# What does an $\oplus$-sign in the superscript mean?

I've come across an expression $M^{\oplus n_i}$ in an article and I have not seen this before? What does is mean? The whole expression looks like this: $$\large M=M_1^{\oplus n_1}\oplus M_2^{\oplus n_2} \oplus\ldots\oplus M_r^{\oplus n_r}$$

• Apparently, $$M_i^{\oplus n_i}:= \underbrace{M_1\oplus M_1\oplus\ldots\oplus M_1}_{n_1\;\text{times}}$$ Commented Jun 6, 2014 at 14:29
• Well.... almost, of course on your LHS you're using $i$, so you should have $i$ on the RHS as well, not $1$. Commented Jun 6, 2014 at 14:32
• @don It seems quite apparently to someone who is familiar whith the notation but for someone who first encounters it, it is usually unwise to just assume things, 'cause it can lead you off the tracks completely if you happen to think wrong. Anyway, thanks a lot for the answer, sincerely. Commented Jun 6, 2014 at 14:39
• Oh dear! How wrong can we be when we don't take into account that we "read" and not see the other person! Not at all, @gebruiker: that "apparently" meant that I think that "apparently" that's what that article's author meant with that notation, based on my personal experience, and not that your question was silly at all ... Commented Jun 6, 2014 at 14:55
• In german, this misunderstanding wouldn't have happened ("anscheinend" $\neq$ "offenbar", instead of apparently = apparently). Commented Jun 6, 2014 at 15:02

$M^{\bigoplus n}$ means a direct sum of $n$ copies of $M$. Another common notation is $M^{(n)}$.
Likewise, $M^{\bigotimes n}$ means a tensor product of $n$ copies of $M$, etc.
• And, if you can believe it, I think I've even seen $M^{\times n}$, which seems a little silly to me, to mean a normal Cartesian product of $n$ copies of $M$ (instead of the usual $M^n$).
• Certainly I can believe this because $M^{\times n}$ is used in my thesis (whereas $M^n$ abbreviates the tensor power $M^{\otimes n}$. This is really common when working with line bundles for example ...) Commented Jun 6, 2014 at 14:54