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Munkres book on Manifolds constructs a wedge product by defining the following sum on $f$ (an alternating $k$-tensor on $V$) and $g$ (an alternating $l$-tensor on $V$): $$(f \wedge g)(v_1,...,v_{k+l}) = \frac{1}{k!l!}\sum_{\sigma} \text{sgn }\sigma f(v_{\sigma(1)},...,v_{\sigma(k)})g(v_{\sigma(k+1)},...v_{\sigma(k+l)})$$

Here $\sigma$ is a permutation on $k+l$ distinct elements.

He goes on to justify why there is the factor $\frac{1}{k!l!}$. I see that there are some permutations that will permute the first set of vectors $\{v_1,...v_k\}$ and the second set of vectors $\{v_{l+1},...v_{l+k}\}$ amongst themselves. I don't understand what will happen with the sign though. Munkres says:

because $f$ and $g$ are alternating tensors, the values of $f$ and $g$ change by being multiplied by the same sign

I'm having trouble seeing this. Knowing whether the permutation on $f$ is even or odd doesn't determine the evenness/oddness of $g$ does it?

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The point is this: If you have a permutation $\sigma = \tau\times\pi$, where $\tau$ permutes $1,\dots,k$ and $\pi$ permutes $k+1,\dots,k+\ell$, then $\text{sgn}\,\sigma = (\text{sgn}\,\tau)(\text{sgn}\,\pi)$, and \begin{align*} \text{sgn}\,\sigma f(&v_{\sigma(1)},\dots,v_{\sigma(k)})g(v_{\sigma(k+1)},\dots,v_{\sigma(k+\ell)}) \\ &=\text{sgn}\,\sigma f(v_{\tau(1)},\dots,v_{\tau(k)})g(v_{\pi(k+1)},\dots,v_{\pi(k+\ell)}) \\ &= \big(\text{sgn}\,\sigma\,\text{sgn}\,\tau\,\text{sgn}\,\pi\big)f(v_1,\dots,v_k)g(v_{k+1},\dots,v_{k+\ell})\\ &= f(v_1,\dots,v_k)g(v_{k+1},\dots,v_{k+\ell}). \end{align*} And of course there are $k!\ell!$ such terms.

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  • $\begingroup$ There is an alternate definition of wedge product which involves summing over only the $(k, l)$-shuffling permutations, instead of all $(k+l)$-permutations. This means that we no longer require the $1\over k!l!$ factor. But I don't quite have the proof of the equivalence of these two definitions. I am not sure if this is the right place to bring this up, but I asked a question about this a couple of months ago and it did not receive any attention. It would be great if you could look at it. Thanks. $\endgroup$ – feynhat Jul 1 at 17:28
  • $\begingroup$ @feynhat Looks like you're done. A good reference comparing the definitions in careful fashion is Warner's classic book. $\endgroup$ – Ted Shifrin Jul 1 at 17:47

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