# Explicitly writing out a differential 2-form

In Tu's An Introduction to Manifolds, one question asks:

At each point $p\in \mathbb{R}^3$, define a bilinear function $\omega_p$ on $T_p(\mathbb{R}^3)$ by: $$\omega_p(\underline{a},\underline{b})=\omega_p((a^1,a^2,a^3),(b^1,b^2,b^3))=p^3(a^1b^2-a^2b^1)$$ For tangent vetors $\underline{a},\underline{b}\in T_p(\mathbb{R}^3)$, where $p^3$ is the third component of $\underline{p}=(p^1,p^2,p^3)$. Since $\omega_p$ is an alternating bilinear function on $T_p(\mathbb{R}^3)$, $\omega$ is a 2-form on $\mathbb{R}^3$. Write $\omega$ in terms of the standard basis $dx^i\wedge dx^j$ at each point.

I understand that we write this as $\omega=a_{ij}dx^i\wedge dx^j$, with $a_{ij}=\omega(e_i,e_j)$ where $e_1,e_2,e_3$ span $T_p(\mathbb{R})$. With this I find that all constants vanish apart from $a_{12}$ and $a_{21}$, which lead to: $\omega=p^3dx\wedge dy-p^3dy\wedge dx=2p^3dx\wedge dy$. In the solutions however, since an alternating function of two arguments is completely determined by its actions on $w(e_{k},e_{l}),k<l$, Tu sums only over $i<j$ leading to $\omega=p^3dx\wedge dy$.

My question is, I thought that whether or not a multilinear function is alternating, you should be able to characterise it by feeding it all possible combinations of basis elements. But it seems in this case that leads to a different answer. Why is this?

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Does it fail? As far as I can see, $\omega(e_1,e_1) = \omega(e_2,e_2) = \omega(e_3,e_3) = 0$, $\omega(e_1,e_2) = -\omega(e_2,e_1) = p^3$, $\omega(e_1,e_3)=-\omega(e_3,e_1)=0$, $\omega(e_2,e_3)=-\omega(e_3,e_2)=0$. So why do you think it fails? – celtschk Jul 14 '13 at 9:24
What I am confused about is that I get two different answers depending on whether I sum over $i<j$ or all $i,j$, as the latter leads to an extra factor of $2$. For general bilinear functions we sum over all $i,j$ (I think), so why does that fail just because the function is alternating? – Ruvi Lecamwasam Jul 14 '13 at 9:32
The expression for $\omega$ should be $\omega = \sum_{i < j} a_{ij} dx^i \wedge dx^j$ with $a_{ij} = \omega(e_i, e_j)$ because it is $e_i \wedge e_j$ that correspond to $dx^i \wedge dx^j$, which span $\Lambda^2(T_p(\mathbb R^3))$. (The space of alternating bilinear functions is smaller (dimension-wise) than the space of bilinear functions.) – Tunococ Jul 14 '13 at 9:33
@AndrewLedesma: If you defined it with $\wedge$ it automatically is alternating. If you defined it with $\otimes$ you have to test it, but then you sum over all $ij$ anyway: $\sum_{ij}\omega(e_i,e_j) dx^i\color{red}\otimes dx^j = \sum_{i\color{blue}\le j}\omega(e_i,e_j)dx^i \color{blue}{\wedge} dx^j$. – celtschk Jul 14 '13 at 9:44
Note: The equality sign in my last comment of course only applies if it is an alternating form, of course. – celtschk Jul 14 '13 at 9:51

Note that $(a\wedge b)_{ij} = 2a_{[i}b_{j]}$ so in particular the 12 component is $a_1b_2-a_2b_1$. Therefore only one of these is needed to reconstruct both the $12$ and $21$ components. Put another way, you are not building the two form not out of usual matrices like $$\pmatrix{0 & 1 \\ 0& 0}$$ but instead ones like $$\pmatrix{0 & 1 \\ -1& 0}$$
The reason this differs from the usual bilinear approach is that we expand two-forms in terms of this latter basis $\mathrm dx_1\wedge \mathrm d x_2=- \mathrm dx_2\wedge \mathrm d x_1$ rather than the more familiar $\mathrm dx_1\otimes \mathrm d x_2 \neq \pm\mathrm dx_2\otimes \mathrm d x_1$