# How do I apply smooth $p$-forms to vector fields?

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.

• 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. – Zhen Lin May 5 '12 at 18:42
• 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 $q$) ? – 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 $$\sigma_{i_{\tau (1)} \ldots u_{\tau (p)}} = \textrm{sgn}(\tau) \, \sigma_{i_1 \ldots i_p}$$

• really helpful, thanks. – 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 $$dx\wedge dy= a_1b_2- a_2b_1$$

• 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? – Paul Slevin May 5 '12 at 18:11
• 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..... – zapkm May 5 '12 at 18:20

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