Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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

1 Answer 1

up vote 4 down vote accepted

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$

share|cite|improve this answer
Thanks for that, from your and celtschk's answers I see where I was going wrong. – Ruvi Lecamwasam Jul 14 '13 at 10:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.