1
$\begingroup$

Let $M$ be a Riemannian manifold defined by a nonsingular algebraic variety.

I wonder if $M$ has a bounded sectional curvature?

Thanks in advance.

  • 1
    Did you try any examples? I tried $(z, z^n)$ and they all have bounded cuvratures. – 2017-02-11
  • 0
    I am not an expert on differential geometry. I studied this example $M(\tau) =\{W \in [0,1]^{m \times m} | \frac{1}{m^3}tr(W^3)=\tau\}$. I conjecture that $M(\tau)$ has negative curvature for every $\tau \in (0,1)$. My question is more related to find literature that have addressed this type of problem. – 2017-02-11
  • 0
    Can you be more specific: Algebraic varieties do not come equipped with some god-given Riemannian metrics. For instance, do you mean a complex affine subvariety with Riemannian metric induced from the standard metric $C^n$? – 2017-02-11
  • 0
    Because $M(\tau)$ is non singular since the normal vector on $M(\tau)$ never is zero. Hence $M(\tau)$ it is a differentiable manifold and itis possible to define a Riemannian metric..Maybe it is better to define $M(\tau)$ as $M(\tau) = \{W \in \mathbb{R}^{m \times m } | \frac{1}{m^2} tr(W^3) = \tau \}$ – 2017-02-12
  • 0
    OK: What I can infer from your comment is that by a "nonsingular algebraic variety" you mean a subset of $R^N$ given by a system of algebraic equations which is a smooth submanifold. With this definition, the conjecture about bounded curvature (of the induced Riemannian metric) is false. As for the specific subset you mentioned, I do not know, it requires some rather unpleasant computations. – 2017-02-12
  • 0
    Yes, there are ugly computations for tackling the problem directly, but I am looking for alternative approaches. By the way one question: I think that every compact Riemannian manifold has a bounded curvature?. – 2017-02-13
  • 0
    Yes, it does. This follows from continuity of the sectional curvature as a function on the 2-plane bundle over the manifold. – 2017-02-13
  • 0
    @MoisheCohen Can you give some examples of algebraic varieties with unbounded curvatures? – 2017-02-14
  • 1
    @JohnMa: $x(y^2+z^2+w^2)=1$. – 2017-02-14
  • 0
    @JohnMa: Thanks for your comment. – 2017-02-14

0 Answers 0