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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 in $TM$ 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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1  
Have you tried expressing $\frac{Dv}{ds}(0)$ with help of Christoffels symbols for some local chart? –  Olivier Bégassat Jul 19 '11 at 22:53
    
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. –  Chu Jul 19 '11 at 23:24
    
@Chu but does it not give you an exact expression that is obviously bilinear? That's certainly what I'd expect. –  Olivier Bégassat Jul 20 '11 at 1:19
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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. –  Chu Jul 20 '11 at 2:01
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@John the one on the right may also be degenerate. Consider the case that $p(t)$ is geodesic, $v(t)$ is its unit tangent. –  Willie Wong Jul 20 '11 at 8:44

1 Answer 1

up vote 7 down vote accepted

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.
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That's a very nice explanation. Thanks. –  John M Jul 20 '11 at 13:39
    
This is an awesome explanation! thank you very much! –  Chu Jul 21 '11 at 0:02
    
how does this construction extend two to a pair of $v\in T_pM$ and $w\in TqM$ for different points $p,q$ in $M$?? –  janmarqz Jan 12 at 3:26
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@janmarqz: eh, what? If two elements of the tangent bundle are not based in the same point, they cannot be compared. –  Willie Wong Jan 16 at 9:55
    
@WillieWong Yes, you are right. The source to my confusion is the use of the word metric in the concept of Riemannian structure. –  janmarqz Jan 16 at 18:44

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