# Big Greeks and commutation

Does a sum or product symbol, $\Sigma$ or $\Pi$, imply an ordering?

Clearly if $\mathbf{x}_i$ is a matrix then:

$$\prod_{i=0}^{n} \mathbf{x}_i$$

depends on the order of the multiplication. But, even if one accepts that it has a sequence, it is not clear if it should mean $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ or $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$.

A similar question, is there a "big" wedge product convention?

$$\overset{n}{\underset{i=0}{\Huge\wedge}} \;{}^{\Large{\mathbf{x}_i} \;=\; \mathbf{x}_0 \wedge \mathbf{x}_1 \;\cdots \mathbf{x}_{n-1}\; \wedge \mathbf{x}_{n}}$$

• @MJD does it work like I said? – Lucas May 22 '13 at 16:00
• In Carter's Simple Groups books he explicitly mentions that it does not matter (for his purposes) what order is taken. This might be because this is important in group theory, but it might aso be because he did not believe there was a completely standard order. – Jack Schmidt May 22 '13 at 16:02
• To avoid confusion for non-commutative operations, just state specifically the ordering you are using in the proof/paper. – recursion.ninja May 22 '13 at 21:58

I think that even if it's not written explicitly anywhere, the $\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n$ convention is the most predictable and sensible.

I've never seen the distinction made explicit, since in most circumstances the operation involved is commutative.

I did see somewhere on m.SE someone suggest $\mathbf{x}_i\prod_{i=1}^n$ to denote $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$, but that may have been with tongue in cheek...

If I wanted $\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0$ I would write it as $$\prod_{i=0}^{n} \mathbf{x}_{n-i}$$

• While you have my +1, note that this trick won't generalize well to infinite products. Although one could always write those as explicit limits, e.g. $\dotsb x_2 x_1 x_0 = \lim_{n\to\infty}\prod_{i=0}^n x_{n-i}$. – Ilmari Karonen May 22 '13 at 21:22

If your elements commute with eachother, then there is no need for an ordering in the case of finite sums/products. In the non-commutative case things are more complicated.

Anyhow, IMO there is no need for an ordering if the sum/product doesn't depend on the order. And this covers many non-commutative cases too. Otherwise, it is clear that one should explain the order.

If you write $_{i=1}^n$ by convention the order is understood to be $1,2,.., n$.

• Even when the elements commute, the order might be important if the index set is infinite. – MJD May 22 '13 at 16:07
• @MJD But them technically $\Sigma$ and $\Prod$ are not a sum or a product symbol, they are limits.... – N. S. May 22 '13 at 16:08
• @N.S.: Yes, but the standard definition of $\prod_{i=1}^{\infty} x_i$ in real analysis is $\lim_{N \rightarrow \infty} \prod_{i=1}^N x_i$. This does in general depend on the ordering -- not of the finite sequence $x_1,\ldots,x_N$ but of the infinite sequence $\{x_i\}_{i=1}^{\infty}$ -- so your remark that there is no need for an ordering in the commutative case seems a little misleading. – Pete L. Clark May 22 '13 at 17:12

I always thought $\prod_{i=1}^n x_i$ as a notational shortcut for $x_1\cdot x_2\cdot\dots\cdot x_n$, but now that you make me think about it, I don't recall having seen the formal definition anywhere.