Can We Choose a Coordinate System for a Levi-Civita Connection on Manifolds?

[Orodruin]

What's the connection? Do you just assume that parallel transport of a vector $V$ leaves the components $V^x$ and $V^y$ unchanged?

You assume that your defined basis is parallel, i.e., if your basis is $\{X,Y\}$, then $\nabla_V X = \nabla_V Y = 0$ for all $V$ (which is what it means for $X$ and $Y$ to be parallel fields). In effect, if you write a vector $V = V^x X + V^y Y$, then yes, $V^x$ and $V^y$ constant along a line would lead to the vector being parallel transported along it. This completely defines a flat connection on the manifold (of course, some manifolds do not allow flat connections because they do not allow you to define vector fields that are linearly independent everywhere). It also gives you sufficient information to uniquely compute the connection coefficients. In particular, if $X = X^a \partial_a$, then
$$
0 = \nabla_a X = [(\partial_a X^b) + \Gamma_{ac}^b X^c]\partial_b,
$$
implying that $\Gamma_{ac}^b X^c = - \partial_a X^b$ with the corresponding relation for each of your basis vectors. You therefore have $N^2$ equations per basis vector (one for each combination of $a$ and $b$), leading to a total of $N^3$ relations to determine the general $N^3$ connection coefficients in whatever coordinate system you have chosen. Note that, hardly surprising, if you choose the coordinate basis of some coordinate patch, the connection coefficients all vanish as the components are constants.

I want to be very clear with the fact that this is just a connection on the manifold. In general there will be other possible connections. Some will be metric compatible, some will be torsion free, and only one will be both (the Levi-Civita connection). The connection defined by the procedure above will generally not be either metric compatible or torsion free (but it will be flat and it might be metric compatible or torsion free, depending on how you chose your fields).

 

[Orodruin]

Just to give a concrete example (other than the compass connection on a sphere, which I worked out in my book), consider the orthonormal basis
$$
X = \partial_r, \quad Y = \frac{1}{r}\partial_\phi
$$
on the plane in polar coordinates with the origin excluded. Requiring that $\nabla_a X = \nabla_a Y = 0$ leads to the relations
\begin{align*}
0 &= \nabla_a X = \nabla_a \partial_r = \Gamma_{ar}^b\partial_b, \\
\quad 0 &= \nabla_a Y = \nabla_a[(1/r)\partial_\phi] = [\partial_a (1/r)] \partial_\phi + \frac{1}{r}\Gamma_{a\phi}^b \partial_b.
\end{align*}
The first relation yields $\Gamma_{ar}^b = 0$ and the second with $a = \phi$ yields $\Gamma_{\phi\phi}^b = 0$. For $a = r$, the second relation becomes
$$
\Gamma_{r\phi}^b \partial_b = - r \frac{\partial(1/r)}{\partial r} \partial_\phi = \frac{1}{r}\partial_\phi,
$$
from which we can read off $\Gamma_{r\phi}^r = 0$ and $\Gamma_{r\phi}^\phi = 1/r$.

Edit: We can also explicitly check that this connection is indeed metric compatible. We generally have
$$
\nabla_a g_{bc} = \partial_a g_{bc} - \Gamma_{ab}^d g_{dc} - \Gamma_{ac}^d g_{bd}.
$$
Since the only non-zero connection coefficient is $\Gamma_{r\phi}^\phi$, we can write this for $a = \phi$ as
$$
\nabla_\phi g_{bc} = \partial_\phi g_{bc} = 0,
$$
since none of the metric components depend on $\phi$. For $a = r$ we instead obtain
$$
\nabla_r g_{bc} = \partial_r g_{bc} - \frac{1}{r}(\delta_b^\phi g_{\phi c} + \delta_c^\phi g_{b\phi}).
$$
For the off-diagonals, e.g., $b = r$, $c = \phi$, we obtain
$$
\nabla_r g_{r\phi} = \partial_r 0 - \frac{1}{r} (0 r^2 + 1\cdot 0) = 0.
$$
For the diagonals, we have
$$
\nabla_r g_{rr} = \partial_r 1 - \frac{1}{r}(0+0) = 0 \quad \mbox{and} \quad
\nabla_r g_{\phi\phi} = \partial_r r^2 - \frac{1}{r}(r^2 + r^2) = 2r - 2r = 0.
$$
Consequently, the connection is compatible with the standard metric in the plane.

 

[TeethWhitener]

That's not quite right. Think about the surface of the Earth. On every point except the North and South pole, you can define a basis by the following:

• Let $e_y$ be a unit vector pointing North.
• Let $e_x$ be a unit vector tangent to the surface of the Earth which is perpendicular to $e_y$ (so that $e_x$ is 90 degrees clockwise away from $e_y$, looking down on the Earth)

This gives you a basis for each point on the Earth (well, except two---nobody lives there, anyway). But it doesn't define the connection. What you need to define the connection is a way to relate the tangent spaces of nearby points. If you have two points $A$ and $B$ that are close by, you need a way to rewrite a vector $V$ defined at point $A$ to a corresponding vector $V'$ defined at point $B$.

I think your idea for defining the connection by just giving a basis at each point might have been to say: Define $V'$ at $B$ by:

• $V'^x = V^x$
• $V'^y = V^y$

That's simple enough, but it can't possibly work on a curved surface.

You do not need the coordinate system, but if you have a set of basis vector fields, then you can define a connection by defining how the connection acts on those fields. In particular, you could define those fields to be parallel. However, there is no guarantee that this connection will have particular properties, such as being the Levi-Civita connection.

Thank you both for your answers. I’m just trying to feel out an intuition for the connection, per the definition in post 6:
$$\nabla_{\mu}e_{\nu} = \Gamma^{\rho}_{\mu\nu}e_{\rho}$$
Since the connection tells you how the basis vectors change from point to point on the manifold, I was thinking that specifying a basis at every point would give the connection a fortiori. Of course it’s not necessarily the Levi-Civita connection, but as long as it meets some minimal criterion of smoothness, would it still be a connection?

 

[Orodruin]

Thank you both for your answers. I’m just trying to feel out an intuition for the connection, per the definition in post 6:
$$\nabla_{\mu}e_{\nu} = \Gamma^{\rho}_{\mu\nu}e_{\rho}$$

Note that what I wrote down there was the connection coefficients in the basis $e_\mu$ and what I computed in the previous post was the connection coefficients in the holonomic basis for the polar system.

Since the connection tells you how the basis vectors change from point to point on the manifold, I was thinking that specifying a basis at every point would give the connection a fortiori. Of course it’s not necessarily the Levi-Civita connection, but as long as it meets some minimal criterion of smoothness, would it still be a connection?

As I discussed above, it is one way of specifying the connection to define how it acts on a set of vector fields that are independent everywhere. If you know this you can write any vector field as a linear combination of those (with coefficients that are functions) and then apply the product rule. Of course, you need not specify that your basis is parallel, it is also fine to use a basis that is not parallel and specify how the connection acts on it.

 

[martinbn]

Another example:

Let $M=\mathbb R^3$ be the Euclidean space with the usual metric and usual coordinates $(x,y,z)$, for convenience number them $(x_1, x_2, x_3)$. Then consider the connection with Christoffel symbols

$\Gamma_{12}^3=\Gamma_{23}^1=\Gamma_{31}^2=1, \quad\Gamma_{21}^3=\Gamma_{32}^1=\Gamma_{13}^2=-1$

All other symbols are zero. Then it is obviously not symmetric, but it is compatible with the metric (and has minimizing geodesics). This is an exercise in Lee's book.