Let $V$ be a vector space and $\omega\in \otimes^k V$. There are $2$ ways (at least) of thinking about $\omega\otimes \omega$.

1) We may think of $\otimes^k V$ as a vector space $W$, and $\omega\otimes \omega$ as a member of $W\otimes W$.
2) We may think of $\omega\otimes \omega$ as a member of $\otimes^{2k}V$.

The two interpretations are "same" because $W\otimes W$ is naturally isomorphic to $\otimes^{2k} V$.

However, the situation is a bit different when talking about "wedging".

Let $\eta\in \Lambda^k V$. We want to wonder about $\eta\wedge \eta$.

1) Let $W=\Lambda^k V$ and think of $\eta\wedge \eta$ as a member of $\Lambda^2 W$. Then $\eta\wedge \eta=0$ by super-commutativity of the wedge-product.

2) Think of $\eta\wedge\eta$ as a member of $\Lambda^{2k}V$. Then $\eta\wedge \eta$ may not be $0$.

Perhaps this confusion would not arise if we write $\wedge_V$ rather that $\wedge$, for when wedging we must remember the base space. Moreover, there is no such thing as taking the wedge product of two vector spaces, thought we can talk about tensor product of two vector spaces.

Admittedly, my mind is not completely clear here. Can somebody throw some more light on the different behaviours of tensoring and wedging.

  • $\begingroup$ I'm not quite sure what the question is, but one nice thing about $\otimes$ is that is makes the category of vector spaces to a (symmetric) monoidal category, i.e. $\otimes$ is "associative" (and "commutative"). As you noticed, $\wedge$ is not even defined for two different vector spaces, and so is best used as a (graded-commutative) associative product on $\bigwedge V$. $\endgroup$ – user8268 Dec 30 '15 at 19:00

This is not really a direct answer but is just too long for a comment.

One curious fact about the wedge construction is that $\bigwedge^n V$ can be (functorially) realized either as a subspace of $\bigotimes^n V$ or as a quotient. (These realizations are canonically isomorphic when the characteristic of the underlying field is $0$ or greater than $n$, but otherwise are non-canonically isomorphic.)

Although the quotient construction tends to be more natural, it's often useful to think about the subspace construction. The little wedge symbol means different things, depending on your construction.

In the quotient construction, $\bigwedge^n V$ is the quotient of $\bigotimes^n V$ by the subspace generated by symbols with repeated vectors, and the symbol $v_1 \wedge \ldots \wedge v_n$ means "the image of $v_1 \otimes \ldots \otimes v_n$ under the quotient map."

In the subspace construction, on the other hand, $\bigwedge^n V$ is the subspace of $\bigotimes^n V$ on which $S_n$ acts via the sign character, and the symbol $v_1 \wedge \ldots \wedge v_n$ means either (depending on your convention and on whether $n! = 0$ in your field) $$ \sum_{\sigma \in S_n} (-1)^{\text{sign}(\sigma)} v_{\sigma(1)}\otimes \ldots \otimes v_{\sigma(n)}$$ or $$ \frac{1}{n!} \sum_{\sigma \in S_n} (-1)^{\text{sign}(\sigma)} v_{\sigma(1)}\otimes \ldots \otimes v_{\sigma(n)}. $$ (The second convention has the advantage that under the natural map "subspace intepretation to quotient interpretation" is compatible with the notation, and the disadvantage that it's only available in characteristic prime to $n!$).

Now let's think about $\bigwedge^2 \bigwedge^n V$ versus $\bigwedge^{2n} V$ in terms of the subspace interpretation. The former is the subspace of $\bigotimes^2 \bigwedge^n V$ on which $S_2$ acts by the sign character, which is the subspace of $\bigotimes^2 \bigotimes^n V$ on which $S_2$ and $S_n$ act independently via the sign character. Under the natural "unravelling map" $$ \bigotimes^2(\bigotimes^n V) \to \bigotimes^{2n} V $$ we get the subspace of $\bigotimes^{2n} V$ on which $(S_n \times S_n) \rtimes S_2 < S_{2n}$ acts via the product of the sign characters. But this is a weaker, and different, demand than demanding that all of $S_{2n}$ act via the sign character, and so this subspace is bigger.

(EDIT: see comments for more details.) In terms of representation theory, your observation could be restated as follows: Write $G = (S_n \times S_n) \rtimes S_2.$ Then there is a natural embedding $G \hookrightarrow S_{2n}$. Now for a field $k$ there is a unique character $G \to k^\times$ which restricts to the sign character each $S_n$ and to the sign character on $S_2$. There is also a character $G \to k^\times$ using the embedding $G \hookrightarrow S_{2n}$ followed by the sign character of $S_{2n}$. These characters aren't the same.

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    $\begingroup$ While I like this approach, I am not sure about the action of $S_n$ of $\bigotimes^2 \bigotimes^n V$: If we just let act $S_n$ on both copies of $\bigotimes^n V$ simultaneously via the sign representation (i.e. if we take the usual tensor product of representations), then it acts trivially on $\bigotimes^2 \bigotimes^n V$ because the sings cancel out. I feel like we should instead let $S_n \times S_n$ act on $\bigotimes^2 \bigotimes^n V$, each copy of $S_n$ on the corresponding copy of $\bigotimes^n V$. $\endgroup$ – Jendrik Stelzner Dec 30 '15 at 21:07
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    $\begingroup$ Thank you. This does make things clearer. One think I am yet not able to see is how do we get the semidirect product $S_n\rtimes S_2$. $\endgroup$ – caffeinemachine Dec 31 '15 at 3:19
  • $\begingroup$ @JendrikStelzner Oops, the above two comments are correct. It is $(S_n \times S_n) \rtimes S_2$ that is acting, not $S_n \rtimes S_2$ as I erroneously stated (and here it is clear what the semidirect product structure is). Then the requirement is that both $S_n$ factors individually act via the sign character, and that $S_2$ does as well. Sorry! Will edit it into the answer when I have more energy. $\endgroup$ – hunter Dec 31 '15 at 3:32
  • $\begingroup$ (edit to above comment: I am confused about what the requirement is now!) $\endgroup$ – hunter Dec 31 '15 at 3:39
  • $\begingroup$ Right. Each $S_n$ has to act via the sign character, and so does $S_2$. This is not the same as demanding that they act via the restriction of the sign character from the ambient $S_{2n}$ to $(S_n \times S_n) \rtimes S_2$, which is maybe a group-theoretic restatement of the phenomenon OP was observing in the first place. $\endgroup$ – hunter Dec 31 '15 at 3:50

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