Exterior algebra of a direct sum

Statement

Let $V$ and $W$ be modules over a commutative ring. There is a canonical isomorphism of graded algebras

\Lambda(V\oplus W)\cong \Lambda V\,\widehat\otimes\,\Lambda W,

where $\widehat\otimes$ is the graded tensor product. Under this isomorphism,

(v,0)\longmapsto v\otimes1, \qquad (0,w)\longmapsto1\otimes w.

The multiplication on the right uses the Koszul sign rule

(a\otimes b)(a'\otimes b') =(-1)^{\deg(b)\deg(a')}(a\wedge a')\otimes(b\wedge b')

for homogeneous elements.

Degree-by-degree form

Taking the homogeneous part of degree $k$ gives a canonical decomposition

\Lambda^k(V\oplus W) \cong \bigoplus_{p+q=k}\Lambda^pV\otimes\Lambda^qW.

Explicitly, the summand indexed by $(p,q)$ maps

(v_1\wedge\cdots\wedge v_p)\otimes (w_1\wedge\cdots\wedge w_q)

to the wedge of the corresponding vectors in $V\oplus W$ , with all $V$ -vectors written before the $W$ -vectors. The graded sign rule makes this prescription independent of how products are regrouped.

Equivariance

If a group or Lie algebra acts on both $V$ and $W$ , the direct sum carries the diagonal action and the displayed isomorphisms are equivariant. In terms of exterior-power representations,

\Lambda^k(V\oplus W) \cong \bigoplus_{p+q=k}\Lambda^pV\otimes\Lambda^qW

is therefore an isomorphism of representations.

Setting $k=\dim V+\dim W$ recovers the determinant identity

\det(V\oplus W)\cong\det(V)\otimes\det(W).

Why the tensor product must be graded

Using the ordinary ungraded tensor-product multiplication would make elements from $V$ commute with elements from $W$ . In $\Lambda(V\oplus W)$ they anticommute in degree one, so the Koszul sign is essential.

References

Nicolas Bourbaki, Algebra I: Chapters 1–3, Springer, 1989, Chapter III. Publisher record.
Christian Kassel, Quantum Groups, Springer, 1995, Chapter XI, §1. Publisher record.