From Local Graphs to Tangent Directions
The Local Graph Theorem for a Regular Level Set shows that near a regular point, a level set can be written as a graph. A graph has directions in which one can move while staying on it. These directions form its tangent space. The derivative of the defining map identifies them: at a regular point, they are exactly the directions that the derivative sends to zero.
We will define tangent vectors using differentiable curves, which makes the idea of “moving along the level set” precise. The resulting description connects the local graph from the previous tutorial to a useful linear calculation. It also distinguishes regular points, where the calculation works exactly, from singular points, where the derivative can give too many candidate directions.
The curve in the definition passes through \(a\) at parameter value zero and remains in \(M\) for parameter values near zero. Its velocity \(\gamma'(0)\) records the instantaneous direction of travel. We allow any continuously differentiable curve, not just a straight line: most tangent directions to a curved level set do not give straight lines lying in the set.
The Tangent Space Is the Derivative's Kernel
Let \(F:U\to\mathbb{R}^k\) be continuously differentiable on an open set \(U\subseteq\mathbb{R}^n\), and let \(a\in U\) satisfy \(F(a)=c\). Suppose \(a\) is a regular point, so \(DF(a)\) has rank \(k\). The kernel \(\ker DF(a)\) consists of all vectors \(v\) such that \(DF(a)v=0\). The next theorem shows that this linear space describes precisely the possible curve velocities in the level set.
Proof. Since \(DF(a)\) has rank \(k\), some \(k\)-by-\(k\) block of its columns is invertible. Reorder coordinates if needed, writing points as \((x,y)\in\mathbb{R}^{n-k}\times\mathbb{R}^k\), so that \(D_yF(a)\) is invertible. By the Local Graph Theorem for a Regular Level Set, there are neighborhoods of the coordinates \(a_x,a_y\) and a continuously differentiable function \(\phi\) such that, near \(a\), the level set is exactly the graph \(y=\phi(x)\). In particular, \(\phi(a_x)=a_y\).
First take \(v\in T_aM\), and choose a curve \(\gamma(t)=(x(t),y(t))\) in \(M\) with \(\gamma(0)=a\) and \(\gamma'(0)=v\). By continuity, for all sufficiently small \(t\) this curve lies in the neighborhood where the graph description holds. There \(y(t)=\phi(x(t))\). Differentiating at zero gives $$ y'(0)=D\phi(a_x)x'(0). $$ Also, \(F(\gamma(t))=c\) near zero. The chain rule therefore gives $$ DF(a)v=\left.\frac{d}{dt}F(\gamma(t))\right|_{t=0}=0. $$ Thus \(v\in\ker DF(a)\).
For the reverse inclusion, differentiate the identity \(F(x,\phi(x))=c\) at \(x=a_x\). For every \(u\in\mathbb{R}^{n-k}\), the chain rule gives $$ D_xF(a)u+D_yF(a)D\phi(a_x)u=0. $$ Now take any \((u,w)\in\ker DF(a)\). Then $$ D_xF(a)u+D_yF(a)w=0. $$ Subtracting the two equations and using the invertibility of \(D_yF(a)\) shows \(w=D\phi(a_x)u\). The curve $$ \gamma(t)=(a_x+tu,\phi(a_x+tu)) $$ is defined for sufficiently small \(t\), lies in the level set, and satisfies \(\gamma(0)=a\) and \(\gamma'(0)=(u,D\phi(a_x)u)=(u,w)\). Hence \((u,w)\in T_aM\), proving \(\ker DF(a)\subseteq T_aM\). The two sets are equal. Finally, the Rank-Nullity Theorem gives \(\dim\ker DF(a)=n-k\), since \(DF(a)\) has rank \(k\). \(\square\)
The proof uses both parts of the local graph description. Curves in the level set must follow the graph, so their velocities satisfy the derivative constraint. Conversely, every vector satisfying that constraint is the velocity of a curve obtained by moving in the graph's free coordinates. This converse is what ensures that the kernel does not merely contain the tangent space: at a regular point, it equals it.
Worked Example: The Tangent Plane to an Ellipsoid
Let \(F:\mathbb{R}^3\to\mathbb{R}\) be \(F(x,y,z)=x^2+2y^2+3z^2\), and consider the level \(M=F^{-1}(\{6\})\) at \(a=(1,1,1)\). Substitution verifies that \(F(a)=1+2+3=6\). The gradient, which represents the derivative of a scalar-valued function, is \(\nabla F(x,y,z)=(2x,4y,6z)\). At \(a\), it is \((2,4,6)\), which is nonzero, so \(DF(a)\) has rank \(1\) and \(a\) is regular.
The Tangent Space to a Regular Level Set Theorem says that \(T_aM\) is the set of vectors \((u,v,w)\) satisfying $$ DF(a)(u,v,w)=2u+4v+6w=0. $$ Dividing by \(2\), the tangent plane through \(a\) has direction vectors satisfying \(u+2v+3w=0\). For example, \((1,-1,0)\) and \((3,0,-1)\) satisfy this equation, since \(1+2(-1)+3(0)=0\) and \(3+2(0)+3(-1)=0\). They are linearly independent, so they span the two-dimensional tangent space. The corresponding affine tangent plane is \(a+T_aM\), or \(x+2y+3z=6\).
Normal Spaces and Constraint Equations
A vector perpendicular to every tangent direction is called a normal vector. The collection of all such vectors is the orthogonal complement of the tangent space, denoted \((T_aM)^\perp\). For a level set defined by several equations, each row of the Jacobian gives a normal vector: tangent directions must make the first-order change in every component of \(F\) equal to zero.
Proof. By the preceding theorem, \(T_aM=\ker DF(a)\). Write \(A=DF(a)\). Every vector in \(\operatorname{im}A^T\) has the form \(A^Tq\) for some \(q\in\mathbb{R}^k\). If \(v\in\ker A\), then $$ (A^Tq)\cdot v=q\cdot(Av)=q\cdot 0=0. $$ Thus \(\operatorname{im}A^T\subseteq(\ker A)^\perp\). Since \(A\) has rank \(k\), its transpose also has rank \(k\), so \(\operatorname{im}A^T\) has dimension \(k\). By the Rank-Nullity Theorem, \(\ker A\) has dimension \(n-k\), and therefore \((\ker A)^\perp\) has dimension \(k\). The inclusion is between two subspaces of the same finite dimension, so they are equal. The vectors in \(\operatorname{im}A^T\) are exactly the linear combinations of the rows of \(A\). When \(k=1\), that row is the gradient \(\nabla F(a)\), proving the final claim. \(\square\)
Worked Example: Tangent Direction from Two Constraints
Define \(F:\mathbb{R}^3\to\mathbb{R}^2\) by \(F(x,y,z)=(x+2y-z,\ 2x-y+z)\). At \(a=(0,0,0)\), \(F(a)=(0,0)\). The derivative matrix at every point is $$ DF(a)= \begin{pmatrix} 1&2&-1\\ 2&-1&1 \end{pmatrix}. $$ Its rows are not scalar multiples, so it has rank \(2\). The point is regular, and the tangent space to the zero level set is the kernel of this matrix.
For a direction \((u,v,w)\), the kernel equations are \(u+2v-w=0\) and \(2u-v+w=0\). Adding them gives \(3u+v=0\), so \(v=-3u\). Substituting into the first equation gives \(u-6u-w=0\), hence \(w=-5u\). Therefore $$ T_aM=\{t(1,-3,-5):t\in\mathbb{R}\}. $$ The vector \((1,-3,-5)\) is indeed in the kernel: \(1+2(-3)-(-5)=0\), and \(2(1)-(-3)+(-5)=0\). The two rows of \(DF(a)\) are normal vectors, and each has dot product zero with \((1,-3,-5)\).
Why Regularity Matters
The equality \(T_aM=\ker DF(a)\) depends on the regularity hypothesis. For any differentiable curve \(\gamma\) in the level set through \(a\), the chain rule still gives \(DF(a)\gamma'(0)=0\). Thus at a point where the derivative is not onto, the kernel still contains every curve velocity. But without regularity, there may be kernel vectors that no curve in the level set can realize.
Worked Example: A Kernel That Is Too Large at a Singular Point
Let \(F:\mathbb{R}^2\to\mathbb{R}\) be \(F(x,y)=x^3\), and take the zero level. Since \(x^3=0\) exactly when \(x=0\), the level set is the vertical line \(M=\{(0,y):y\in\mathbb{R}\}\). At \(a=(0,0)\), every curve in \(M\) has first coordinate identically zero. Its velocity therefore has first coordinate zero. Conversely, for every \(s\in\mathbb{R}\), the curve \(\gamma(t)=(0,st)\) lies in \(M\) and has velocity \((0,s)\). Consequently, $$ T_aM=\{(0,s):s\in\mathbb{R}\}. $$
However, \(DF(x,y)=(3x^2,0)\), so \(DF(a)=(0,0)\) and \(\ker DF(a)=\mathbb{R}^2\). For instance, \((1,0)\) lies in this kernel, but it cannot be the velocity at the origin of a curve lying in \(M\), because every point in \(M\) has first coordinate zero. The derivative kernel is therefore larger than the tangent space here. This example shows why the regular-point hypothesis cannot be omitted.
For regular level sets, the tangent-space calculation is a direct way to translate equations into geometry. Given \(F(a)=c\), compute \(DF(a)\), check that its rank is \(k\), and solve \(DF(a)v=0\) to find tangent directions. The rows of that same matrix span the normal space. If the rank check fails, the kernel equation remains a necessary condition on curve velocities, but it need not describe all and only the tangent vectors. At such a point, the level set must be examined more directly.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- How is a tangent vector to a set defined using a continuously differentiable curve?
- Why does the chain rule imply that every tangent vector to a level set belongs to the kernel of its derivative?
- At a regular point of a level set in \(\mathbb{R}^n\) defined by \(k\) equations, what is the dimension of the tangent space?
- How can the rows of the Jacobian be used to find normal vectors?
- In the example \(F(x,y)=x^3\), why is the kernel of \(DF(0,0)\) larger than the tangent space to the zero level set?