The matrix specify an algernating $k$-tensor on $V$, and dim$\bigwedge^k(V^*)=1$

Question

The end of the hint "The matrix specify an algernating $k$-tensor on $V$, and dim$\bigwedge^k(V^*)=1$" does not make sense to me.

In my not very assured understanding, the $k$-tensor $\bigwedge^k(V^*)$ eats $k$ vectors and gives a real number, like the determinant operator does, therefor the dimension is 1. Is this correct? Then how can I proceed?

Also, I am award of this theorem but not sure if {$\phi_1, \ldots, \phi_k$} is a basis.

P.S. Definition of Tensor Product Thank you~

Answer 1

Let us consider a finite dimensional vector space $V$ over a field $\mathbb K$ of characteristic zero (we can take $\mathbb K=\mathbb R$ or $\mathbb C$, for example). Let $k$ be the dimension of $V$: when dealing with vectors in $V^{\otimes k}$ or ${V^{*}}^{\otimes k}$ we refer to $k$ as the "weight" of the tensor. This terminology is standard in homological algebra.

To reconstruct the result you want to prove, we need to consider a couple of maps. The first one is the alternating map

$$Alt^{k}_{*}: \wedge^{k}V^{*}\rightarrow Alt^k(V^{*})\subset V^{*}\otimes\dots\otimes V^{*},$$

with

$$Alt^{k}_{*}(\omega_1\wedge\dots\wedge \omega_k):=\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi} \omega_{\pi(1)}\otimes\dots\otimes\omega_{\pi(k)},$$

denoting by $Alt^{k}(V^{*}):=Im(Alt^{k}_{*})$ the linear subspace of $V^{*}\otimes\dots\otimes V^{*}$ ($k$-times) consisting of all "alternating" tensors of weight $k$. $Alt^{k}_{*}$ is an isomorphism.

The second map we need to consider is

$$\varphi:(V^{*}\otimes\dots\otimes V^{*})\otimes(V\otimes\dots\otimes V)\rightarrow\mathbb K,$$

where

$$\varphi(\omega_{1}\otimes\dots\otimes\omega_{k}\otimes v_{1}\otimes\dots\otimes v_{k}):=\prod_{r=1}^k\omega_r(v_r).$$

$\varphi$ is the (I d not discuss unicity here) multilinear extension of the duality $V^{*}\otimes V\rightarrow \mathbb K$ to the components of weight $k$ of the tensor algebras

$$ T(V^{*})=\bigoplus_{k\geq 0} {V^{*}}^{\otimes k} $$

$$T(V)=\bigoplus_{k\geq 0} V^{\otimes k} $$

In summary, we want to characterize the map (always focusing on weight $k$ tensors)

$$\varphi\circ(Alt^{k}_{*}\otimes 1): \wedge^{k}V^{*}\otimes (V\otimes\dots\otimes V )\rightarrow \mathbb K.$$

The above map gives us the formula of the "pairing" in your question.

It follows that, choosing a basis $\{e_i\}$ for $V$ and the dual basis $\{\omega_i\}$ for $V^{*}$ the composition

$$\varphi(Alt_{*}(\omega_1\wedge\dots\wedge\omega_k)\otimes(e_1\otimes \dots\otimes e_k))=\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi}\varphi( \omega_{\pi(1)}(e_1)\otimes\dots\otimes\omega_{\pi(k)}(e_k))=\\ =\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi}\prod_{i=1}^k\omega_{\pi(i)}(e_i)=\frac{1}{k!}\det(A), $$

where $A$ is the $k\otimes k$ matrix with entries $A_{ij}:=\omega_{i}(e_j)$.

Considering dependent vectors $\omega_{*}$ results in having a zero determinant.

Answer 2

This is how I understood Avitus' proof. My version is much less involved - not involving $\varphi$ and the basis of $V^*$, $\omega_i$s. Don't feel right about this...

If $T \in \wedge^p(V^*)$ and $S \in \wedge^q(V^*)$, then the definition of their wedge product is $$T \wedge S := \text{Alt}(T \otimes S) \in \wedge^{p+1}(V^*)$$

Meanwhile, we have that $$\text{Alt}(T) = \frac{1}{p!}\sum_{\pi \in S_p}(-1)^{\pi} T^\pi.$$

Hence, for this problem, we have that \begin{eqnarray*} \phi_1 \wedge \cdots \wedge \phi_k (v_1, \cdots, v_k) & =& \text{Alt}(\phi_1 \otimes \cdots \otimes \phi_k (v_1, \cdots, v_k))\\ & = & \frac{1}{k!}\sum_{\pi \in S_k}(-1)^{\pi} T^\pi(v_1, \cdots, v_k)\\ & = &\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi} \phi_{\pi(1)}\otimes\dots\otimes\phi_{\pi(k)}(v_1, \cdots, v_k)\\ & = &\frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi} \prod_{i=1}^k\phi_{\pi(i)}(v_i). \end{eqnarray*}

Without loss of generality, we choose a basis $\{e_i\}$ for $V$ %and the dual basis $\{\omega_i\}$ for $V^{*}$ . Then \begin{eqnarray*} \phi_1 \wedge \cdots \wedge \phi_k (e_1, \cdots, e_k) & =& \frac{1}{k!}\sum_{\pi\in S_k}(-1)^{\pi} \prod_{i=1}^k\phi_{\pi(i)}(e_i)\\ &=& \frac{1}{k!}\det(A). \end{eqnarray*} where $A$ is the $k\otimes k$ matrix with entries $A_{ij}=\phi_{\pi(i)}(e_i)$.

Considering dependent vectors $\omega_{*}$ results in having a zero determinant.