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.
Best Answer
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).