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If $F= a+ k_1 b + k_2 b^2/ a$ , where $a$ is the Riemannian metric , $a = \sqrt{a_{ij} y^i y^j}$ and $b$ is the 1-form $b = b_i y^i$, then find the fundamental form $g_{ij}$ ?? I need the steps for $g_{ij}$?

Kortlek
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SAB
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  • I do not understand your notation; usually the Finsler function $F$ is a function on the tangent bundle of the manifold $M$, i.e. for all $x\in M$, $F_x$ takes vectors in the tangent space $T_xM$ and returns scalars (see for example http://en.wikipedia.org/wiki/Finsler_manifold). I do not understand the formula $F=a+k_1b+k_2b^2/a$: $a$ is the Riemannian metric, while $b$ is a 1-form (which implies $b^2=0$)...what am I missing here? – Avitus Aug 11 '13 at 15:33
  • If F is a Minkowski Norm, Could you find the g_ij ? Here, k1 and k2 are constants. – SAB Aug 13 '13 at 13:09

1 Answers1

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The fundamental tensor of a Finsler metric is its Hessian in the "vertical" direction. That is to say,

$$ g_{v}(u,w) = \frac12 \frac{\partial^2}{\partial s \partial t} F(x, v + su + tw)^2 $$

where $v,u,w\in T_x M$.

Suppose, as you indicated, that

$$ F(x,v) = \sqrt{a_{ij}(x) v^i v^j} + k_1 b_i(x) v^i + k_2 \frac{(b_i v^i)^2}{\sqrt{a_{ij}v^i v^j}} $$

we have that

$$ F(x,v)^2 = a_{ij}(x) v^i v^j + (k_1^2+2k_2) (b_i(x) v^i)^2 + k_2 \frac{(b_i v^i)^4}{a_{ij} v^i v^j} + 2 \sqrt{a_{ij}(x) v^i v^j} k_1 b_k v^k + 2 k_1 k_2 \frac{(b_k v^k)^3}{\sqrt{a_{ij} v^i v^j}} $$ we can directly compute its fundamental tensor.

The first two terms contribute to $g_v(u,w)$

$$ a_{ij} u^i w^j + (k_1^2 + 2 k_2)(b_i u^i)(b_j w^j) $$

the remaining three terms require a bit more computation but can be dealt with using the standard rules of multivariable calculus.

Willie Wong
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