There exist various, though certainly equivalent, ways of introducing the wedge product. I am trying to grasp one particular method. This approach implies that we first define $\,\wedge\,$ for unit forms, $$ e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_r}\;\equiv\;r!\,\left[\,e^{i_1}\,\otimes\;.\,.\,.\;\otimes\,e^{i_r}\,\right] \quad,\qquad $$ and then employ the simple theorem $$ \left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right]\;=\; \left[\;\; \left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_k}\,\right]\;\otimes\;\left[\,e^{j_{k+1}}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right] \;\;\right] $$ as a means to extend the definition of $\,\wedge\,$ to arbitrary-order skew forms: $$ \omega^{k+l}\,=\;\omega^k\wedge\omega^l\,\equiv\;\frac{(k+l)!}{k!\,\;l!}\;\left[\,\omega^k\,\otimes\,\omega^l\,\right]\;\;. $$ As ever, $\,\;[\;.\;.\;.\;]\;\,$ denotes alternation.

To ask my questions, I shall now provide the said development in mode detail.

As agreed, we start out with the DEFINITION: $$ e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_r}\;=\;r!\,\;\left[\,e^{i_1}\,\otimes\,\, .\,.\,.\,\,\otimes\,e^{i_r}\,\right] \,\;. $$ Equivalently, $$ e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_r}\;=\;\sum_{j_1\,.\,.\,.\,j_r}\delta^{i_1\,.\,.\,.\,i_r}_{j_1\,.\,.\,.\,j_r}\;\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_r}~~. $$

Within my notation, $\,\sum_{j_1 ... j_k}\,$ implies summation over $\,$all$\,$ values taken by each index $\,j_s\,=\,1\,,\,.\,.\,.\,,\,n\,$, with $\,$no$\,$ additional condition (like $\,j_1\,<\,.\,.\,.\,<\,j_k\,$).

Now we recall that a skew form is a form which is equal to its own alternation, and also recall that alternation of a sum equals the sum of alternations of its summands. This renders us: $$ \omega^r(x_1\,,\,.\,.\,.\,,\,x_r)\;=\;\left[\;\omega^r(x_1\,,\,.\,.\,.\,,\,x_r)\;\right] $$ $$ =\;\sum_{j_1\,.\,.\,.\,j_k}^n\,\omega^r_{j_1\;.\;.\;.\;j_r}\;\left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_k}\,\right]\;\;(x_1\,,\,.\,.\,.\,,\,x_r) $$ $$ =\;\frac{1}{r!}\,\sum_{j_1\,.\,.\,.\,j_k}^n\,\omega^r_{j_1\;.\;.\;.\;j_r}\; e^{j_1}\,\wedge .\,.\,. \wedge\,e^{j_r}\;\;\;\,(x_1\,,\,.\,.\,.\,,\,x_r)\;\;, $$ whence we see that all $\,e^{j_1}\,\wedge .\,.\,. \wedge\,e^{j_r}\,$ make a basis.

Our next step is to use the well known THEOREM: $$ \left[\;\; \left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_k}\,\right]\;\;\left[\,e^{j_{k+1}}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right] \;\;\right]\;=\; \left[\,e^{j_1}\otimes\, .\,.\,.\,\otimes\,e^{j_{k+l}}\,\right]\;\;, $$ which can be rewritten as $$ \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\;=\; \frac{\textstyle 1}{\textstyle (k+l)!}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}}\,\;. $$

To extend the wedge-product definition to arbitrary skew forms, we can multiply the above equality by some $\;\omega_{i_1\;.\;.\;.\;i_k}\;\,\omega_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;$ and sum over all indices: $$ \nonumber \left[\;\;\left(\,\;\frac{\textstyle 1}{\textstyle k!}\;\,\omega_{i_1\;.\;.\;.\;i_k}\;\,e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} \,\;\right)\,\otimes\, \left(\,\;\frac{\textstyle 1}{\textstyle l!}\;\,\,\omega_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} \,\right)\;\;\right]\qquad\qquad\\ \label{9} $$ $$ =\;\frac{\textstyle 1}{\textstyle (k+l)!}\;\;\omega_{i_1\;.\;.\;.\;i_k}\;\,\omega_{i_{k+1}\;.\;.\;.\;i_{k+l}}\;\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}}\;\,. \qquad(1) $$ At this point, we introduce $\,\omega^k\,$ and $\,\omega^l\,$ as $$ \omega^k\;=\;\frac{1}{k!}\;\sum_{j_1\,.\,.\,.\,j_k}^n\,\omega_{j_1\;.\;.\;.\;j_k}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k} $$ $$ \omega^l\;=\;\frac{1}{l!}\;\sum_{j_{k+1}\,.\,.\,.\,j_{k+l}}^n\,\omega_{j_{k+1}\;.\;.\;.\;j_{k+l}}\;e^{i_{k+1}}\,\wedge .\,.\,. \wedge\,e^{i_{k+l}} $$ and also introduce $\;\omega^k\wedge\omega^l\;$ as $$ \omega^k\wedge\omega^l\;\equiv\;\frac{1}{(k+l)!}\;\sum_{j_1\,.\,.\,.\,j_{k+l}}^n\,\omega_{j_1\;.\;.\;.\;j_{k+l}}\;e^{i_1}\,\wedge .\,.\,. \wedge\,e^{i_k}\quad.\;\;\quad $$ Be mindful that in the latter expression the matrix $\,\omega_{j_1\;.\;.\;.\;j_{k+l}}\,$ is not yet defined.

If we choose to set $$ \omega_{j_1\;.\;.\;.\;j_{k+l}}\;\equiv\;\omega_{j_1\;.\;.\;.\;j_k}\;\omega_{j_{k+1}\;.\;.\;.\;j_{k+l}}\;\;,\qquad (2) $$ then equality (1) becomes $$ \left[\;\omega^k\,\otimes\,\omega^l\;\right]\;=\; \omega^k\wedge\omega^l\,\;.\qquad(3) $$

If, however, we set $$ \omega_{j_1\;.\;.\;.\;j_{k+l}}\;\equiv\;\frac{(k+l)!}{k!\;l!}\;\omega_{j_1\;.\;.\;.\;j_k}\;\omega_{j_{k+1}\;.\;.\;,\;j_{k+l}}\;\;.\qquad(4) $$ then equality (1) will read as $$ \left[\;\omega^k\,\otimes\,\omega^l\;\right]\;=\; \frac{k!\;l!}{(k+l)!}\; \omega^k\wedge\omega^l\;\,.\qquad (5) $$

Now, my questions:

(a) Is my overall logic consistent? Did I not mess up things in equation (1) ?

(b) In the literature, they traditionally choose convention (4 - 5), not (2 - 3). Why??

At first glance, choice (4) looks artificial --- but I may be missing some big motivation behind it.

  • $\begingroup$ There's an ambiguity in the notation that gets to the heart of the question: Does $\sum_{j_1 \cdots j_k}^n$ denote the sum over all ordered $k$-tuples of indices in $\{1, \ldots, n\}$ (or, equivalently, ordered $k$-tuples of distinct indices in $\{1, \ldots, n\}$), or the sum over all ordered $k$-tuples of indices in $\{1, \ldots, n\}$ satisfying $j_1 < \cdots < j_k$? (Only one of these choices is actually reasonable here.) $\endgroup$ – Travis Feb 16 at 0:52
  • 1
    $\begingroup$ In any case, Lee's Introduction to Smooth Manifolds, $\S$ 12, tidily addresses exactly this issue, and it discusses the advantages of the two conventions for $\wedge$, which it calls the "determinant convention" and the "Alt convention". In particular, this treatment avoids index expressions in $\omega$. $\endgroup$ – Travis Feb 16 at 0:54
  • $\begingroup$ Dear Travis, many thanks! In my notation $\sum$ implies summation over all values taken by each index $\,j_s\,=\,1,\,,\,.\,.\,.\,n\,$, with no additional conditions (like $\,j_1\,<\,.\,.\,.\,<\,j_k\,$). Within this convention, is my line of reasoning correct? $\endgroup$ – Michael_1812 Feb 16 at 1:06

Your Answer

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

Browse other questions tagged or ask your own question.