Let $M$ be a smooth manifold. My definition of a smooth $p$-form is a map section $\omega: M \rightarrow \Lambda^p TM^*$, i.e. if $q \in M$ is contained in a chart $U$ with co-ordinates $x_1, \ldots, x_n$, then $$ \omega (q) = \sum_{i_1 < \ldots < i_p}a_{i_1\ldots i_p}(q) \operatorname{d}x_{i_1}|_q \wedge \ldots \wedge \operatorname{d} x_{i_p}|_q$$ or in other words we can write $\omega$ locally as $$ \omega = \sum_{i_1 < \ldots < i_p}a_{i_1\ldots i_p} \operatorname{d}x_{i_1} \wedge \ldots \wedge \operatorname{d} x_{i_p}$$ where each $a_{i_1\ldots i_p} \in C^{\infty}(U)$.

But my question is what does it mean for me to apply this to vector fields? For example if I have a 2-form $\sigma$, what does $\sigma (X,Y)$ mean? Does it take a value in $\mathbb{R}$? How do I calculate it? Any help would be appreciated.

  • 2
    $\begingroup$ By definition (or at least, any good definition), an element of $\Lambda^p T^*_x M$ corresponds to an alternating multilinear map $T_x M \times \cdots \times T_x M \to \mathbb{R}$. So just apply the $p$-form pointwise to the vector field. $\endgroup$ – Zhen Lin May 5 '12 at 18:42
  • $\begingroup$ But I'm not sure how to see the corresponding map. Say $M = \mathbb{R}^2$ and $\omega = \operatorname{d} x \wedge \operatorname{d} y$, a 2-form. Let $X = a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y}$, and $Y = c\frac{\partial}{\partial x} + d\frac{\partial}{\partial y}$. Then how do I know what the corresponding map $(\operatorname{d} x \wedge \operatorname{d} y)_q (X_q, Y_q)$ is (at some point $q$) ? $\endgroup$ – Paul Slevin May 5 '12 at 20:37

Let $(x^1, \ldots, x^n)$ be local coordinates on a manifold $M$. Since we are working locally, we may assume that these are global coordinates. Let us write $\partial_1, \ldots, \partial_n$ for the vector fields corresponding to these coordinates; this is a global frame (i.e. a trivialisation) of the tangent bundle $T M$. The differential 1-forms $d x^1, \ldots, d x^n$ are defined at first to be the duals of $\partial_1, \ldots, \partial_n$, so that $$d x^i (\partial_j) = \delta^i_j = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \ne j \end{cases}$$ and this holds pointwise.

Now, we define what $d x^{i_1} \wedge \cdots \wedge d x^{i_p}$ does. It is completely determined by the following: $$(d x^{i_1} \wedge \cdots \wedge d x^{i_p}) ( \partial_{j_1}, \ldots, \partial_{j_p} ) = \begin{cases} +1 & \text{if } (i_1, \ldots, i_p) \text{ is an even permutation of } (j_1, \ldots, j_p) \\ -1 & \text{if } (i_1, \ldots, i_p) \text{ is an odd permutation of } (j_1, \ldots, j_p) \\ 0 & \text{otherwise} \end{cases}$$ This can be extended by linearity to all differential $p$-forms. Note that this gives an embedding of $\Lambda^p T^* M$ into $(T^* M)^{\otimes p}$ by $$d x^{i_1} \wedge \cdots \wedge d x^{i_p} \mapsto \sum_{\tau \in S_p} \textrm{sgn}(\sigma) \, d x^{i_{\tau (1)}} \otimes \cdots \otimes d x^{i_{\tau (p)}}$$ and in the case $p = 2$ this amounts to $dx^{i_1} \wedge dx^{i_2} \mapsto dx^{i_1} \otimes dx^{i_2} - dx^{i_2} \otimes dx^{i_1}$ as others have said.

If $\sigma$ is a 2-form, then $\sigma = \sum \sigma_{i j} \, dx^i \wedge dx^j$ for some smooth functions $\sigma_{i j}$... except what they are depends on the convention you use. Physicists typically use the convention where $\sigma_{i j}$ is defined as $-\sigma_{j,i}$ for $i \ge j$, so that $$\sigma = \sum_{i < j} \sigma_{i,j} \, dx^i \wedge dx^j = \frac{1}{2} \sum_{i, j} \sigma_{i j} \, dx^i \wedge dx^j$$ Note that this is compatible with the identification of $dx^i \wedge dx^j$ with $dx^i \otimes dx^j - dx^j \otimes dx^i$. I'll use this convention here. Let $X = \sum X^i \partial_i$ and $Y = \sum Y^j \partial_j$. Then, $$\sigma (X, Y) = \frac{1}{2} \sum_{i, j, k, \ell} \sigma_{i j} X^k Y^\ell \, (dx^i \wedge dx^j) (\partial_k, \partial_\ell) = \frac{1}{2} \sum_{i, j} \sigma_{i j} (X^i Y^j - X^j Y^i)$$ but since $\sigma_{i j} = - \sigma_{j i}$, we can simplify the RHS to $$\sigma (X, Y) = \sum_{i, j} \sigma_{i j} X^i Y^j$$

More generally, if $\sigma$ is a $p$-form with $$\sigma = \frac{1}{p !} \sum_{i_1, \ldots, i_p} \sigma_{i_1 \ldots i_p} \, dx^{i_1} \wedge \cdots \wedge dx^{i_p}$$ and $X_1, \ldots, X_p$ are vector fields with $X_r = \sum_j X_r^j \, \partial_j$, we have $$\sigma (X_1, \ldots, X_p) = \sum_{i_1, \ldots, i_p} \sigma_{i_1 \ldots i_p} X^{i_1} \cdots X^{i_p}$$ where we have assumed that $\sigma_{i_1 \ldots i_p}$ is a totally antisymmetric in the sense that $$\sigma_{i_{\tau (1)} \ldots u_{\tau (p)}} = \textrm{sgn}(\tau) \, \sigma_{i_1 \ldots i_p}$$

  • $\begingroup$ really helpful, thanks. $\endgroup$ – Paul Slevin May 5 '12 at 21:26

We know that $$dx\wedge dy= dx\otimes dy -dy\otimes dx$$

If $v= a_1\frac{\partial}{\partial x}+ a_2\frac{\partial}{\partial y}$ and $w=b_1\frac{\partial}{\partial x}+b_2\frac{\partial}{\partial y}$ are two vectors then we have $$dx\wedge dy= dx\otimes dy(v,w) -dy\otimes dx(v,w)$$ Also we have $$dx\otimes dy(a_1\frac{\partial}{\partial x}+ a_2\frac{\partial}{\partial y},b_1\frac{\partial}{\partial x}+ b_2\frac{\partial}{\partial y})= a_1b_2$$

Hence we have $$dx\wedge dy= a_1b_2- a_2b_1$$

  • $\begingroup$ Sorry but I am confused, why does $dx \wedge dy = dx \otimes dy - dy \otimes dx$? The tensor product $dx \otimes dy$ lives in $TM^* \otimes TM^*$ and the wedge product $dx \wedge dy$ lives in $ \Lambda^2 TM^* \otimes TM^*$, right? $\endgroup$ – Paul Slevin May 5 '12 at 18:11
  • $\begingroup$ yes you are right.... But where does $\wedge^2T_pM^*\otimes T_oM^*$ lies?? Actually wedge product of two element from $T_p^*M$ is defined in terms of alternating tensor with some constant multiplication..... So basically later one is subspace of previous one.. but pointwise..... $\endgroup$ – zapkm May 5 '12 at 18:20

Yes, $\sigma(X,Y)$ would be a smooth scalar field (or, commonly known as a smooth function).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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