What does the dual parallel transport, or conjugate affine connection, actually do?

Question

I'm reading both Frank Nielsen's An Elementary Introduction to Information Geometry as well as Amari's Information geometry and its applications, and confused by something they use (I'll focus on AEITIG here).

He first introduces the concept of an affine connection $\nabla$, which is basically a covariant derivative on vector fields, so if we have vector fields $X, Y$, it tells us how to calculate $\nabla_X Y$, i.e., how $Y$ changes as we move in the $X$ direction, at each point. If we have the basis $(\partial_i) = (\frac{\partial}{\partial x_i})$ for a smooth chart $(U, x)$, then exactly what the connection does is defined in terms of the Christoffel symbols:

$\nabla_{\partial_i} \partial_j = \Gamma_{ij}^k \partial_k$

This all makes sense to me. Then, from what I understand, if we want to transport a vector from the tangent plane at one point to another, we could do some sort of integration of that derivative over the path taken, and that would give the final vector.

He then introduces (section 3) the concept of a conjugate connection $\nabla^\ast$ that's conjugate to a given connection $\nabla$. He mentions a few properties of it, like an identity that must be satisfied for it to be conjugate, but never what it actually does or how to figure it out. Here's that identity:

What I don't understand is the following:

1. What does the conjugate connection actually do to a vector? I.e., for the regular connection, it's clear that it's saying how it's changing the basis vectors.

2. Given a connection $\nabla$, how does one actually find the conjugate $\nabla^\ast$?

I have a couple guesses but they're only that. My guess for (1) is that if the primal connection tells us how the basis vectors change, $\nabla_{\partial_i} \partial_j$, then maybe the conjugate one tells us how the reciprocal basis vectors $\partial^i$ change, i.e., $\nabla_{\partial^i} \partial^j$. This is somewhat complicated by the fact that (like I asked about in my previous question) his dual basis isn't the covector dual basis $dx_i$ I'm used to from earlier differential geometry, it's the dual (covector) basis with the "sharp" operator applied to it, making it a vector again, such that we have two ways of expressing the same vector.

For (2), I'm even less sure, but he notes that if we have a pair of conjugate connections $\nabla, \nabla^\ast$ that are coupled with the metric connection $g$, then the mean connection must equal the Levi-Civita connection, so

$\nabla^{LC} = \overline \nabla = \frac{\nabla + \nabla^\ast}{2}$

Since we can figure out the Christoffel symbols of the LC connection from the metric $g$, it seems like given the symbols defining $\nabla$, we could figure out an explicit form for $\nabla^\ast$. Is this right, or is there an easier way?

Answer 1

Let $(e_i)$ be a local orthonormal frame (so that $\left< e_i, e_j \right> = \delta_{ij}$). Write $\nabla_{e_i} e_j = \Gamma_{ij}^k e_k$ and $\nabla^{*}_{e_i} e_j = \tilde{\Gamma}_{ij}^k e_k$. Plugging $X = e_i, Y = e_j, Z = e_k$ into the defining equation of the conjugate connection, we get that $$ 0 = e_i \left< e_j, e_k \right> = \left< \nabla_{e_i} e_j, e_k \right> + \left< e_j, \nabla^{*}_{e_i} e_k \right> = \Gamma_{ij}^k + \tilde{\Gamma}_{ik}^j. $$ Hence, if you know the coefficients of $\nabla$ with respect to an orthonormal frame you can easily compute the coefficients of $\nabla^{*}$ with respect to the same frame by the formula $\tilde{\Gamma}_{ik}^j = -\Gamma_{ij}^k$. In particular, you also see that $\nabla^{**} = \nabla$ as expected.

Working with orthonormal frame allows you also to see that the conjugate connection $\nabla^{*}$ is in fact equal to the dual connection $\nabla^{\sharp}$ if you identify vectors and covectors using the metric. More explicitly, let $e^i$ be the dual basis of the frame $(e_i)$. Since $(e_i)$ is orthonormal, the dual frame $(e^i)$ is identified with the frame $(e_i)$ under the musical isomorphisms. The dual connection $\nabla^{\sharp}$ is defined by the similar equation $$ X \{ \omega, Z \} = \{\nabla^{\sharp}_X(\omega), Z \} + \{ \omega, \nabla_X Z \} $$ where $X,Z$ are vectors fields, $\omega$ is a covector field and $\{ \cdot, \cdot \}$ is the natural pairing (i.e $\{ \omega, Z \} = \omega(Z)$). Plugging in $X = e_i, Z = e_j$ and $\omega = e^k$ and using the fact that $\{ \omega, Z \} = e^k(e_j) = \delta^k_j$ is constant, one immediately gets that $$ 0 = \{ \nabla^{\sharp}_{e_i}(e^k), e_j \} + \{ e^k, \nabla_{e_i} e_j \} = \{ \nabla^{\sharp}_{e_i}(e^k), e_j \} + \Gamma_{ij}^k. $$ Hence $$\nabla^{\sharp}_{e_i}(e^k) = \{ \nabla^{\sharp}_{e_i}(e_k), e_j \}e^j = -\Gamma_{ij}^k e^j $$ so the coefficients of the dual connection with respect to the coframe $e^i$ are also given by the same formula.

Answer 2

This is a partial answer.

Take two connections $\nabla,\nabla^*$. Each connection has its own Christoffel symbols $$\begin{align} \nabla_{\partial_i}\partial_j &= \Gamma^r_{ij}\partial_r \\ \nabla^*_{\partial_i}\partial_j &= \Gamma^{*r}_{ij}\partial_r \\ \end{align}$$ which determine it completely.

Now take a connection $\nabla$ and let $\nabla^*$ be its conjugate with respect to $g$. In components, the equation $$X(g(Y,Z)) = g(\nabla_XY,Z)+g(Y,\nabla^*_XZ) \tag 1$$ looks like $$\partial_i g_{jk} = g_{rk}\Gamma^r_{ij}+g_{jr}\Gamma^{*r}_{ik}$$ so you can solve for $\Gamma^*$ as $$g_{jr}\Gamma^{*r}_{ik} = \partial_i g_{jk} - g_{rk}\Gamma^r_{ij}$$ $$\Gamma^{*s}_{ik} = g^{js}(\partial_i g_{jk} - g_{rk}\Gamma^r_{ij}).$$

Now from the defining equation $(1)$ you can show that the mean connection $\bar\nabla=\frac 1 2 (\nabla+\nabla^*)$ is metric compatible, i.e. $\bar\nabla g=0$. What is not so clear to me is that $\bar\nabla$ has zero torsion. I'll think about it.