Suppose $dx_i$ is the dual basis in $R^n$ so that $$dx_i (e_j) = \delta_{ij}.$$ It makes sense to me how 1-forms work: a 1-form evaluated at a point gives some linear functional, which takes tangent vectors to scalars. But for k-forms evaluated at points, I'm not exactly sure how a wedge product of linear functionals takes a wedge product of vectors (a k-parallelepiped) to a scalar.

How does $ dx_1 \wedge dx_2 $ act on $ e_1 \wedge e_2 $? How does, say, $ dx_1 \wedge dx_3 $ act on $ e_1 \wedge e_2 $? And in general how does an element of $$\wedge^k(R^{n*}) \simeq \wedge^k(R^{n})^*$$ (is that isomorphism correct?) act on an element of $\wedge^k(R^n)$?

Finally, how does this relate to the Jacobian in Rudin's definition of the action of a k-form on a k-surface?

For reference: Rudin says a k-form in $E$ ($\subset R^n$) is a function $\omega$, symbolically represented by the sum $$\omega = \sum{a_{i_1 \cdots i_k}(x) dx_{i_1}\wedge \cdots \wedge dx_{i_k}}$$

(where the indices $i_1, \ldots, i_k$ range independently from 1 to n), which assigns to each k-surface $\phi$ in $E$ a number $\omega(\phi) = \int_{\phi}{\omega}$ according to the rule $$\int_{\phi}{\omega}= \int_{D}{\sum{a_{i_1 \dots i_k}(\phi(u)) \frac{\partial (x_{i_1}, \ldots, x_{i_k})}{\partial (u_1, \ldots, u_k)} du}},$$

where $D$ is the (compact) domain of $\phi$ (for instance the k-cell $[0,1]^k$) and $\frac{\partial (x_{i_1}, \ldots, x_{i_k})}{\partial (u_1, \ldots, u_k)}$ is the Jacobian.


1 Answer 1


The wedge $\omega_1 \wedge \ldots \wedge \omega_k$ acts on $v_1 \wedge \ldots \wedge v_k$ as $$ \omega_1 \wedge \ldots \wedge \omega_k \cdot v_1 \wedge \ldots \wedge v_k = \det ( \omega_i (v_j) )_{1\le i,j\le k}$$

Sometime normalizing factors involving factorials do appear, because the spaces of alternating covariant and contravariant tensors are considered inside the space of usual tensors. But more or less, up to a factor, this is the pairing.

According to this definition, $dx_1 \wedge dx_2 \cdot e_1 \wedge e_2$ is the determinant of the $2\times 2$ identity matrix, that is $1$. And $dx_1 \wedge dx_3 \cdot e_1 \wedge e_2$ is zero, because it's the det of a matrix with zero second row.

This thing with determinants appear in other situations. For instance, we have the following formula for usual vectors in $3$ space ( the dot product of some cross products)

$$v_1\times v_2 \cdot w_1 \times w_2= \det (v_i \cdot w_j)_{1\le i,j, \le 2}$$

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    $\begingroup$ How can that be derived from the fact that for ordinary tensors $$f^*_1 \otimes \cdots \otimes f^*_k (v_1 \otimes \cdots \otimes v_k) = f^*_1(v_1)\cdots f^*_k(v_k)?$$ $\endgroup$
    – user217664
    Dec 26, 2015 at 0:07
  • $\begingroup$ @agrasin: Aha, in fact $f_1\wedge \ldots \wedge f_k$ can be viewed as the antisymmetrization of $f_1\otimes \ldots \otimes f_k$ ( now some factors might appear), and so can $v_1 \wedge \ldots \wedge v_k$ . Check what the antisymmetrization of a tensor is, there are some variants ( bc of the normalizing factors) $\endgroup$
    – orangeskid
    Dec 26, 2015 at 0:11

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