# Exterior derivative of a 2-form

I want to prove that the exterior derivative of a 2-form in $\mathbb{R}^n$: $${\alpha = \sum a_{ij} dx_i \wedge dx_j}$$ is: $${d \alpha = \sum (\frac{\partial a_{ij}}{\partial x_k} + \frac{\partial a_{jk}}{\partial x_i} - \frac{\partial a_{ik}}{\partial x_j})dx_i \wedge dx_j \wedge dx_k.}$$ The thing is that I'm stucked after taking the differential of the function $a_{ij}$ and then multiplying the $dx_k$ term from the 1-form that comes out from ${da_{ij} = \sum \frac{\partial a_{ij}}{\partial x_k}dx_k}$. The answer seems very pleasant (permutating the indices $i,j,k$), but I missed the simple algebraic step that takes to it.

• Use $$dx^i \wedge dx^j = - dx^j \wedge dx^i$$ – jimbo Jan 18 '15 at 21:21

I think the explanation comes from what is hidden, namely the range of the summation indices. Because of the anticommutativity $$\alpha=\sum_{i<j}a_{ij}\,dx^i\wedge dx^j.$$ When you take the exterior derivative it comes out as (I switch temporarily to indices $r,s,t$ - bear with me for a moment) \begin{aligned} d\alpha&=\sum_{r<s}(da_{rs})\,dx^r\wedge dx^s\\ &=\sum_{r<s}\left(\sum_t\frac{\partial a_{rs}}{\partial x^t}\,dx^t\right)\,dx^r\wedge dx^s\\ &=\sum_{r<s}\sum_t\frac{\partial a_{rs}}{\partial x^t}\,dx^t\wedge dx^r\wedge dx^s. \end{aligned} At this point the problem is to list the various ways that we get a term $dx^i\wedge dx^j\wedge dx^k$ with $i<j<k$ to emerge here. We know that $r$ and $s$ are in correct order, but $t$ can be anything. This leaves us with three alternatives:
• $t=i$, $r=j$, $s=k$, this happens when $t<r<s$
• $r=i$, $t=j$, $s=k$, this happens when $r<t<s$
• $r=i$, $s=j$, $t=k$, this happens when $r<s<t$
The anticommutativity rule introduces, respectively, $0$, $1$ or $2$ sign changes, as we need to move $dx^t$ to its correct place. So in the second case we also get a minus sign. The terms coming from $t=r$ or $t=s$ obviously vanish.
This accounts for the three partial derivatives in your sum together with the correct signs. In the end we get, listing the terms in order described above, $$d\alpha=\sum_{i<j<k}\left(\frac{\partial a_{jk}}{\partial x^i}-\frac{\partial a_{ik}}{\partial x^j}+\frac{\partial a_{ij}}{\partial x^k}\right)\,dx^i\wedge dx^j\wedge dx^k.$$
• Thanks, my real trouble was in turning the sum in $k$ inside the brackets in a sum only in crescent indices. Your explanation was very helpful. – B. Chinaski Jan 19 '15 at 2:32