0 (Warmup). Let \(V\) be a real vector space of dimension \(n\), \(\alpha _1,\alpha _2\ldots \alpha _k,\beta _j\in \cM ^1(V)\) and \(\lambda ,\mu \in \R \). Show that \(\alpha _1\alpha _2\cdots (\lambda \alpha _j+\mu \beta _j)\alpha _{j+1}\cdots \alpha _k=\lambda \,\alpha _1\alpha _2\cdots \alpha _j\alpha _{j+1}\cdots \alpha _k+\mu \, \alpha _1\alpha _2\cdots \beta _j\alpha _{j+1}\cdots \alpha _k\) in \(\cM ^k(V)\) and deduce that

\begin{multline*} \alpha _1\wedge \alpha _2\wedge \cdots \wedge (\lambda \alpha _j+\mu \beta _j)\wedge \alpha _{j+1}\wedge \cdots \wedge \alpha _k\\ =\lambda \,\alpha _1\wedge \alpha _2\wedge \cdots \wedge \alpha _j\wedge \alpha _{j+1}\wedge \cdots \wedge \alpha _k+\mu \, \alpha _1\wedge \alpha _2\wedge \cdots \wedge \beta _j\wedge \alpha _{j+1}\wedge \cdots \wedge \alpha _k. \end{multline*}

[Solution: Evaluating the left hand side of the first identity on \(v_1,v_2,\ldots v_k\in V\) (using the definition) gives

\[ \alpha _1(v_1)\alpha _2(v_2)\cdots (\lambda \alpha _j+\mu \beta _j)(v_j)\alpha _{j+1}(v_{j+1})\cdots \alpha _k(v_k)\in \R \]

Now \((\lambda \alpha _j+\mu \beta _j)(v_j)=\lambda \,\alpha _j(v_j)+\mu \,\beta _j(v_j)\) (pointwise operations) so the result follows by the distributive law. To obtain the second identity, apply \(\alt \) to both sides of the first identity, and use that \(\alt \colon \cM ^k(V)\to \Alt ^k(V)\) is linear: \(\alt (\lambda \alpha +\mu \beta )=\sum _{\sigma \in S_k}\sgn (\sigma )\sigma \cdot (\lambda \alpha +\mu \beta )\) and it is easy to see that \(\sigma \cdot (\lambda \alpha +\mu \beta )=\lambda \, \sigma \cdot \alpha +\mu \,\sigma \cdot \beta \) (evaluate both sides on \(v_1,v_2,\ldots v_k\)).]

1. Let \(V\) be a real vector space of dimension \(n\).

• (i) Let \(v_1, \ldots , v_k \in V\) and \(\alpha _1, \ldots , \alpha _k \in V^*\). Let \(A \in M_{k,k}(\R )\) be the matrix with \(A_{ij} = \alpha _i(v_j)\). Show that \((\alpha _1 \wedge \cdots \wedge \alpha _k)(v_1, \ldots , v_k) = \det A . \)

  [Hint: Compare the result of evaluating the full antisymmetrisation of \(\alpha _1\cdots \alpha _k \in \mathcal {M}^k(V)\) on \(v_1, \ldots , v_k\) with the sum formula for the determinant of a matrix.]

• (ii) Let \(\phi : V \to V\) be a linear operator. Show that for any \(\alpha \in \Alt ^n(V)\),

  \[ \phi ^* \alpha = (\det \phi ) \alpha \in \Alt ^n (V) \]

  [Hint: Recall that \(\mathrm {det} \, \phi \) means the determinant of the matrix \(A\) representing \(\phi \) with respect to any basis \(e_1,e_2,\ldots e_n\) of \(V\). Use the dual basis \(\epsilon _1,\epsilon _2,\ldots \epsilon _n\) to write the entries \(A_{ij}\) in the form of part (i), and note that \(\epsilon _1\wedge \epsilon _2\wedge \ldots \wedge \epsilon _n\) is a basis for \(\Alt ^n(V)\).]

2. Let \(e_1, \ldots , e_5 \in \R ^5\) be the standard basis, and let \(\epsilon _1, \ldots , \epsilon _5 \in (\R ^5)^*\) be the dual basis. Let

\[ \alpha = 3\epsilon _1 \wedge \epsilon _3 + \epsilon _2 \wedge (7\epsilon _3 - 2\epsilon _5) \in \Alt ^2(\R ^5) . \]

• (i) Evaluate \(\alpha (e_1 + 2e_3, e_3 + e_4) \in \R \).

• (ii) Express \(\alpha \wedge (2\epsilon _1 + \epsilon _2 -3 \epsilon _4) \in \Alt ^3 (\R ^5)\) in terms of the standard basis \(\epsilon _1 {\wedge } \epsilon _2 {\wedge } \epsilon _3\), \(\epsilon _1 {\wedge } \epsilon _2 {\wedge } \epsilon _4\), …, \(\epsilon _3 {\wedge } \epsilon _4 {\wedge } \epsilon _5\).

[Hint: When expanding, bear in mind that \(\epsilon _i \wedge \epsilon _j = -\epsilon _j \wedge \epsilon _i\), and \(\epsilon _i \wedge \epsilon _i = 0\).]

3 (Less essential). For a real inner product space \(V\) and \(v \in V\), define \(v^\flat \in V^*\) to be the linear map \({V \to \R ,} \; {w \mapsto v.w}\). For \(\alpha \in \Alt ^{k+1}(V)\) and \(v \in V\), define the ‘contraction’ \(v \inner \alpha \in \Alt ^k(V)\) by

\[ (v \inner \alpha )(w_1, \ldots , w_k) = \alpha (v, w_1, \ldots , w_k) \]

for all \(w_1, \ldots , w_k \in V\). Show that the cross product on \(\R ^3\) is related to the wedge product on \((\R ^3)^*\) by

\[ (u \times v) \inner {\Det } = u^\flat \wedge v^\flat \in \Alt ^2 (\R ^3) \]

for any \(u, v \in \R ^3\).

[Hint: Express \(u\) and \(v\) in terms of the standard basis, i.e., write \(u = u_1 e_1 + u_2 e_2 + u_3 e_3\) etc., and find the components of both sides of the equation with respect to the standard basis \(\epsilon _1 \wedge \epsilon _2, \epsilon _1 \wedge \epsilon _3, \epsilon _2 \wedge \epsilon _3\) of \(\mathrm {Alt}^2\mathbb {R}^3\).]

4. Let \(\phi : V \to W\) be a linear map between real vector spaces. Show that

• (i) \(\phi ^* \alt (\alpha ) = \alt (\phi ^*\alpha ) \in \Alt ^k(V)\) for any \(\alpha \in \mathcal {M}^k(W)\).

  [Hint: First check that \(\sigma \cdot (\phi ^*\alpha ) = \phi ^*(\sigma \cdot \alpha ) \in \mathcal {M}^k(V) \) for any \(\alpha \in \mathcal {M}^k(W)\) and \(\sigma \in S_k\).]

• (ii) \(\phi ^*(\alpha _1\wedge \cdots \wedge \alpha _k) = (\phi ^*\alpha _1)\wedge \cdots \wedge (\phi ^*\alpha _k)\in \Alt ^k(V)\) for any \(\alpha _1,\ldots , \alpha _k \in \Alt ^1(W)\).

  [Hint: First show \(\phi ^*(\alpha _1\cdots \alpha _k)=(\phi ^*\alpha _1)\cdots (\phi ^*\alpha _k)\) by evaluating both sides on \(v_1,\ldots v_k\).]

5. Let \(V\) and \(W\) be vector spaces with bases \(v_1, v_2, v_3\) and \(w_1, w_2, w_3, w_4\) respectively. Let \(\phi : V \to W\) be the linear map represented with respect to these bases by

\[ \pmat {2 & 0 & -3 \\ 1 & 6 & 0 \\ 0 & 1 & -1 \\ 1 & 0 & 5} \]

Let \(\epsilon _1, \epsilon _2, \epsilon _3 \in V^*\) and \(\delta _1, \delta _2, \delta _3, \delta _4 \in W^*\) denote the dual bases to the given bases. Compute

\[ \phi ^*(3\delta _1 \wedge \delta _3 + \delta _2 \wedge \delta _4) \in \Alt ^2(V) \]

in terms of the standard basis \(\epsilon _i\wedge \epsilon _j:i<j\) for \(\Alt ^2(V)\).

[Hint: What matrix represents \(\phi ^* : W^* \to V^*\) with respect to the bases \(\delta _i\) and \(\epsilon _j\)? There is a reason why the dual map \(\phi ^*\) is sometimes called the "transpose" of \(\phi \)! You should find, for instance, that \(\phi ^*\delta _1 = 2\epsilon _1 - 3\epsilon _3\).]