10
$\begingroup$

I have a question about Do Carmo notion of horizontal vector (page 79). So he defines natural metric on $TM$ of manifold $M$. Now he chooses vector $V\in T_{(p,v)}(TM)$ and calls $V$ horizontal vector if it is orthogonal to fiber $\pi^{-1}(p)$ under metric of $TM$ (where $\pi :TM\to M$ is natural projection.

What I am confused is that $V$ and $\pi^{-1}(p)$ do not live in a same space, so metric on $TM$ can not receive as input an element of $\pi^{-1}(p)$. Can someone clarify me how should this be understood?

  • 5
    If $A$ is an affine space, and $x \in A$, then there is a natural identification of $A$ with $T_xA$. The space $\pi^{-1}(p)$ is a submanifold of $TM$. Thus the tangent space at $(p,v) \in TM$ is the direct sum of the tangent space $T_{(p,v)}(\pi^{-1}(p))$ and a 'horizontal' vector space. The former we can identify with $\pi^{-1}(p)$ itself, according to the first sentence of this comment.2012-10-10

1 Answers 1

3

Consider a curve in $TM$ : $c(0)=p,\ c'(0)=X,\ (c(t), tv+x ) $

whose tangent is $ (X, v) $

Vertical Component : We want to define a metric $G$ on $TM$. If $u_i$ is a coordinate for $(M,g)$ where $g_{ij} = g(\partial_{u_i} , \partial_{u_i}), $ then if $c(t)=p$ i.e., $X=0$, $ G_{p,x} ( (0,v=v_i\partial_{u_i}), (0,w=w_i\partial_{u_i}) )=v_i w_j g_{ij} (p) $

Horizontal Component : If $Z$ is a parallel along $c$ with $c(0)=p,\ Z(0)=x,$ then $ 0= (\nabla_X Z)^i = X(Z^i) + X^j Z^k \Gamma_{jk}^i $

Define a curve $ (c(t), Z(t)) $ whose tangent is $ (X,X(Z))= (X, - X^j Z^k \Gamma_{jk}^i \partial_{u_i} ) = (X, - X^j x^k \Gamma_{jk}^i\partial_{u_i} ) $

Decomposition : Then we have orthogonal decomposition : $ (X,v) = (X, - X^j x^k \Gamma_{jk}^i ) + (0, v +X^j x^k \Gamma_{jk}^i ) $

That is $ |(X, - X^j x^k \Gamma_{jk}^i )|^2_{G_{p,x}} = |X|_{g_p}^2 $

$|(X,v)|^2_{G_{p,x}} = |X|^2 + g_{im} ( v^i +X^j x^k \Gamma_{jk}^i )( v^m +X^j x^k \Gamma_{jk}^m) $