direct

1.1L2L3L5algebra

For homogeneous $a \in A_{d}$ the map $r_{a} (x) : = x a$ sends $(A {d})_{e} = A_{e - d}$ into $A_{e}$ because the multiplication of the graded algebra is homogeneous, so $r_{a}$ is degree-zero; it is left $A$ -linear because $r_{a} (x^{'} x) = (x^{'} x) a = x^{'} (x a) = x^{'} r_{a} (x)$ by associativity of $A$ ; here $A {d}$ carries the same underlying left $A$ -module as $A$ .

2.1step 1.1L1L3L9L11

The composite $m \cdot a : = F (r_{a}) θ_{A, d}^{- 1} (m)$ is well defined: $F (r_{a}) : F (A {d}) \to F (A) = M$ is degree-zero $B$ -linear by step 1.1 and [L3], and $θ_{A, d}^{- 1} : M {d} \to F (A {d})$ is a degree-zero $B$ -linear isomorphism by [L1]; hence $m \mapsto m \cdot a$ is a degree-zero $B$ -linear map $M {d} \to M$ , so for $m \in (M {d})_{f} = M_{f - d}$ one has $m \cdot a \in M_{f}$ , which is the homogeneity statement $M_{f - d} A_{d} \subseteq M_{f}$ , and the map is additive and $k$ -linear in $m$ because $B$ -linear maps of graded $B$ -modules are $k$ -linear [L9, L11].

3.1step 2.1L1L6L8L9

For homogeneous $a, a^{'}$ of the same degree the identity $r_{a + a^{'}} = r_{a} + r_{a^{'}}$ holds pointwise, so $F (r_{a + a^{'}}) = F (r_{a}) + F (r_{a^{'}})$ by additivity [L6] and $m \cdot (a + a^{'}) = m \cdot a + m \cdot a^{'}$ ; for $λ \in k$ one has $r_{λ a} = λ r_{a}$ pointwise, so $F (r_{λ a}) = λ F (r_{a})$ by $k$ -linearity [L8] and $m \cdot (λ a) = λ F (r_{a}) θ_{A, d}^{- 1} (m) = F (r_{a}) θ_{A, d}^{- 1} (λ m) = λ (m \cdot a)$ , using that $θ^{- 1}$ and $F (r_{a})$ are $k$ -linear; hence the prescription extends by the finite homogeneous decomposition $a = \sum_{d} a_{d}$ to a well-defined pairing $M \times A \to M$ that is additive and $k$ -linear in each variable.

3.2step 2.1L1L2L7

Unit: $r_{1} = {i d}_{A}$ as a map $A {0} = A \to A$ and $θ_{A, 0} = 1_{F (A)}$ by [L1], so $m \cdot 1 = F ({i d}_{A}) θ_{A, 0}^{- 1} (m) = 1_{M} (m) = m$ by [L7].

3.3step 2.1L1L2L6L7algebra

Associativity: for homogeneous $a \in A_{d}$ , $b \in A_{e}$ one has $r_{a b} = r_{b} \circ (r_{a} {e})$ as maps $A {d + e} \to A$ , both sending $x$ to $x a b$ ; hence $F (r_{a b}) = F (r_{b}) F (r_{a} {e})$ by [L7]. The identity $θ_{A, e}^{- 1} F (r_{a}) θ_{A, d}^{- 1} = F (r_{a} {e}) θ_{A, d + e}^{- 1}$ of underlying maps from $M$ to $F (A {e})$ follows from the naturality of $θ$ at $r_{a}$ with parameter $e$ , $θ_{A, e} F (r_{a} {e}) = (F (r_{a}) {e}) θ_{A {d}, e}$ , from the cocycle $θ_{A, d + e} = (θ_{A, d} {e}) θ_{A {d}, e}$ and from the fact that shifting a morphism leaves the underlying map unchanged [L2], since then $(F (r_{a}) {e}) (θ_{A, d} {e})^{- 1} = F (r_{a}) θ_{A, d}^{- 1}$ as functions. Substituting into $(m \cdot a) \cdot b = F (r_{b}) θ_{A, e}^{- 1} (F (r_{a}) θ_{A, d}^{- 1} (m))$ gives $(m \cdot a) \cdot b = F (r_{a b}) θ_{A, d + e}^{- 1} (m) = m \cdot (a b)$ .

4.1step 2.1step 3.1step 3.2step 3.3L3L4L8L9L10L11

The left $B$ -action commutes with the reconstructed right action, $b (m \cdot a) = (b m) \cdot a$ , because $F (r_{a})$ and $θ_{A, d}^{- 1}$ are $B$ -linear; the induced $k$ -actions agree because for $t \in k$ one has $η_{A} (t) \in A_{0}$ , $r_{η_{A} (t)} = t {i d}_{A}$ , $θ_{A, 0} = 1$ and therefore $m \cdot η_{A} (t) = F (t {i d}_{A}) (m) = t m$ by $k$ -linearity [L8], while $η_{B} (t) m = t m$ ; steps 3.1, 3.2 and 3.3 give additivity in both variables, the unit law and associativity, so $M$ is a left $B$ -module and a right $A$ -module with commuting actions and common central $k$ -action, homogeneous under both; by [L3, L4] $M$ is a graded $(B, A)$ -bimodule.

5.1step 4.1L1L2L5algebra∎

Every map used is $F$ applied to the canonical maps $r_{a}$ , shifted by the canonical comparisons and the canonical identifications of the shift functor, so no basis, generator or element is selected and no choice is used; and the shifts are essential: a degree-zero $A$ -linear endomorphism $φ$ of $A$ satisfies $φ (x) = x φ (1_{A})$ with $φ (1_{A}) \in A_{0}$ , so the endomorphisms of $A$ alone recover only the actions of degree $0$ , whereas the maps $r_{a}$ with $a$ of degree $d \neq 0$ have source $A {d}$ and enter $M$ only through the comparisons $θ_{A, d}$ .

Depends on

Used by

Dependency tree · two levels

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