Hermitian musical isomorphism

complex-geometryhermitian-matricesriemannian-geometry

I have encountered a problem regarding the musical isomorphism in its complex or Hermitian form, having a very different appearance under two different conventions. I ask for a revision of the exposition (maybe I get something wrong) and a possible fix for the ambiguity stated at the end.

Let $(V,h)$ be a Hermitian finite-dimensional complex vector space. Here I use the mathematician's (rather than the physicist's) convention: $h$ is linear in its 1st entry and conjugate linear in its 2nd.

$$
h(\alpha v, w) = h(v,\overline{\alpha}w) = \alpha h(v,w)
$$

Now fixing bases for $V=\langle e_1,\dots,e_d\rangle$ and the natural dual basis for $V^*=\text{Hom}_{\mathbb{C}}(V,\mathbb{C}) = \langle e^1,\dots,e^d \rangle$ satisfying $e^j(e_i) = \delta^j_i$. We define the Riesz map $F:V\rightarrow V^*$ by
$$
(Fv)(w) = h(w,v)
$$
Depending on the author's one defines the matrix of the Hermitian form $h$ w.r.t. the basis $\{e_i\}$ by either

  1. Convention 1: $h_{ij} = h(e_j,e_i)$. This convention is used in Hermitian geometry by authors like Wells.
  2. Convention 2: $h_{ij} = h(e_i,e_j)$. This convention is used by Kobayashi and Huybrechts.

I am looking for a closed expression of the induced Hermitian metric on the dual space $V^*$, namely $h^*$. This Hermitian metric is defined so that the above Riesz map is actually an isometry, i.e.
$$
h^*(Fv,Fw) = h(v,w)
$$
One subtle fact is that the Riesz map is not a complex isomorphism, but a conjugate linear one, and $h^*$ presumably plays a role in proving that $F$ is actually surjective.

The problem

Under different conventions I would get that the matrix expression for $h^*$ is either

  1. Convention 1: $h^* = (h^t)^{-1}$.
  2. Convention 2: $h^* = (\overline{h})^{-1} h (\overline{h})^{-1}$

Now this might seem like a minor inconvenience of the conventions, but in particular for the field that I am studying (connections on Hermitian vector bundles), this has a important consequence in relating the connection and curvature on a vector bundle to the ones on the dual vector bundle.

Best Answer

I have found that the imposition for the Riesz map to be an isometry must be wrong. Should we require that $$ h^*(Fv,Fw) = h(v,w) $$ we immediately run into contradiction by rescaling $v\mapsto \alpha v$ for $\alpha \in \mathbb{C}$ and imposing linearity in the first argument. Since the Riesz map is conjugate-linear: $$ h^*(F(\alpha v),Fw)=h(\alpha v, w) \implies \overline{\alpha} h^*(Fv,Fw) = \alpha h(v,w) $$ which indeed can't hold for arbitrary $v,w\in V$.

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