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Hello! in Little Spivak, p. 79, we find this:

We will therefore define a new product, the wedge product $\omega\wedge\eta\in > \Lambda^{k+\ell}(V)$ by $$ > \omega\wedge\eta = > \frac{(k+\ell)!}{k!\ell!}\mathrm{Alt}(\omega\otimes\eta). > $$

It is ambiguous whether Spivak means for this to be defined for all $\omega\in\mathcal{J}^k(V),\eta\in\mathcal{J}^{\ell}(V)$ or only for $\omega\in\Lambda^k(V),\eta\in\Lambda^{\ell}(V)$. ($\mathcal{J}^n(V)$ is the set of $n$-tensors over vector space $V$, $\Lambda^n(V)$ is the set of alternating $n$-tensors over $V$.) Certainly the definition would make sense either way.

Which one is it?

In short: Is the wedge product applied to general tensors, or only to tensors which are already antisymmetric?

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The wedge product is applied to a pair of tensors that are each already antisymmetric (not "already symmetric", which for the moment is what you've written at the end). I have no idea what your abbreviation "TL/DR" means. –  KConrad Mar 31 '13 at 17:36
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@KConrad: Too Long; Didn't Read. I'm never sure if writing it is condescending or accommodating. –  Ryan Reich Mar 31 '13 at 18:45
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Thank you. I fixed the "symmetric" bit to "antisymmetric" and changed "TL/DR" to "In short". –  Anonymous Mar 31 '13 at 21:19
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1 Answer

Without having the book handy, I can't say for sure what Spivak means by this. As KConrad points out in his comment, this is usually applied to tensors that are already alternating, and it turns the vector space $$ \Lambda(V) := \bigoplus_{k=0}^\infty \Lambda^k(V) $$ into an associative algebra, called the exterior algebra of $V$.

I would argue that this is sort of an unnatural way to view the exterior algebra. See this question, where this viewpoint is discussed. The gist of it is the following: the exterior algebra is most naturally viewed as the quotient algebra $T(V)/\langle x \otimes y + y \otimes x \mid x,y \in V\rangle$, where $T(V)$ is the tensor algebra of $V$. The description above of $\Lambda(V)$ as the direct sum of the antisymmetric tensors means that we are embedding the exterior algebra as a subspace of the tensor algebra $T(V)$ rather than a quotient. It is not a sub-algebra, because the concatenation $\omega \otimes \eta$ of two antisymmetric tensors is not antisymmetric. So the definition you/Spivak give of the wedge product is the way to express the natural multiplication of the exterior algebra in terms of operations in the tensor algebra. Applying $\mathrm{Alt}$ makes it antisymmetric again, and the factorials are a fudge factor required to make it associative.

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Although the exterior powers are arguably better thought of as a quotient space, and that way of treating them is essential for exterior powers of modules over a general commutative ring, my impression (as an outsider) is that in differential geometry and physics it is still common to regard the exterior powers as subspaces of antisymmetric tensors. –  KConrad Apr 1 '13 at 1:21
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