A counterexample on nilpotent endomorphisms in triangulated categories

Matthias Künzer

September 16, 2009

Abstract

We give a counterexample to an assertion on nilpotent endomorphisms of degree 3 in triangulated categories. Roughly speaking, it is in general not possible to lower the nilpotency degree by passing to a certain cone.

1 Nilpotency degree 2 in triangulated categories

Let C be a Verdier triangulated category.

Suppose given X ∈ Ob C and d ∈ End_{C} X such that d^{2} = 0. Let X ^{\overline{}^{d}} X ^{\overline{}^{f}} C ^{\overline{}^{g}} X^{+1} be a (distinguished) triangle. M. Breuning made use of the following endomorphism of triangles.

\xymatrix{X \ar[r]^{d} \ar[d]^{0} & X \ar[r]^{f} \ar[d]^{d} & C \ar[r]^{g} \ar[d]^{0} & X^{+1} \ar[d]^{0} \\ X \ar[r]^{d} & X \ar[r]^{f} & C \ar[r]^{g} & X^{+1}}

So the pair (0, d) can be extended to an endomorphism of triangles (0, d, e) such that e = e^{1} = 0.

2 Nilpotency degree 3 in abelian categories

Let A be an abelian category.

Suppose given X ∈ Ob A and d ∈ End_{A} X such that d^{3} = 0. Let f be a cokernel of d. We obtain an endomorphism of sequences.

[figure: X; d^{2}; 0; X; f; d; C; e; X; d^{2}; X; f; C]

Now fe^{2} = d^{2}f = 0, whence e^{2} = 0. So the pair (0, d) can be (uniquely) extended to an endomorphism of sequences (0, d, e) such that e^{2} = 0.

3 Nilpotency degree 3 in triangulated categories

Let C be a Verdier triangulated category. Suppose given X ∈ Ob C and d ∈ End_{C} X such that d^{3} = 0. Let X ^{\overline{}^{d}} X ^{\overline{}^{f}} C ^{\overline{}^{g}} X^{+1} be a triangle. There exists C ^{\overline{}^{e}} C fitting into an endomorphism of triangles as follows.

\xymatrix{X \ar[r]^{d^{2}} \ar[d]^{0} & X \ar[r]^{f} \ar[d]^{d} & C \ar[r]^{g} \ar[d]^{e} & X^{+1} \ar[d]^{0} \\ X \ar[r]^{d^{2}} & X \ar[r]^{f} & C \ar[r]^{g} & X^{+1}}

Now, eg = 0, whence e = bf for some C ^{\overline{}^{b}} X. Hence e^{3} = bfe^{2} = bd^{2}f = 0. Motivated by §1 and §2, we consider the following Assertion. There exists an endomorphism C ^{\overline{}^{ẽ}} C with ẽ^{2} = 0 such that (0, d, ẽ) is an endomorphism of triangles. Counterexample. Let C = \underline{Z/64 -mod} ; cf. [1]. Let (X ^{\overline{}^{d}} X) := (Z/8 ^{\overline{}^{2}} Z/8). Suppose given the following endomorphism

\xymatrix{Z/8 \ar[r]^{4} \ar[d]^{0} & Z/8 \ar[r]^{^{(^{1 2})}} \ar[d]^{2} & Z/4 ⊕ Z/16 \ar[r]^{^{(\begin{matrix}−2 \\ 1 \end{matrix})}} \ar[d]|{ẽ= (\begin{matrix}a & 4b \\ c & d \end{matrix})} & Z/8 \ar[d]^{0} \\ Z/8 \ar[r]^{4} & Z/8 \ar[r]^{_{(_{1 2})}} & Z/4 ⊕ Z/16 \ar[r]^{_{(\begin{matrix}−2 \\ 1 \end{matrix})}} & Z/8}

of triangles, in which a, b, c, d ∈ Z/4. The middle quadrangle yields (^{a+2c 4b+2d}) = (^{2 4}), as morphisms from Z/8 to Z/4 ⊕ Z/16. So a + 2c ≡_{4} 2 and 4b + 2d ≡_{8} 4. Hence a = 2 − 2c and d = 2 − 2b. Thus ẽ = (\begin{matrix}2−2c & 4b \\ c & 2−2b \end{matrix}). Now the right hand side quadrangle yields (\begin{matrix}−4+4b+4c \\ 2−2b−2c \end{matrix}) = (\begin{matrix}0 \\ 0 \end{matrix}) as morphisms from Z/4 ⊕ Z/16 to Z/8. So b + c ≡_{2} 1. We calculate ẽ^{2} = (\begin{matrix}0 & 8b(b+c) \\ 2c(b+c) & 0 \end{matrix}) = (\begin{matrix}0 & 8b \\ 2c & 0 \end{matrix}), which is nonzero because of b + c ≡_{2} 1.

References

[1] Künzer, M., Nonisomorphic Verdier octahedra on the same base, J. Homotopy and Related Structures 4(1), p. 7–38, 2009.

MSC2000: 18E30.