how (dx)^2 affect a formula

Question

I am given α = xdx − ydy, β = zdxdy + xdydz, and required to compute αβ. After distributions and reorderings of differentials I get: $$xz(dx^2dy)+x^2(dxdydz)+yz(dy^2dx)-yx(dy^2dz)$$ Since I am not told that forms are in any dimension of spaces, do I need to apply $$dx_i^2=0$$, so that it remains the second term above? Also, how do you compute dα in this case? I assume the d operator is distributed into xdx and ydy respectively, but I am confused on how to compute those.

Answer 1

Since forms are antisymmetric, it will always be the case that $\alpha\wedge\alpha = 0$. In particular, $dx_i\wedge dx_i = dx_i^2 = 0$ for all $i$. So the answer to your first question is, "yes."

To compute $d\alpha$, note first that the $d$ operator obeys a Leibniz rule: $$ d(\omega\wedge\eta) = (d\omega)\wedge\eta + (-1)^\cdot \omega\wedge (d\eta) $$ where the $\cdot$ is a number depending on the order of $\omega$ (I think it's $|\omega|+1$). So in particular if $\eta$ is already a coboundary, you will have $$ d(\omega\wedge\eta) = (d\omega)\wedge\eta $$ so the answer to your second question is that the differential is computed by distributing the $d$ operator to the function and then multiplying out the $dx_i$s.

(I'll let you work out $d\alpha$. In the case of $\beta$, we have $$ d\beta = d(zdxdy) + d(xdydz) = dzdxdy + dxdydz = 2 dxdydz. $$ Note that I transposed twice in the last equality to get $dz$ to the end of the first term; if I had transposed once, the two terms would have canceled.)