Why Symmetry Helps
A function can have many stationary points, and solving its full system of partial-derivative equations may obscure useful structure. If the function is unchanged by reversing a coordinate or exchanging two variables, its critical points and Hessians often inherit that symmetry.
This structure can reduce the work: it can identify points that must be stationary, show that other stationary points occur in related groups, and split a Hessian test into simpler directions. But symmetry does not by itself guarantee that every stationary point lies on a symmetry axis or plane.
Throughout, assume \(f\) has continuous second partial derivatives near the points under discussion. A stationary point is a point where every first partial derivative is zero. Its Hessian is the matrix of second partial derivatives.
Coordinate Evenness and Symmetry Planes
If \(f\) is even in \(x\), then \(f(-x,y,z)=f(x,y,z)\). Differentiating with respect to \(x\) shows that \(f_x\) is odd in \(x\), so \(f_x(0,y,z)=0\). Thus the plane \(x=0\) automatically satisfies one of the three stationary-point equations. A stationary point on that plane still has to satisfy \(f_y=0\) and \(f_z=0\).
Evenness also restricts the Hessian on the fixed plane. The mixed partial derivatives \(f_{xy}\) and \(f_{xz}\) are odd in \(x\), so both vanish when \(x=0\). The Hessian then separates the \(x\)-direction from the \(y,z\)-directions.
This is a useful way to locate candidates and simplify their classification. It does not say that all stationary points have \(x=0\): points with \(x\ne 0\) may also solve the full gradient system.
Variable Exchange and Related Critical Points
Suppose \(f(x,y,z)=f(y,x,z)\). Exchanging \(x\) and \(y\) leaves the function unchanged. If \((a,b,c)\) is stationary, then \((b,a,c)\) is also stationary. Their Hessians have the same curvature classification, with the \(x\)- and \(y\)-directions exchanged.
At a point fixed by the exchange, where \(x=y\), the first partial derivatives satisfy \(f_x=f_y\). The Hessian has the form
The direction \((1,-1,0)\), which changes \(x-y\) while leaving \(x+y\) unchanged, separates from the other two directions. The remaining directions lie in the plane \(x=y\). This converts a three-dimensional Hessian test into one test in the exchange-antisymmetric direction and a two-dimensional test in the exchange-symmetric plane.
Why the Hessian Inherits Symmetry
The structure above follows from differentiating the symmetry relation. For an orthogonal linear symmetry represented by a matrix \(\mathbf{P}\), suppose \(f(\mathbf{P}\mathbf{x})=f(\mathbf{x})\). The chain rule gives
Because \(\mathbf{P}\) is invertible, a zero gradient at \(\mathbf{x}\) implies a zero gradient at \(\mathbf{P}\mathbf{x}\). Differentiating a second time gives
At a fixed point, \(\mathbf{P}\mathbf{x}=\mathbf{x}\), so the Hessian satisfies \(\mathbf{P}^{\top}\mathbf{H}(\mathbf{x})\mathbf{P}=\mathbf{H}(\mathbf{x})\). In the variable-exchange case, \(\mathbf{P}\) swaps the first two coordinates. The vector \((1,-1,0)\) changes sign under this swap, while vectors of the form \((u,u,v)\) remain fixed. The Hessian therefore cannot mix these two types of direction: its quadratic form separates into the antisymmetric direction and the symmetric plane. This is the reason the block structure simplifies classification.
Worked Example: Evenness Reveals Critical-Point Families
Worked Example: Evenness Reveals Critical-Point Families
Consider
The function is even in each coordinate. Differentiate:
The equation \(f_z=0\) gives \(z=0\). The other equations give \(x=0\) or \(2x^2+y^2=2\), and \(y=0\) or \(x^2+2y^2=2\). If \(x=0\), then \(y=0\) or \(y^2=1\). If \(y=0\), then \(x=0\) or \(x^2=1\). When both are nonzero, solve
Subtracting the equations gives \(x^2-y^2=0\), so \(x^2=y^2\). Substitution then gives \(3x^2=2\), hence \(x^2=y^2=2/3\). The complete list is the origin, the four axis points \((\pm1,0,0),(0,\pm1,0)\), and the four points with \(x=\pm\sqrt{2/3}\), \(y=\pm\sqrt{2/3}\), \(z=0\).
The Hessian is
At the origin it is \(\operatorname{diag}(-4,-4,2)\), which is indefinite, so the origin is a saddle. At \((1,0,0)\), for example, it is \(\operatorname{diag}(8,-2,2)\), also indefinite; the other axis points have the same classification by the coordinate symmetries.
At a point with \(x^2=y^2=2/3\), the two diagonal entries of the \(x,y\) block are \(16/3\), and its off-diagonal entry is \(4xy=\pm 8/3\). The eigenvalues of that block are \(16/3+8/3=8\) and \(16/3-8/3=8/3\). The remaining eigenvalue is \(2\). All are positive, so each of these four points is a local minimum.
As a check on the candidate equations, at \(x^2=y^2=2/3\), each bracket in \(f_x\) and \(f_y\) equals \(2(2/3)+2/3-2=0\); also \(z=0\), so all three first partial derivatives vanish. The sign choices occur in symmetry-related groups, which lets one classification apply to the entire group.
Worked Example: A Symmetric Hessian Splits into Two Tests
Worked Example: A Symmetric Hessian Splits into Two Tests
Consider the function, unchanged when \(x\) and \(y\) are exchanged,
Its gradient equations are
Subtracting the first two equations gives \(x-y=0\), so set \(x=y=t\). The first and third equations become \(3t+z=2\) and \(2t+4z=0\). The second gives \(t=-2z\); substituting into the first yields \(-5z=2\). Therefore \(z=-2/5\) and \(x=y=4/5\). Substitution checks the equations: \(8/5+4/5-2/5-2=0\) in each of the first two, and \(8/5-8/5=0\) in the third.
The Hessian is constant:
In the antisymmetric direction \((1,-1,0)\), the eigenvalue is \(1\). In the symmetric plane, use the orthonormal directions \((1,1,0)/\sqrt{2}\) and \((0,0,1)\). The corresponding block is
For coordinates \(u,v\) in this plane, its quadratic form is
This is positive for every nonzero pair \((u,v)\), and the antisymmetric eigenvalue is positive as well. Thus the Hessian is positive definite, and the stationary point is a strict local minimum.
Worked Example: Exchange Symmetry Does Not Force the Diagonal
Worked Example: Exchange Symmetry Does Not Force the Diagonal
Let \(s=x+y\), \(d=x-y\), and define
Exchanging \(x\) and \(y\) keeps \(s\) fixed and changes \(d\) to \(-d\), so \(f\) is unchanged. Since \(s,d\) are invertible linear coordinates, a point is stationary in \(x,y\) exactly when both derivatives with respect to \(s,d\) vanish. These are
Thus \(s\in\{0,-1,1\}\) and \(d\in\{0,-2,2\}\), giving nine stationary points. Recover their original coordinates using \(x=(s+d)/2\), \(y=(s-d)/2\). For example, \(s=1,d=2\) gives \((x,y)=(3/2,-1/2)\), which is not on the diagonal. Exchanging the variables gives the distinct point \((-1/2,3/2)\).
In the \(s,d\) coordinates, the second derivative matrix is diagonal, with entries \(12s^2-4\) and \(12d^2-16\). At \(s=0,d=0\), both entries are negative, so the origin is a strict local maximum. At \(s=\pm1,d=\pm2\), both are positive, so the four corresponding points are strict local minima. At the other four combinations, one entry is positive and the other negative, so those points are saddles.
The classification is valid in the original coordinates as well: the change from \((x,y)\) to \((s,d)\) is invertible, so it preserves whether the quadratic form has positive and negative directions. This example shows why it is unsafe to search only on \(x=y\): symmetry pairs off-diagonal stationary points, but does not rule them out.
Using the Structure Carefully
Symmetry is most useful when it guides a complete search rather than replacing one. For coordinate evenness, inspect the fixed plane where the corresponding coordinate is zero, but also solve for possible stationary points away from it. For variable exchange, exploit the diagonal to simplify equations when appropriate, while checking for off-diagonal solutions. Once candidates are found, use the Hessian structure only at points where the relevant symmetry actually fixes the point.
For a three-variable Hessian, positive definiteness gives a strict local minimum and negative definiteness gives a strict local maximum. An indefinite Hessian gives a saddle. If the Hessian is only semidefinite, its second-derivative test may be inconclusive; symmetry does not remove the need for further analysis in that case.
Check Your Understanding
Use the symmetry and Hessian ideas above to answer these questions.
- If \(f\) is even in \(x\), which first and mixed second partial derivatives must vanish on \(x=0\)?
- If \(f(x,y,z)=f(y,x,z)\) and \((a,b,c)\) is stationary, what other point must be stationary?
- Why does the Hessian at a point with \(x=y\) separate the direction \((1,-1,0)\) from the plane \(x=y\)?
- Does variable-exchange symmetry imply that every stationary point has \(x=y\)? Explain.
- What additional analysis may be needed if a symmetry-reduced Hessian is semidefinite?