Problem 1: Let V be a vector space, let ω1 , ω2 , ω3 ∈ Alt1 V , and let v1 , v2 , v3 ∈ V . Using only
the definition of ∧, work out an explicit formula for (ω1 ∧ ω2 ∧ ω3 )(v1 , v2 , v3 ).
Solution: Brute force computation. Use
P that for a permutation τ of {1, 2, 3}, sign(τ ) = +1 if
and only if τ is cyclic. The answer is τ ∈S3 sign(τ )ω1 (vτ (1) )ω2 (vτ (2) )ω3 (vτ (3) ).
Problem 2: Let {ωi }i=1,...,4 with ωi (v) = v · ei denote the standard basis of Alt1 R4 . Define
two subspaces Alt2+ R4 , Alt2− R4 ⊂ Alt2 R4 as follows:
Alt2+ R4 = span(ω1 ∧ ω2 + ω3 ∧ ω4 , ω1 ∧ ω3 − ω2 ∧ ω4 , ω1 ∧ ω4 + ω2 ∧ ω3 ),
Alt2− R4 = span(ω1 ∧ ω2 − ω3 ∧ ω4 , ω1 ∧ ω3 + ω2 ∧ ω4 , ω1 ∧ ω4 − ω2 ∧ ω3 ).
(a) Prove that Alt2 R4 = Alt2+ R4 ⊕ Alt2− R4 .
(b) Prove that if we write ω ∈ Alt2 R4 as
combination of the 6 basis forms above, with
Pa3 linear
1
2
3
i 2
coefficients λ± , λ± , λ± , then ω ∧ ω = 2( i=1 [(λ+ ) − (λi− )2 ])ω1 ∧ ω2 ∧ ω3 ∧ ω4 .
Solution: (a) We know that ωi ∧ ωj (1 6 i < j 6 4) form a basis of Alt2 R4 . But clearly each
of these 6 forms can be written as a linear combination of the 6 forms given in the problem,
e.g. ω1 ∧ ω2 = 21 (ω1 ∧ ω2 + ω3 ∧ ω4 ) + 12 (ω1 ∧ ω2 − ω3 ∧ ω4 ). Thus, the 6 forms given in the
problem form a basis of Alt2 R4 as well, and the claim is obvious from this.
(b) It suffices to check that if ω is one of the given basis forms of Alt2± R4 , then ω ∧ ω = ±2
(ω1 ∧ ω2 ∧ ω3 ∧ ω4 ), and if η 6= ω is another one of these basis forms, then ω ∧ η = 0. For this,
write down one or two examples and claim that the rest follows by symmetry.
Problem 3 (Exercises 8.3 and 8.4 in the book): Let v : Rn → Rn be differentiable. We
can view v as a vector field by ‘attaching’ the vector v(x) to the point x, for every x ∈ Rn . Let
ω = dx1 ∧ . . . ∧ dxn ∈ Ωn Rn . Then clearly d(v y ω) = f ω for some function f : Rn → R. Find
f in terms of v. Hint: See Problem 3 on Homework 4 and the solution I posted.
P
. . . ∧ dxı̂ ∧ . . . ∧ dxn ),
Solution: From the solution to HW4, #3, v y ω = ni=1 (−1)i−1 v i (dx1 ∧P
∂g
j
where v(x) = (v 1 (x), . . . , v n (x)). Thus, applying d and using that dg = nj=1 ∂x
j dx ,
!
n
n
X
X
∂v i j
i−1
d(v y ω) =
(−1)
dx ∧ dx1 ∧ . . . ∧ dxı̂ ∧ . . . ∧ dxn
j
∂x
i=1
j=1
=
n
X
i=1
=
∂v i
∧ dxi ∧ dx1 ∧ . . . ∧ dxı̂ ∧ . . . ∧ dxn
i
∂x
!
(−1)i−1
n
X
∂v i
i=1
∂xi
(dx1 ∧ . . . ∧ dxn ) = f ω.
Bonus: Show that the linear map associated with a symmetric 3 × 3 matrix is the orthogonal
projection map onto some line through the origin in R3 if and
ifthe matrix entries satisfy
a only
11 a12 a13 a21
a11 a12
a11 a13
a22 a23
22 a23 = 0.
a11 + a22 + a33 = 1, | a21 a22 | + | a31 a33 | + | a32 a33 | = 0, and a31 aa32
a33
Solution: Any symmetric matrix A has an orthonormal basis of eigenvectors. Then clearly
the associated linear map is the orthogonal projection map onto some line through the origin if
and only if A has eigenvalues 0, 0, 1, or equivalently if det(tI − A) = t2 (t − 1) = t3 − t2 . On the
other hand, we can calculate det(tI − A) explicitly in terms of the entries of A; cf. the solution
to HW4, #4, given in class. Comparing coefficients yields the result.