# Sum notation, $\wedge$-product

I never really understood how you would evaluate sums, which have a summation like this:

$\sum_{\mu_1<\dotso<\mu_k}\omega(e_{\mu_1},\dotso, e_{\mu_k})\delta^{\mu_1}\wedge\dotso\wedge\delta^{\mu_k}$

What confuses me, is the notation $\mu_1<\dotso<\mu_k$ How would this sum actually look like, when you right it out?

I think it is best with a simpler example of the sum.

$\sum_{\mu_1<\mu_2<\mu_3}\omega(e_{\mu_1},e_{\mu_2}, e_{\mu_3})\delta^{\mu_1}\wedge\delta^{\mu_2}\wedge\delta^{\mu_3}$

I just found an easier form of this notation here: https://en.wikipedia.org/wiki/Summation#Identities

But I am not sure how to adapt it to the sum above.

I'm not totally sure I know the context of your problem but I think the thing with the $\mu_i$'s is probably just so the wedge products of the $\delta^i$'s form a basis.

If the superscripts are not in ascending order, you can just anticommute the elements and put them in order. Including all ordering a would be redundant and you could combine several with this trick.

This is the case in exterior algebra and Clifford algebra anyway.

• Thanks you, I thought the context of the problem is not necessary, be cause you get across this notation alot. Edit: What is about this notation: $\sum_{i<j<k}a_{i,j,k}$
– Siu
Sep 29, 2016 at 3:09
• @Siu what about it ? It's just adding an extra constraint to the indices. What's unclear? Sep 29, 2016 at 3:14
• I do not know, how I could calculate this, when I had to.
– Siu
Sep 29, 2016 at 3:17