Big Greeks and commutation

Question

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?

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

Answer 1

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...

Answer 2

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

Answer 3

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.

Answer 4

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 $1,2,.., n$.