Question

Let \(f: S^{2} \rightarrow R^{1}\) be a smooth function on the sphere all of whose critical points are simple (i.e., the second differential at each critical point is nonde. generate). Prove that \(m_{0}-m_{1}+m_{2}=2\), where \(m_{i}\) is the number of critical points whose negative index of inertia of the second differential is \(\mathrm{i}\).

Short answer

Based on the step-by-step solution provided, complete the following short answer: Prove that for a smooth function \(f: S^{2} \rightarrow R^{1}\) with simple critical points, the number of critical points with a negative index of inertia of the second differential equals to \(i\) is given by \(m_0 - m_1 + m_2 = 2\). Solution: First, we find an expression for the index of inertia of the second differential at a critical point by considering the Hessian matrix. Then, we use the Euler-Poincaré characteristic formula to relate the number of critical points with the index of inertia to the Euler characteristic of the manifold. Finally, we show that because all critical points of \(f\) are simple, the index of inertia of the Hessian matrix at critical points can only take the values 0, 1 or 2, leading to the required result: \(m_0 - m_1 + m_2 = 2\).

Step-by-step solution

Find an expression for the index of inertia

To find the index of inertia of the second differential at a critical point, we need to determine the Hessian matrix of the function \(f\) at a critical point \((x, y, z)\) on the unit sphere \(S^2\). The Hessian matrix \(H\) is given by: $$H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^2 f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z^2} \end{bmatrix}$$ Since \((x, y, z)\) is on the unit sphere, we have the constraint \(x^2 + y^2 + z^2 = 1\). To incorporate this condition, we introduce a Lagrange multiplier \(\lambda\) and consider the Hessian matrix of the Lagrangian function \(L = f(x, y, z) - \lambda(x^2 + y^2 + z^2)\) instead: $$H_L = \begin{bmatrix} \frac{\partial^2 L}{\partial x^2} & \frac{\partial^2 L}{\partial x \partial y} & \frac{\partial^2 L}{\partial x \partial z} \\ \frac{\partial^2 L}{\partial y \partial x} & \frac{\partial^2 L}{\partial y^2} & \frac{\partial^2 L}{\partial y \partial z} \\ \frac{\partial^2 L}{\partial z \partial x} & \frac{\partial^2 L}{\partial z \partial y} & \frac{\partial^2 L}{\partial z^2} \end{bmatrix}$$ The index of inertia is the number of negative eigenvalues of the Hessian matrix \(H_L\).

Use the Euler-Poincaré characteristic formula

The Euler-Poincaré characteristic formula for a smooth function on a 2-dimensional manifold is given by: $$\sum_{i=0}^{n} (-1)^ia_i = \chi(M)$$ where \(n = 2\) in our case (as \(S^2\) is a 2-dimensional sphere), \(a_i\) is the number of critical points with an index of inertia \(i\), and \(\chi(M)\) is the Euler characteristic of the manifold \(M\) (which is 2 for the sphere \(S^2\)). In our case, the formula becomes: $$m_0 - m_1 + m_2 = 2$$

Prove the required condition

Since all critical points of \(f\) are simple, it implies that the Hessian matrix \(H_L\) at all critical points of \(f\) is non-degenerate. This means all eigenvalues of \(H_L\) are nonzero at those critical points. Since the rank of \(H_L\) is equal to 3 (it is a 3x3 matrix), there can be at most 3 non-zero eigenvalues. Moreover, since \(H_L\) corresponds to a second differential on the sphere, it must be either positive or negative definite, or have a signature (+,-,-) or (-,+,+). Thus, the index of inertia of the Hessian matrix \(H_L\) at critical points can only take the values 0, 1 or 2. Now by substituting the required values into the Euler-Poincaré formula, we arrive at the desired result: $$m_0 - m_1 + m_2 = 2$$ This proves the given condition for the function \(f: S^{2} \rightarrow R^{1}\) under the provided constraints.

Key concepts

Understanding the Hessian Matrix

When exploring critical points of a function, the Hessian matrix becomes a vital tool. It's essentially a square matrix of second-order partial derivatives. Imagine you have a function like the one in our exercise, defined on the surface of a sphere, and you want to understand the nature of its critical points. At any such point, the Hessian matrix will capture the curvature of the function in various directions.

Constructing this matrix for a spherical function requires a special touch: since points lie on the sphere, we work under the constraint that the sum of the squares of the coordinates equals one, enforcing this with a Lagrange multiplier. The altered function with the multiplier included is the Lagrangian, and it's the second derivatives of this, not the original function, that constitute our Hessian matrix here.

Why do we care about the Hessian? Because it tells us whether a critical point is a peak, a valley, or a saddle point. These are determined by the eigenvalues of the Hessian: all positive eigenvalues mean we're at a peak, all negative mean a valley, and a mix implies a saddle point. However, for our sphere function, we must also consider the inherent constraints of the spherical surface, which add a layer of complexity.

Deciphering the Index of Inertia

The index of inertia is somewhat like an identity badge for a critical point—it identifies whether the point is a local maximum, minimum, or a saddle point based on the sign of the eigenvalues of the Hessian matrix. In simpler terms, if we consider the number of negative eigenvalues, we get the index of inertia. For our spherical function, this can be 0, 1, or 2, corresponding to a local minimum, saddle point, and local maximum, respectively.

The beauty of simple critical points, as stated in our exercise, is that they provide a clear separation between these types. With non-degenerate second differentials, there's no ambiguity, and each critical point garners a distinct index.

Practical Importance

Why wrestle with the index of inertia? Because in physics, it's tied to stability—objects want to rest at minimum energy points, so knowing the nature of critical points can predict physical outcomes. Similarly, in optimization problems, we hunt for minima, so distinguishing these points guides efficient solution paths.

The Euler-Poincaré Characteristic

The Euler-Poincaré characteristic is a grand unifier in topology, providing a bridge between geometry and algebra. In essence, it's a number that describes a shape's structure, counting features like vertices, edges, and faces. For 2-dimensional surfaces like the sphere (\( S^2 \)) in our exercise, this characteristic is incredibly special—it equals 2, no matter how you distort the sphere.

Our exercise turns this concept into an elegant formula that relates to the critical points of a function. By assigning alternating signs to counts of critical points based on their index of inertia and summing them, we arrive at this characteristic. It's an assertion of balance among the types of critical points, and it stands true for any simple function on a sphere.

What's astonishing is how this abstract mathematical invariant, stemming from the terrain of topology, manifests in the analysis of a function's critical points on a geometric shape. It joins the study of shapes with the calculus of variations and optimization, showcasing the pervasive interconnectivity of different areas in math.