research33 min studyLesson 913 in Mathematics

Retractions, vector transports and exponential maps

Separate feasible first-order updates from geodesic motion and check the length behavior of transported tangent vectors.

On this page

What you’ll understand

  • State the defining local conditions of a smooth retraction.
  • Distinguish first-order agreement from second-order agreement with the exponential map.
  • Compare projected vector transport with parallel transport on a sphere.

A tangent step is not generally a feasible point. Retractions turn such steps into manifold points using inexpensive smooth maps, while the exponential map follows the chosen metric’s geodesics. Both agree at first order, but their higher-order behavior and cost can differ. Algorithms that compare gradients or search directions from different tangent spaces also need a transport rule. A transport can be convenient and smooth without preserving lengths, so convergence proofs must use the properties actually supplied rather than treating every transport as parallel translation.

Retractions preserve the initial point and velocity

A smooth retraction is defined on a neighborhood of the zero section of the tangent bundle. At each x, its local map Rₓ sends zero to x and has derivative at zero equal to the identity on TₓM. Thus the curve t↦Rₓ(tη) starts at x with velocity η. These conditions let objective pullbacks reproduce first derivatives correctly. They do not assert that the curve is a geodesic, that a finite displacement has unchanged length, or that the map is globally defined for arbitrarily large tangent vectors.

The exponential map also satisfies the retraction conditions, but is defined by solving a geodesic initial-value problem. On a complete manifold it is defined for every tangent vector, although it need not be globally injective. A second-order retraction additionally has zero intrinsic acceleration at time zero along every radial tangent curve. In smooth local coordinates it agrees with the exponential through second order, giving an order-three local difference. A generic first-order retraction need only have an order-two difference. The adjective second-order refers to covariant acceleration, not merely to normalizing a point twice.

Rx(0)=x,DRx(0)[η]=η,DdtddtRx(tη)∣t=0=0for a second-order retractionR_x(0)=x,\qquad DR_x(0)[\eta]=\eta,\qquad\left.\frac{D}{dt}\frac{d}{dt}R_x(t\eta)\right|_{t=0}=0\quad\text{for a second-order retraction}
The first two identities define a retraction locally. The third uses the chosen metric’s covariant acceleration and is an additional condition.↔ Scroll wide equations horizontally.

Sphere normalization is feasible but not the exponential

On the unit sphere with its induced metric, normalizing x+η gives a smooth retraction for every tangent η, since x and η are orthogonal and their sum cannot vanish. Its radial curve has initial ambient acceleration −∥η∥²x, which is normal, hence zero intrinsic acceleration. It is therefore second order. The exact exponential uses sine and cosine instead. Their Taylor expansions agree through degree two, but normalization uses an angular displacement arctan∥η∥ rather than ∥η∥ for a finite step.

A retraction can be first order without being second order. On a circle at (1,0), the curve with angle t+t² has the correct initial tangent velocity but a nonzero tangential acceleration at zero. This difference affects a nonstationary objective’s second-order pullback model. Choosing a retraction should consequently be part of the algorithm specification, especially for Newton or trust-region methods. First-order methods can often use either order, provided their required smoothness and step-size bounds hold on the actual retraction domain.

Rx(η)=x+η1+∥η∥2,Exp⁡x(η)=cos⁡(r)x+sin⁡(r)rη,r=∥η∥,Tx→yξ=(I−yy⊤)ξR_x(\eta)=\frac{x+\eta}{\sqrt{1+\|\eta\|^2}},\qquad\operatorname{Exp}_x(\eta)=\cos(r)x+\frac{\sin(r)}r\eta,\quad r=\|\eta\|,\qquad\mathcal T_{x\to y}\xi=(I-yy^\top)\xi
The exponential formula uses the continuous sine-over-r value at r=0. The last expression is a convenient projected transport into TᵧS; it is generally not isometric.↔ Scroll wide equations horizontally.

Transport rules require their own assumptions

A vector transport associated with a retraction maps an old tangent vector into the tangent space at the new point, is linear in that vector, and agrees with the identity at a zero step, with suitable smoothness. Projection onto the new tangent space supplies such a local construction for embedded manifolds. It can shrink a vector or even lose invertibility at a large step. Parallel transport along a specified curve instead preserves the metric inner product through a covariant differential equation. The two constructions should not be conflated.

Conjugate-gradient recurrences, quasi-Newton updates and momentum methods compare tangent data across iterates. Their analyses may require bounded transport norms, metric compatibility, an inverse transport, or explicit scaling to avoid undesirable expansion. These properties must be checked for the selected transport and step range. A valid retraction by itself does not imply them. Conversely, insisting on exact geodesics and parallel transport can impose unnecessary cost when a proof only needs a controlled local approximation. The correct choice balances geometric accuracy with the assumptions used by the algorithm.

Normalize a step and compare two transports

On the unit circle take x=(1,0), η=(0,1/2) and ξ=(0,1). Compute the normalized update y, its angle, the projection transport of ξ, and parallel transport along the shorter circular arc.

  1. Normalization gives y=(2,1)/√5, with angle arctan(1/2), approximately 0.463648. The exponential at η instead has angle 1/2, so these are different finite updates.

  2. Since yᵀξ=1/√5, projected transport is ξ−y/√5=(−2/5,4/5), with norm 2/√5.

  3. The positively oriented unit tangent at y is (−1,2)/√5. Parallel transport of ξ along the shorter arc is this vector, of norm one; projection transport is its multiple 2/√5.

The result

Both transported vectors are tangent at y, but only the parallel-transported one preserves ξ’s length. The normalized update is second order without being the exact exponential.

Explore the model

Interactive lab

Interactive lab · vectors

Quantify the shrinking caused by tangent projection

Pause and Reset, then set angle=30°,magnitude=1 to locate a new circle point y. Project the old vertical unit tangent from (1,0) onto TᵧS: predict norm cos30°=√3/2, whereas parallel transport has norm one. For a fresh comparison Reset and set angle=60°,magnitude=1; predict projected norm 1/2. The displayed vector locates y; it does not animate the transport equation or claim projected transport is isometric.

Quantify the shrinking caused by tangent projectionPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA smooth tangent transport can lose length, with the loss depending on the finite step.

Check your understanding

Which extra condition makes a smooth retraction second order for a chosen Riemannian metric?

Try it yourself

At (1,0) on the unit circle consider R(t)=(cos(t+t²),sin(t+t²)). Compute velocity and ambient acceleration at zero, then decide whether this radial update is second order. For f(x,y)=y compare its second pullback derivative with the intrinsic Hessian at the base point.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate the angle twice and project acceleration onto the vertical tangent line.

Compare with the solution

The initial velocity is (0,1), so the curve satisfies the first-order retraction condition locally. Its ambient acceleration is (−1,2), whose tangent projection is (0,2), not zero. Thus this radial map is not second order. The pullback f(R(t)) has second derivative two at zero. For the linear ambient objective y, the sphere Hessian along the unit vertical tangent at (1,0) is zero, so the mismatch is exactly the gradient paired with the nonzero tangent acceleration.

Take it with you

  • Retractions preserve initial point and tangent velocity, while the exponential follows metric geodesics.
  • Second-order retractions require zero covariant acceleration at zero.
  • Projection transport is generally not parallel transport and may change lengths.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.