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Let $M$ be a closed manifold equipped with an affine connection $\nabla$. Let $X \in \Gamma(TM)$, and suppose $X$ is a parallel vector field, i.e. $\nabla X=0$.

Is it true that the flow of $X$ preserves the connection? i.e. let $\phi_t$ be the flow of $X$. Is it true that $\phi_t^* \nabla=\nabla$?

Here, for a given diffeomorphism $\phi:M \to M$, I define $\phi^* \nabla$ by requiring

$$ \phi_*\big((\phi^* \nabla)_XY\big)= \nabla_{\phi_* X}\phi_* Y,$$ where $\phi_*$ is the pushforward operation.

(I think this is the 'right' definition).

What about the converse, that is suppose that $\phi_t^* \nabla=\nabla$. Is $X$ parallel?

I am not sure how to compute the 'variational derivative' $\frac{d}{dt}|_{t=0}\phi_t^* \nabla$.

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The quantity you call the "variational derivative" is just the Lie derivative of $\nabla$ along $X$: This is the tensor field $\mathcal L_X \nabla: TM \to T^*M \otimes TM$ characterized by $$(\mathcal L_X \nabla)(Y) = [\mathcal L_X, \nabla] Y = \mathcal L_X (\nabla Y) - \nabla \mathcal L_X Y ,$$ which measures the failure of $\mathcal L_X$ to commute with $\nabla$, that is, the infinitesimal failure of the flow of $X$ to preserve $\nabla$. (For more see these expository notes of Mike Eastwood, which are ultimately concerned with infinitesimal projective symmetries, that is, vector fields whose flow preserves the projective class of a connection.) It's straightforward to verify that this quantity really is tensorial (in $Y$).

If $\nabla$ is torsion-free, which we henceforth assume, we may express the Lie derivative of a vector field in terms of $\nabla$, and in fact in this case the object in $T^*M \otimes T^*M \otimes TM$ obtained by dualizing is actually in $S^2 T^*M \otimes TM$ (this dualized object is the infinitesimal version of the difference tensor between two connections). A straightforward calculation gives that $$(\mathcal L_X \nabla)_{ab}{}^c = \nabla_a \nabla_b X^c + \color{red}{R_{da}{}^c{}_b X^d} ,$$ where $R$ is the curvature tensor.

In the case that $X$ is parallel, the first term on the r.h.s. vanishes by definition, and so the preservation of $\nabla$ by the flow of $X$ is governed (at least in the torsion-free case) precisely by the vanishing of $\color{red}{R_{da}{}^c{}_b X^d}$. In general this quantity need not be zero:

Example Consider the (torsion-free) connection $\nabla$ on $\Bbb R^2$ characterized by $$\nabla_{\partial_1} \partial_1 = x^2 \partial_1$$ and $\nabla_{\partial_i} \partial_j = 0$ for all other $(i, j)$. We can read off immediately that $\partial_2$ is parallel w.r.t. $\nabla$, but computing gives that $R = - dx^1 \wedge dx^2 \otimes \partial_1 \otimes dx^1$, so $\mathcal L_X \nabla = dx^1 \otimes \partial_1 \otimes dx^1 \neq 0$.

On the other hand, if $\nabla$ is the Levi-Civita connection of some metric, the symmetries of a metric curvature tensor allow us to rewrite $\color{red}{R_{da}{}^c{}_b X^d}$ as $g^{ce} g_{af} R_{be}{}^f{}_d X^d$, but this vanishes by the definition of curvature and the condition that $X$ is parallel. In other words, if we specialize to Levi-Civita connections, the answer to the question is always positive.

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  • $\begingroup$ Thanks. Your answer is very interesting. I now see that in hindsight, $\nabla X=0$ cannot contain enough information to conclude $L_X \nabla=0$, not even in the torsion-free case. (If $\dim M=n$, $\nabla X=0$ is $n^2$ equations while $L_X \nabla=0$ is something like $n^3$ equations-$n^3$ in the general case, and $\frac{n^2(n+1)}{2}$ in the torsion-free case). Just to make sure, your calculations were in abstract index notation, right? Also, do you have more references on Lie derivatives of connections? $\endgroup$ – Asaf Shachar Jan 25 '18 at 9:08
  • $\begingroup$ You're welcome, I'm glad you found it useful. Yes, every expression in my answer in which indices appear use abstract notation index. I don't have more references handy, and in fact offhand I don't know anywhere in the literature where the above formula appears, though it's straightforward to derive by hand. $\endgroup$ – Travis Willse Jan 25 '18 at 15:37
  • $\begingroup$ Even if you don't know this formula, it's still easy to see that the claim is true for Levi-Civita connections, by the way: Lowering an index of $\nabla X = 0$ and symmetrizing gives $\nabla_{(a} X_{b)} = 0$, but it's straightforward to check, using the formula for the Lie derivative of a tensor in terms of a torsion-free connection, that the left-hand side here is exactly $(\mathcal L_X g)_{ab}$, so $\mathcal L_X g = 0$, that is, the flow of $X$ preserves $g$ (i.e., $X$ is a Killing field). Since $\nabla$ is constructed invariantly out of $g$, we must have $\mathcal L_X \nabla = 0$, too. $\endgroup$ – Travis Willse Jan 25 '18 at 15:42
  • $\begingroup$ Thanks. I guess you referred here to the formula $(L_V g)(X,Y)=g(\nabla_X V,Y)+g(X,\nabla_Y V)$ which holds for the Levi-Civita connection? (I really should learn how to work better with coodinates expressions). BTW, I am guessing that in the general case of torsion-free connection, the converse also doesn't hold? that is, $\phi_t^* \nabla=\nabla$ that does not imply $X$ is parallel. What do you think? $\endgroup$ – Asaf Shachar Jan 25 '18 at 17:29
  • $\begingroup$ Correct, and in fact the converse doesn't even hold for the flat connection on $\Bbb R^n$: In that case, the vector space of vector fields preserving $\nabla$ has dimension $n^2+n$ (as in that case the components of $\mathcal L_X \nabla$ are just the second coordinate derivatives of the components $X^i$ of $X$, all w.r.t. standard coordinates), but any parallel vector field on a connected affine manifold $(M,\nabla)$ is determined via parallel transport by its value at a single point, so the dimension of the space of parallel vector fields is always $\leq\dim T_x M=\dim M$ for any $x\in M$. $\endgroup$ – Travis Willse Jan 26 '18 at 17:54

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