2
$\begingroup$

Basically a mathematically flavoured duplicate of https://physics.stackexchange.com/questions/368455/the-inverse-vielbein :

I have derived an identity for the inverse of bundle isometries. 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 Manifold $M$ and vector-bundles $E$, $E'$ equipped with pseudo-riemannian bundle-metrics $g$, $\eta$ of the same signature. A bundle isometry is then a bundle-isomorphism $e: E \rightarrow E'$ with $e^*\eta = g$ and $(e^{-1})^*g = \eta$.

I then claim, that the inverse of $e$ 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 $E$ 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 $(e^{-1})^{\mu}_{~i} = e_{i}^{~\mu} = e^{j}_{~\nu} \eta_{ji} g^{\mu \nu}$ where I used Einstein sum convention.

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.

$\endgroup$
1
  • 1
    $\begingroup$ 1. You should correct the typos: $TM$ is $E'$. 2. Other than that the statement is a fancy way to say that for an orthogonal matrix $A$, $A^{-1}=A^T$. $\endgroup$ Nov 12, 2017 at 15:28

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.