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(doCarmo, Riemannian Geometry, p.56, Q2)

I want to prove that the Levi-Civita connection $\nabla$ is given by $$ (\nabla_X Y)(p) = \frac{d}{dt} \Big(P_{c,t_0,t}^{-1}(Y(c(t)) \Big) \Big|_{t=t_0}, $$ where $p \in M$, $c \colon I \to M$ is an integral curve of $X$ through $p$, and $P_{c,t_0,t} \colon T_{c(t_0)}M \to T_{c(t)}M$ is the parallel transport along $c$, from $t_0$ to $t$.

My approach is to use the uniqueness of the Levi-Civita connection (a theorem proved elsewhere in the textbook) and show that the RHS satisfies all of its properties, i.e.

  1. It is an affine connection,
  2. It is symmetric,
  3. It is compatible with the metric.

However, for the first part, I am stuck on proving that $$ \nabla_{fX + gY}Z = f \nabla_X Z + g \nabla_Y Z. $$ So far, I have the following

$$ f \nabla_X Z = f \Big( \frac{d}{dt} \Big( P_{c_X,t_0,t}^{-1}(Z(c_X(t)) \Big) \Big|_{t=t_0} \Big), $$

$$ g \nabla_Y Z = g \Big( \frac{d}{dt} \Big( P_{c_Y,t_0,t}^{-1}(Z(c_Y(t)) \Big) \Big|_{t=t_0} \Big), $$

$$ \nabla_{fX + gY}Z = \frac{d}{dt} \Big( P_{c,t_0,t}^{-1}(Z(c(t)) \Big) \Big|_{t=t_0}, $$ where $$ c_X (t_0) = c_Y (t_0) = c(t_0) = p, $$ $$ \frac{d c_X}{dt} = X(c_X(t)), $$ $$ \frac{d c_Y}{dt} = Y(c_Y(t)), $$ $$ \frac{d c}{dt} = fX(c(t)) + gY(c(t)). $$

I'm sure the solution is something simple like working in local coordinates but I'm having trouble so any direction would be appreciated.

Andrew Whelan
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1 Answers1

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You can probably make your idea work but it won't be easy. The reason is that your formula that recovers the connection from the parallel transport is true not only for the Levi-Civita connection but also for arbitrary connections. This means that in order to identify the right hand side as the Levi-Civita connection, you will need to understand what makes the parallel transport of the Levi-Civita connection special compared with the parallel transport of a general connection. The compatibility with the metric is easy - this implies that parallel transport is an isometry. However, to understand how the symmetry affects the parallel transport is much more delicate (see here for example).

A much less painful way to solve the exercise is to use the notion of a parallel frame along $c$ (which if I remember correctly is introduced in one of the other exercises). Namely, pick some basis of $\xi_1(p), \dots, \xi_n(p)$ of $T_pM$ and extend it by parallel transport to a frame $(\xi_1, \dots, \xi_n)$ of vectors fields along $c$. Now, write the restriction of $Y$ to $c(t)$ as $Y = Y^i(t) \xi_i(c(t))$ (summation convention is in use) and note that

$$ (\nabla_X Y)(c(t)) = \frac{DY(c(t))}{dt} = \dot{Y}^i(t) \xi_i(c(t)) + Y^i(t) \frac{D\xi_i(t)}{dt} = \dot{Y}^i(t) \xi_i (c(t)) $$

which means that the covariant derivative relative to the frame $\xi_i$ is given simply by the regular derivative. Then,

$$ \frac{d}{dt} \left( P_{c,t_0, t}^{-1}(Y(c(t)) \right)|_{t = t_0} = \frac{d}{dt} \left( P_{c,t_0, t}^{-1}(Y^i(t) \xi_i(c(t))) \right)|_{t = t_0} \\ = \frac{d}{dt} \left( Y^i(t) \xi_i(p) \right)|_{t = t_0} = \dot{Y}^i(t_0) \xi_i(p) = (\nabla_X Y)(p).$$

levap
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  • This answers my question beautifully - thank you. However, I'm still curious as to how I'd go about proving the property I mentioned and seeing that the symmetry is the only hard part. – Andrew Whelan Nov 02 '16 at 17:22
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    @AndrewWhelan: I think the best approach is to prove that the expression only depends on the tangent vector defined by $c$ at $c(t_0)$. That is, you can use any curve you like with the same tangent vector and it doesn't have to be an interval curve of $X$. To do that I see no other way than to describe the parallel transport operator in local coordinates and once you do that, you can immediately prove the equality anyway by getting an explicit formula for the expression which agrees with the covariant derivative. – levap Nov 02 '16 at 20:50
  • @levap Sorry for being too late, hope you don't mind answering a small question. How did you justify the fact that $Y^i(t)$ is differentiable? As far as I know, $\xi_i(c(t))$ need not come from a chart so we cannot just apply the definition of $Y$ being smooth, can we? – BigbearZzz Jan 10 '18 at 20:51
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    @BigbearZzz: When you restrict the vector field $Y$ to the curve $c$ the resulting object $Y \circ c$ is a smooth vector field along a curve, or, in other words, a section of the pullback bundle. Now you can show that if you represent a smooth section of some vector bundle locally with respect to a smooth frame, the coefficients will be smooth functions. In your case, the smooth frame is $\xi_i(c(t))$ so it amount to showing that the result of parallel transport is a smooth vector field along the curve. All those facts are shown in most textbooks. – levap Jan 14 '18 at 04:41
  • Is $P_{c,t_0,t}$ $C^\infty(M)$-linear? – Gomes93 May 08 '19 at 19:56
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    @AndreGomes: For fixed $t,t_0$, the map $P_{c,t_0,t}$ is a linear map from $T_{c(t_0)}$ to $T_{c(t)}$ and this is what I use in the proof. It doesn't really make sense to say that $P_{c,t_0,t}$ is linear over $C^{\infty}(M)$ because it is defined only given a specific curve $c$ and then on a tangent space at a specific point. – levap May 09 '19 at 16:06