# What is the motivation for wanting to remove redundant terms that arise from the wedge product of two multilinear functions?

I am currently working through Loring Tu's "An Introduction to Manifolds," specifically the section in which the exterior algebra of multicovectors is introduced, and I am having trouble grasping the motivation for a need to get rid of the redundant terms in the following formula: $$f\wedge g= \frac{1}{k!l!}\sum_{\sigma \in S_{k+l}}sgn(\sigma)f(v_{\sigma(1)}, ..., v_{\sigma(k)})g(v_{\sigma(k+1)},...,v_{\sigma(k+l)})$$

In this instance $f$ is an alternating $k$-linear function and $g$ is an alternating $l$-linear function, both defined on the same vector space.

The beginning of the section mentions that we want to define a product so that the product of two alternating multilinear functions is also alternating, but in keeping the terms it doesn't seem to ruin that property for the wedge product. I feel like there is something obvious I am missing here. Is there some other reason for being concerned with these redundant terms?

Edit: For example, if $f$ is bilinear and $g$ is linear, then we consider their alternating tensor product map:

$$A(f\otimes g)=\sum_{\sigma \in S_3}f(v_{\sigma(1)},v_{\sigma(2)})g(v_{\sigma(3)})$$ evaluated at $(v_1, v_2, v_3)$ which gives us:

$f(v_1,v_2)g(v_3)-f(v_1,v_3)g(v_2)+ f(v_2,v_3)g(v_1) -f(v_2,v_1)g(v_3)+ f(v_3,v_1)g(v_2)-f(v_3,v_2)g(v_1)$

Now since $f$ is alternating, we can see that we have repeated values in this sum. To avoid these repeated values he divides by $l!$ and $k!$ (which in this case is 2! and 1!) to compensate for these repetitions in our sum.

• Which terms are you thinking of that could be kept? I think the idea is that $f \otimes g$ is not alternating, but $f \wedge g$ is. – littleO Apr 24 '16 at 21:38
• So in the text they define the alternating operator $A$ which takes a $k$-linear function $f$ on a vector space $V$ and makes it alternating via $A(f)=\sum_{\sigma \in S_{k}}sgn(\sigma)f(v_{\sigma(1)}, ..., v_{\sigma(k)})$. This makes no mention of the properties of $f$ other than being $k$-linear. We can just take $A(f\otimes g)$ and this will be alternating, so I don't see the need to get rid of the extra terms. – Brandon Thomas Van Over Apr 24 '16 at 21:46
• you should take a look at https://en.wikipedia.org/wiki/Exterior_algebra – reuns Apr 24 '16 at 22:12
• What specific part of the page are you referring to? – Brandon Thomas Van Over Apr 24 '16 at 22:21
• Even with your comment I don't understand what extra terms you're getting rid of. What is Tu doing that you don't like, precisely, and what would you rather he be doing, precisely? – user98602 Apr 24 '16 at 22:29