There are various equivalent ways of introducing the wedge product. One possibility is to first define $\,\wedge\,$ for basis forms, $$ e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_r}\;\equiv\;r!\,\left[\,e^{i_1}\,\otimes\;.\,.\,.\;\otimes\,e^{i_r}\,\right] \quad,\qquad $$ (where [...] = Alt = $\frac{\textstyle 1}{\textstyle k!}\sum \,.\,.\,.\,$), and then to employ the simple theorem $$ \left[\;\; \left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_k}\,\right]\;\otimes\;\left[\,e^{j_{k+1}}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right] \;\;\right] \;=\;\left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right] $$ as a means to extend the definition of $\,\wedge\,$ to arbitrary skew forms.

To this end, we rewrite the above equality as $$ \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\;=\; \frac{\textstyle 1}{\textstyle (k+l)!}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} ~\qquad $$ and then as $$ \frac{(k+l)!}{k!\;l!}\;\left[\;\;\left(\,\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\;=\; e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} ~\;\;.\qquad\qquad $$ We now see that, if we extend the definition of $\,\wedge\,$ to arbitrary exterior forms as $$ \omega^k\,\wedge\,\omega^l\;\equiv\;\frac{(k+l)!}{k!\;l!}\;\left[\;\omega^k\,\otimes\,\omega^l\;\right]\;\; $$ and apply it to $\;\left(\,\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\;$ and $\;\left(\,\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;$, we end up with $$ \left(\,\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\wedge\, \left(\,\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;=\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}}\;\,. $$ This serves as a motivation to introduce the factor of $\,{(k+l)!}/(k!\,l!)\,$ in the above definition. The purpose of these factorials is to make the operation $\,\wedge\,$ associative.

Consider three forms of the orders $\,k\,$, $\,l\,$, and $\,k+l\;$: $$ \alpha^k\;=\;\frac{1}{k!}\;\sum_{i_1\,.\,.\,.\,i_k}^n\,\alpha_{i_1\;.\;.\;.\;i_k}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \;=\;\sum_{i_1\,<\,.\,.\,.\,<\,i_k}^n \,\alpha_{i_1\;.\;.\;.\;i_k}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \quad, $$ $$ \beta^l\;=\;\frac{1}{l!}\;\sum_{i_{k+1}\,.\,.\,.\,i_{k+l}}^n\,\beta_{i_1\;.\;.\;.\;i_l}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \;=\;\sum_{i_{k+1}\,<\,.\,.\,.\,<\,i_{k+l}}^n\,\beta_{i_1\;.\;.\;.\;i_l}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}}\quad, $$ $$ \omega^{k+l}=\frac{1}{(k+l)!}\sum_{i_1\,.\,.\,.\,i_{k+l}}^n\,\omega_{i_1\;.\;.\;.\;i_{k+l}}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k}\;=\sum_{i_1\,<\,.\,.\,.\,<\,i_{k+l}}^n\,\omega_{i_1\;.\;.\;.\;i_{k+l}}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \;\;. $$

Now, the Question:

If we assume that $$ \omega^{k+l}\,=\,\alpha^k\wedge\beta^l\;\;, $$ how to employ the definition $$ \omega^{k+l}\,=\;\alpha^k\wedge\beta^l\,\equiv\;\frac{(k+l)!}{k!\,\;l!}\;\left[\,\alpha^k\,\otimes\,\beta^l\,\right] $$ to express the components $\;\omega_{i_1\;.\;.\;.\;i_{k+l}}\;$ via $\;\alpha_{i_1\;.\;.\;.\;i_k}\;$ and $\;\beta_{i_1\;.\;.\;.\;i_l}\;\;?$

My first attempt was to start out with $$ \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\;=\; \frac{\textstyle 1}{\textstyle (k+l)!}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} ~\;\;, $$ multiply both sides with some $\;\alpha_{i_1\;.\;.\;.\;i_k}\;\,\beta_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;$: $$ \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,\alpha_{i_1\;.\;.\;.\;i_k}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,\,\beta_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right] $$ $$ =\;\frac{\textstyle 1}{\textstyle (k+l)!}\;\;\alpha_{i_1\;.\;.\;.\;i_k}\;\,\beta_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \qquad $$ and sum over all indices on both sides. Comparing the outcome with the definition $$ \left[\;\alpha^k\,\otimes\,\beta^l\;\right]\;=\; \frac{k!\;l!}{(k+l)!}\; \alpha^k\wedge\beta^l\;\;, $$ thus I arrived at $$ \omega_{j_1\;.\;.\;.\;j_{k+l}}\;\equiv\;\frac{(k+l)!}{k!\;l!}\;\alpha_{j_1\;.\;.\;.\;j_k}\;\beta_{j_{k+1}\;.\;.\;,\;j_{k+l}}\;\;.\qquad $$ This result has the correct number of terms, $\;{(k+l)!}/({k!\;l!})\;$. However, all of these come out equal, while in reality they must contain the alternating factor, and the right answer must be $$ \omega_{j_1\;.\;.\;.\;j_{k+l}}\;\equiv\;\sum_{i_1\;.\,.\,.\;i_k}\sum_{i_{k+1}\,.\,.\,.\,i_{k+l}}\delta^{i_1\;.\,.\,.\;i_{k+l}}_{j_1\;.\;.\;.\;j_{k+l}}\; \alpha_{i_1\;.\;.\;.\;i_k}\;\beta_{i_{k+1}\;.\;.\;,\;i_{k+l}}\;\;.\qquad $$ It is not, however, clear to me how to derive it.

  • $\begingroup$ What do you mean by "unit" form? The construction of an exterior power does not need anything like "unit" vectors, so explain the meaning of that terminology. And what are your references on the construction of exterior powers? $\endgroup$ – KCd Mar 16 at 15:11
  • $\begingroup$ I'd suggest that if you're getting confused, look at a concrete example in low dimensions so you're not dealing with a huge number of indices. Try the 2nd exterior power of a 3 or 4-dimensional space, for instance. $\endgroup$ – KCd Mar 16 at 15:13
  • $\begingroup$ We physicists are usually taught that, if a linear space has a basis e_j, then the unit forms (unit covectors) make a basis in the dual covector space: e^i(e_j) = \delta^i_j . I am sure the exterior algebra can be explained also without reference to a basis. However, for physicists and applied mathematicians this old-fashioned language is easier. $\endgroup$ – Michael_1812 Mar 16 at 15:56
  • $\begingroup$ The notion of dual vectors does not require anything be a "unit" vector: every basis has a dual basis. Anyway, look at a low-dimensional example to clear up any issues about indices. $\endgroup$ – KCd Mar 16 at 18:01
  • $\begingroup$ Agreed. I keep using the term "unit" by inertia. Borrowed from naive a primitive courses for physicists. Will correct now. $\endgroup$ – Michael_1812 Mar 16 at 18:22

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