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I know that the metric tensor can not always be formulated as a Hessian, but sometimes it can. Can you help me to understand what the special conditions are under which the metric tensor is a Hessian of the first fundamental form? (Note: Einstein notation follows)

If the first fundamental form of some vectors of same length u and v is $$I(u,v) = <u,v> = \vec u^T g \vec v = u^i v^j g_{ij} = u^i v^j <x_i , x_j>, $$ then why doesn't the Hessian of the first fundamental form yield the same quantities as the metric tensor?

$$H_{ij} I(u,v) = \frac{\delta}{\delta x_i} \frac{\delta}{\delta x_j} I(u,v) = u^k v^l \frac{\delta}{\delta x_i} \frac{\delta}{\delta x_j} <x_k , x_l> \ne <x_i , x_j> $$

Can you also provide an equation that shows the relationship between the metric tensor and the Hessian of the first fundamental form?

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  • $\begingroup$ What do you mean by the "first fundamental form? Or its Hessian? As far as I know, the term "first fundamental form" is used mainly for a submanifold of Euclidean space and refers to the induced Riemannian metric on the submanifold. $\endgroup$
    – Deane Yang
    Commented Sep 5, 2014 at 16:40
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    $\begingroup$ AFAIK the "metric tensor" is the "first fundamental form". These are synonyms. A metric is a symmetric positive bilinear form on the tangent space. The Hessian of a function is a bilinear form. To take a Hessian you need a connection lying around, because otherwise how can you talk about changes in the derivative at nearby points? Sometimes you can find a function whose Hessian is the actual metric. For instance, the metric on the Poincare disk (IIRC) is the hessian of a function. $\endgroup$
    – Steven Gubkin
    Commented Sep 5, 2014 at 16:57
  • $\begingroup$ If you are in the setting of a submanifold of Euclidean space, as Deane Yang suggests, then the connection you are using to define you Hessian is the Levi-Civita connection. $\endgroup$
    – Steven Gubkin
    Commented Sep 5, 2014 at 17:19
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    $\begingroup$ You may also be interested in this question and answer: mathoverflow.net/questions/123633/… $\endgroup$
    – Steven Gubkin
    Commented Sep 5, 2014 at 18:27
  • $\begingroup$ Steven, the link you posted is very helpful. I don't think I'm dealing with a submanifold of Euclidean space. General relativity is formulated in non-Euclidean space. Also, I thought the first fundamental form was the entire order 2 polynomial formed as the sum of x^i x^j g_ij on all i and j. Is the first fundamental form just the metric tensor? What do you call that order 2 polynomial? $\endgroup$
    – Heather
    Commented Sep 5, 2014 at 22:18

2 Answers 2

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The first fundamental form is just the metric. If you express it in Riemannian normal coordinates (the Riemannian exponential map centered at $x_0$), the the first derivative of this vanishes at $x_0$, and the Hessian at $x_0$ (which exists since the first derivative vanishes) is the purely covariant form of the Riemann curvature tensor. This is the way how Riemann found his tensor in his price essay for the Paris academy:

  • Commentatio mathematica, qua respondere tentatur quaestioni ab Illma Academia Parisiensi propositae: ``Trouver quel doit être l'état calorifique d'un corps solide homogène indéfeni pour qu'un système de courbes isothermes, à un instant donné, restent isothermes après un temps quelconque, de telle sorte que la température d'un point puisse s'exprimer en fonction du temps et de deux autres variables indépendantes.'' (1861, Gesammelte Mathematische Werke, Zweite Auflage, 391-404). (pdf)

A very good explanation in English of Riemann's work is contained in:

  • Michael Spival: A comprehensive introduction to differential geometry, Volume 2. Publish or Perish, Inc.
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I'm not sure what you mean by $x^i$ when you write $g_{ij}x^ix^j$. If they are local co-ordinates, then this quantity is co-ordinate-dependent and has no geometric meaning or name. If on the other hand, they are the components of a tangent vector, this quadratic polynomial makes sense and defines a function on the tangent bundle. This function is indeed the metric itself and could be called the first fundamental form.

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