# Definition The dual basis is a [[basis]] $\mathcal{B}^*$ of the [[Dual Space]] $V^*$ corresponding to a basis $\mathcal{B}$ of the [[vector space]] $V$. It is defined via the relation $\tilde{e}^i(\pmb{e}_j) = \delta^i_{\,j}$ and $\mathcal{B}^* = \{e^j\}$ and $\mathcal{B} = \{e_i\}$. Note that for the dual basis to make sense, the basis of the vector space $V$ has to be *fully* specified. # Differential Geometry Let's discuss the same thing in the language of differential geometry. In the [[cotangent space]] $T_P^*$ of a [[manifold]] $M$ at a point $P \in M$, where $M$ is an $n$-[[Dimension|dimensional]] [[manifold]], any $n$ linearly independent [[One-Form|one-forms]] constitute a [[basis]]. However, once a [[basis]] $\{\pmb{e}_i\}$ is chosen for the [[tangent space]] $T_P$, 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 $j^{\text{th}}$ 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: $ \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. # Coordinate Dual Basis This discussion extends to [[One-Form Field|one-form fields]]. If the set of [[Vector Field|vector fields]] $\{\pmb{e}_i\}$ is a [[basis]] at every point in some region $U$ of $M$, then the fields $\{\tilde{\omega}^j\}$ are likewise a set of one-form fields that defines a basis at all points of $U$. A [[coordinate chart|coordinate system]] on $U$, $\{x^i\}$, 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.