Exercise 0.18

Question

Show that $S^1 \ast S^1 = S^3$, and more generally $S^m \ast S^n = S^{m+n+1}$.

Amit Awasthi · Accepted Answer

\begin{proof}
  We first want to show $S^n = \operatorname{\raisebox{-0.6ex}{\scalebox{2.5}{$\ast$}}}_{i=0}^n S^0$. This trivially holds for $n=0$, and so we consider the inductive case. Recall $S^{n+1} = S(S^n) = S\left( \operatorname{\raisebox{-0.6ex}{\scalebox{2.5}{$\ast$}}}_{i=0}^n S^0 \right)$, and so $X \ast S^0 = S(X)$ implies $S^{n+1} = \operatorname{\raisebox{-0.6ex}{\scalebox{2.5}{$\ast$}}}_{i=0}^{n+1} S^0$. Thus,
  \begin{equation*}
    S^m \ast S^n = \left( \operatorname{\raisebox{-0.6ex}{\scalebox{2.5}{$\ast$}}}_{i=0}^{m} S^0 \right) \ast \left( \operatorname{\raisebox{-0.6ex}{\scalebox{2.5}{$\ast$}}}_{k=0}^{n} S^0 \right) = \operatorname{\raisebox{-0.6ex}{\scalebox{2.5}{$\ast$}}}_{i=0}^{m+n+1} S^0 = S^{m+n+1}
  \end{equation*}
  since $\ast$ is associative as remarked on p.~9. In particular, $S^1 \ast S^1 = S^3$.
\end{proof}

Exercise 0.18

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Exercise 0.18

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