# Geometric interpretation of horizontal and vertical lift of vector field

In many References such as D.E. Blair, Riemannian Geometry of Contact and Symplectic Manifolds chapter 9, and Differential Geometric Structures By Walter A. Poor Page 54; the horizontal and vertical lift(space) of a vector field on $M$, $X\in\Gamma(TM)$, are defined as follows:

If $X$ is a vector field on $M$, its vertical lift $X^V$ on $TM$ is the vector field defined by $X^V\omega = \omega(X)\circ \pi$, where $\omega$ is a 1-form on $M$, which on the left side of this equation is regarded as a function on $TM$. For an affine connection $\nabla$ on $M$, the horizontal lift $X^H$ of $X$ is defined by $X^H\omega = \nabla_X\omega$.

The span of the horizontal lifts at $t ∈ TM$ is called the horizontal subspace of $T_tTM$.

The other approach is as follows:

It is well-known that the tangent space to $TM$ at $(x, u)$ splits into the direct sum of the vertical subspace $VTM_{(x,u)}=ker\pi_*|_{(x,u)}$ and the horizontal subspace $HTM_{(x,u)}$ with respect to $\nabla$ $$TTM=HTM\oplus VTM.$$ For $X\in T_xM$, there exists a unique vector $X^h$ at the point $(x, u)\in TM$ such that $X^h\in HTM_{(x,u)}$ and $\pi_*(H^h) = X$. $X^h$ is called the horizontal lift of $X$ to $(x, u)$. There is also a unique vector $X^v$ at the point $(x, u)$ such that $X^v\in VTM_{(x,u)}$ and $X.(df) = Xf$ for all functions $f$ on $M$. $X^v$ is called the vertical lift of $X$ to $(x, u)$.

My problems are:

1. How can I split every vector field in $TTM$ into horizontal and vertical part?
2. What is the geometric interpretation of horizontal and vertical spaces?
3. why the tangent sphere bundle and tangent bundle is different in the sense of horizontal and vertical part?

It seems that after solving the question I can to prove the following identities: $$[X^v,Y^v]=0,\quad dX(Y)=Y^h+(\nabla_YX)^v\quad X,Y\in\Gamma(TM).$$

Thanks.

• First, you should notice that no metrix is needed, but only the covariant derivative (connection). Moreover, the construction works essentially the same for arbitrary vector bundles with the necessary small changes. Then you can find this in many books on differential geometry. A more recent one is e.g. the nice book of Peter Michor. – Stefan Waldmann Aug 1 '16 at 6:21

To every vector bundle $\pi: E \to M$ (in your case $E = TM$) we have an associated exact sequence of vector bundles (sometimes called the Atiyah sequence, at least in the principal bundle case): $$0 \to V E \to TE \to \pi^* TM \to 0$$ Here $VE$ denotes the bundle of vertical tangent bundles. Now there are three ways to split an exact sequence:
• Write down an isomorphism between the middle term and the sum of the terms on the right- and left-hand side. In our case, this corresponds to the decomposition $TE = VE \oplus HE$.
• Split on the left, i.e. give a map $TE \to VE$. This is the connection form $\omega$.
• Split on the right, i.e. specify a map $\pi^* TM \to TE$. This corresponds to lifting a tangent vector from $M$ to $E$.
So now it should be pretty clear how to translate between the different viewpoints (modulo some natural isomorphisms). For example, every tangent vector $X \in T_e E$ can be written as a sum $X^v + X^h$, where $X^v = \omega(X)$ and $X^h$ is the horizontal lift of some vector $Y \in T_{\pi(e)}M$.
• Is the $\pi^* TM \simeq HTM$? By this approach, How can I prove the equations $[X^v,Y^v]=0,\quad dX(Y)=Y^h+(\nabla_YX)^v\quad X,Y\in\Gamma(TM).$ – C.F.G Aug 3 '16 at 9:51
• Yes, $\pi^{\ast} TM \simeq HE$ which is just another way of saying the exact sequence splits. You have $VE$ is the kernel of the map $TE \to \pi^{\ast}TM$ and $HE \simeq TE/VE \simeq \pi^{\ast}TM$. But there are many ways to realise the isomorphism $HE \simeq \pi^{\ast} TM$ and these correspond to choosing a connection. This is much the same as say $\mathbb{R}^2 \simeq \mathbb{R}(1, 0) \oplus \mathbb{R} (0, 1) \simeq \mathbb{R} (1, 0) \oplus \mathbb{R} (1, 1)$ etc. The choice of connection corresponds to a choice of complementary subspace $HE$ to $VE \subset TE$. – Paul Bryan Sep 16 '16 at 23:03