I'm trying to grasp the notation and concept of wedge products(, and tensors as well).

In my lecture notes, the following expansion/notation for a $(n,r)$-tensor is used:

In a basis $\left\{ \partial_{a_1} \otimes...\otimes\partial_{a_n}\otimes\mathrm{d}x^{b_1} \otimes...\otimes \mathrm{d}x^{b_r} \right\}$, also noting that $\alpha = \alpha^{\mu_i} \partial_{\mu_i}$, $\beta = \beta_{\nu_j} \mathrm{d}x^{\nu_j}$ \begin{align} T &= T^{a_1,...,a_n}_{b_1,...,b_r} \left\{ \partial_{a_1} \otimes...\otimes\partial_{a_n}\otimes\mathrm{d}x^{b_1} \otimes...\otimes \mathrm{d}x^{b_r} \right\} \\ T(\alpha_{1},...,\alpha_{n},\beta_{1},...,\beta_{r}) &= T^{a_1,...,a_n}_{b_1,...,b_r} \alpha_{1}^{\mu_1} \cdots \alpha_{n}^{\mu_n}\beta_{1,\nu_1} \cdots \beta_{r,\nu_r} \end{align} Where I find the last line quite confusing, actually. I guess on the right are the component written out on a specific basis. On the left... there's just a list of the arguments of the tensor (but the arguments itself can have components, can't they?).

The basis for $r$-forms (i.e. differentiable forms i.e. completely antisymmetric covariant tensor of rank $r$) is $\{ \mathrm{d}x^{b_1} \wedge ... \wedge \mathrm{d}x^{b_r} \}$ where $\mathrm{d}x^{b_1} \wedge ... \wedge \mathrm{d}x^{b_r} = \sum_{\sigma \in Perm} sgn(\sigma) \mathrm{d}x^{\sigma(b_1)} \otimes ... \otimes \mathrm{d}x^{\sigma(b_r)} $

Now an $r$-form $\omega$ can be expanded as \begin{align} \omega &= \frac{1}{r!} \omega_{b_1,...,b_r} \mathrm{d}x^{b_1} \wedge ... \wedge \mathrm{d}x^{b_r} \\ \end{align} And now the wedge product between the $n$-form $\omega$ and $r$-form $\upsilon$ is defined (still, with the relatively confusing notation) as \begin{align} (\omega \wedge \upsilon)(\beta_1,...,\beta_{n+r}) = \sum_{\sigma \in Perm} sgn(\sigma) \omega(\beta_{\sigma(1)},...,\beta_{\sigma(n)}) \cdot \upsilon (\beta_{\sigma(1)},...,\beta_{\sigma(r)}). \end{align}

On to the questions.

  1. What does it mean for $\omega$:s coefficients to be antisymmetric? Say, if in some orthonormal basis $\omega = (\omega_{1},\omega_{2})$, this just means that $\omega_1 = - \omega_2$?

  2. What is the basis of $\omega \wedge \upsilon$? If $\omega$ has a basis $\{ e_{\omega} \}$ and $\upsilon$ has a basis $\{ e_{\upsilon} \}$, then is the basis $e_{\omega} \otimes e_{\upsilon}$ or $e_{\omega} \wedge e_{\upsilon}$? EDIT: wolfram's wedge product page says that tha basis is indeed $e_{\omega} \wedge e_{\upsilon}$.

  3. I don't understand the concept very well, as I don't see why for even $r$-forms can be true that $\omega \wedge \omega \neq 0$? As they both have the same basis, so that when I take the direct product (see question 2), I have the same index $b_i$ always appear twice (and at least $\mathrm{d}x^{b_1} \wedge ... \wedge \mathrm{d}x^{b_r} = 0$ if the same $b_i$ appears twice). EDIT: My problem here has to do with tensor product, clearly, since wolfram's wedge product page gives an example of this. If we have a 4-form, it's basis should be $\left\{ e_1 \otimes ... \otimes e_4 \right\}$, so what does it mean to have a 4-form like $\omega = e_1 \otimes e_2 + e_3 \otimes e_4$? How do you write that in the basis $e_1 \otimes e_2 \otimes e_3 \otimes e_4$?

  4. Would the definition be different for tensors of type $(r,0)$ instead of $(0,r)$ (i.e. for contravariant tensors)? (This is the least important question.)

  • $\begingroup$ the arguments $\alpha$'s and $\beta$'s in your $T$ are misplaced. $\endgroup$ – janmarqz Nov 14 '14 at 0:49

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