Disclaimer: First, being a software engineer, I did not properly study the subject besides quickly reading some wikipedia pages, but the few things I found are fascinating. Second, sorry for this unappealing block of text and these probably not so clear questions.

Edit 2: After way longer that one can forgive, I realized, the $f$ function I was considering in the text below is not at all a real valued function onto the manifold but the implicit function defining the manifold. Sorry about the complete mix-up, not sure if what is remaining makes any sense. I will probably ask a new question rather than re-edit this one, if I have more specific points in mind.

I have some visual intuition about the fact that given a riemannian manifold $M$ and a differentiable real valued function $f$ on $M$, if $f$ is "smooth enough" around its critical points, the Jacobian of the gradient (aka. Hessian) will not be "perturbed" by $f$ and so we will be in fact able to measure the riemannian curvature.

But my visual intuition also tells me that around the other non-critical points, $f$ is only skewing the curvature measure and that this "skewing" should be exactly accounted for by the forms dual to the gradients. If we "subtract" these forms to f we should get a smooth function with critical points everywhere and hopefully be able to measure the curvature on all $M$.

Although I realize this "subtraction" makes no sense as we would subtract $0$ everywhere. No, we need to do, is only need to subtract $0$ at a given point and as we roll the tangent spaces we should update the form but keep track of our given point in the new tangent space... and then I browsing wikipedia realized what I needed was called a connection (between tangent spaces).

I also found the fundamental theorem of riemannian geometry which state that given a metric there is a unique torsion free connection. And visually it seems quite clear that the torsion of the chosen connection should indeed perturb a measure of the curvature. So I guess there is no hope for my first dream of being able to measure the Riemanian curvature with for only tools the Hessian and the gradient of a smooth enough real valued function, is it ?

But with the same reasoning, I would conjecture that given a smooth (it appears to be the right adjective there) manifold, a connection and a smooth enough real valued function $f$ we should be able to build a Riemanian metric. The curvature at a given point should be measured by rolling tangent spaces around it subtracting the value of the gradients-dual forms to $f$ and taking the Hessian. Is such a construction possible ?

Sorry if the question is not exactly clear, I was quite excited by these findings and wanted to share them because, first, I might be partially/completely wrong and second, you might give me some other interesting insight on this great topic.

Edit: After thinking a bit more about it I realized I was not clear about what the hessian was. Because, it is in fact defined in terms of the Levi-Civita connection since you need to derive the gradient while "moving" it between tangent spaces. So you need (and have) a connection from the start. So I guess all the insight in the text above could be summarized as: Given a riemanian manifold $(M, g)$ and a smooth (enough) real-valued function $f$ on $M$.

Have we, $g(\theta) = \nabla\nabla(f - df(\theta)^{*})(\theta)$ ?

where $\nabla$ is the Levi-Civita connection and $df(\theta)^{*}$ is the form dual to the differential of $f$ at $\theta$. But I may have missed something again and this formula might very well be wrong...

  • 1
    $\begingroup$ There are over a thousand questions being asked at MSE every day so there is certainly no shortage of room for extra questions, so you may do better to subdivide this into a number of separate questions (you can start by limiting this particular question to the first three paragraphs). $\endgroup$ – Mikhail Katz Aug 24 '17 at 15:01

This is not a complete answer to all your questions, but is at least somewhat relevant.

It seems like you're asking how to construct a metric from functions. What comes to my mind, which you may find interesting, is the concept of Finsler manifolds.

For a manifold $M$, with tangent bundle $TM$, define a function (called the fundamental tensor) $F$ on $TM$. Assume $F$ is positively homogeneous of degree 1, positive for all non-zero tangent vectors, and $$ \frac{\partial (F(x,\dot{x})^2)}{\partial \dot{x}^i\partial \dot{x}^j}\xi^i\xi^j > 0 \text{ if } \xi^k\ne 0 $$ i.e. it is positive definite.

Then $F$ defines a norm on $T_xM$ with line element length $ds = F(x,dx)$, and thus a metric as well.

A Riemannian metric is a special case of this. In local coordinates,
$$\tag{1} F(x,dx)=\sqrt{g_{ij}(x)\, dx^i dx^j} $$ so in this case $$ \frac{\partial (F(x,\dot{x})^2)}{\partial \dot{x}^i\partial \dot{x}^j}\xi^i\xi^j= g_{ij}(x)\xi^i\xi^j $$ which is exactly the metric on the tangent space we are familiar with.

In other words, to define a metric from a function (at least locally), you need a rather special one. In general, it can depend on $TM$, but in the case that it is of the special form in (1), it is Riemannian. In such a case, you are basically constructing a function whose mixed partials (wrt to the tangent vectors' components) form the metric tensor. Notice the similarity to a Hessian.

(Sorry if I misinterpreted your question!)

  • $\begingroup$ That is completely related to my question. Thanks ! I don't have much time to delve into it now but I will let you know if I have some questions/remarks. $\endgroup$ – matovitch Aug 24 '17 at 18:34

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