My trouble is best described by the following diagram:

$$ \begin{array}{ccccc} \mathrm{Alt}^k V &\stackrel{\sim}{\rightarrow}& (\Lambda^k V)^* &\stackrel{\sim}{\rightarrow}& \Lambda^k V^* \cr i \downarrow &&&& \downarrow \mathrm{Sk}\cr \mathrm{Mult}^k V &\stackrel{\sim}{\leftarrow} & (\otimes^k V)^* & \stackrel{\sim}{\leftarrow} & \otimes^k V^* \end{array} $$ The problem is that this diagram is not commutative but let me explain the terminology first.

Here $\mathrm{Alt}^k V$ and $\mathrm{Mult}^k V$ are the spaces of alternating and multilinear $k$-forms, respectively, on the vector space $V$. All the horizontal isomorphisms are canonical. The left vertical arrow is the inclusion of the alternating forms in the multilinear ones. The only "questionable" arrow is the right-hand vertical one ($\mathrm{Sk}$, following the notation in Birkhoff-MacLane "Algebra", Section XVI.10). It is given as an extension of the following alternating map

$$ (*)\quad (v_1^*, \ldots, v_k^*) \mapsto {1\over k!} \sum_{\sigma\in S_k} (-1)^{\sigma} v_{\sigma(1)}^* \otimes \cdots \otimes v_{\sigma(k)}^*. $$

If the characteristic is zero (which I assume) then Sk is an inclusion. There are two good things about this inclusion: First, it is a linear inversion of the canonical projection modulo the graded ideal generated by squares of elements of grade 1, i.e. of $\otimes^k V^* \rightarrow \Lambda^k V^*$.

Second, if $\mathrm{Sk'}$ is a map $\otimes^k V^* \rightarrow \otimes^k V^* $ which is again obtained by extending a multilinear map (*), then we have $$ Sk(a \wedge b) = Sk'(Sk(a)\otimes Sk(b)) $$ (i.e. to learn what $a\wedge b$ is you map both to tensors via $Sk$ and then antisymmetrize their tensor product in the tensor algebra).

So the above argument suggests that $Sk$ is somewhat canonical as well. However, here is a strange situation. Suppose that $e_1, \ldots, e_n$ are the basis of $V$. Then consider the alternating 2-form that operates on $V\times V$ as follows: $$ f(v_1, v_2) = e_1^*(v_1) e_2^*(v_2) - e_1^*(v_2) e_2^*(v_1) $$ Its image in $\Lambda^k V^*$ is $e_1^* \wedge e_2^*$, and thus under $Sk$ it gets mapped to $$ {1\over 2} (e_1^* \otimes e_2^* - e_2^* \otimes e_1^*) $$ Therefore applying the other two bottom isomorphisms we arrive at a multilinear form that operates on $V\times V$ as follows: $$ g(v_1, v_2) = {1\over 2} (e_1^*(v_1) e_2^*(v_2) - e_1^*(v_2) e_2^*(v_1)) $$

Clearly $g\neq f$ and this is precisely the non-commutativity I was talking about.

Can somebody explain if I made a mistake somewhere? And if not, why then so many physicists happily use "skew-symmetric tensors" and refuse to use "differential forms" and still arrive to the very same answers never loosing coefficient like $1\over 2$?

Thanks in advance! This looks really puzzling to me and I know this is too easy for MO, but I am in a situation much different from the rest of MO having zero mathematicians around to ask such silly questions to. Again, thanks for reading!

**Added later:** As Andrew and Georges point out it is easy to make the diagram commute by either redefining the $\mathrm{Sk}$ without the ${1\over k!}$ or by changing $(\Lambda^k V)^* \rightarrow \Lambda^k V^* $ from the $\mathrm{det}$-map to ${1\over k!}\mathrm{det}$. Let me explain why I think why either approach is confusing.

First, redefining the $\mathrm{Sk}$ map as Georges suggests revokes its first property: namely it is no longer a right inverse of the projection $\otimes^k V \to \Lambda^k V$. On the other hand, map $(\Lambda^k V)^* \rightarrow \Lambda^k V^* $ determines what we call a wedge-product in the graded algebra $\mathrm{Alt}^* V$ (since the wedge-product in $\Lambda^* V^* $ canonically comes from $\otimes^* V^* $ via projection). Therefore, if we are to redefine the meaning of $(\Lambda^k V)^* \rightarrow \Lambda^k V^* $ as Andrew proposes, then we have to agree that now

$$ (* *)\quad (dx \wedge dy) (\partial_x, \partial_y) = {1\over 2},$$ which I think many will find somewhat weird. (Although, it seems things like Stokes theorem do not depend on the agreement (**), right?)

To sum up: if we agree what $\wedge$ means in $\mathrm{Alt}^k$ then this determines the definition of $\mathrm{Sk}$. And thus with the usual definition of $\wedge$ in $\mathrm{Alt}^k$ we unfortunatelly obtain the $\mathrm{Sk}$ which is not the right-inverse of the projection. Am I correct in this?

`$(V^*)^{\otimes 2} \neq (V^{\otimes 2})^*$`

. +1 for the question, BTW: these conventions matter in mathematics, too, and it's easy to be off by $k!$. – Theo Johnson-Freyd Sep 1 '10 at 18:03viasomething else. This can be seen by the fact that in the determinant map (for example), there's noa priorireason why either of the terms has to be alternating. 2: $Alt^k(V)$ already has a product, it doesn't need one imposed from any isomorphism. 3:Keep those diagrams separate! After a little more reflection, I think that the problem is that $Sk_{V^*}$ is not $p_V^*$. I recommend that you try to understand that statement. – Andrew Stacey Sep 1 '10 at 18:55