Definition of Metric dual

Question

I'm reading Nadir Jeevanjee's An Introduction to Tensors and Group Theory for Physicists (2ed). In section 2.7, he discusses the relation between dual spaces and non-degenerate Hermitian forms. In it, he says the following:

Given a non-degenerate Hermitian form $(\cdot|\cdot)$ on a finite-dimensional vector space $V$, we can associate with any $v\in V$ a dual vector $\tilde v \in V^*$ defined by \begin{equation} \tilde v(w)\equiv (v|w). \end{equation} This defines a very important map \begin{align} L:V&\rightarrow V^* \\ v&\mapsto\tilde v, \end{align} ... We'll sometimes write $\tilde v$ as $L(v)$ or $(v|\cdot)$, and refer it to as the metric dual of $v$. Now, $L$ is conjugate-linear since for $v=cx+z$, where $v,x,z \in V$, \begin{equation} L(v) = (v|\cdot) = (cx+z|\cdot) = \bar c(x|\cdot)+(z|\cdot) = \bar c L(x)+L(z), \end{equation} ... A word of warning here: as a map from $V \rightarrow V^*$, $L$ is conjugate linear, but any $L(v)$ in the range of $L$ is a dual vector, hence is a fully linear map from $V\rightarrow C$.

($C$ is the set of scalars over which $V$ is defined.)

Why is this dual vector called 'metric' dual? What does he mean by 'fully' linear map? Can anyone please explain the warning mentioned above more elaborately?

From what he has said earlier in the book and from what I understand: The non-degenerate Hermitian form $(\cdot|\cdot)$ is linear in its second argument and conjugate-linear in its first argument. If the vector space in which it is defined is a real vector space, then the Hermicity condition [$(v|w)=\overline {(w|v)}$] reduces to a Symmetric one [$(v|w)=(w|v)$], and the hermitian form becomes linear in both its arguments (bilinear) and is said to be a metric. (If it is also positive-definite, then it is said to be an inner product.)

Is the metric dual called 'metric' using this context? If yes, please explain how.

Answer 1

Metric dual is not called 'metric'; typically if any words are omitted when talking about these things then 'metric' would be omitted and we'd talk just about the dual. However, that would not be a particularly good idea in this context.

$V^*$ is traditionally the collection of all continuous linear functions (called functionals) that map from $V$ to the base field, which seems to be ${\mathbb C}$ for you. So any element $v^* \in V^*$ is a function $v^*:V \rightarrow {\mathbb C}$. When the author refers to 'fully linear' maps they are talking about these functionals. $V^*$ is called the dual of $V$, or topological dual to distinguish it from the algebraic dual $V^\sharp$ which is the collection of all linear functions (continuous or not) from $V$ to ${\mathbb C}$.

The metric dual then is distinct from both the topological and algebraic duals which is why it would not be a very good idea to refer to it just as 'the dual'. For people used to topological duals (especially) conjugate-linearity would be unexpected, so the warning is there to draw attention to the fact that the metric dual doesn't behave quite the same way as you might expect from the choice of name.

Answer 2

Why is this dual vector called 'metric' dual?

That seems unclear. It is worth noting that simply referring to the "dual" of a vector $v$ is ambiguous; producing a functional corresponding to $v$ requires some kind of structure (such as a non-degenerate Hermitian form) since there is no canonical isomorphism between $V$ and $V^*$.

What does he mean by 'fully' linear map?

We say that $f$ is fully linear if we have $f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)$ for vectors $x,y$ and scalars $\alpha,\beta$. We say that $f$ is conjugate linear if we have $f(\alpha x + \beta y) = \bar \alpha f(x) + \bar \beta f(y)$. The map $L$ (which produces a map in $V^*$) is conugate-linear, but the maps in $V^*$ are themselves fully-linear maps.

Is the metric dual called 'metric' using this context?

It is not. There is no clear connection between the usage of the term "metric" in these two contexts.