0
$\begingroup$

Given a closed Riemannian manifold $(X,g)$ and let $p\colon TX\to X$ be the usual projection, the paper I'm reading asserts that the Levi-Civita connection induces a splitting $T(TX)= H(TX)\oplus V(TX)\cong p^* TX\oplus p^*TX$, so that $$ \begin{aligned}\Omega^\bullet(TX) &= \{ \Gamma(H(TX))\to C^\infty(TX)\}\otimes_{C^\infty(TX)} \{\Gamma(V(TX))\to C^\infty(TX) \} \\ &\cong p^*\Omega^\bullet (X)\otimes_{C^\infty(TX)} p^*\Omega^\bullet(X). \end{aligned} $$

I want to investigate this isomorphism explicitly, in the sense that I can compute the images of the local coordinate functions on $X$ under the two maps

$$ \phi_1\colon C^\infty(X)\xrightarrow{d}\Omega^1(X)\xrightarrow{p^*} p^*\Omega^\bullet(X) \cong \{\Gamma(V(TX))\to C^\infty(TX) \} \hookrightarrow \Omega^\bullet(TX),$$

$$\phi_2\colon C^\infty(X)\xrightarrow{d}\Omega^1(X)\xrightarrow{p^*} p^*\Omega^\bullet(X) \cong \{\Gamma(H(TX))\to C^\infty(TX) \} \hookrightarrow \Omega^\bullet(TX). $$

But I failed finding any material that could tell me how the horizontal vector bundle is splitted out using the Levi-Civita connection, without which I cannot do any computation.

Also, I found that $V(TX) = \ker ( d p\colon T(TX)\to TX)$ on Wikipedia, but I'm not quite sure how the isomorphism $V(TX)\cong p^*TX$ is given. The isomorphism that I came up with is the following:

Let $x_1,\cdots,x_n$ be the coordinate functions of a coordinate neighborhood $U$ of $X$, then $x_1,\cdots,x_n,y_1:=d x_1,\cdots, y_n:=dx_n$ are the coordinates of $p^{-1}(U)\cong U\times\mathbb{R}^n$ of $TX$. Then the fibres of $V(TX)|_{p^{-1}(U)}$ are exactly the spans of $\frac{\partial}{\partial y_i}$'s. Since the fibres of $p^*TX|_{p^{-1}(U)}$ are the spans of $\frac{\partial}{\partial x_i}$'s, the isomorphism is given by $\frac{\partial}{\partial x_i}\mapsto \frac{\partial}{\partial y_i} $ along each fibre.

So my questions are:

  1. How the horizontal vector bundle is constructed using the Levi-Civita connection?
  2. Is the above isomorphism the correct one?
  3. What are the images of the local coordinate functions on $X$ under the maps $\phi_1$ and $\phi_2$ given above? (Probably I will be able to do this by myself with (1) and (2) solved, but I'm putting it here anyway, since it is the ultimate goal of mine)

To be honest, I'm not familiar to Riemannian geometry, so it would be the best if there exists a self-contained material that could provide a throughout treatment to this matter.

Thanks in advance for any help.


EDIT: I thought it over and a new question (4) arises: See, a choice of coordinates $x_1,\cdots,x_n$ of a neighborhood $U$ of $X$ induces coordinates $x_1,\cdots,x_n,y_1:=d x_1,\cdots, y_n:=dx_n$ of $p^{-1}(U)$ of $TX$, which gives a basis $\frac{\partial }{\partial x_1},\cdots,\frac{\partial }{\partial x_n},\frac{\partial }{\partial y_1},\cdots,\frac{\partial }{\partial y_n}$ of $T(TX)|_{p^{-1}(U)}$. While the span of $\frac{\partial }{\partial y_i}$’s gives $V(TX)| _{p^{-1}(U)} $, why isn’t $H(TX)|_{p^{-1}(U)}$ simply the span of $\frac{\partial}{\partial x_i}$’s? I don’t see any trouble gluing these choices of $H(TX)|_{p^{-1}(U)}$‘s up as $U$ varies, which seems to have no difference with $p^*TX$. Am I missing something in this construction, or is there any other reason to define $H(TX)$ differently? This approach looks canonical — not even a Riemannian metric is required.

$\endgroup$

1 Answer 1

2
$\begingroup$

I am just answering to (1) and (2): The horizontal bundle is defined as follows. Consider a point $v\in T_pX\subset TX,$ and $w\in T_pm$. Consider a curve $\gamma$ through $p$ with $\gamma'(0)=w.$ Consider the parallel transport $V$ through $v$ along $\gamma,$ i.e. the curve $V\colon (-\epsilon, \epsilon)\to TX$ satisfying $$\nabla_\gamma' V=0\quad\text{and}\quad V(0)=v.$$ Then, $V'(0)\in T_vTX$ is per Definition horizontal, i.e. $$V'(0)\in H_v(TX).$$ Of course, one has to show that this gives a well-defined bundle, which is isomorphic to $p^*TX.$ The later isomorphism is given by $$w\mapsto V'(0),$$ with inverse $$V'(0)\mapsto d_vp(V'(0)).$$

(2) seems correct. There is the following invariant way describing the isomorphism: Take $v\in T_pX\subset TX,$ and $w\in T_pX.$ Then, $$t\in(-\epsilon,\epsilon)\mapsto v+t w\in T_pX$$ is a vertical curve $\gamma_w$. The isomorphism then is $$w\in T_{p(v)} X\mapsto \gamma_w'(0)\in V_vV.$$

$\endgroup$
1
  • $\begingroup$ Thanks! But I'm a little bit confused by what I've found: since $T(TX)=H(TX)\oplus V(TX)$ and $V(TX)$ lives in the kernel of $dp$, I find that the composition $T(TX)\twoheadrightarrow H(TX)\cong p^*TX$ is exactly the map $\widetilde{dp} :T(TX)\to p^*TX$ induced by $dp:T(TX)\to TX$ and the universal property of the pullback bundle. So essentially the connection plays no role for my "ultimate goal" -- am I missing something, or is this really the case? $\endgroup$
    – Shana
    May 15 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.