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In the linear algebra book that I'm reading, the exterior product is defined explicitly as $a \wedge b := a \otimes b - b \otimes a$. Obviously this satisfies the required properties of the exterior product, but the definition seems a bit ad hoc. A better definition seems to be this:

Consider a vector space $V$ and let $R$ be the subspace of $V \otimes V$ spanned by elements of the form $a \otimes b + b \otimes a$. Then $V \wedge V := (V \otimes V) / R$.

It's clear that these are conceptually equivalent definitions, but I'm not sure how we can explicitly derive the form of the exterior product from the above definition (and this explicit form seems to be important for further applications).

Also, if I interpret the above definition in terms of an equivalence relation on $V \otimes V$, I am effectively saying that for every $a \wedge b \in V \wedge V$, $a \wedge b + b \wedge a \sim 0$, so $a \wedge b \sim -b \wedge a$, which is exactly what I want.

However, if I want to find the exact form of $a \wedge b$, I think it'd be more convenient to interpret it as a coset; $a \wedge b = a \otimes b + R$, but now it is completely unclear to me how we could get $a \wedge b = -b \wedge a$.

To elaborate a bit further, I'm also asking how one could come up with this definition $a \wedge b = a \otimes b - b \otimes a$. Perhaps it is somewhat reasonable to guess in this two-dimensional case, but I'm interested in the general process, since, for example, we also have

$$a \wedge b \wedge c = a \otimes b \otimes c - b \otimes a \otimes c + c \otimes a \otimes b - c \otimes b \otimes a + b \otimes c \otimes a - a \otimes c \otimes b,$$

which is definitely not easy to guess. I'm hoping the "process" of the derivation will generalize to the $n$-dimensional case.

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    $\begingroup$ Isn’t $a\otimes b+R=(a\otimes b+b\otimes a)-b\otimes a+R=-b\otimes a+R$? $\endgroup$
    – user1551
    Aug 23, 2023 at 15:16
  • $\begingroup$ @user1551 Oops, true. Thanks. Still not sure how to derive the general form, though. $\endgroup$
    – zaq
    Aug 23, 2023 at 15:27
  • $\begingroup$ Why is working with a quotient vector space "better" than working with explicit elements of the vector space? For universal mapping purposes, I might concede the point, but ... $\endgroup$ Aug 23, 2023 at 16:27
  • $\begingroup$ @TedShifrin It's not "better" for computational purposes, but I prefer it because it's much more intuitive. If you think how we construct the space $V \otimes V$, it's by creating a quotient space $(V \times V)/ P$, where $P$ is the space spanned by the properties of the tensor product we want. It makes intuitive sense to construct the exterior product similarly and it is easier to construct it this way than to guess a gnarly expression, as is e.g. the computational definition for $a \otimes b \otimes c$ I've given above. $\endgroup$
    – zaq
    Aug 25, 2023 at 15:33
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    $\begingroup$ It’s not weird at all. You act by the permutation group and use the sign of the permutation as the coefficient. If you use subscripts $1,\dots,n$ instead of different letters, you can write the formula easily with a summation symbol. $\endgroup$ Aug 25, 2023 at 15:51

2 Answers 2

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$ \newcommand\Ext{\mathop{\textstyle\bigwedge}} \newcommand\MVects[1]{\mathop{\textstyle\bigwedge^{\mkern-2mu#1}}} \newcommand\Tensor{\mathop{\textstyle\bigotimes}} \newcommand\MTensors[1]{\mathop{\textstyle\bigotimes^{#1}}} \newcommand\alt{\mathop{\mathrm{alt}}\nolimits} $Let $V$ be an $n$-dimensional $K$-vector space. This is just the "inverse" of the projection map $\pi : \Tensor V\to \Ext V$, with $\Tensor V$ the tensor algebra and $\Ext V$ the exterior algebra.

If you're concerned with "better" definitions, then rather than something like $R$ we should be considering something like $v\otimes v$. This is because in characteristic 2 we have $1 = -1$ and so $v\otimes w + w\otimes v = v\otimes w - w\otimes v$, meaning that your $R$ will actually give us a symmetric power rather than an exterior power. That being said, this is not as important as it may sound because there is no injective algebra homomorphism $\Ext V \to \Tensor V$ in characteristic $\leq n$. So in characteristic 2 we're restricted to 1D vector spaces which is... really boring. We will from here on assume a field characteristic $0$, though this is stricter than necessary.

The tensor algebra is characterized as the unique-up-to-isomorphism associative algebra with the following universal propery

  • Let $A$ be an associative $K$-algebra and $f : V \to A$ be linear. Then $f$ extends uniquely to an algebra homomorphism $f' : \Tensor V \to A$ such that $$ f'(v_1\otimes\dotsb\otimes v_k) = f(v_1)\dotsb f(v_k) $$ for all $k$ and $v_1,\dotsc, v_k \in V$.

The tensor algebra can be explicitly constructed as $$ \Tensor V = \bigoplus_{i=0}^\infty(\MTensors iV) $$ where $\MTensors iV$ is the $i^{\text{th}}$ tensor power and $\MTensors0V = K$ and $\MTensors1V = V$, and we give it the obvious product suggested by the tensor powers.

The exterior algebra is similarly characterized by a universal property:

  • Let $A$ be an associative $K$-algebra and $f : V \to A$ be linear with $f(v)^2 = 0$ for all $v \in V$. Then $f$ extends uniquely to an algebra homomorphism $f' : \Ext V \to A$ such that $$ f'(v_1\wedge\dotsb\wedge v_k) = f(v_1)\dotsb f(v_k) $$ for all $k$ and $v_1,\dotsc,v_k \in V$.

Now consider the ideal of $\Tensor V$ $$ I = \langle v\otimes v \;:\; v \in V\rangle $$ generated by tensors of the form $v\otimes v$. It is easy to confirm that $\Tensor V/I$ satisfies the above universal property by construction; we take this as our definition of $\Ext V$. It also follows that $\Ext V$ is a direct sum of exterior powers $\MVects iV = \MTensors iV/I$ $$ \Ext V = \bigoplus_{i=0}^n\MVects iV $$ and is in fact graded on them. This quotient construction naturally gives us a projection homomorphism $\pi : \Tensor V \to \Ext V$. We identify $\pi(V)$ with $V$ and $\pi(K)$ with $K$.

The alternation property $v\wedge v = 0$ means that $v\wedge w = -w\wedge v$ for $v, w \in V$; in particular define $\alt_0 : \Tensor V \to \Tensor V$ by $$ \alt_0(v_1\otimes\dotsb\otimes v_k) = \sum_{\sigma\in S_k}\mathrm{sgn}(\sigma)v_{\sigma(1)}\otimes\dotsb\otimes v_{\sigma(k)} $$ where $S_k$ is the set of all $k$-permutations, and the image $\alt_0\Tensor V$ is called the space of alternating tensors. Then evidently for any $X \in \MTensors kV$ $$ \pi(\alt_0(X)) = k!\,\pi(X). $$ So define the normalized alternation map $\alt : \Tensor V \to \alt_0\Tensor V$ by $$ \alt(X) = \frac1{k!}\alt_0(X),\quad X \in \MTensors kV $$ (this is where we start having issues with characteristic!) which now satisfies the equality $$ \pi(\alt(X)) = \pi(X),\quad X \in \Tensor V. $$ So $$ I = \ker(\pi) = \ker(\pi\circ\alt) = \alt^{-1}(\ker\pi) \supseteq \ker(\alt). $$ It is straightforward to show from the definition of $\alt$ that $I \subseteq \ker(\alt)$, so $\ker(\alt) = I$. Also, $\alt$ is itself a projection $$ \alt(\alt(X)) = \alt(X),\quad X \in \Tensor V. $$ Finally, we can make $\alt$ into a homomorphism; we need an associative product $\bullet$ on $\alt_0\Tensor V$ such that $$ \alt(v_1\otimes\dotsb\otimes v_k) = v_1\bullet\dotsb\bullet v_k $$ noting that $\alt(v) = v$ for all $v \in V$. What you'll find is that the LHS actually defines such a product.

So we have an algebra homomorphism $\alt : \Tensor V \to (\bullet,\, \alt_0\Tensor V)$ which is surjective and whose kernel is $I$; thus there is an algebra isomorphism $$ \varphi : \Ext V \cong (\bullet,\, \alt_0\Tensor V). $$ The product $\bullet$ is usually also denoted $\wedge$ and gives an exterior product on alternating tensors defined by $$ v_1\wedge\dotsb\wedge v_k = \alt(v_1\otimes\dotsb\otimes v_k) = \frac1{k!}\sum_{\sigma\in S_k}\mathrm{sgn}(\sigma)v_{\sigma(1)}\otimes\dotsb\otimes v_{\sigma(k)}. $$ It is trivial at this point that $\varphi$ is an inverse of $\pi$: $$ \pi(\varphi(X)) = X,\quad X \in \Ext V. $$


Answering zaq's first comment below:

My second to last display equation gives $a\wedge b = \tfrac12(a\otimes b - b\otimes a)$ (I was missing a normalization and have edited). You're entirely right that if we are only concerned with exterior powers then you could choose any prefactor. So we have to look past exterior powers and look at the exterior algebra and reason about how these exterior powers combine with each other; in particular we want $\wedge$ to be an associative product. Only certain prefactors are compatible with associativity; if $X$ is a $p$-vector and $Y$ is a $q$-vector and $$X\wedge Y = k_{pq}\mathrm{alt}_0(X\otimes Y),$$ then to get associativity we must have $$k_{p(q+r)}k_{qr}(q+r)! = k_{pq}k_{(p+q)r}(p+q)!$$ for all $p,q,r$. My construction gives $k_{pq} = 1/(p+q)!$. Another common solution is $k'_{pq} = 1/p!q!$, and there are even more. These two are special, however.

Once we have a wedge product defined by such a set of coefficients $k$, we can define $$\mathrm{alt}_k(v_1\otimes\dotsb\otimes v_p) = v_1\wedge\dotsb\wedge v_p$$ using the $k$-wedge product. My construction shows that the $\mathrm{alt} = \mathrm{alt}_k$ we get from $k_{pq} = 1/(p+q)!$ is special: it is the only one which is compatible with the projection $\pi : {\bigotimes}V \to {\bigwedge}V$, and also the only one which is a linear projection $\mathrm{alt}\circ\mathrm{alt} = \mathrm{alt}$.

Once we commit to embedding ${\bigwedge}V$ in ${\bigotimes}V$ using this $\mathrm{alt}$ and the corresponding wedge product, then it can be shown (through quite some effort) that $\mathrm{alt}_{k'}$ with $k'_{pq} = 1/p!q!$ and its corresponding wedge product is the natural way to embed ${\bigwedge}V^*$ into ${\bigotimes}V^*$.

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  • $\begingroup$ A lot of what you've written goes over my head. I've spent some time trying to understand and I think I got some of it, but this is still unclear to me: how does this show that my (or your) definition of $\wedge$ is equivalent to $a \wedge b := a \otimes b - b \otimes a$? I mean, does it follow that we can not have $a \wedge b$ equal anything else from our definitions? Certainly it is still possible that $a \wedge b = k(a \otimes b - b \otimes a)$ for some $k \in K$, but is it necessary that this holds or are there other isomorphic characterizations? $\endgroup$
    – zaq
    Aug 26, 2023 at 21:08
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    $\begingroup$ @zaq I added a reply to the end of my answer. $\endgroup$ Aug 27, 2023 at 1:03
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For everything to be rigorous, you can try to prove the following. Let $W_1=span\{a\otimes b-b\otimes a:a,b\in V\}\subset V\otimes V$ and $W_2=V\otimes V/R$ ($R$ is as defined as in the question). Show this mapping is well-defined and is an isomorphism:

$$ T:W_1\to W_2,a\otimes b-b\otimes a\mapsto a\otimes b+R. $$


Update:

Some idea on how to derive the specific construct: I think the intuition is similar to constructing symmetric polynomials by symmetrization, i.e., adding all symmetric variants of a known monomial. For example, if you know $x_1x_2^2$ belongs to a symmetric polynomial $p$, then $p$ is at least $x_1x_2^2+x_1^2x_2$.

For wedge products, the symmetry is in fact antisymmetry. Suppose you somehow know $a\otimes b$ is a term in $a\wedge b$, then by antisymmetry $-b\otimes a$ must also be a term. For $k$-forms, suppose you know $v_1\otimes v_2\otimes\cdots\otimes v_n$ is a term in $v_1\wedge v_2\wedge\cdots\wedge v_n$, then all terms $sign(\sigma)v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)}$ also belong to it. Here, $\sigma$ ranges over all permutation on $\{1,\cdots,n\}$. This gives the construct: $$ v_1\wedge\cdots\wedge v_n:=\sum_{\sigma}sign(\sigma)v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(n)}. $$

Note that this is an observation, not a proof. You'd still need to prove the isomorphism as above.

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  • $\begingroup$ Sure, something like this has to be true in order for the definitions to be equivalent. But this still runs into the problem of seeming ad hoc. Is there any intuitive reason why we would even think of this as being an isomorphism? $\endgroup$
    – zaq
    Aug 23, 2023 at 15:30
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    $\begingroup$ @zaq Are you asking how can one think of this specific construct $a\otimes b-b\otimes a$ when one only knows the general definition? $\endgroup$
    – trisct
    Aug 23, 2023 at 15:31
  • $\begingroup$ Yes. And also how to prove it, but you've already answered that. I edited my post to elaborate a bit. Sorry for the confusion. $\endgroup$
    – zaq
    Aug 23, 2023 at 15:43
  • $\begingroup$ @zaq I made an edit explaining the possible intuition behind the construct. $\endgroup$
    – trisct
    Aug 23, 2023 at 16:55

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