research31 min studyLesson 911 in Mathematics

Embedded constraints and tangent projections

Derive feasible first-order directions from regular constraints and compute projections only in the metric for which they are valid.

On this page

What you’ll understand

  • Use a full-rank constraint derivative to identify a smooth tangent space.
  • Compute the Euclidean orthogonal projector onto a regular level-set tangent space.
  • Distinguish tangent feasibility from remaining on the constraint after a finite step.

A smooth equality constraint often defines a manifold whose tangent spaces contain the feasible infinitesimal directions. Optimization can then move through these spaces instead of repeatedly solving a full constrained subproblem. The construction requires regularity and a declared metric: a kernel formula can fail at singular constraints, and Euclidean projection is not automatically the gradient for another Riemannian metric. Throughout this module manifolds are finite dimensional and smooth, objectives have the differentiability stated in each result, and matrix examples are real unless explicitly noted.

Regular level sets provide local geometry

Let h map an open part of Rⁿ smoothly into Rᵐ, and suppose its derivative has full row rank at every point of the zero set under consideration. The regular level-set theorem gives an embedded manifold of dimension n−m. Differentiating h along any feasible smooth curve shows that its velocity lies in the kernel of Dh. Regularity supplies the converse locally: every vector in that kernel is a tangent velocity. Merely counting equations without checking their independent derivatives can give the wrong dimension.

The rank assumption belongs to the chosen representation as well as to the feasible set. For h(t)=t² at zero, the feasible set is a smooth zero-dimensional singleton, but Dh vanishes and its kernel is the whole line. The kernel formula therefore fails for that singular description. For h(x,y)=xy, the zero set has a crossing and is not even a manifold near the origin. These examples explain why a failed constraint qualification is a warning to inspect the geometry rather than a license to treat every derivative-kernel direction as feasible.

M=h−1(0),rank⁡Dh(x)=m ⟹ TxM=ker⁡Dh(x),dim⁡M=n−m\mathcal M=h^{-1}(0),\quad\operatorname{rank}Dh(x)=m\ \Longrightarrow\ T_x\mathcal M=\ker Dh(x),\qquad\dim\mathcal M=n-m
The full-row-rank condition makes the differentiated constraints a complete characterization of tangent directions, locally on the regular level set.↔ Scroll wide equations horizontally.

Projection removes the normal component

With the Euclidean inner product, the normal space of a regular level set is the range of Dh(x) transposed. Minimizing the squared distance from an ambient vector z to the derivative kernel gives a small normal equation. Full row rank makes Dh times its transpose invertible, producing the displayed orthogonal projector. This is a linear projection at one point, not a projection of the point itself onto the nonlinear feasible set. Its output and the removed component are orthogonal in the declared Euclidean metric.

For the unit sphere, use h(x)=(∥x∥²−1)/2. Its derivative is x transposed, so projection subtracts the radial component. For the Stiefel constraint XᵀX=I, tangent matrices satisfy Xᵀη+ηᵀX=0, and the Frobenius-orthogonal projection subtracts X times the symmetric part of XᵀZ. These formulas exploit the constraints’ structure and avoid forming a large explicit Jacobian. Their metric qualification matters: weighted or canonical metrics generally require different gradient or projection formulas.

Px=I−Dh(x)⊤(Dh(x)Dh(x)⊤)−1Dh(x),Pxspherez=z−x(x⊤z),PXStZ=Z−Xsym⁡(X⊤Z)P_x=I-Dh(x)^\top\bigl(Dh(x)Dh(x)^\top\bigr)^{-1}Dh(x),\qquad P_x^{\mathrm{sphere}}z=z-x(x^\top z),\qquad P_X^{\mathrm{St}}Z=Z-X\operatorname{sym}(X^\top Z)
These are Euclidean or Frobenius orthogonal projections; sym(A)=(A+Aᵀ)/2. The sphere formula assumes ∥x∥=1 and the Stiefel formula assumes XᵀX=I.↔ Scroll wide equations horizontally.

A tangent direction needs a finite-step mechanism

For a tangent η, the straight ambient point x+tη satisfies the constraints only to first order. On a sphere its squared norm becomes one plus t²∥η∥², so every nonzero finite step leaves the sphere. A retraction will repair this by mapping a tangent displacement back to the manifold while preserving its initial velocity. Projecting an ambient gradient and then adding it without such a feasible update is therefore not yet a complete manifold algorithm. The derivative constraint controls velocity rather than all higher-order terms.

Computationally, tangent projection should be checked by its defining identities: the output satisfies the differentiated constraint, projection is idempotent, and the residual is normal. These tests reveal transposes, missing symmetric factors and inappropriate metrics. They do not certify a global algorithmic rate, because convergence additionally depends on step choice, objective regularity and the update map. Starting with the correct tangent geometry isolates those later assumptions instead of mixing a geometric error with a line-search or curvature issue.

A nonradial ellipsoid normal

On h(x)=x₁²+4x₂²+x₃²−1=0, take x=(3/5,2/5,0) and z=(2,1,−1). Compute the Euclidean tangent projection of z.

  1. The point is feasible because 9/25+16/25=1. The constraint gradient is (6/5,16/5,0), proportional to n=(3,8,0), with squared norm nᵀn=73.

  2. The normal coefficient in the n direction is (nᵀz)/(nᵀn)=14/73. Subtracting it gives Pz=(104/73,−39/73,−1).

  3. The dot product with n is (312−312)/73=0. The removed vector is (42/73,112/73,0), parallel to n, verifying both tangent feasibility and normality of the residual.

The result

The projection is (104/73,−39/73,−1), whose squared norm is 242/73. The ellipsoid normal is not the same direction as the position vector.

Explore the model

Interactive lab

Interactive lab · vectors

Locate a sphere point before computing its tangent component

Pause and Reset, then set angle=60°,magnitude=1. The point is x=(1/2,√3/2); for ambient z=(1,0), predict radial coefficient xᵀz=1/2 and tangent projection (3/4,−√3/4). For a fresh comparison Reset and set angle=0°,magnitude=1; predict the tangent projection of the same z is zero. The instrument displays the unit point and its coordinate projections; the tangent projection is the stated algebraic calculation, not a separately rendered solver.

Locate a sphere point before computing its tangent componentPlayback 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 tangent projector depends on the current feasible point, even when the ambient direction is unchanged.

Check your understanding

What is needed to conclude Tₓh⁻¹(0)=ker Dh(x) from the regular level-set theorem?

Try it yourself

Take X=[e₁,e₂] in St(3,2) and Z=[[1,2],[3,4],[5,6]]. Compute the Frobenius tangent projection and verify the tangent condition.

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

Give me a hint

Find sym(XᵀZ), subtract X times it, and inspect the top two rows of the result.

Compare with the solution

Here XᵀZ=[[1,2],[3,4]], whose symmetric part is [[1,5/2],[5/2,4]]. Thus the projection is [[0,−1/2],[1/2,0],[5,6]]. Its product XᵀP is the skew-symmetric top block [[0,−1/2],[1/2,0]], so XᵀP+PᵀX=0. The removed matrix is X times a symmetric matrix and is normal in the Frobenius metric. The tangent space has dimension 3·2−2·3/2=3.

Take it with you

  • Regularity, not equation counting alone, justifies a tangent kernel formula.
  • Orthogonal projection formulas are tied to their Euclidean or Frobenius metric.
  • A tangent displacement is infinitesimally feasible but needs a retraction for a finite feasible update.

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