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The Exterior Algebra and the Wedge Product

Force antisymmetry onto the tensor product and you get the wedge product. The exterior powers it generates explain determinants, oriented volume, and why dimension n has exactly n+1 graded pieces.

Antisymmetry as a design choice

The wedge product v ^ w is what you get from v tensor w after demanding antisymmetry: v ^ w = -(w ^ v), and therefore v ^ v = 0. Build all such products and you get the exterior algebra of V. It is the quotient of the full tensor algebra by the relations that kill symmetric parts.

Exterior powers and their dimensions

The degree-k piece is the k-th exterior power Lambda^k V, spanned by wedges e_{i1} ^ ... ^ e_{ik} with strictly increasing indices. Its dimension is C(n,k) = n choose k. So Lambda^0 V = scalars, Lambda^1 V = V, and Lambda^n V is one-dimensional. Above degree n everything is zero, because you would have to repeat an index.

For n = dim V = 3, basis e1,e2,e3:
  Lambda^0 V : {1}                         dim C(3,0)=1
  Lambda^1 V : {e1, e2, e3}                 dim C(3,1)=3
  Lambda^2 V : {e1^e2, e1^e3, e2^e3}        dim C(3,2)=3
  Lambda^3 V : {e1^e2^e3}                   dim C(3,3)=1
  Lambda^4 V : {0}                          (must repeat an index)

Total dim of exterior algebra = sum_k C(n,k) = 2^n  (here 1+3+3+1 = 8).

Reorder cost = sign of the permutation:
  e2 ^ e1 ^ e3 = -(e1 ^ e2 ^ e3),   e2 ^ e3 ^ e1 = +(e1 ^ e2 ^ e3).
Exterior powers of a 3-D space; dims are binomial coefficients summing to 2^n.

Why this is the determinant

Lambda^n V is one-dimensional, so a linear map A induces multiplication by a single scalar on it: (Av_1) ^ ... ^ (Av_n) = (det A) * (v_1 ^ ... ^ v_n). That scalar is the determinant. The alternating, multilinear, normalized rules you memorized in Vol I are not axioms pulled from a hat — they are exactly the rules of the top exterior power.