Question

I'm not too comfortable with "why" questions, but one

One good reason for the ubiquity of the exterior algebra construction is that it has nice universality basic properties . As Ben mentioned, the exterior algebra of a (super)vector space which if made precise will uniquely define it):

It is the symmetric algebra on a parity-shifted version of that vector space. The exterior algebra functor is therefore the left adjoint from vector spaces to the functor that forgets the multiplicative structure in a commutative ring, and switches parity(strictly) supercommutative algebras.If R is a commutative ring in supervector spaces, then there is a natural isomorphism between the space of linear maps from V

Direct sums are taken to R[1] and the space of commutative ring homomorphisms from the exterior algebra tensor products of V to Ralgebras.

I can give a very incomplete description of the distinguished nature of fermions

It plays well with base change and bosons in physics. In quantum mechanicsdescent, for each type of particlei.e., there is a complex vector space of single particle states. States are in general considered equivalent if they differ one can make exterior algebra bundles by a nonzero complex scalargluing.If we wish

It takes a line to consider two an odd line plus an even line (or more) particles, the possible states lie used to great effect in a tensor product of Torsten's answer here - note the two (or more) single-particle state spaceslast paragraph comparing exterior with symmetric cases).However, multiple copies of the same particle do not have distinct identities, so when we switch two identical particles,

Bonus properties in the resulting state vector should differ only by multiplication by finite rank case include:

There is a nonzero complex numbermultiplicative determinant subfunctor to invertible objects (read: graded lines).If we write the switching operation as a suitably well-behaved pair of noncolliding trajectories, we describe

It yields a path of states in the tensor product spaceHopf algebra.If space is at least 3 dimensional and simply connected, then all choices of such paths are homotopy equivalent, and any switching operation on

You have a set of identical particles depends only on perfect "Hodge dual" pairing valued in the underlying permutation.

In this determinant (and pretty much any) setting, symmetric and antisymmetric behavior is distinguished as pointed out by the fact that any nontrivial symmetric group has only two possible one-dimensional complex representations. Bosons correspond to the trivial representation - switching does nothing to the stateMarc Nieper-Wißkirchen).Fermions correspond to the sign representation - switching multiplies by -1. The space of states for all finite collections of identical bosonic particles (called the bosonic Fock space) is therefore the symmetric algebra on the single-particle space, and the fermionic Fock space is

Regarding the exterior algebra.

When it comes to pointing out ways the exterior algebra is useful when the symmetric algebra is not, I think most all of the applications you listed revolve around the finite rank properties - in particular, the distinguished nature of the determinant as a canonical one-dimensional tensor. The only one-dimensional symmetric tensor is the trivial one, which carries no information. Any attempt to make things like volumesboth multilinear and nontrivial , cup products, or Hodge stars requires an orientation, which can be viewed as a determinant (see e.g., my answer here).

As Wikipedia mentions, exterior algebras satisfy a universal property: for any linear map from a vector space $V$ to an associative algebra $A$ landing in the square zero subspace, there is a unique algebra homomorphism from $\bigwedge V$ to $A$ making a certain triangular diagram commute. This yields a description of the exterior algebra functor as a left adjoint to a forgetful functor from strictly supercommutative algebras. If 2 is invertible, then it is equivalent to the parity-shifted symmetric algebra functor, but in general, it represents a genuinely different functor. In particular, the fact that determinants exist in characteristic 2 is an indication that the exterior algebra is more important than the shifted symmetric algebra.

I should emphasize that supersymmetry is different from the existence of fermions. In short, fermions are just odd fields that transform a certain way under ordinary spacetime symmetries, but supersymmetry is the specification of additional odd symmetries of spacetime. This is substantially more exotic: theories containing fermionic particles (like electrons) can exist without supersymmetry, and sign considerationsin fact existed happily for about 50 years before supersymmetry was hatched.