- Dual Basis vs Metric Dual

1. Definition of Metric Dual #

$L : V \to V^*, \quad v \mapsto \tilde{v}$, where $\tilde{v}(w) = (v|w)$, $( \cdot | \cdot )$ is a non-degenerate Hermitian form, and $\tilde{e}_i = L(e_i)$ is called the metric dual of $e_i$. While the term “metric” is primarily used in physics for Euclidean or Minkowski spaces,

mathematically, a complex vector space equipped with a non-degenerate Hermitian form is also a metric space, which may explain the use of the term “metric” as a modifier here.

In other words, it seems that the metric dual refers to the dual defined using a positive-definite non-degenerate Hermitian form (complex inner product) or Minkowski metric for a vector space equipped with such forms.

2. $e^i$ vs. $\tilde{e_i}=L(e_i)$ #

Since the dual space exists even without a positive-definite non-degenerate Hermitian form or Minkowski metric, we can consider the relationship between the basis of the dual space of such a vector space and the metric dual.

The dual basis $e^i$ satisfies $e^i(e_j) = \delta^i_j$, and $\sub{\tilde{e}}{i}(e_j) = (e_i|e_j)$. Therefore, if $(e_i \mid e_j) = \sub{\delta}{ij}$ holds, then $\sub{\tilde{e}}{i} = e^i$. In other words, when the basis is orthonormal and satisfies $(e_i \mid e_i) = +1$, the dual basis and the metric dual are the same.

Consequently, we can understand that the Minkowski space is not a vector space where the dual basis and the metric dual are the same.

3. Additional Note #

Although this content may not be particularly important, understanding the difference between the dual basis of a vector space and the metric dual of an inner product space or metric space has its own significance. Moreover, in a complex inner product space, the conjugate-linear map $L$ can be used to define the concept of the Hermitian adjoint of a matrix.