Isomorphisms between $\mathbb{R}^3$ and $\left( \mathbb{R}^3\right)^*$

Question

Consider the oriented euclidean space $\mathbb{R}^3$. Every vector $A \in \mathbb{R}^3$ determines a $1$-form $ω^1_A$ by $ω^1_A(ξ)=(A,ξ), ξ\in \mathbb{R}^3$ (scalar product) and a $2$-form $ω^2_A$ by \begin{equation} ω^2_A(A,ξ_1,ξ_2)=(A,ξ_1,ξ_2) \text{*triple scalar product*} \end{equation} Show that the maps $A \to ω^1_A$ and $A \to ω^2_A$ establish isomoprhisms of the linear space $\mathbb{R}^3$ of vectors $A$ with the linear spaces of $1$-forms on $\mathbb{R}^3$ and the $2$-forms on $\mathbb{R}^3$. Then choose an orthonormal oriented coordinate system $(x_1,x_2,x_3)$ on $\mathbb{R}^3$ and show that: \begin{align} & ω^1_Α=A_1 x_1+A_2 x_2+A_3 x_3 \\ & ω^2_A=A_1 x_2\wedge x_3+A_2 x_1\wedge x_3+A_3 x_1\wedge x_2 \end{align}

I do understand the last two relations, since it can be proved that the set of linear coordinates $\{x_1,...,x_n\}=B$ is a basis for $\left( \mathbb{R}^n \right)^*$, therefore every $1$-form can be written as a linear combination of the elements of the basis, and moreover, I have already proved that every $2$-form can be written as a linear combination of the exterior product of elements of $B$, therefore like a linear combination of $1$-forms , exactly like $ω^2=\sum_{i<j}^n α_{ij}x_i \wedge x_j$.

I am not able to prove the first part, concerning the isomorphisms, plus I cant really see why does it have to be an orthonormal coordinate system. I am guessing just orthogonality will do fine as far as linear independece is concerned.

How can I prove the first part? Thanks.

Answer 1

There is no need for orientations or bases to show that the two given (linear!) maps $$\nu_1:\ A\ \longmapsto\ \omega_A^1\qquad\text{ and }\qquad \nu_2:\ A\ \longmapsto\ \omega_A^2,$$ are isomorphisms. The spaces of $1$-forms and $2$-forms are of dimensions $\tbinom{3}{1}=3$ and $\tbinom{3}{2}=3$, respectively. So to show that $\nu_1$ and $\nu_2$ are isomorphisms, it suffices to show that they are injective.

1. If $A\in\ker\nu_1$ then $\omega_A^1=0$, so in particular $$\langle A,A\rangle=\omega_A^1(A)=0,$$ which shows that $A=0$ and hence $\nu_1$ is injective.

2. If $A\in\ker\nu_2$ then $\omega_A^2=0$, so for all $\xi_1,\xi_2\in\Bbb{R}^2$ $$\det(A,\xi_1,\xi_2)=\omega_A^2(\xi_1,\xi_2)=0,$$ which shows that $A=0$ and hence $\nu_2$ is injective.

For the relations to hold you don't need $\Bbb{R}^3$ to be oriented and you don't need $(x_1,x_2,x_3)$ to be an orthonormal basis for $\Bbb{R}^3$. What you do need is for $(x_1,x_2,x_3)$ to be the dual basis to whatever basis $(e_1,e_2,e_3)$ you choose for $\Bbb{R}^n$: If for all $A\in\Bbb{R}^3$ you have $$\omega_A^1=A_1x_1+A_2x_2+A_3x_3,$$ then in particular for $A=e_i$ you get $\omega_{e_i}^1=x_i$ where $\omega_{e_i}^1(\xi)=\langle e_i,\xi\rangle$ for all $\xi\in\Bbb{R}^3$. This means precisely that $(x_1,x_2,x_3)$ is the dual basis to $(e_1,e_2,e_3)$.

When $(x_1,x_2,x_3)$ is the dual basis to $(e_1,e_2,e_3)$ then verifying the two relations is a matter of checking that they hold on any basis for $\Bbb{R}^3$. This is easiest for $(e_1,e_2,e_3)$; for $A=e_1$ we get \begin{eqnarray*} A_1x_1(\xi)+A_2x_2(\xi)+A_3x_3(\xi)=x_1(\xi) &=&\langle e_1,\xi\rangle=\omega_{e_1}^1(\xi).\\ &\ &\\ A_1x_2\wedge x_3(\xi_1,\xi_2)+A_2x_1\wedge x_3(\xi_1,\xi_2)+A_3x_1\wedge x_2(\xi_1,\xi_2) &=&x_2\wedge x_3(\xi_1,\xi_2)\\ &=&\begin{vmatrix}x_2(\xi_1)&x_3(\xi_1)\\ x_2(\xi_2)&x_3(\xi_2)\end{vmatrix}\\ &=&\begin{vmatrix} 1&0&0\\ x_1(\xi_1)&x_2(\xi_1)&x_3(\xi_1)\\ x_2(\xi_2)&x_2(\xi_2)&x_3(\xi_2) \end{vmatrix}\\ &=&\omega_{e_1}^2(\xi_1,\xi_2), \end{eqnarray*} for all $\xi,\xi_1,\xi_2\in\Bbb{R}^3$, and the same argument works for $e_2$ and $e_3$, of course.