# Why does $\phi^*_{-t}$ move coordinates in the opposite way?

Let $$\phi_t: M \mapsto M$$ be a one-parameter group of diffeomorphisms defined on some manifold $$M$$. It then follows that a tensor field $$T^{b_1 \dots b_k}_{a_1 \dots a_l}$$ can be pushed forward along $$\phi_t$$ as follows:

$$\phi^*_tT^{b_1 \dots b_k}_{a_1 \dots a_l}(\omega_1)_{b_1}\dots (v_1)^{a_1}\dots = T^{b_1 \dots b_k}_{a_1 \dots a_l} (\phi_{t*}\omega_1)_{b_1}\dots ((\phi_t^*)^{-1}v_1)^{a_1}\dots$$

In other words, the vectors and covectors inserted into the tensor are pulled back so that the tensor can be pushed in the other direction. This of course requires the inverse map $$(\phi^*)^{-1}$$ to exist, which it does for diffeomorphisms.

If $$v$$ is a vector field generating $$\phi_t$$, then the lie derivative with respect to $$v$$ is given by:

$$\mathcal{L}_v T^{b_1 \dots b_k}_{a_1 \dots a_l} = \lim_{t \to 0}\frac{\phi_{-t}^*T^{b_1 \dots b_k}_{a_1 \dots a_l} - T^{b_1 \dots b_k}_{a_1 \dots a_l}}{t}$$

In my book (Wald's General Relativity), the components of the lie derivative are worked out in a coordinate system adapted to $$v$$ such that $$v^a = \left( \frac{\partial}{\partial x} \right)^a$$. First, it is said that if the map $$\phi_t$$ transforms coordinates $$(x^1, \dots , x^n)$$ to $$(y^1, \dots , y^n)$$ in some chosen coordinate system, then the matrix corresponding to $$\phi_t$$ is the jacobian $$\partial y^\mu / \partial x^\nu$$. This part makes sense and I managed to work it out on my own.

However, in the next step, it is said that $$\phi_{-t}$$ corresponds to a coordinate transformation $$(x^1, x^2, \dots , x^n) \to (x^1 + t, x^2, \dots , x^n)$$. This part does not make any sense to me. If $$v^a = \left( \frac{\partial}{\partial x} \right)^a$$, then $$\phi_t$$ acting on a point in $$M$$ should be moved in the $$\left( \frac{\partial}{\partial x} \right)^a$$-direction if $$t > 0$$. To me, this sounds like a point with $$x^1$$-coordinate $$x^1$$ will be translated increasingly to $$x^1 + t$$. Now, if we instead consider $$\phi_{-t}$$, then wouldn't it be translated in the other direction, i.e. its $$x^1$$-coordinate becomes $$x^1 - t$$? Apparently this is not the case.

Where does this line of thinking go wrong? At first I thought I must have misinterpreted what was meant by points being moved (for instance, maybe somehow you can see it as points staying the same and the coordinate system being moved in the other direction), however it just seems like thinking of it as points being moved by the action of $$\phi$$ seems more in line with the description used in the book.

Edit: Below is the quote from the book

To analyze the action of $$\mathcal{L}_v$$ on an arbitrary tensor field, it is helpful to introduce a coordinate system on $$M$$ where the parameter $$t$$ along the integral curves of $$v^a$$ is chosen as one of the coordinates $$x^1$$, so that $$v^a = (\partial / \partial x^1)^a$$ (This always can be done locally in any region where $$v^a \neq 0$$.) This action of $$\phi_{-t}$$ then corresponds to the coordinate transformation $$x^1 \to x^1 + t$$, with $$x^2, \dots , x^n$$ held fixed. From the parenthetical remark below equation (C.1.1), we have $$(\phi^*)^\mu_\nu = \delta^\mu_\nu$$ and hence, the coordinate basis components of $$\phi_{-t}^* T^{a_1 \dots a_k}_{b_1 \dots b_l}$$ at the point $$p$$ whose coordinates are $$(x^1, \dots, x^n)$$ are

$$(\phi^*_{-t} T^{\mu_1 \dots \mu_k}_{\nu_1 \dots \nu_l})(x^1, \dots, x^n) = T^{\mu_1 \dots \mu_k}_{\nu_1 \dots \nu_l}(x^1 + t, x^2, \dots, x^n)$$

• This does not make sense. I think that "in the next step", some additional assumption or definition is being made because $\phi(t,x^1, x^2, \dots , x^n)= (x^1 + t, x^2, \dots , x^n)$ is a very specific flow on $M$, namely, the flow of the vector field $X=\frac{\partial}{\partial x^1}.$ This is so because, by definition, we have $X_{\phi^p(t)}=\phi^p{'}(t)$, which unwinds to the system of differential equations $x^1{'}(t)=1;\ x^i{'}(t)=0: 2\le i\le n.$ Dec 30, 2022 at 2:15
• @Matematleta Thanks! I agree that it does not make any sense the way it's phrased right now. I don't know if it helps, but I added the exact quote from the book. Are you able to decipher what they mean?
– Max
Dec 30, 2022 at 10:28
• I find the terminology confusing. I guess $v^a$ (which I will call $X_a$) is a vector in the tangent space to $M$ at $a,\ T_aM$. By the flowbox lemma, we can arrange it so that $X_a=\frac{\partial}{\partial x^1}$ in some coordinates $(x^i)$ in a neighborhood $U\ni a.$ Hence, the flow, $\phi ^a(t)$ of $X$ starting at $a$ satisfies $\phi ^a(0)=a$ and $\phi^p{'}(t)=X_{\phi^a(t)}$ and we are back to my previous comment. Dec 30, 2022 at 15:12
• If the author's intent is to explain how the Lie derivative works, I suggest going over the ideas first for $\mathscr L_V W$, the Lie derivative of $W$ along $V$ where $W$ and $V$ are vector fields. The extension to arbitrary tensor fields is almost immediate. Dec 30, 2022 at 15:14
• I have given a coordinate-free explanation of the Lie derivative for vector fields in the answer section. The analogue for tensor fields uses the same idea. The advantage is that once you have the formula, you can do the calculation in any coordinates you want. But if my answer is not useful to you, please let me know so I can delete it. Dec 30, 2022 at 17:14

If we are given two vector fields $$V,W\in \mathfrak{X}(M),$$ then $$\mathscr L_VW$$ is basically the directional derivative of $$W$$ vector field in the direction specified by the flow of $$V.$$ In Euclidean space, if we start at a point $$p$$ and advance to $$p+tv$$ after some time $$t$$ in the direction of a fixed point $$v$$, then we can compare the vectors $$W_{p+tv}$$ and $$W_p$$ only because these vectors are canonically identified with $$\mathbb R^n$$ itself. Another way of looking a this is that we can always move $$W_{p+tv}$$ back so that it starts "at the same place" as $$W_p$$, then we calculate the directional derivative. This is done in vector analysis courses on $$\mathbb R^n$$ without even saying so. The vectors are simply "slid around" as we please. But on a manifold, this makes no sense. If $$p$$ and $$q$$ are distinct points on a manifold $$M$$, then there is no reason to expect that $$T_pM$$ and $$T_qM$$ have any relation to each other, so vectors in one of them cannot be compared to vectors in the other.
So, to get around this problem, we use the flow of $$V$$ to push the vector $$W_{p+vt}$$ back to $$T_pM$$ and then we take derivative. The flow of $$V,\ \theta:(-\epsilon,\epsilon)\times U\to M$$ a diffeomorphism for each $$t$$ at least for some time in a neighborhood $$U\ni p$$ that satisfies $$\theta(p, 0)=\theta_0(p)=\theta^p(0)=p.$$ Now, $$\theta_t(p)\in M$$ for each $$t\in (-\epsilon,\epsilon).$$ Fix such a $$t.$$ We want to compare $$W_{\theta^p(t)}$$ with $$W_p$$ but they are in different tangent spaces.
But as the pushforward $$(\theta_t)_*$$ is a linear map from $$T_pM$$ into $$T_{\theta(p,t)}M,$$ its inverse $$(\theta_{-t})_*$$ maps $$T_{\theta(p,t)}M$$ back to $$T_pM$$, so $$(\theta_{-t})_*(W_{\theta^p(t)})$$ is a vector in $$T_p$$ so $$(\theta_{-t})_*(W_{\theta^p(t)})-W_p$$ makes sense. Now, we can define $$\mathscr L_V W=\underset{t\to 0}\lim \frac{(\theta_{-t})_*(W_{\theta^p(t)})-W_p}{t}$$ and call this the Lie derivative.