# 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.

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 $$v,w\in V$$.