TITLE: Classifying triangulated structures on a graded category
QUESTION [14 upvotes]: I know of several results to the effect that two triangulated categories are equivalent categories (usually one coming from algebra and one coming from topology).  However, it's never been clear to me how to classify the possible different triangulated structures lifting a given "graded" category (a category enriched in graded abelian groups).
For example, how many different triangulated structures can there be with underlying category the derived category of Z-modules?  All objects in this category are equivalent to direct sums of Z-modules concentrated in different degrees.
As another example, we have the category of "Z/2-graded abelian groups", which is equivalent to the derived category of differential graded modules over a Laurent polynomial ring on a generator in degree 2.  There's another, equivalent-looking, homotopy category of modules over the periodic complex K-theory spectrum KU.  These have equivalent underlying categories but I have never been clear on whether the triangulated structures are the same.

REPLY [3 votes]: In 2002, Paul Balmer wrote a nice two page note answering the same question.
He sent it to me because I had asked the same question in my paper ``The 
additivity of traces in triangulated categories'', where I had recalled the
standard fact (well-known in topological examples since the 1960's) that a 
triangulation usually differs from its negative.  I just looked, and it 
seems Paul never published his note.  I'll mostly use his notation.
Fix a category $\mathcal K$ with a triangulation $\mathcal T$ that differs 
from its negative $-\mathcal T$.  We will never change the suspension $\Sigma$.  His 
first example is trivial.  For any $n\geq 1$, the $n$-fold product $\mathcal{K}^n$
with the obvious suspension has at least $2^n$ distinct triangulations, choosing 
$\mathcal T$ or $-\mathcal T$ on each factor.  
Define a global automorphism $\alpha$ of $(\mathcal K,\Sigma)$ to be an automorphism of
the identity functor on $\mathcal K$ that commutes with $\Sigma$.  It is given by 
isomorphisms $\alpha\colon A\to A$ for all objects $A$ of $\mathcal K$ such that $\alpha f = f\alpha$
for all maps $f\colon A\to B$ in $\mathcal K$ and $\Sigma \alpha$ is the $\alpha$ for $\Sigma A$.
The simplest example is to take $\mathcal K$ to be (any version of) the derived category 
of a commutative ring $R$ and let $\alpha(u)\colon A\to A$ be multiplication by a unit $u\in R$.
For a global automorphism $\alpha$ of $\mathcal K$, define $\mathcal{T}_{\alpha}$ to be the
class of all triangles $(f,g,h)$ such that $(f\alpha,g,h)$ is a trangle in $\mathcal T$.  Then
$(\mathcal K,\mathcal{T}_{\alpha})$ is a triangulated category. 
Now return to the derived category $\mathcal K$ of a commutative ring $R$. Let $u$ be a unit of $R$.
Suppose there is an element $r$ such that $r$ is not a zero divisor and $1-u\notin rR$. 
Then the triangulations $\mathcal T$ and $\mathcal{T}_{\alpha(u)}$ are different. If 
we take $u=-1$, this very often proves that $\mathcal T \neq -\mathcal T$.  Now let 
$R = k[x]$ be a polynomial ring over a commutative ring $k$ with infinitely many units.
consider the units $u$ other than $1$ in $k\subset R$ and let $r=x$.  We have a 
triangulation $\mathcal{T}_{\alpha(u)}$ for each such $u$.  Clearly 
$(\mathcal {T}_{\alpha(u)})_{\alpha(v)} = \mathcal {T}_{\alpha(uv)}$.  Thus, if $\mathcal{T}_{\alpha(u)} = \mathcal{T}_{\alpha(v)}$,
then $\mathcal{T}= \mathcal{T}_{\alpha(u^{-1}v)}$.  Therefore $\mathcal K$ has infinitely
many distinct triangulations.