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? 



Alternating forms as skew-symmetric tensors: some inconsistency?
I can't speak to what is actually *used*, particularly what is used by physicists! However, I can try to shed some light on the diagram and the maps in question. In actual fact, there are two diagrams here and you are conflating them. This, simply put, is the source of the confusion. Let me expand (at a bit more length than I intended!) on that.



Firstly, there are too many maps flying around and some are more canonical than others. The most canonical is the identification of $(\bigotimes^k V)^\*$ with $\operatorname{Mult}^k(V)$ since this is by (one of the) definition(s) of the tensor product. So let us start with that. The inclusion $\operatorname{Alt}^k(V) \to \operatorname{Mult}^k(V)$ is probably next in line since it is the inclusion of a subspace. After that, I'd put the map $\bigotimes^k V^\* \to (\bigotimes^k V)^\*$. So, so far we have a diagram:




$$
\begin{array}{ccccc} \operatorname{Alt}^k V \\
 i \downarrow \\
 \operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
 \end{array}
$$




That the horizontal maps are isomorphisms is nice, but only holds for finite dimensional vector spaces so I'm not going to write in the fact that they are isomorphisms. I want to emphasise what's really canonical and what's not.




Now let us consider $(\Lambda^k V)^\*$. We appear to have a canonical map from this to $\operatorname{Alt}^k(V)$ but in fact, we don't. We have a canonical map from this to $(\bigotimes^k V)^\*$ given by:




$$
f(v\_1 \otimes \cdots \otimes v\_k) = f(v\_1 \wedge \cdots \wedge v\_k)
$$




This is dual to the projection map $\bigotimes^k V \to \Lambda^k V$. That projection map is pretty canonical as we usually define $\Lambda^k V$ as a quotient of $\bigotimes^k V$. Taking its dual is a natural thing to do, so this also appears on my list of "canonical maps". Now when we go "down" and "across" we happen to end up in the subspace $\operatorname{Alt}^k(V)$ so we can add a horizontal arrow $(\Lambda^k V)^\* \to \operatorname{Alt}^k(V)$ if we like, but the new map that we add by doing this is one step removed from the really canonical maps so I'm going to leave it out at this stage.




Now we come to $\Lambda^k V^\*$. This is, as for $\Lambda^k V$, defined as a quotient of the tensor product. So we have a projection $\bigotimes^k V^\* \to \Lambda^k V^\*$. This, again, is pretty canonical. So our "canonical" diagram looks like this:




$$
\begin{array}{ccccc} \operatorname{Alt}^k V && (\Lambda^k V)^\* && \Lambda^k V^\* \cr
 i \downarrow &&{p\_V}^\* \downarrow&& \uparrow p\_{V^\*}\cr
 \operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
 \end{array}
$$




At this point, an obvious question is as to whether or not we can fill in the gaps. I've already said that we can in the top-left. Can we in the top-right? That is, is there a map $\Lambda^k V^\* \to (\Lambda^k V)^\*$ making the diagram commute? (Thinking about infinite dimensions says that this is the correct direction.) The answer is: (drum roll) No. And the reason is quite simply that we start in $\bigotimes^k V^\*$ and can choose any element there as our starting point, but would want to end up in the *alternating* part of $(\bigotimes^k V)^\*$.




Okay, now we throw in the Alternator (probably time for another drum roll). The Alternator does what it says on the tin: it alternates stuff. But we have to be careful and ensure that we only apply it to stuff that can genuinely be alternated. So we have an alternator: $\operatorname{Alt} \colon \operatorname{Mult}^k(V) \to \operatorname{Alt}^k(V)$ given by




$$
\operatorname{Alt}(f)(v\_1,\dotsc,v\_k) = \frac{1}{k!} \sum (-1)^{\sigma} f(v\_{1\sigma}, \dotsc, v\_{k\sigma})
$$




The $1/k!$ is to make this a left inverse of the inclusion $\operatorname{Alt}^k(V) \to \operatorname{Mult}^k(V)$.




We also have an alternator $\Lambda^k V \to \bigotimes^k V$ given by:




$$
v\_1 \wedge \dotsb \wedge v\_k \mapsto \frac{1}{k!} \sum (-1)^{\sigma} v\_{1\sigma} \otimes \dotsb v\_{k\sigma}
$$




Again, the multiplier is chosen to ensure that this is a right inverse of the projection map. This is your $Sk$ map. Putting these into a diagram, we get:




$$
\begin{array}{ccccc} \operatorname{Alt}^k V && (\Lambda^k V)^\* && \Lambda^k V^\* \cr
 \operatorname{Alt} \uparrow &&{Sk\_V}^\* \uparrow&& \downarrow Sk\_{V^\*}\cr
 \operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
 \end{array}
$$




Again, the obvious question is: can we fill in the gaps? We can fill in the first one. Indeed, the same filler map works in this diagram as in the last. That was the map $\alpha \colon (\Lambda^k V)^\* \to \operatorname{Alt}^k(V)$ with the property that $i \alpha = \eta {p\_V}^\*$ (where $\eta \colon (\bigotimes^k V)^\* \to \operatorname{Mult}^k(V)$ is the isomorphism). So:




$$
i \alpha (Sk\_V)^\* = \eta {p\_V}^\*(Sk\_V)^\* = \eta (Sk\_V p\_V)^\* = \eta\;\text{and}\; i \operatorname{Alt} \eta = \eta
$$




Thus, as $i$ is an injection, $\alpha (Sk\_V)^\* = \operatorname{Alt} \eta$.




But it's the other gap that's more interesting. Now we *can* fill it in. And the "filler" map is laid out for us already: it's simply follow-the-arrows. If we work it out in detail, it's the following map:




$$
\begin{aligned}
f\_1 \wedge \dotsb \wedge f\_k \mapsto \Big((v\_1 \wedge \dotsb \wedge v\_k) \mapsto & Sk\_{V^\*}(f\_1 \wedge \dotsb \wedge f\_k) \big( {Sk\_V}^\*(v\_1 \wedge \dotsb \wedge v\_k)\big)\Big) \\
&= \frac{1}{k!} \frac{1}{k!} \sum\_\sigma \sum\_\tau (-1)^{\sigma} (-1)^{\tau} f\_{1\sigma}(v\_{1\tau}) \dotsb f\_{k\sigma}(v\_{k\tau})
\end{aligned}
$$




This simplifies considerably by rewriting $f\_{j\sigma}(v\_{j\tau})$ as $f\_{j\rho}(v\_j)$. Then we end up with $k!$ of each term, so we get:




$$
(f\_1 \wedge \dotsb \wedge f\_k)(v\_1 \wedge \dotsb \wedge v\_k) = \frac{1}{k!} \operatorname{det}(f\_i(v\_j))
$$




But notice the factor of $1/k!$ in this!




So to make that right-hand rectangle commute, one of the maps has to have a factor of $1/k!$ in it. It doesn't have to be the top one, but that's the most obvious one since if you modify one of the $Sk$s then you ought to modify the other one - though there's no reason to do so, and in fact this might be what's going on: the physicists are keeping one of the $Sk$s as it is and defining the other one to be suitably scaled so that the upper map is the determinant map. But that's speculation, returning to reality we have a diagram:




$$
\begin{array}{ccccc} \operatorname{Alt}^k V && (\Lambda^k V)^\* &\stackrel{\frac{1}{k!}\operatorname{det}}{\leftarrow} & \Lambda^k V^\* \cr
 \operatorname{Alt} \uparrow &&{Sk\_V}^\* \uparrow&& \downarrow Sk\_{V^\*}\cr
 \operatorname{Mult}^k V &\leftarrow & (\otimes^k V)^\* & \leftarrow & \otimes^k V^\*
 \end{array}
$$




Finally, let's compare this to your original diagram. The key thing to notice is that in my diagrams, I have two vertical maps in one direction and one in the other. In your diagram, you have two vertical maps in the same direction (and are missing the third). But whichever of my diagrams you prefer, one of your maps is going in the wrong direction.
So, in conclusion, the mistake is that your diagram isn't supposed to commute. Rather, there's two commuting diagrams there with some maps from one diagram and some from another.




(I have a feeling that I haven't really answered the question. This was what I wrote out when trying to make sense of the question rather than towards an answer. But I hope that it helps clarify the issue for you.)



