# Notes on the pullback connection

$\newcommand{\lie}{\mathcal{L}} \newcommand{\cd}{\nabla} \newcommand{\ad}{\mathop{\rm ad}\nolimits}$

In an effort to keep myself organized, I decided I should type up some notes I have laying around. These calculations are not enough to be a journal article (or even a note on the arXiv), but slightly non-trivial, so I don’t want to lose them.

## Definitions

This note is on a construction called the pullback connection. Consider a manifold $\mathcal{M},$ and let there be a one-parameter family of diffeomorphisms $\varphi_{t}:\mathcal{M}\to \mathcal{M}.$ The one-parameter family is of course a Lie group under composition, and $\varphi_{s}\circ \varphi_{t} = \varphi_{s+t}.$ The diffeomorphism induces a number of maps, e.g. the pullback map $\varphi_{t}^{*}$ and the pushforward map $\varphi_{t*},$ sometimes written $d\varphi_{t}.$ While the pullback and pushforward are defined for general maps between manifolds, since this one is a diffeomorphism from the manifold to itself, the pullback and pushforward are inverses, $\varphi_{t*} = \varphi_{-t}^{*}.$

Let $v^{a}$ be the vector field tangent to the one-dimensional orbits of $\varphi.$ Recall that the Lie derivative is defined as \begin{align} \label{eq:lie-deriv-def} \lie_{v} T = \frac{d}{dt} \Big[ \varphi_{t}^{*} T \Big]_{t=0} \,. \end{align} Conversely, we can think of the pullback as exponentiation of the Lie derivative, $\varphi_{t}^{*}=\exp[t\lie_{v}]$.

Now let there be a connection $\cd$ on this manifold, so that for any vector field $X\in \mathcal{X}(\mathcal{M}),$ $\cd_{X}$ is a map from the space of tensor fields of rank $(p,q)$ to the same space; that $\cd_{X}$ satisfies the Leibniz property, and so on; and in particular that it only involves $X$ but not derivatives of $X$ (one example of such a connection is of course the Levi-Civita connection of a metric, but we will get to this later).

Given the connection $\cd$ and a diffeomorphism $\varphi_{t},$ there is a new connection we can construct: the pullback connection, $(\varphi_{t}^{*}\cd).$ We define it as follows: \begin{align} \label{eq:pullback-connection-defn} (\varphi_{t}^{*}\cd)_{X} T \equiv \varphi_{t}^{*} \left[ \cd_{d\varphi_{t}(X)} \left( \varphi_{t*} T \right) \right] \,. \end{align} In words, this definition is: take tensor $T$ and push it forward along the map; take its derivative with your given connection $\cd$, but in the direction given by the pushed-forward version of $X$; then take the whole result and pull it back.

As we know, the space of connections is an affine space, and there are connection coefficients $C^{c}{}_{ab}$ which are the difference between two particular connections. A natural question would be: given $\cd$ and $\varphi_{t},$ what is the difference $(\varphi_{t}^{*}\cd)-\cd?$

## For a Killing vector field

I will address the general case below, but first specialize to the case when $\cd$ is the Levi-Civita connection of some metric $g$, and $\varphi_{t}$ is an isometry of $g$, i.e. that \begin{align} \label{eq:g-isometry} g = \varphi_{t} g \,. \end{align} In this case, we see that \begin{align} \label{eq:pullback-conn-of-g-vanishes} (\varphi_{t}^{*}\cd)_{X} g = 0 \quad \forall X\,. \end{align} This is a direct application of the definition of the pullback connection in Eq. $\eqref{eq:pullback-connection-defn}$ and the isometry condition Eq. $\eqref{eq:g-isometry}.$ But since the Levi-Civita connection is unique, this means that \begin{align} \label{eq:isometry-conns-agree} \varphi_{t}\text{ is an isometry of }g \qquad\Longrightarrow\qquad (\varphi_{t}^{*}\cd) = \cd \,. \end{align}

Since the two connections agree, we have the equality \begin{align} \cd_{X} T = \varphi_{-t}^{*} \left[ \cd_{\varphi_{t}^{*}(X)} \left( \varphi_{t}^{*} T \right) \right] \end{align} for all $X,T$ (the astute reader will have noticed that we have flipped the sign of $t\to -t$ in order to make later calculations prettier). Now we will find the infinitesimal version of this identity. Apply the inverse $\varphi_{t}^{*}$ to both sides of this equality, differentiate both sides with respect to $t$, and then evaluate at $t=0$: \begin{align} \frac{d}{dt} \Big[ \varphi_{t}^{*} \cd_{X} T \Big]_{t=0} &= \frac{d}{dt} \Big[ \cd_{\varphi_{t}^{*}(X)} \left( \varphi_{t}^{*} T \right) \Big]_{t=0} \,, \\ \lie_{v} \left( \cd_{X} T \right) &= (\cd_{X} \lie_{v} T) + \cd_{(\lie_{v}X)} T \,. \end{align} This can be elucidated by expanding $\cd_{X}$ in abstract index notation (still suppressing indices on $T$), \begin{align} \lie_{v} \left( X^{a}\cd_{a} T \right) &= X^{a}(\cd_{a} \lie_{v} T) + (\lie_{v}X)^{a} \cd_{a} T \,, \\ (\lie_{v} X)^{a} \cd_{a} T + X^{a} \lie_{v} \cd_{a} T &= X^{a}(\cd_{a} \lie_{v} T) + (\lie_{v}X)^{a} \cd_{a} T \,. \end{align} Now cancel the $\lie_{v}X$ terms on both sides. Noting that $X^{a}$ is arbitrary, we have derived an identity \begin{align} \lie_{v}\cd_{a} T = \cd_{a}\lie_{v} T \end{align} when $\cd$ is the Levi-Civita connection of $g$ and $v$ is a Killing vector field.

Note: another way to say this is

\begin{align} [ \lie_v, \cd_a ] T = 0 \,. \end{align}

In other words, when $v$ is a Killing vector field of the metric $g$, $\lie_v$ commutes with the Levi-Civita connection $\cd$ of $g$.

## General connection coefficients for $(\varphi_{t}^{*}\cd)-\cd$

Let us further study the difference $(\varphi_{t}^{*}\cd)-\cd,$ and we will lift the restriction that $\varphi_{t}$ is an isometry of $g.$ First, we need some general properties of the difference of these connections. Much of this follows Sec. 3.1 of Wald.1

All connections agree when acting on scalar functions, so we have

\begin{align} \label{eq:diff-agree-scalar} (\varphi_{t}^{*}\cd)_{a}f-\cd_{a}f = 0 \end{align}

for all scalar fields $f$. Next, examine their difference when acting on the product of a scalar field $f$ and a one-form field $\omega_{b}$. Using the Leibniz rule, we find

\begin{align} \label{eq:diff-on-f-omega} (\varphi_{t}^{*}\cd)_{a}(f\omega_{b})-\cd_{a}(f\omega_{b}) = f \left[ (\varphi_{t}^{*}\cd)_{a}\omega_{b}-\cd_{a}\omega_{b} \right] \,. \end{align}

This tells us that the difference in fact only depends on the value of $\omega_{b}$ at each point, and not on the derivative of $\omega_{b}$: consider a point $p$ where $f(p)=1$ but the gradient of $f$ is arbitrarily large; still at that point, $(\varphi_{t}^{*}\cd)_{a}(f\omega_{b})-\cd_{a}(f\omega_{b})$ agrees with $(\varphi_{t}^{*}\cd)_{a}\omega_{b}-\cd_{a}\omega_{b}$. There is a more rigorous argument given in 1, but the conclusion is the same: $(\varphi_{t}^{*}\cd)_{a}-\cd_{a}$ is just a linear transformation of its argument at each point on the manifold. Therefore, there is a tensor field $C(t)^{c}{}_{ab}$ (the connection coefficients) such that

\begin{align} \label{eq:diff-def-of-C} (\varphi_{t}^{*}\cd)_{a}\omega_{b}-\cd_{a}\omega_{b} = C(t)^{c}{}_{ab} \omega_{c} \,. \end{align}

This is extended by linearity to vectors and tensors in the usual way, with $C(t)^{c}{}_{ab}$ correcting each down index with a plus sign and each up index with a minus sign. For example, when acting on the metric, we have

\begin{align} \label{eq:diff-on-g} (\varphi_{t}^{*}\cd)_{a}g_{bc}-\cd_{a}g_{bc} = C(t)^{d}{}_{ab} g_{dc} + C(t)^{d}{}_{ac} g_{bd} \,. \end{align}

The first property of $C(t)^{c}{}_{ab}$ we can establish is that it is symmetric in its lower two indices. To show this, let the one-form be the gradient of a scalar, $\omega_{b}=\cd_{b}f = (\varphi_{t}^{*}\cd)_{b}f$. Then we have

\begin{align} \label{eq:C-is-symm} (\varphi_{t}^{*}\cd)_{a}(\varphi_{t}^{*}\cd)_{b} f - \cd_{a} \cd_{b} f = C(t)^{c}{}_{ab} \cd_{c} f \,. \end{align}

Since $\cd_{a}$ is (by assumption) torsion-free, then so is $(\varphi_{t}^{*}\cd)_{a}$. Therefore, both terms on the left hand side of Eq. \eqref{eq:C-is-symm} are symmetric, and so is their difference. Thus $C(t)^{c}{}_{ab} = C(t)^{c}{}_{(ab)}$ is symmetric in its lower two indices.

Now to make progress, we will expand $C(t)$ as a power series in $t$,

\begin{align} \label{eq:C-power-series-first-few} C(t)^{c}{}_{ab} &= C^{(0)c}{}_{ab} + t C^{(1)c}{}_{ab} + \ldots \\ \label{eq:C-power-series} C(t)^{c}{}_{ab} &= \sum_{k=0}^{\infty} \frac{t^{k}}{k!}C^{(k)c}{}_{ab} \,, \\ C^{(k)c}{}_{ab} &\equiv \frac{d^{k}}{dt^{k}} C(t)^{c}{}_{ab} \Big|_{t=0} \,. \end{align}

Clearly $C^{(0)c}{}_{ab} = 0,$ since $\varphi_{0}$ is the identity map, so $(\varphi_{0}^{*}\cd)=\cd.$

Now order-by-order, the connection coefficients’ expansion $C^{(k)c}{}_{ab}$ can be computed by taking $k$ derivatives of Eq. \eqref{eq:diff-on-g} (with $X^{a}$ contracted into the $a$ slot). We will perform the expansion for $k=1$:

\begin{align} \frac{d}{dt} \left[ (\varphi_{t}^{*}\cd)_{X}g_{bc} \right] &= \frac{d}{dt} \left[ X^{a}C(t){}_{cab} + X^{a}C(t){}_{bac} \right] \\ \frac{d}{dt} \left[ \varphi_{t}^{*} \left( \cd_{(\varphi_{-t}^{*}X)} \varphi_{-t}^{*} g_{bc} \right) \right] &= X^{a} \left( C^{(1)}_{cab} + C^{(1)}_{bac} \right) \,. \end{align}

There are three terms on the left, one from each of the pullbacks. These are

\begin{align} \lie_{v}( \cd_{X} g_{bc}) + \cd_{(\lie_{-v}X)} g_{bc} + \cd_{X} \lie_{-v} g_{bc} &= X^{a} \left( C^{(1)}_{cab} + C^{(1)}_{bac} \right) \\ - X^{a} \cd_{a} \lie_{v} g_{bc} &= X^{a} \left( C^{(1)}_{cab} + C^{(1)}_{bac} \right) \\ \label{eq:cd-lie-g-is-C-plus-C} - \cd_{a} \lie_{v} g_{bc} &= C^{(1)}_{cab} + C^{(1)}_{bac} \,. \end{align}

Now we can do the following index gymnastics. Take Eq. \eqref{eq:cd-lie-g-is-C-plus-C} with the given index order, on the left hand side, $abc$; add to it the same expression, but with the indices in the order $bac$; and then subtract from it the same expression but with the indices in the order $cab$. Then using the symmetry on the last two indices of $C$, we can solve for

\begin{align} \label{eq:C1-down-sol} C^{(1)}_{cab} = \frac{-1}{2} \Big( \cd_{a} \lie_{v} g_{bc} + \cd_{b} \lie_{v} g_{ac} - \cd_{c} \lie_{v} g_{ab} \Big) \,. \end{align}

Note: Again if $v$ is a Killing vector field of $g$, this immediately vanishes. This says that to linear order in $t$, $(\varphi_t^*\cd)$ agrees with $\cd$.

## Expansion of pullback connection

Here we are going to present an explicit formula in terms of Lie derivatives for the expansion of the pullback connection. I think this is not completely rigorous, but it seems satisfactory to me, and I checked it up to high order in Mathematica.

First, let us re-write the definition of the pullback connection Eq. \eqref{eq:pullback-connection-defn} as

\begin{align} (\varphi_{t}^{*}\cd)_{X} T &{}\equiv \varphi_{t}^{*} \left[ \cd_{d\varphi_{t}(X)} \left( \varphi_{t*} T \right) \right] \\ &{}= \varphi_{t}^{*} \left[ (\varphi_{-t}^{*} X)^{a} \cd_{a} \left( \varphi_{-t}^{*} T \right) \right] \,. \end{align}

Here, if the left hand side is evaluated at point $p$, then on the right hand side, the up/down indices $a$ are in the spaces $T_{\varphi(p)}\mathcal{M}$ and $T_{\varphi(p)}^{*}\mathcal{M}$. Now we are about to do something which a classical differential geometer might find illegal, but it seems fine to me. Recall that the pullback/pushforward commute with tensor product, so $\varphi(A\otimes B) = \varphi(A)\otimes\varphi(B)$. We apply this to the right hand side, yielding

\begin{align} (\varphi_{t}^{*}\cd)_{X} T &{}= X^{a} \varphi_{t}^{*} \left[ \cd_{a} \left( \varphi_{-t}^{*} T \right) \right] \,. \end{align}

There is no pullback/pushforward acting on $X^{a}$ on the RHS because the pullback and pushforward have inverted each other. Now this notation looks quite bad, because the up index on $X^{a}$ is in $T_{p}\mathcal{M}$. Inside the square brackets of the pullback, the down index on $\cd_{a}$ is very much in the space $T_{\varphi(p)}^{*}\mathcal{M}$. However, this index is then pulled back to live in $T_{p}^{*}\mathcal{M}$, so it may be correctly contracted with $X^{a}$. I’m not sure if this is just a shortcoming of the abstract index notation, or if there is a deeper issue to understand here.

We continue being cavalier physicist and replace $\varphi_{t}^{*}$ with $\exp[t\lie_{v}]$. Therefore, we have an expression for the pullback connection as

\begin{align} \label{eq:pullback-conn-as-exp-cd-exp} (\varphi_{t}^{*}\cd)_{a} T = \exp[t\lie_{v}] \cd_{a}\exp[-t\lie_{v}] T \,. \end{align}

Through a standard formula in the theory of Lie algebras, we can formally re-express this as

\begin{align} (\varphi_{t}^{*}\cd)_{a} T ={}& (e^{t[\lie_{v},-]} \cd_{a}) T = (e^{t \ad_{\lie_{v}} } \cd_{a}) T \\ (\varphi_{t}^{*}\cd)_{a} T ={}& \sum_{k=0}^{\infty} \frac{t^{k}}{k!} \left((\ad_{\lie_{v}})^{k} \cd_{a}\right) T \,, \end{align}

where $[-,-]$ is the commutator, $\ad_{\lie_{v}}\cd_{a} = [\lie_{v},\cd_{a}]$ is the adjoint action of $\lie_{v}$ on $\cd_{a}$, and

\begin{align} \label{eq:ad-k-cd} (\ad_{\lie_{v}})^{k} \cd_{c} = \underbrace{[\lie_{v},[\lie_{v},\cdots [\lie_{v}}_{ k\text{ Lie derivative commutators}} , \cd_{c} ]]] \,. \end{align}

Through explicit calculation in the xTensor package for Mathematica, I have confirmed that this expression agrees with the geometric definition up to 12th order in $t$ (this calculation takes only half a second for 6th order, but scales very badly with order because of two pushforwards and one pullback; and thus 12th order takes almost 7 minutes on my laptop).

This calculation allows us to find an explicit expression for $C^{(k)}_{cab}$ as defined in Eq. \eqref{eq:C-power-series}. We follow the same approach as above, by acting with $(\varphi_{t}^{*}\cd)_{a}-\cd_{a}$ on $g_{bc}$. Now each side is expanded in a power series and powers of $t^{k}$ are equated. Again we play the index gymnastics game and take the combination of equations with indices in the orders $(abc)+(bac)-(cab)$. This allows us to find

\begin{align} \label{eq:C-k-explicit} 2 C^{(k)}_{cab} = \left((\ad_{\lie_{v}})^{k} \cd_{a}\right) g_{bc} + \left((\ad_{\lie_{v}})^{k} \cd_{b}\right) g_{ac} - \left((\ad_{\lie_{v}})^{k} \cd_{c}\right) g_{ab} \,. \end{align}

Note: Again inspect the case when $v$ is a Killing vector field of $g$. Since $[\lie_v, \cd_a] = 0$, we recover the fact that $(\varphi_t^*\cd)$ and $\cd$ agree to all orders of $t$.