# Do any non-combinatorial proofs of the elementary properties of wedge products exist?

The wege product, an operation defined between two alternating tensors, has a number of elementary properties such as associativity, distributivity, etc. There are many proofs of these properties e.g., see Analysis on Manifolds by Mukres or Topology, Geometry and Guage Fields by Naber. These proofs differ in the details but all ultimately rely on combinatorial arguments that require a considerable amount of algebra and index juggling. Once you have worked your way through the details you don't really feel like you've gained any insight into the essence of the matter. Motivated by these considerations, I am led to ask the following question:

Is anyone aware of alternate proofs of the basic properties of the wedge product that don't rely on combinatorial arguments?

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I personally like the following approach.

Define the tensor algebra $T^{\circ}(V)=\bigoplus V^{\otimes n} = k\oplus V \oplus V\otimes V +\cdots$. Because there natural isomorphisms $V^{\otimes m}\otimes V^{\otimes n} \cong V^{\otimes m+n}$, and tensor products distributes over direct sums, we have a natural multiplication on $T^{\circ}(V)$. Because tensor products are associative, this multiplication is associative. Because we have a natural isomorphism $k\otimes_k V \cong V$, the multiplication is unital. Further more, because tensor products (over $k$) are $k$-linear, all the maps in sight are $k$-linear. Thus, $T^{\circ}(V)$ is a noncommutative, associative, unital $k$-algebra.

There is a grading on $T^{\circ}(V)$ where $V^{\otimes n}$ has degree $n$, and so $T^{\circ}(V)$ is graded. If we quotient out by a homogeneous ideal, the result will still be a noncommutative, graded, associative, unital $k$-algebra. So let's quotient out by the (two sided) ideal generated by all elements of the form $v\otimes v$ where $v\in V$. We call the quotient ring the exterior algebra of $V$ and denote it $\bigwedge^{\circ}(V)$. You can actually define the product to be the wedge product, and you get all the properties (except skew symmetry) from the properties of $T^{\circ}(V)$. Skew symmetry comes from the defining relations in the ideal we quotiented by.

So we can get all the basic properties of the wedge product without passing to combinatorics. Of course, you still have to link this definition of wedge product with whatever other definition you are using, and since I don't know your definition, I can't really say if there will be any combinatorics involved in that.

Unfortunately, we have replaced combinatorics (which, while messy, is elementary) with tensor products and graded rings, which are slightly higher level. In particular, this is probably not an appropriate definition for an undergraduate class in calculus on manifolds, at least one that doesn't presuppose some algebra.

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This is how John Lee does it in "Introduction to Smooth Manifolds", the book I learned from, and I think this method is very transparent. +1 –  Jeff Jun 10 '11 at 3:34
@Jeff: That makes me wish I had read the book more carefully, or at least used it as a primary reference. It is quite a wonderful book, and the few times I used it as a reference, it greatly clarified things for me. Though on the subject of great books, I also recommend the book "Differential Topology" by Guillemin and Pollack –  Aaron Jun 10 '11 at 3:40
Very elegant approach. Thank you! –  ItsNotObvious Jun 10 '11 at 11:37