16
$\begingroup$

Let $(M,g)$ be a Riemannian manifold with $\text{dim}(M)=n$. Then, there is a "natural" metric $\tilde{g}$ on the tangent bundle $TM$, so that $(TM,\tilde{g})$ is a Riemannian manifold, called the Sasaki metric, where a line element is written (with local coordinates of $TM$ given by $(x,v)$): $$ d\sigma^2 = g_{ij}\,dx^idx^j + g_{ij}\, Dv^iDv^j $$ where $D$ represents covariant differentiation: $$ Dv^i = dv^i + \Gamma_{jk}^iv^jdx^k $$ In components, letting indices range over $1$ to $n$, this is: \begin{align} \tilde{g}_{jk} &= g_{jk} + g_{\alpha \gamma}\Gamma_{\mu j}^\alpha\Gamma_{\eta k}^\gamma v^\mu v^\eta =: g_{jk}+A_{jk} \\ \tilde{g}_{j(n+k)} &= g_{kd}\Gamma^d_{\lambda j}v^\lambda =: B_{jk}\\ \tilde{g}_{(n+j)(n+k)} &= g_{jk} \end{align} Or, as a matrix: $$ \tilde{g} = \begin{bmatrix} g+A & B \\ B^T & g \end{bmatrix} $$

Question: intuitively speaking, why is this "natural"?

I am aware of other "natural" metrics on the tangent bundle; this question is specifically about this one, and the geometric intuition for why it is a good choice of metric. I can't seem to picture it.

Related: [1], [2], [3]

$\endgroup$
8
$\begingroup$

TL;DR: The connection gives us a way to canonically decompose $TTM$ as the direct sum of two copies of $TM$ (the "horizontal" and "vertical" bundles), so we just give each copy the Riemannian metric and declare the direct sum to be orthonormal. Long version:

A Riemannian metric on $TM$ is a (smoothly varying) choice of inner product on the double tangent space $T_v TM$ for each $v \in TM$. Since $\pi : TM \to M$ is a vector bundle over $M$, each $T_v TM$ has as a subspace the vertical tangent space $V_v TM$, which consists of the velocity vectors of curves in the vector space $T_{\pi(v)} M$, and thus can be canonically identified with $T_v M$. The Levi-Civita connection of $(M,g)$ provides a canonical horizontal subspace $H_v TM$, which consists of the velocity vectors of curves $(\gamma(t), V(t)) \in TM$ such that $v= (\gamma(0),V(0))$ and $\nabla_{\dot \gamma} V = 0.$

The upshot of all this is that we have a direct sum decomposition $TTM = VTM \oplus HTM$, with canonical isomorphisms $V_v TM \simeq T_{\pi(v)} M$ (described earlier) and $H_v TM \simeq T_{\pi(v)} M$ (by sending the velocity of $(\gamma,V)$ to the velocity of $\gamma$). If this isn't intuitive, think about the Euclidean case - if you have a tangent vector $v$ to $p= \pi(v) \in \mathbb R^n$, then the directions you can move it decouple in to one copy of $R^n$ for the motion of the basepoint and another copy for the motion of the vector.

The Sasaki metric can then be naturally defined by declaring $V_vTM$ and $H_vTM$ to be orthogonal, with the metric on each factor just being the pullback of $g$ from $T_{\pi(v)}M$ via the canonical isomorphisms.

This construction works for any vector bundle $E$ (over a Riemannian manifold $M$) equipped with a fibre metric and compatible connection: the vertical tangent spaces take the fibre metric from $E$, while the horizontal spaces (as defined by the connection) take the metric from $TM$. I have seen this general construction called the Kaluza-Klein metric.

$\endgroup$
  • $\begingroup$ Thank you for this, especially the third paragraph. I was wondering if you could mention a little more on the horizontal/vertical subspace decomposition of $TTM$, in particular it's not really intuitively clear to me how/why the "velocity vectors" in $V_vTM$ and $H_vTM$ have to be orthogonal or cover $TTM$. $\endgroup$ – user3658307 Jul 28 '17 at 17:13
  • 1
    $\begingroup$ @user3658307: they span $TTM$ thanks to the definition of a connection: motion of a tangent vector in any direction can be achieved by simultaneously parallel transporting in some direction in the base (the horizontal component) and adjusting the vector independently of this (the vertical component). We define the Sasaki metric so that these components are orthogonal. $\endgroup$ – Anthony Carapetis Jul 28 '17 at 23:14

Your Answer

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

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