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
    $\begingroup$ 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. $\endgroup$
    – yasmar
    Oct 10, 2012 at 5:21

1 Answer 1


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

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


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