# Is the tangent bundle of the top space the direct sum of the pull back of tangent bundle and relative tangent bundle?

Let $Y_1$ and $Y_2$ be two real compact differentiable manifolds such that $\pi:Y_2\to Y_1$ is an $S^1$ bundle and $p:Y_2\to S^1$ is a bundle with fibre $Y_1$. Let $TY_i$ be the tangent bundle of $Y_i$. Let $\pi^*TY_1$ be the pull back of the $TY_1$ to $Y_2$. There is a natural map from $TY_2\to \pi^*TY_1$. Let $T_vY_2$ denote the kernel of this map. $T_vY_2$ is the relative tangent bundle of $Y_2$ wrt $Y_1$. Let $p^*TS^1$ be the pull back of the tangent bundle of $S^1$ to $Y_2$ (which is the trivial line bundle).

Can we say that $TY_2\cong T_vY_2\oplus p^*TS^1$?

From this answer I thought that I can conclude that $TY_2\cong T_vY_2\oplus \pi^*TY_1$ (Is that right?). But I'm not sure if I can proceed from here of if this helps me in any way.

Thank you.

The relative tangent bundle consists of vectors that are tangent to the fibers and so $T_vY_2$ is a one-dimensional vector bundle and so it $p^{*}(TS^1)$ (being the pullback of a one-dimensional bundle). Hence, it is not possible for $TY_2 \cong T_vY_2 \oplus p^{*}(TS^1)$ unless $Y_2$ has dimension two. If $Y_2$ is two dimensional, then $Y_1$ must be $S^1$ and then the isomorphism does hold because of the identity

$$TY_2 \cong T_V Y_2 \oplus \pi^{*} TY_1$$

and since both $\pi^{*} TY_1$ and $p^{*} TS^1$ are trivial line bundles.

• No matter what the dimension of $Y_2$, and $Y_1$, the dimension of $T_vY_2$ is always one?
– R_D
Aug 31 '16 at 2:00
• Yep. The local model of a fiber bundle is of the form $U \times F \rightarrow U$ and so the differential is onto with kernel of dimension $\dim F$ (in your case, one). Aug 31 '16 at 2:03
• I got it now. Thank you!
– R_D
Aug 31 '16 at 2:06