Math 593 Assignment # 10 Solutions

Bryden R. Cais

December 12, 2002

1. Let V be a vector space of dimension n over a field k.

(a) Define

ϕ : V^{∗} × ⋀^{n} V ⟶ ⋀^{n−1} V

by

ϕ(f, v_{1} ∧ ⋯ ∧ v_{n}) = ∑_{i=1}^{n}(−1)^{i}f(v_{i})v_{1} ∧ ⋯ ∧ v̂_{i} ∧ ⋯ ∧ v_{n},

where v̂_{i} denotes the omission of v_{i} from the wedge product. We first check that ϕ is well defined. For this, it is enough to show that if x ∈ ⋀^{n} V is 0 then ϕ(f, x) = 0 for any f ∈ V^{∗}. We restrict ourselves to the case that x = v_{1} ∧ ⋯ ∧ v_{n} and v_{1} = v_{j} for 1 < j ≤ n. The general case is not much more difficult. We may permute v_{j} so that it is adjacent to v_{1}, and we only change the sign of our wedge product. Thus, without loss of generality, suppose that j = 2. Then

ϕ(f, v_{1} ∧ v_{2} ∧ ⋯ ∧ v_{n}) = f(v_{1})v_{2} ∧ ⋯ ∧ v_{n} − f(v_{2})v_{1} ∧ v_{3} ∧ ⋯ ∧ v_{n}
= f(v_{1})v_{2} ∧ ⋯ ∧ v_{n} − f(v_{1})v_{2} ∧ ⋯ ∧ v_{n}
= 0,

where we have used the fact that v_{1} ∧ ⋯ ∧ v̂_{i} ∧ ⋯ ∧ v_{n} = 0 for i ≠ 1, 2. It is straightforward to verify that ϕ is bilinear, so that we have a well defined bilinear map ϕ. This induces a linear map

φ : V^{∗} ⊗ ⋀^{n} V ⟶ ⋀^{n−1} V.

We claim that φ is surjective. Let e_{i} be a basis for V. Then it suffices to show that the basis elements

w_{i} = e_{1} ∧ ⋯ ∧ ê_{i} ∧ ⋯ ∧ e_{n}

of ⋀^{n−1} V are in the image of φ. Define f_{i} ∈ V^{∗} by

f_{i}(e_{j}) = \begin{cases}1 & if i = j \\ 0 & otherwise \end{cases} .

Then we have

φ(f_{i} ⊗ e_{1} ∧ ⋯ ∧ e_{n}) = (−1)^{i}w_{i}.

It follows that φ is surjective. Since ⋀^{n} V is one dimensional, and V^{∗} is (non-canonically) isomorphic to V, and since ⋀^{n−1} V is n dimensional, it follows that

dim(V^{∗} ⊗ ⋀^{n} V) = dim( ⋀^{n−1} V),

so that since φ is surjective, it must also be injective. Hence φ is an isomorphism, as desired.

(b) Suppose k < n and let ω ∈ ⋀^{k} V. Suppose that ω ≠ 0. Pick a basis e_{1}, …, e_{n} of V and let

ω = ∑_{i_{1}<...<i_{k}} a\overline{i}e_{i_{1}} ∧ ⋯ ∧ e_{i_{k}}.

Since ω is nonzero, we must have a\overline{i} ≠ 0 for some \overline{i} = (i_{1}, i_{2}, …, i_{k}). Let

η = ∧_{l∉{i_{1},i_{2},...i_{k}}}e_{l}.

Then we have

η ∧ ω = a\overline{i} ≠ 0,

so that, in particular,

ω ∧ e_{l} ≠ 0

for any l ∉ {i_{1}, …, i_{k}}. It follows that if ω ∧ v = 0 ∈ ⋀^{k+1} V for all v ∈ V then ω = 0.

2. (a) Define

S^{k}ρ : G ⟶ GL(S^{k}V)

and

⋀^{k} ρ : G ⟶ GL(⋀^{k} V)

by

S^{k}ρ(g)(v_{1} ⊗ ⋯ ⊗ v_{k}) = ρ(g)v_{1} ⊗ ⋯ ⊗ ρ(g)v_{k}

and

⋀^{k} ρ(g)(v_{1} ∧ ⋯ ∧ v_{k}) = ρ(g)v_{1} ∧ ⋯ ∧ ρ(g)v_{k},

respectively, and extend by linearity to all of S^{k}V, ⋀^{k} V. It is straightforward to verify that these maps are well defined (i.e. that they respect the nontrivial identities in S^{k}V and ⋀^{k} V). We show that S^{k}ρ is a group homomorphism. Indeed, we have

S^{k}ρ(gh)v_{1} ∧ ⋯ ∧ v_{n} = ρ(gh)v_{1} ⊗ ⋯ ⊗ ρ(gh)v_{k}
= ρ(g)ρ(h)v_{1} ⊗ ⋯ ⊗ ρ(g)ρ(h)v_{k}
= S^{k}ρ(g)(ρ(h)v_{1} ⊗ ⋯ ⊗ ρ(h)v_{k})
= S^{k}ρ(g) ∘ S^{k}ρ(h)(v_{1} ⊗ ⋯ ⊗ v_{k}),

since ρ is a group homomorphism. The same verification works identically for ⋀^{k} ρ. Observe that

S^{k} id = ⋀^{k} id = id .

Thus we see that the image of S^{k}ρ is indeed in GL(S^{k}V) since

S^{k}ρ(g^{−1}) ∘ S^{k}ρ(g) = id .

Similarly for ⋀^{k} ρ. It follows that S^{k}ρ, ⋀^{k} ρ are representations.

(b) Let dim V = n and realize ρ(g) ∈ GL(V) by n × n matrices. We have seen that

⋀^{n} ρ(g) = det(ρ(g)),

so that the one dimensional representation on ⋀^{n} V ≃ k given by ⋀ ρ is multiplication by the determinant of ρ(g) (e.g. by the 1 × 1 matrix det(ρ(g))).

(c) Let e_{1}, e_{2} be the standard basis of ℂ^{2} as a ℂ vector space and let

s = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ GL_{2}(ℂ)

. Then we easily have

s(e_{1}) = ae_{1} + ce_{2}
s(e_{2}) = be_{1} + de_{2}.

It follows that

S^{2} id(s)(e_{1} ⊗ e_{1}) = (ae_{1} + ce_{2}) ⊗ (ae_{1} + ce_{2})  = a^{2}e_{1} ⊗ e_{1} + 2ace_{1} ⊗ e_{2} + c^{2}e_{2} ⊗ e_{2}
S^{2} id(s)(e_{1} ⊗ e_{2}) = (ae_{1} + ce_{2}) ⊗ (be_{1} + de_{2})= abe_{1} ⊗ e_{1} + (ad + bc)e_{1} ⊗ e_{2} + cde_{2} ⊗ e_{2}
S^{2} id(s)(e_{2} ⊗ e_{2}) = (be_{1} + de_{2}) ⊗ (be_{1} + de_{2})  = b^{2}e_{1} ⊗ e_{1} + 2bde_{1} ⊗ e_{2} + d^{2}e_{2} ⊗ e_{2}.

Thus, in the basis e_{1} ⊗ e_{1}, e_{1} ⊗ e_{2}, e_{2} ⊗ e_{2} of S^{2}ℂ^{2} the matrix of S^{2} id(s) is given by

(\begin{matrix}a^{2} & ab & b^{2} \\ 2ac & ad + bc & 2bd \\ c^{2} & cd & d^{2} \end{matrix}) .

3. (a) Recall that

ker g = {v ∈ V : g(x, v) = 0 for all x ∈ V

while

ker L_{g} = {v ∈ V : L_{g}(v) ≡ 0} = {v ∈ V : L_{g}(v)(x) = g(x, v) = 0 for all x ∈ V.

Thus it is easy to see that ker L_{g} = ker g. Since dim V = n it follows from the rank nullity theorem that

rank(L_{g}) + dim ker L_{g} = rank(g) + dim ker g = n,

as required. Moreover, we know that g is nondegenerate if and only if ker g = {0}, that is, iff dim ker g = 0, if and only if rank(g) = n.

(b) It suffices to show that there exist symmetric n × n matrices of any rank 0 ≤ r ≤ n. This is clear,

since

M = (\begin{matrix}I_{r} \\ & 0_{n−r} \end{matrix})

clearly has rank r and is symmetric. Here I_{r} is the square r × r identity matrix and 0_{n−r} is the square matrix of size n − r consisting of all zeroes. Now let g be an alternating form on V and set N = ker g. Then h = g|W is nondegenerate and skew, where

W = V/N

and rank(h) = rank(g) = dim W.

Now h^{T} = −h so that

det(h) = det(h^{T}) = det(−h) = (−1)^{dim W} det(h)

from which it follows that dim W = rank g is even since det h ≠ 0. As above, it is enough to show that there exist skew symmetric n× matrices of any given even rank. Let

M = (\begin{matrix}T \\ & T \\ & & . & . & . \\ & & & & & 0_{n−2r} \end{matrix})

where

T = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix})

and there are r T ’s along the diagonal. This matrix clearly has rank 2r and is skew.

(c) Let W ⊂ V be a codimension 1 subspace and suppose that g is a nondegenerate symmetric form

on V. Then

L_{g} : V → V^{∗}

is an invertible matrix. Moreover, since dim W = n − 1 we see that

dim(L_{g}(W)) = n − 1

where L_{g}(W) ⊂ V^{∗}. The rank of g|W is the rank of L_{g|W}, which is the dimension of

L_{g}(W) ∩ W^{∗}.

Since W^{∗} also has dimension n − 1 and V has dimension n, the intersection L_{g}(W) ∩ W^{∗} has dimension at least n − 2. Thus, the possibilities for the rank of g|W are

n − 1, n − 2.

It is not difficult to find examples where each possibility occurs. Indeed, The rank(g_{W}) = n − 1 if and only if L_{g}(W) = W^{∗}.

(d) Set

g(v, w) = ∑_{i=1}^{n} v_{i}w_{n−i+1}

and let W = {v ∈ V : v_{i} = 0 for 1 ≤ i ≤ m}. Then it is easy to see that g is symmetric and g(w_{1}, w_{2}) = 0 for any w_{1}, w_{2} ∈ W but that for any w ≠ 0 ∈ W there exists v ∈ V \ W such that g(w, v) ≠ 0. It follows that g has rank m. Similarly, the nondegenerate skew symmetric form

g(v, w) =∼_{i=1}^{m} (v_{i}w_{n−i+1} − v_{n−i+1}w_{i})

has the m dimensional isotropic subspace

W = {v ∈ V : v_{i} = 0 for 1 ≤ i ≤ m}

. Finally, it is not difficult to see that the usual dot product on ℝ^{n} has no isotropic subspaces other than {0}. This follows from the fact that the only solution to

∑_{i=1}^{n} v_{i}^{2}

over ℝ is v = 0.