I have derived an identity for the inverse vielbein. But somehow, since this is quite an fundamental question, and I did no found any reference which even bother on this problem, I ask you if my argumentation is correct:

Consider a smooth pseudo-riemannian Manifold $M$ with metric $g$ and a vector-bundle $E$ equipped with a pseudo-riemannian bundle-metric $\eta$ of the same signature. I then call a vielbein a bundle-isomorphism $e: TM \rightarrow E$ with $e^*\eta = g$ and $(e^{-1})^*g = \eta$.

I then claim, that the inverse vielbein is given by $e^{-1} = \sharp \circ e^* \circ \flat$.

Now I prove this:

Let $v \in TM$, which implies that $e(v) \in E$.

Then $\flat e(v) = \eta(e(v), \cdot)$.

Therefore $e^* [\flat e(v)] = e^* [\eta(e(v), \cdot)] = \eta(e(v), e(\cdot))$.

Now $\sharp [\eta(e(v), e(\cdot))] = \sharp [(e^*\eta)(v, \cdot)] = \sharp g(v, \cdot) = v$.

Hence $\sharp \circ e^* \circ \flat \circ e = id.$. Smoothness follows since $\sharp, e^*, \flat$ are smooth, which completes the proof.

Now for local frames $X_\mu$ of $TM$ and $S_i$ of $E$ and $e(X_\mu) := e^{i}_{~\mu}S_i$ and $e^{-1}(S_i) = (e^{-1})^{\mu}_{~i} X_\mu$ this corresponds to $(e^{-1})^{\mu}_{~i} = e_{i}^{~\mu} = e^{j}_{~\nu} \eta_{ji} g^{\mu \nu}$.

Is there any mistake? Basically this says, that we get for any bundle isometry the inverse in terms of the musical isomorphisms "for free", which should be a result stated somewhere, but I did not find any reference.


The statement that, in local frames $(e^{-1})^\mu{}_i= e^j{}_\nu\eta_{ij}g^{\mu\nu}$ can be stated as $\eta_{ij}e^i{}_\mu e^j{}_\nu = g_{\mu\nu}$, which is well known since, as is made obvious by this form, this corresponds directly to $e^*\eta = g$. If one seeks to calculate the inverse, $e^{-1}$, your form is certainly helpful, but in my view it follows almost trivially from $e^*\eta = g$, and is, I think used without much motivation. I think this is one of those situations where the use of abstract index notation makes work with the musical isometries seem redundant to a physicist, while to a mathematician (or someone so inclined) their existence should not be thus trivialized. That is to say, the physicists that I have talked to (although I am sure there are plenty who disagree) would think "why do all that work (although this was a rather short proof), when you can just write it down in index notation see that it follows?" While a mathematician would probably think something along the lines of "the reason you can do that at all is because of the musical isometries, and the argument should be made, not assumed." Indeed, the crucial part of your argument is $$ e^*(\flat e) = e^*\eta, $$ which, once you write it out, as one always does in index notation becomes trivial $$ \eta_{ij}e^i{}_\mu e^j{}_\nu = \eta_{ij}e^i{}_\mu e^j{}_\nu. $$

Anyway, the answer is that you have made no mistake, and the result holds.


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