Suppose that $A$ is a real, $n \times n$ symmetric matrix. Define the map $F: \mathbf{R}^n \to \mathbf{R}$ by $x \mapsto x^T A x$.

Question: For which matrices $A$ is the map $f = F|_{S^{n-1}}$ convex?

Let us give an interpretation for convexity in the sense of the question above.

Definition: Let $M$ be a Riemannian manifold. A map $f: M \to \mathbf{R}$ is convex, provided that for every geodesic $\gamma: [0, 1] \to M$, and every $t \in [0, 1]$, $$ f(\gamma(t)) \leq t f(\gamma(0)) + (1-t) f(\gamma(1)).$$

Let us make three remarks now.

  • First, this definition implies that $f \circ \gamma$ is a convex map for all geodesics $\gamma$ (check this via restricting $\gamma$, yielding yet another geodesic).
  • Secondly, when $M = \mathbf{R}^n$ with the usual flat metric, this definition reduces to $$ f(x_t) \leq t f(x_0) + (1-t) f(x_1), \qquad \mbox{for every $x_0, x_1 \in \mathbf{R}^n$}, $$ where above $x_t:= tx_0 + (1-t)x_1$, $t\in [0, 1]$. In other words, the usual definition of convexity for maps $\mathbf{R}^n \to \mathbf{R}$.

  • Finally, if $S^{n-1}$ is replaced by $\mathbf{R}^n$ above, then the answer to the question is simply for $A$ nonnegative definite.


The notion of convexity you are using is unnatural for functions defined on compact (connected) Riemannian manifolds: With this definition every convex function is constant. Indeed, suppose that $f: M\to {\mathbb R}$ is a convex function on a compact connected Riemannian manifold. Then $f$ is necessarily convex and, hence, attains its maximum at some point $p$. But for every $q\in M$ there is a geodesic $c$ in $M$ containing $q$ and $p$ as its interior point. Thus, the convex function $f|c$ is nonconstant. But a nonconstant convex function on an interval cannot attain its maximum at an interior point. Hence, $f(p)=f(q)$ and, therefore, $f$ is constant. (A small modification of this proof works even if you assume that $f$ is convex only on all distance-minimizing geodesics.)

Applied in the setting of your question, you have the bilinear form $\langle x, y\rangle=x^TAy$ on ${\mathbb R}^n$. Either the corresponding quadratic form $q(x)=f(x)$ is identically zero (which implies that $A=0$), or after multiplying the matrix $A$ by a scalar, we get $q(x)=1$ for all $x\in S^{n-1}$. But a bilinear form is uniquely determined by its quadratic form: $$ \langle x, y\rangle= \frac{1}{2}(q(x+y) - q(x) -q(y)). $$ Hence, $\langle x, y\rangle$ equals the standard dot product. In other words, $A=I$.

| cite | improve this answer | |
  • $\begingroup$ So is there a "natural definition" of convexity for functions defined on compact Riemannian manifolds? $\endgroup$ – Adam Chalumeau Jul 16 '19 at 9:07
  • 1
    $\begingroup$ @AdamChalumeau: Only for locally defined functions. $\endgroup$ – Moishe Kohan Jul 16 '19 at 10:04
  • $\begingroup$ On such manifolds what is the local definition? $\endgroup$ – Drew Brady Jul 16 '19 at 14:29
  • $\begingroup$ @DrewBrady: The same as you gave. Just the function is defined on some open subset in your manifold, not on the entire manifold. The "right" setting for the definition of convexity of a globally defined function is that $M$ is simply-connected complete manifold of nonpositive curvature. Then there is a large supply of convex functions such as distance functions to convex subsets, Busemann functions, etc. $\endgroup$ – Moishe Kohan Jul 16 '19 at 17:29

I provide a partial answer below, but I would be interested to see what people think.

Claim: If $A$ is not a multiple of the identity, then $f$ is not convex.

  1. Let $\lambda_i$, $e_i$ denote the (real) eigenvalue-eigenvector pairs for $A$, for $i = 1, 2, \dots, n$.

  2. Suppose that $\lambda_i < \lambda_j$ for some distinct $i, j \in \{1, 2, \dots, n\}$. Let us consider the geodesics with images lying in $E_{ij} := \mathrm{span}\{e_i, e_j\} \cap S^{n-1}$.

  3. For $x \in E_{ij}$, write $x = c_i e_i + c_j e_j$, for $c_i^2 + c_j^2 = 1$, and note that $f|_{E_{ij}}(x) = \lambda_i c_i^2 + \lambda_j c_j^2$. We may reparameterize this as $f|_{E_{ij}}(\theta) = \lambda_i \cos^2(\theta) + \lambda_j \sin^2(\theta)$.

  4. Geodesics $\gamma$ with images in $E_{ij}$ correspond to connected intervals $I \subset [0, 2\pi)$ (of length less than $\pi$). Hence such maps $f \circ \gamma$ are restrictions of $f_{E_{ij}}$ to intervals.

  5. There exists a connected interval $I$ of length less than $\pi$ on which $\cos^2(\theta)$ is neither convex nor concave.
  6. Hence, if $\lambda_i < \lambda_j$, $f \circ \gamma$ is not convex for some $\gamma$ whose image lies in $E_{ij}$. Thus $f$ is not convex.
  7. If no such pair $i, j$ exists, then $A = cI$ for some $c \in \mathbf{R}$.
| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.