Lemma 13.36.1. Let $\mathcal{T}$ be a triangulated category. Let $E$ be an object of $\mathcal{T}$. For $n \geq 1$ we have

For $n, n' \geq 1$ we have $\langle E \rangle _{n + n'} = smd(\langle E \rangle _ n \star \langle E \rangle _{n'})$.

Lemma 13.36.1. Let $\mathcal{T}$ be a triangulated category. Let $E$ be an object of $\mathcal{T}$. For $n \geq 1$ we have

\[ \langle E \rangle _ n = smd(\langle E \rangle _1 \star \ldots \star \langle E \rangle _1) = smd({\langle E \rangle _1}^{\star n}) = \bigcup \nolimits _{m \geq 1} smd(add(E[-m, m])^{\star n}) \]

For $n, n' \geq 1$ we have $\langle E \rangle _{n + n'} = smd(\langle E \rangle _ n \star \langle E \rangle _{n'})$.

**Proof.**
The left equality in the displayed formula follows from Lemmas 13.35.1 and 13.35.2 and induction. The middle equality is a matter of notation. Since $\langle E \rangle _1 = smd(add(E[-\infty , \infty ])])$ and since $E[-\infty , \infty ] = \bigcup _{m \geq 1} E[-m, m]$ we see from Remark 13.35.6 and Lemma 13.35.2 that we get the equality on the right. Then the final statement follows from the remark and the corresponding statement of Lemma 13.35.4.
$\square$

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