TITLE: Alternating forms as skew-symmetric tensors: some inconsistency?
QUESTION [10 upvotes]: 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?

REPLY [7 votes]: Dear Paul, first of all let me congratulate you for the extremely clear formulation of your interesting question (which is not silly at all, contrary to what you say): +1.
The source of your trouble is the identification $Sk$: it is not the correct one.Why the cocksure statement? Because Laurent Schwartz wrote it! (In his book Les Tenseurs Hermann, 1975). Talking of (an analagon) of $Sk$, he writes on page 61:"il se trouve que cette identification n'est pas la bonne" (it happens that this identification is not the right one).
He recommends, for a vector space $E$,  the embedding $\Sigma k:\Lambda ^k E\to \otimes^k E$ where $\Sigma k=k! Sk$ (in your notation).The image of $\Sigma k$ is exactly the subspace $ A^k E \subset\otimes ^k E$ consisting of antisymmetric tensors This is valid in all characteristics $\neq 2$ and has  the crucial advantage that it preserves products: for $\alpha \in  \Lambda ^k E$ and
$\beta \in  \Lambda ^l E$, he proves $\Sigma (\alpha \wedge \beta)=\Sigma (\alpha)\otimes _a 
\Sigma (\beta)$, where $\otimes_a$ is the antisymmetric product defined on antisymmetric tensors and yielding antisymmetric tensors (This product is defined with shuffle permutations).The calculation is on page 104.
I find it very satisfying that Schwartz's point of view solves your trouble, without having to resort to ad hoc trickery. By the way, the book I mentioned is in French but many distinguished anglophones on this site have claimed energetically that mathematical French is no problem for an English-speaking person. I'll let you be the judge!
COMPLEMENT In case somebody is interested, here is the formula for the antisymmetric product I alluded to. Let $\alpha \in A^k E \subset \otimes ^k E$ and $\beta \in A^l E \subset \otimes ^l E$ be antisymmetric tensors. Their antisymmetric product is the antisymmetric tensor $\alpha \otimes_a \beta \in A^{k+l }E \subset \otimes ^{k+l} E$ defined by
$\alpha \otimes _a \beta=\sum sgn(s) s\bullet(\alpha \otimes \beta)$
(the sum is over shuffle permutations $s$,  i.e. permutations with   $s_1 < \ldots < s_k$ and $s_{k+1} < \ldots < s_{k+l}$ The bullet denotes the action of the symmetric group on tensors.) 
Of course Schwartz emphasizes that it is not a great idea to use antisymmetric tensors, which form a subspace of the tensor product: it is far better to use exterior products which are a quotient of said tensor product.