^^ [[Schutz - 2 - Differentiable Manifolds and Tensors|2. Differentiable Manifolds and Tensors]] << [[Schutz - 2.19 - The Gradient and the Pictorial Representation of a One-Form|2.19 The Gradient and the Pictorial Representation of a One-Form]]| [[Schutz - 2.21 - Index Notation|2.21 Index Notation]] >> # Basis One-Forms and Components of One-Forms In the [[Schutz - 2.16 - One-Forms#^0fa8ac|cotangent space]] at a point $P \in M, T_P^*$, where $\text{dim} M = n$, any $n$ linearly independent one-forms constitute a basis. However, once a basis $\{\pmb{e}_i\}$ is chosen for $T_P$, this induces a preferred basis for $T_P^*$. > **The Dual Basis** > Pick a basis for $T_P$, $\{\pmb{e}_i\}$. For $\pmb{V} \in T_P$, the *dual basis* is defined as: > $\tilde{\omega}^i(\pmb{V}) = V^i$ > This is a linear function in the argument $\pmb{V}$. Since $\pmb{e}_j$ has only the $j^th$ component non-zero, then: > $\tilde{\omega}^i(\pmb{e}_j) = \delta^i_{\,j}$ > This constitutes the definition of $\tilde{\omega}^i$. One must know *all* the basis vectors $\{\pmb{e}_j\}$ to define any of the $\{\tilde{\omega}^i\}$, any change in any of the basis vectors changes the entire dual basis. To show that the $\{\tilde{\omega}^i\}$ do form a basis, consider an arbitrary $\tilde{q} \in T^*_P$ acting on an arbitrary $\pmb{V} \in T_P$: $ \begin{align} \tilde{q}(\pmb{V}) &= \tilde{q}\left(\sum_j V^j \pmb{e}_j\right),\\ &= \sum_j \tilde{q}(V^j \pmb{e_j}),\\ &= \sum_j V^j\tilde{q}(\pmb{e_j}),\\ &= \sum_j \tilde{\omega}^j(\pmb{V})\tilde{q}(\pmb{e_j}),\\ &= \sum_j q_j \tilde{\omega}^j(\pmb{V}),\\ \end{align} $ where we have defined $q_j = \tilde{q}(\pmb{e}_j)$. These are known as the components of $q$ on the basis dual to $\{\pmb{e}_j\}$. Since $\pmb{V}$ is arbitrary and a one-form is defined by its values on vectors, it follows that: ^a65b2c $ \tilde{q} = \sum_j q_j \tilde{\omega}^j $ And thus the $\{\tilde{\omega}^j\}$ is indeed a basis (there's only $n$ of them, and an arbitrary $\tilde{q} \in T_P^*$ is a linear combination of them). This also shows that the $\{q_j\}$ are the components of $\tilde{q}$ in the dual basis in the ordinary sense. Thus, if we know the components of $\tilde{q}$,$\{q_j\}$, and the components of $\pmb{V}$, $V^j$, then we can write the action of the one-form on the vector as: $ \tilde{q}(\pmb{V}) = \sum_j V^j q_j, $ which follows from the derivation above. One can see that the notion of [[Schutz - 2.19 - The Gradient and the Pictorial Representation of a One-Form#^89c474|differentiability of one-form fields]] can be defined in terms of components as well. That is, $\tilde{q}$ is a differentiable $C^\infty$ one-form field if and only if its components $\{q^i\}$ associated with a $C^\infty$ basis for vector fields are $C^\infty$ [[Schutz - 2.8 - Basis Vectors and Basis Vector Fields#^af0f66|differentiable]] functions. ^d5f715 This extends to [[Schutz - 2.19 - The Gradient and the Pictorial Representation of a One-Form#^8740b5|one-form fields]]. If the set of vector fields $\{\pmb{e}_i\}$ is a basis at every point in some region $U$ of $M$, then the fields $\{\tilde{\omega}^j\}$ is likewise a set of one-form fields that defines a basis at all points of $U$. A coordinate system on U, $\{x^\}$ defines a natural basis for vector fields $\{\partial/\partial x^i\}$. It also defines a natural set of $n$ one-forms, the gradients $\{\tilde{d}x^i\}$. These one-forms are in fact a basis dual to the coordinate basis vectors: $ \tilde{x}x^i(\partial/\partial x^i) \equiv \partial x^i /\partial x^i = \delta^i_{\,j} $ Thus, the two bases defined by the coordinates $\{x^i\}$, $\{\partial/\partial x^i\}$ and $\{\tilde{d}x^i\}$ are dual to one another.