I'm trying to solve the following problem (from do Carmo's Riemannian Geometry). particularly I'm having trouble proving that the inner product defined is bilinear.

Problem. It is possible to define a Riemannian metric in the tangent bundle $TM$ of a Riemannian manifold $M$ in the following manner. Let $(p,v)\in TM$ and $V,W$ be tangent vectors to $TM$ at $(p,v).$ Choose curves in $TM$

$$\begin{align*}& \alpha: t \rightarrow (p(t), v(t))\\ & \beta: t \rightarrow (q(s), w(s))\end{align*}$$

with $p(0)=q(0)=p$, $v(0)=w(0)=v$ and $V=\alpha'(0)$, $W=\beta'(0)$. Define an inner product on $TM$ by

$$\langle V,W \rangle_{(p,v)} = \left\langle d\pi(V), d\pi(W) \right\rangle_{p} + \left\langle \frac{Dv}{dt}(0) , \frac{Dw}{ds}(0) \right \rangle_{p},$$

where $\pi:TM \rightarrow M$. Prove this is a well-defined Riemannian metric on $TM$.

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    $\begingroup$ Have you tried expressing $\frac{Dv}{ds}(0)$ with help of Christoffels symbols for some local chart? $\endgroup$ – Olivier Bégassat Jul 19 '11 at 22:53
  • $\begingroup$ Yep, but maybe I'm missing some keypoint when doing it or something. I couldn't see beyond just, using the bilinearity of $< , >_{p}$ when doing so. $\endgroup$ – Sak Jul 19 '11 at 23:24
  • $\begingroup$ @Chu but does it not give you an exact expression that is obviously bilinear? That's certainly what I'd expect. $\endgroup$ – Olivier Bégassat Jul 20 '11 at 1:19
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    $\begingroup$ I'm not sure but, I think he one on the left is not a riemannian metric by itself because since $d\pi$ is not necessarily inyective, it might result degenerate. $\endgroup$ – Sak Jul 20 '11 at 2:01
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    $\begingroup$ @John the one on the right may also be degenerate. Consider the case that $p(t)$ is geodesic, $v(t)$ is its unit tangent. $\endgroup$ – Willie Wong Jul 20 '11 at 8:44

The two terms you wrote down are roughly the horizontal and vertical parts of the metric. Roughly speaking, the first part gives the metric for the first factor $T_pM$ of $T_{p,v}TM$. The second part gives the metric for the second factor $T_vT_pM$.

By the definition of the projection operator $\pi$ and the definition of the covariant derivative on $(M,\langle\cdot\rangle)$, it is clear that the expression you wrote down is coordinate independent. There are several things to check to make sure that it is a metric

  • It is positive definite
  • It is bilinear
  • It is tensorial

Since we already have coordinate independence, it is most convenient to work over a fixed coordinate system.

Let $\{x^1,\ldots,x^n\}$ be a system of coordinates for $M$; this can be extended to a system of local coordinates $\{x^1,\ldots,x^n;y^1,\ldots,y^n\}$ for $TM$ where $(x,y)$ corresponds to the point $(p,v)\in TM$ with $p$ the point in $M$ specified by $x$, and $v\in T_pM$ given by $\sum y^i\partial/\partial x^i$.

At a fixed point $(p,v)$, an element of $T_{p,v}TM$ can be then described by $$ V = \sum \xi^i \frac{\partial}{\partial x^i} + \sum \zeta^i \frac{\partial}{\partial y^i}$$ with the projection $$ d\pi(V) = \sum \xi^i \frac{\partial}{\partial x^i} $$

Express the curve $\alpha(t) = (p,v)(t)$ in this coordinates we have that the condition $\alpha'(0) = V$ is simply the statement that $\frac{d}{dt}p^i(0) = \xi^i$ and $\frac{d}{dt}v^i(0) = \zeta^i$.

With this we can compute $$ \frac{D}{dt}v^i(0) = \frac{d}{dt} v^i(0) + \Gamma^i_{jk}\left(\frac{d}{dt}p^j(0)\right)\left(v^k(0)\right) $$ using the definition, and where $\Gamma$ is the Christoffel symbol of the Riemannian metric on $M$. This we immediately see to be $$ \frac{D}{dt}v^i(0) = \zeta^i + \Gamma^i_{jk}v^k(0)\xi^j $$ which in fact is a linear map from $T_{p,v}TM$ to $T_pM$.

Now the three properties are easily checked:

  • It is tensorial because the expressions are completely independent of which curve $\alpha$ is chosen, as long as $\alpha'(0)= V$.
  • It is bilinear because $T_{p,v}TM\ni V\mapsto (d\pi(V),\frac{D}{dt}v(0))\in T_pM \oplus T_pM$ is linear, and the Riemannian metric on $M$ is bilinear
  • Since the Riemannian metric induced on $T_pM\oplus T_pM$ is positive definite, to prove that the new object is also positive definite, it suffices to check that the map $V\mapsto (d\pi(V),\frac{D}{dt}v(0))$ is injective. But this is true by direct inspection.
  • $\begingroup$ That's a very nice explanation. Thanks. $\endgroup$ – John M Jul 20 '11 at 13:39
  • $\begingroup$ This is an awesome explanation! thank you very much! $\endgroup$ – Sak Jul 21 '11 at 0:02
  • $\begingroup$ how does this construction extend two to a pair of $v\in T_pM$ and $w\in TqM$ for different points $p,q$ in $M$?? $\endgroup$ – janmarqz Jan 12 '14 at 3:26
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    $\begingroup$ @janmarqz: eh, what? If two elements of the tangent bundle are not based in the same point, they cannot be compared. $\endgroup$ – Willie Wong Jan 16 '14 at 9:55
  • $\begingroup$ @WillieWong Yes, you are right. The source to my confusion is the use of the word metric in the concept of Riemannian structure. $\endgroup$ – janmarqz Jan 16 '14 at 18:44

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