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I am currently reading Calculus on Manifolds by Spivak. In there, it defined wedge product as follows

To determine the dimensions of $\Lambda^k(V)$, we would like a theorem analogous to Theorem 4-1. Of course, if $\omega \in \Lambda^k(V)$ and $\eta \in \Lambda^l(V)$, then $\omega \otimes \eta$ is usually not in $\Lambda^{k+l}(V)$. We will therefore define a new product, the wedge product $\omega \wedge \eta \in \Lambda^{k+l}(V)$ by $$ \omega \wedge \eta=\frac{(k+l) !}{k ! l !} \operatorname{Alt}(\omega \otimes \eta) $$

I am not very sure if here Spivak assumed $\omega \in \Lambda^k(V)$ and $\eta \in \Lambda^l(V)$, i.e. $\omega$ and $\eta$ are the alternating tensors. In the rest of the text, should I always assume whenever Spivak used the $\wedge$ symbol, he means that both operands are alternating tensors? Or in other words, is there any reason to use this definition on tensors that are not alternating?

Later on Spivak defined differential forms with alternating tensors, and I don't think I have experienced the case where $\omega$ and $\eta$ are not alternating.

I'm aware of other definitions of wedge products as the product in exterior algebra. However, I don't know anything about exterior algebra, and it seems to be very time consuming for me to learn about it just for this simple question.

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    $\begingroup$ What do you mean by "does this definition work"? The RHS of the definition of $\omega \wedge \eta$ is also well defined when $\omega$ and $\eta$ are general tensors (not necessarily alternating), if that is what you are asking? $\endgroup$
    – jd27
    Commented Jul 19, 2023 at 17:15
  • $\begingroup$ @jd27 Thank you for pointing this out. I have edited my question. I meant for the rest of the text, can we assume that both operands of $\wedge$ are alternating tensors? $\endgroup$ Commented Jul 19, 2023 at 19:51
  • $\begingroup$ Did you not quote Spivak as "Of course, if $\omega \in \Lambda^k(V)$ ... " etc. ? This should answer your question. BTW: If you are aware of other defintions of wedge product (exterior algebra) I strongly recommend to invest some time in studying those as well. $\endgroup$
    – Kurt G.
    Commented Jul 19, 2023 at 20:08
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    $\begingroup$ Wedge product is only defined when both (all) operands are alternating. No ambiguity. $\endgroup$ Commented Jul 19, 2023 at 21:17
  • $\begingroup$ @TedShifrin Thank you. Would you mind putting also answering this so I can mark the question as answered? $\endgroup$ Commented Jul 19, 2023 at 22:11

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Wedge product is only defined when both (or all) operands are alternating. No ambiguity.

By the way, if you want to see differential forms (and wedge products) done with no tensor products or Alt operator, check out my lectures on YouTube (linked in my profile). They are all based on determinants (the most important alternating tensor).

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  • $\begingroup$ Any chance this is in your book or you have written notes on this? $\endgroup$
    – Deane
    Commented Jul 20, 2023 at 1:03
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    $\begingroup$ @Deane: Sure, it's in the book. I just want to provide a legal free source. :) $\endgroup$ Commented Jul 20, 2023 at 1:39

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