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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?

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Perhaps the discussion at physicsforums.com/showthread.php?p=2025445 is helpful? –  Hans Lundmark Sep 1 '10 at 9:18
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I assume that in addition to working in characteristic $0$, your vector space $V$ is finite-dimensional? Otherwise $(V^*)^{\otimes 2} \neq (V^{\otimes 2})^*$. +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 18:03
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Comments on the addendum: 1: There is no map $(\Lambda^k V)^* \to \Lambda^k V^*$. A map can be constructed in the opposite direction, but it always goes via something else. This can be seen by the fact that in the determinant map (for example), there's no a priori reason 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
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Andrew, sorry for being equivocal. 1. By the map $(\Lambda^k V)^* \rightarrow \Lambda^k V^*$ I mean the inverse of your $\mathrm{det}$ map (which exist since I assume my $V$ to be finite dimensional). 2. You are right that algebra $\mathrm{Alt}^* V$ has a product, but the reason we denote it $\wedge$ is because the map $\mathrm{Alt}^* V$ in the top-row of my diagram is an isomorphism of graded algebras. 3. I am not sure what you mean $\mathrm{Sk}_{V^*}$ and $p^*_V$ go in the opposite directions (and are the inverses of each other). Again, thanks for clarifying discussion! –  Paul Yuryev Sep 2 '10 at 1:30
    
To elaborate on 2: if you change the top-row map then it is natural to change the definition of $\wedge$ in $\mathrm{Alt}^* V$ to preserve the isomorphism of algebras. For this reason I said before that the choice of $(\Lambda^k V)^* \rightarrow \Lambda^k V^*$ implies the choice of $\wedge$ in $\mathrm{Alt}^* V$. BTW, I just realized that convention (**) also affects differentials (i.e. d(x dy) is now a different 2-form) and volumes of Riemannian manifolds (so that the volume of unit 2-sphere becomes 2\pi). –  Paul Yuryev Sep 2 '10 at 1:36
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3 Answers

up vote 12 down vote accepted

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.)

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Dear Andrew, Your answer looks great from all points of view, but I have a tiny typographical problem with it: one line is MUCH longer than all the others. Wouldn't it be possible to break it? –  Pierre-Yves Gaillard Sep 1 '10 at 10:50
    
Pierre: Sure, but you'll have to tell me which line as it all looks fine to me (I get my MathJaX served as MathML which may make a difference to how it looks). –  Andrew Stacey Sep 1 '10 at 11:22
    
(Switched to HTML+CSS to see which it was and tried to fix it. Hope that's acceptable now. The MathML rendering really does look much, much nicer!) –  Andrew Stacey Sep 1 '10 at 11:33
    
Thanks a lot!!! It's MUCH better! (At least to me.) [The 2 lines stick out a little bit, but that's ok. And that's not your fault, that's math's fault: these formulas ARE nasty.] [If this helps, I'm using Firefox (sometimes Safari or Chrome) on a MacBook Pro. Your post looks the same on all three.] –  Pierre-Yves Gaillard Sep 1 '10 at 11:46
    
Pierre: In firefox, right-click on a piece of maths. Select "Settings->Math Renderer->MathML" and it'll all look right again! Slightly more seriously, when trying to break them I realises that actually it might be better expressed as two completely separate lines, but there's a limit to the amount of time I can spend on such matters. –  Andrew Stacey Sep 1 '10 at 12:32
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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.

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The formula for the antisymmetric product also appears in Warner's "Foundations of differentiable manifolds and Lie groups". It's Formula (2) p. 60. You can view it as follows: go to amazon.com/Foundations-Differentiable-Manifolds-Lie-Groups/dp/…, then choose "Click to LOOK INSIDE", then type "shuffle" in the box "Search Inside This Book", then go to p. 60. –  Pierre-Yves Gaillard Sep 1 '10 at 13:26
    
    
Georges, thanks! BTW, Birkhoff-MacLane in Section XVI.10 where they discuss $\mathrm{Sk}$ never use the fact that $\mathrm{char} F=0$ and all their statements hold true without ${1\over k!}$ and under the assumption $\mathrm{char} F \neq 2$ (in particular, they never state that $\mathrm{Sk}$ is the right inverse of the projection). So I think they also hiddenly agree with Schwartz! –  Paul Yuryev Sep 1 '10 at 16:38
    
I think the following holds. Let $K$ be a commutative ring and $V$ a $K$-module. Then there is a unique $K$-linear map $f$ from $\bigwedge V$ to $\bigotimes V$ mapping $v_1\wedge\cdots\wedge v_n$ to the sum over $\sigma\in S_n$ of the $\epsilon_\sigma\,\sigma(v_1\otimes\cdots\otimes v_n)$. Moreover $f(a\wedge b)=f(a)*f(b)$, where $*$ is the "shuffle product" described in Georges's answer. –  Pierre-Yves Gaillard Sep 3 '10 at 4:28
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I think the problem might be with your definition of $\wedge$.

If you look at chapter 7 in volume 1 of Michael Spivak's Differential Geometry, you will see that the way to define $\wedge$ from $\operatorname{Sk}$ (which he calls $\operatorname{Alt}$) involves some combinatorial factors that makes $\wedge$ have all the nice properties, including associativity. Indeed, if $\alpha \in \Lambda^k V^*$ and $\beta \in \Lambda^\ell V^*$, then their wedge product is defined as $$\alpha \wedge \beta = \frac{(k+\ell)!}{k!\ell!} \operatorname{Sk}(\alpha\otimes\beta).$$ In particular, if $\alpha,\beta \in V^*$, one has $$\alpha \wedge \beta = \alpha\otimes\beta - \beta\otimes\alpha$$ and not half of that.

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One can also look at Section 2.10 p. 59 of "Foundations of Differentiable Manifolds and Lie Groups" by Frank W. Warner. –  Pierre-Yves Gaillard Sep 1 '10 at 10:06
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There are two natural definitions of $\wedge$, one with one set of combinatorial factors, one with another, and both are associative. You can use $\alpha \wedge \beta = \operatorname{Sk}(\alpha\otimes \beta)$ if you want, provided you pick the correct embedding $\bigwedge^k \hookrightarrow \bigotimes^k$. The equivalence (in char=$0$) can be seen by applying the map ${\rm T}W = \bigoplus_{k=0}^\infty W^{\otimes k} \to {\rm T}W$ that acts as $k!$ on the $k$th piece. This is not an algebra homomorphism on ${\rm T}W$, but respects $\bigwedge W$, and intertwines its two multiplications. –  Theo Johnson-Freyd Sep 1 '10 at 18:19
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