In this post, I talk about the Sasaki metric.

The Sasaki Metric, Pt. 1

The Sasaki metric is a very useful extension of a metric from a manifold $M$ to its tangent bundle $TM$. I want to use it to construct Cartan’s structural equations, which will be useful in studying Pestov’s integral identities. We’ll be following Gidea and Burns book on differential geometry pretty closely, along with these notes.

The basic idea of the Sasaki metric is easy: we want to decompose vectors into their “vertical” and “horizontal” components. In terms of pictures, this is not super hard to see on a surface (and there are great books out there which actually draw this picture – I’m sadly not much of an artist, so I won’t attempt it). On general manifolds, the picture is a little less clear. Since vertical vectors don’t necessarily need the metric, we’ll start with defining what a vertical vector even is.

Points on the tangent bundle of our surface consist of two bits of information: $\theta \in TM$ is of the form $\theta = (x,v)$, where $x$ is some point in our manifold and $v$ is some vector tangent to our surface based at $x$. To get a tangent vector to this point, we take $\xi \in T_{\theta}TM$. This is achieved by some curve $c : (-\epsilon, \epsilon) \rightarrow TM$. We’ll say this is vertical if $(d\pi)_\theta(\xi) = 0$, where $\pi : TM \rightarrow M$ is the footprint map. Notice that if this is the case, then the curve that defines $\xi$ is of the form $c(t) = (x, v(t))$. In other words, it only moves the vector, not the base point. Thus, at the point $\theta$ we have

$$V_\theta := \ker(d\pi_\theta)$$

gives us the space of vertical vectors.

The goal now is to construct horizontal vectors. In terms of pictures, if the vertical vectors are the ones that remain inside the tangent space at a base, then the horizontal vectors should be vectors which are sort of orthogonal; i.e. they should arise as derivatives of parallel transports. Let’s try to formalize this idea now.

Let $\xi \in T_{\theta} TM$, and let $V(t)$ be a curve through $\theta$ in $TM$ with $\xi = \frac{dV}{dt}(0).$ Thus this is the curve which realizes our vector $\xi$. We can push $V(t)$ to the manifold via the footprint map. Let $h(t) := \pi \circ V(t)$. Then one way to think about $V(t)$ is a vector field along the curve $h(t)$. Recall that, with reference to the metric, we have the covariant derivative of a vector field along a curve, $\frac{D_hV}{dt}(0)$. Now what we’d like to do is define a connection map $K : TTM \rightarrow TM$ by $K_\theta(\xi) = \frac{D_hV}{dt}(0)$. The issue is that this value may depend on our choice of curve $V(t)$. The reality is that this does not! The proof is actually pretty easy, so let’s discuss it briefly. Take a coordinate system on $TM$ $(x_1, \ldots, x_n, v_1, \ldots, v_n)$ centered at $(x,v)$. Let $h(t) = (x_1(t), \ldots, x_n(t))$ on $M$ in local coordinates. Since $V$ a vector field, we can write it as

$$V(t) = \sum_{k=1}^n V_k(t) \frac{\partial}{\partial x_k} \bigg|_{h(t)}.$$

Let’s remember how the covariant derivative works now. Based on the compatibility of the connection, the covariant derivative satisfies

$$\frac{D_h V}{dt} = \nabla_{\frac{dh}{dt}} V.$$

In local coordinates, we can write

$$\frac{dh}{dt} = \sum_{j=1}^n \frac{\partial x_j}{\partial t} \frac{\partial }{\partial x_j}.$$

Thus using the product rule, linearity, and compatibility, we get

$$\frac{D_hV}{dt}(0) = \sum_{k=1}^n \left( \frac{dV_k}{dt}(0) + \sum_{i,j} V_k \frac{dx_i}{dt}(0) \Gamma_{ij}^k\right)\frac{\partial}{\partial x_k}\bigg|_{h(0)},$$

where we recall that the $\Gamma_{ij}^k$ are the Christoffel symbols. Recall that we have $\frac{dx_i}{dt}(0) = v_i$ by our construction of $h$ and $\frac{dV_k}{dt}(0) = \xi_k$, where we write $\xi = (\xi_1, \ldots, \xi_n)$. Substituting those in, we see that everything only depends on local information $\xi$, $\theta$, and the Christoffel symbols. Nothing depends on our choice of map, as desired, and so our connection map is well-defined. We can then define the horizontal vectors to be

$$H_\theta := \ker(K_\theta).$$

Can a vector be both horizontal and vertical? Intuitively, the answer should be no, but based on the above descriptions it’s not so obvious (to me, at least). Take $\xi \in H_\theta \cap V_\theta$. This means that $\xi \in \ker(d\pi_\theta) \cap \ker(K_\theta)$. From our earlier description, we know that this means that the curve that defines $\xi$ must remain entirely inside of one tangent space, $T_xM$, however this means that the parallel transport is not moving. So taking the derivative of how the vector changes along the curve $h(t)$, we get that $\xi$ must be $0$.

The claim now is that $T_\theta(TM) = H_\theta \oplus V_\theta.$ Since they intersect trivially, we need to show that they cover the entire space. To do that, we’ll use a dimension argument. The first claim to check is that $V_\theta \cong T_xM$. This is actually not bad – a vector $\xi \in V_\theta$ is defined by a curve of the form $(x, v(t))$. If we let $w = v'(0)$, $v = v(0)$ (where $(x,v) = \theta)$, then really it’s defined by curves of the form $(x, v + tw)$, and $v \in T_xM$. But this is exactly enough to give us our isomorphism between $T_xM$ and $\xi$, since this means that it’s just determined by vectors $w$. The horizontal vectors are realized as parallel vector fields along geodesics through the point $x$. This of course is in linear bijection with $T_xM$, and so we’re done. Moreover, we can define an isomorphism

$$i_\theta : T_\theta(TM) \rightarrow T_xM \times T_xM, \ \ i_\theta(\xi) := (d\pi_\theta(\xi), K_\theta(\xi)).$$

Now from this, we can use the first isomorphism for vector spaces to realize

$$V_\theta \cong \text{Im}(K_\theta), \ \ K_\theta \cong \text{Im}(d\pi_\theta).$$

So using this, we can define a new metric on $TM$ by declaring these bundles be orthogonal to one another. This is the Sasaki metric, which is denoted by

$$\langle \xi, \eta \rangle = g(d\pi_\theta(\xi), d\pi_\theta(\eta)) + g(K_\theta(\xi), K_\theta(\eta)).$$

The Sasaki Metric, Pt. 2

Now an overflow post explains that there’s actually an even “easier” way of realizing things, which is to use the connection form to realize these things via a split exact sequence. The meat of the subject is still the same, since we have to actually define this connection which was the hard part of what we did above, but it allows us to skip the sort of trivial details of checking isomorphisms and orthogonality conditions. Let’s quickly discuss the abstract notions so that we have those in our toolbox for later.

Recall a principal $G$-bundle (or just principal bundle) is a fiber bundle $\pi : P \rightarrow X$ together with a continuous right action $P \times G \rightarrow P$ such that $G$ preserves the fibers of $P$ (meaning that if $y \in \pi^{-1}(\{x\}),$ then $yg \in \pi^{-1}(\{x\}).$) A very easy example comes from looking at the unit tangent bundle, generally denoted by $SM$ or $T^1M$ (we’ll use $SM$ for ease of typing). This forms a principal $S^1$ bundle, where the circle action is rotation (sometimes this is called a principal $U(1)$ bundle).

A connection on a principle bundle $G \rightarrow P \rightarrow M$ is a smooth choice for each $p \in P$ of a linear map $\sigma_p : T_m M \rightarrow T_pP$ where $m = \pi(p)$ so that the following two conditions hold: (1) $\pi_* \circ \sigma_p = \text{Id},$ (2) $\sigma_{pg} = g_* \sigma_p$. Condition (2) is sometimes called the elevator property.

An equivariant path lifting is a way of associating to each curve $\alpha : I \rightarrow M$ and $p \in \pi^{-1}(\alpha(0))$ a curve $\alpha_p : I' \rightarrow P$ with $I' \subseteq I$ so that $\pi \circ \alpha_p = \alpha$ and $\alpha_{pg}(t) = \alpha_p(t)g$ for all $g \in G$ and $t \in I'$. It is non-trivial, but useful, to note that connections define equivariant path liftings. Conversely, an equivariant path lifting defines a connection (this is actually a much easier direction). So this is another way of identifying connections. This isn’t important in our discussion of the Sasaki metric, but I included it just as an interesting thing to keep in perspective.

So now assume that we have a vector bundle $\pi : E \rightarrow TM$, then letting $VE = \ker(\pi^*)$ and $HE = \text{Im}(\pi^*)$, we get the following exact sequence:

$$0 \rightarrow VE \rightarrow TE \rightarrow HE \rightarrow 0.$$

Recall an exact sequence splits if the middle term can be written as a direct sum of the outer terms, so in this case if we have $TE = HE \oplus VE.$ The splitting lemma tells us the following are equivalent: (1) the sequence splits (2) there is a left split, i.e. a map $\omega : TE \rightarrow VE$ so that $\omega \circ i = \text{Id}$ ($i$ here is the natural injection), (3) there is a right split, i.e. a map $\psi : HE \rightarrow TE$ so that $p \circ \psi = \text{Id}$ (here $p$ denotes the projection). As Tobias points out, (2) is given to us via the connection form $K$ in part 1. So knowing the existence of a connection form tells us the sequence splits, and so gives us the Sasaki metric!

The goal of the next post (whenever it may come) is to use the Sasaki metric to create Cartan’s structural equations.