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A degree-one derivation need not square to zero

Statement refuted

For every graded associative unital $Q$ -algebra $B$ , every degree- $+ 1$ $Q$ -linear map $δ : B \to B$ satisfying $δ (a b) = δ (a) b + (- 1)^{∣ a ∣} a δ (b)$ for homogeneous $a, b$ is square-zero.

Facts & Assumptions

Given: The usual rational field and the free associative graded $Q$ -algebra $B = Q ⟨ x, u, v ⟩$ with $∣ x ∣ = 0$ , $∣ u ∣ = 1$ , and $∣ v ∣ = 2$ , whose multiplication is concatenation of words. Its $Q$ -basis consists of finite words in $x, u, v$ , including the empty word $1$ . Define $δ (x) = u$ , $δ (u) = v$ , and $δ (v) = 0$ ; for a word $w = a_{1} \dots a_{n}$ , define $δ (w) = \sum_{i = 1}^{n} (- 1)^{∣ a_{1} ∣ + \dots + ∣ a_{i - 1} ∣} a_{1} \dots a_{i - 1} δ (a_{i}) a_{i + 1} \dots a_{n},$ with $δ (1) = 0$ , and extend $Q$ -linearly. Refute the statement.

[F1]

The rationals form a field: $Q$ with its usual operations is a field.

[F2]

Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring: Every field is a commutative ring with unit, so $Q$ is an admissible base ring for the DGA definition.

[F3]

Algebras over a commutative ring, central structure maps, and algebra homomorphisms: A $Q$ -algebra has a unital structure map from the commutative base ring $Q$ .

[F4]

Cohomological differential graded algebra: The defining cochain differential satisfies $d^{n + 1} d^{n} = 0$ .

Counterexample

1.1givenF1F2F3constructalgebra

By [F1] and [F2], $Q$ is a commutative ring with unit, and [F3] is the unital base-algebra convention used in the construction of $B$ . The words form a $Q$ -basis, so the displayed formula defines a unique linear map. Every nonzero generator image has degree one higher than its input. For homogeneous words $s, t$ , the summands replacing letters of $s$ give $δ (s) t$ , while those replacing letters of $t$ acquire the extra sign $(- 1)^{∣ s ∣}$ ; hence $δ (s t) = δ (s) t + (- 1)^{∣ s ∣} s δ (t)$ . Bilinearity extends the identity to all homogeneous elements.

2.1step 1.1algebra

Applying the generator rules twice gives $δ^{2} (x) = δ (u) = v$ . The word $v$ is a nonzero basis element of the free algebra, so $δ^{2} (x) \neq 0$ .

3.1F4step 2.1∎

By [F4], square-zero is required for a cohomological DGA differential. Thus $δ$ is a degree-one graded derivation that is not square-zero, refuting the statement.

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