11
$\begingroup$

In the paper ``Morse theory on Hilbert manifolds'' (1963), on page 326, Richard Palais makes a remark that if $\phi\colon V \to W$ is an isometry (of submanifolds of $\mathbb{R}^n$), then this does not imply that the natural lift $\bar{\phi}\colon H_1([0,1];V) \to H_1([0,1];W)$ is an isometry of Hilbert manifolds. There is no explanation of why this is not generally true, though. This remark is puzzling me, since I don't see how this would fail to be true.

A little bit of background details. Palais constructs the Hilbert manifold $H_1([0,1];V)$ non-intrinsically (as compared to the construction in Klingenberg's ``Riemannian geometry'') through the embedding of $V$ into some $\mathbb{R}^n$ and then choosing a metric on $\mathbb{R}^n$ such that $V$ is a totally geodesic submanifold.

So my question is: what is wrong with my argument below that $\bar{\phi}$ is indeed an isometry?

Let $V,W$ be two embedded submanifolds in Euclidean space, so $V,W$ become Riemannian manifolds with metric given by the restriction of the Euclidean metric to their respective tangent bundles. Let $\phi\colon V \to W$ be an isometry.

Now let $\sigma \in H_1([0,1];V)$ and $\lambda,\mu \in H_1([0,1];V)_\sigma$. The metric $g$ on $H_1([0,1];V)$ is defined by $$ g_\sigma(\lambda,\mu) = \int_0^1 \langle \lambda(t), \mu(t) \rangle \,dt $$ with $\langle \,\cdot\,, \,\cdot\, \rangle$ the Euclidean inner product. Thus, with $\bar{\phi}(\sigma) = \phi \circ \sigma$ we get (using $d\bar{\phi}$ as given in Theorem~7 of the paper) $$ \bar{\phi}_*(g)_{\bar{\phi}(\sigma)}(d\bar{\phi}_\sigma(\lambda), d\bar{\phi}_\sigma(\mu)) = \int_0^1 \langle d\phi_{\sigma(t)}(\lambda(t)), d\phi_{\sigma(t)}(\mu(t)) \rangle \,dt. $$ Since $\phi$ was an isometry, this reduces to the metric on $H_1([0,1];W)$.

Note that I interpreted $H_1([0,1];V)$ and $H_1([0,1];W)$ as Riemannian manifolds, where I assumed that the metric is defined by the inner product on the canonical Hilbert spaces on which the tangent spaces are modeled.

$\endgroup$
5
  • $\begingroup$ it seems that Palais meant something else, or made a mistake $\endgroup$ Sep 26, 2013 at 9:26
  • 7
    $\begingroup$ The problem lies in the line above defining $g_\sigma(λ,μ)$. This is the metric for $H_0$, NOT the metric for $H_1$. The correct definition of the metric for $H_1$ uses $⟨λ'(t),μ'(t)⟩$ rather than $⟨λ(t),μ(t)⟩$ (see page 222 of my article where both metrics are defined), and with this change it is fairly obvious why my remark in the cited paper is in fact correct. $\endgroup$ Sep 30, 2013 at 3:20
  • 1
    $\begingroup$ @DickPalais: indeed, thank you for pointing this out. My understanding now is that the derivatives of $\lambda,\mu$ are non-intrinsically defined using the (simple Euclidean) covariant derivative of the ambient space of the embedding. $\endgroup$ Sep 30, 2013 at 8:59
  • $\begingroup$ @JaapEldering: can I convince you to convert your comment above to a self-answer (and get this question off the unanswered list)? $\endgroup$ Jun 1, 2023 at 3:37
  • $\begingroup$ If you feel that's appropriate, sure. $\endgroup$ Jun 15, 2023 at 8:56

1 Answer 1

3
$\begingroup$

Turning the comments by Dick Palais and me into an answer:

The problem with the argument in the question lies in defining $g_\sigma(\lambda,\mu)$ as metric on $H_1([0,1];V)$. It is the metric for $H_0$, not the metric for $H_1$. The correct definition of the metric for $H_1$ uses $\langle \lambda′(t),\mu′(t) \rangle$ rather than $\langle \lambda(t),\mu(t) \rangle$, (see page 222 of the article where both metrics are defined), and with this change it is fairly obvious why the remark in the Palais' paper that the natural lift $\bar{\phi}: H_1([0,1];V) \to H_1([0,1];W)$ is not an isometry of Hilbert manifolds, is in fact correct: the derivatives of $\lambda,\mu$ are non-intrinsically defined using the (simple Euclidean) covariant derivative of the ambient space of the embedding.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.