Monoidal triangulated categories I have a monoidal (not symmetric) triangulated category $(A,\otimes, 1)$ with unit 1.
Define $C$ the localizing subcategory of $A$ generated by the unit 1.

*

*is $(C, \otimes, 1) $ a symmetric monoidal triangulated category?

*Do we have a natural isomorphism $c\otimes a\cong a\otimes c$ for any $a\in A$ and any $c\in C$ ?

 A: The answer to question 2, even with the assumption that $a\in C$, is no in general. It follows that the answer to question 1 is also no.
For a counterexample, one can consider for example a (derived) category of bimodules. Let $k$ be a base commutative ring, and $R$ a (say) flat $k$-algebra. Consider the derived category $A= D(R\otimes_k R^{op})$ of $(R,R)$-bimodules over $k$. This is monoidal under $\otimes^\mathbb L_R$, with unit $R$ viewed as an $R$-bimodule in the canonical way.
An explicit example is given by the polynomial ring $R= k[t]$. In this case, $R\otimes_k R^{op} = k[t_0,t_1]$.
There is $k[t_0,t_1]$-linear extension $0\to k[t]\to k[t_0,t_1]/(t_0-t_1)^2\to k[t]\to 0$, where the rightmost map is the algebra map $t_0,t_1\mapsto t$, while the leftmost map is $P\mapsto (t_0-t_1)P(t_0,t_0)$. One easily checks that this is an extension: first, the rightmost map is clearly a bimodule map. For the leftmost map, one checks that $(t_0-t_1)t_1P(t_0,t_0) = (t_0-t_1)t_0P(t_0,t_0)$. Indeed it suffices to check it when $P = 1$ and then the relation is $t_0t_1-t_1^2 = t_0^2-t_1t_0$ which is a rephrasing of the relation $(t_0-t_1)^2 = 0$.
Second, suppose $P(t,t) = 0$. Then $P$ is divisible by $(t_0-t_1)$, to $P = (t_0-t_1)Q$ for some $Q$, and now $(t_0-t_1)Q = (t_0-t_1)Q(t_0,t_0)$ modulo $(t_0-t_1)^2$, because $Q-Q(t_0,t_0)$ is divisible by $(t_0-t_1)$. It follows that $P$ is in the image of the leftmost map.
This extension corresponds to a morphism $k[t]\to \Sigma k[t]$ in $A$, with fiber $E= k[t_0,t_1]/(t_0-t_1)^2$. In particular, $E$ belongs to $C$.
Another object in $C$ is $k$ where $t_0,t_1$ act trivially. Indeed, it is the mapping cone of $k[t]\xrightarrow{t}k[t]$ (this is a bimodule map, as $t$ is central).
I claim that $E\otimes^\mathbb L_{k[t]}k \not\simeq k\otimes_{k[t]}^\mathbb L E$.
To see this, we first observe that these derived tensor products are underived. The key lemma for this is:
Lemma: Suppose $(t_0-t_1)$ divides $t_0 P$. Then $(t_0-t_1)$ divides $P$.
Proof: The assumption implies that $tP(t,t) = 0$ in $k[t]$. But $t$ is not a zero divisor, so $P(t,t) = 0$ and the conclusion follows.
Now we use this lemma twice : suppose $t_0 P = 0$ in $E$. This means $(t_0-t_1)^2\mid t_0P$ in $k[t_0,t_1]$. It follows that $(t_0-t_1)$ divides $t_0 P$, and thus by the lemma, it divides $P$. Write $P= (t_0-t_1)Q$, and it follows easily that $(t_0-t_1) \mid t_0 Q$, so that by the lemma, $(t_0-t_1)$ divides $Q$, and so $(t_0-t_1)^2$ divides $P$. So $P = 0$ in $E$.
By symmetry, the same holds for $t_1$, and so the action of $t_0$ and $t_1$ on $E$ is injective, which proves the claim that the tensor products are underived.
Now, $E\otimes_{k[t]}k$ is therefore $E/t_1 E\cong k[t_0,t_1]/((t_0-t_1)^2, t_1) = k[t_0,t_1]/(t_0^2,t_1)\cong k[t_0]/(t_0)^2$, on which $t_0$ acts nontrivially while $t_1$ acts trivially. Conversely, $k\otimes_{k[t]}E\cong k[t_1]/(t_1^2)$, on which $t_0$ acts trivially but $t_1$ nontrivially. It follows that they are not isomorphic as bimodules, hence not isomorphic in the derived category either.
