differential-geometryriemannian-geometry
Prove that connection $\nabla $ on a Riemannian manifold $M$ is compatible with metric iff
$$Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ),$$
for every smooth vector fields $X,Y,Z$.
I am confused about how to prove compatibility from this equation. Any help is appreciated it. By my text it should be obvious (Do Carmo).
This is a cultural comment, rather than an answer; but it is a bit long for the comment box, which is why I am writing it here.
My comment is the following: giving a connection, and giving the associated parallel transport, are essentially the same thing, and so it should be possible to establish the formula you are trying to prove by reasonably conceptual, high-level, thinking, rather than mucking about too much with complicated differential equations.
Let me elaborate a little: how would you define the derivative of a vector field along a curve? Well, the idea is that you want to form the usual Newton quotient $$ \dfrac{V\bigl(c(t+\epsilon)\bigr) - V\bigl(c(t)\bigr)}{\epsilon}$$ and then let $\epsilon$ go to zero, to compute the derivative of $V$ at the point $c(t)$.
The only problem is that the subtraction in this formula doesn't make sense, because the tangent vectors being subtracted are based at different points: one is based at $c(t+\epsilon)$ and the other at $c(t)$.
Enter parallel transport: suppose we have an agreed upon way to shift, or transport, tangent vectors along a curve. Then we can apply this so as to transport $V\bigl(c(t+\epsilon)\bigr)$ from $c(t+\epsilon)$ to $c(t)$, and then form the above Newton quotient, and proceed to compute a derivative.
An agreed upon method for shifting vectors along curves is formally referred to as a choice of parallel transport, because if you have such a method, and if you define a vector field along a curve $c(t)$ by choosing some $V_0$ at $c(t_0)$ and then using the agreed upon parallel transport to define $V\bigl(c(t))$ by tranporting $V_0$ along $c$ to $c(t)$, you obtain a vector field along $c(t)$ whose derivative at each point will be zero (by construction!). Thus this vector field does not change along $c(t)$, and so consists of parallel vectors, hence the name parallel tranport. (Of course, the notions of change and parallel are not intrinsict to the vector field; they depend upon our particular choice of parallel transport.)
Of course, our choice of parallel transport should satisfy some axioms (it should be linear; it should be smooth (so that the vector fields along curves that it gives rise to are smooth); and so on). If you look in the right text, you will find these axioms written down.
As we've seen, a choice of parallel transport on our manifold gives a way of differentiating vector fields along curves. But since a tangent vector is just an infinitesimal curve, one sees that we in fact have a way of differentiating a vector field in the direction of a given vector field: extend that vector field to a curve, and then differentiate along that curve.
Thus a choice of parallel transport determines an affine connection, i.e. a way of differentiating vector fields along tangent vectors. Conversely, this is enough information to determine the parallel transport: we parallel transport a tangent vector along a curve by making sure that its derivative (using the given affine connection) at every point of the curve, in the tangent direction of the curve, vanishes.
So we have gone full circle, from parallel transport, to affine connection, back to parallel transport. The text you are reading is spelling out the second half of this circle, but is perhaps a little scanty on details regarding the first half; in fact, the exercise you are trying to solve is exactly about the first half of the circle, and its goal is to verify that you truly are going around in a circle: i.e. that you are ending up where you started from.
There really is something to verify here (i.e. the exercise is not trivial if you have not done it before, and are learning these ideas for the first time), but I hope that the above discussion may help shed some light on its meaning, and also make it seem less daunting (and perhaps more conceptual and less computational) than it might otherwise appear.
As in the problem statement, we let $\overline{\nabla}$ denote the Riemannian connection on $N$, and define $(\nabla_XY)(p) = \pi^\top[\overline{\nabla}_{X^*}Y^*(p)]$, where $\pi^\top\colon TN|_{f(U)} \to TM$ denotes the tangential projection. We aim to show that $\nabla$ is the Riemannian connection on $M$.
By the uniqueness of the Riemannian connection on $M$, it suffices to show that (1) $\nabla$ is a connection, (2) $\nabla$ is compatible with the metric, and (3) $\nabla$ is symmetric.
(1) $\nabla$ is a connection
Let $X_1, X_2, Y_1, Y_2$ be differentiable vector fields on $f(U)$ that extend to differentiable vector fields on an open subset of $N$. We note that $$(\nabla_{X_1 + X_2}Y)(p) = (\overline{\nabla}_{X_1^* + X_2^*}Y^*)(p) = (\overline{\nabla}_{X_1^*}Y^* + \overline{\nabla}_{X_2^*}Y^*)(p)^\top = (\nabla_{X_1}Y)(p) + (\nabla_{X_2}Y)(p)$$ and $$(\nabla_X(Y_1 + Y_2))(p) = (\overline{\nabla}_{X^*}Y_1^* + \overline{\nabla}_{X^*}Y_2^*)(p)^\top = (\nabla_XY_1)(p) + (\nabla_XY_2)(p)$$ and if $h\in C^\infty(f(U))$ is any differentiable function on $f(U)$, then $$\begin{align*} (\nabla_X(hY))(p) & = (\overline{\nabla}_{X^*}(hY^*))(p)^\top \\ & = \left[(\overline{\nabla}_{X^*}h)(p)Y^*(p) + h(p)(\overline{\nabla}_{X^*}Y^*)(p) \right]^\top \\ & = \left[ (Xh)(p)\,Y^*(p)\right]^\top + h(p)(\overline{\nabla}_{X^*}Y^*)(p)^\top \\ & = (Xh)(p)\,Y(p) + h(p)\,(\nabla_XY)(p), \end{align*}$$ which shows that $\nabla$ is a connection.
(2) $\nabla$ is compatible with the metric.
This is a computation:
$$\begin{align*} X_p\langle Y_p, Z_p \rangle_g & = X^*_p\langle Y^*_p, Z^*_p \rangle_{\overline{g}} \\ & = \overline{\nabla}_{X^*}\langle Y^*, Z^* \rangle_{\overline{g}}(p) \\ & = \langle (\overline{\nabla}_{X^*}Y^*)(p), Z^*_p\rangle_{\overline{g}} + \langle Y^*_p, (\overline{\nabla}_{X^*}Z^*)(p)\rangle_{\overline{g}} \\ & = \langle (\overline{\nabla}_{X^*}Y^*)(p)^\top + (\overline{\nabla}_{X^*}Y^*)(p)^\perp, Z_p \rangle_{\overline{g}} + \langle Y_p, (\overline{\nabla}_{X^*}Z^*)(p)^\top + (\overline{\nabla}_{X^*}Z^*)(p)^\perp \rangle_{\overline{g}} \\ & = \langle (\nabla_XY)(p), Z_p \rangle_g + \langle Y_p, (\nabla_XZ)(p)\rangle_g, \end{align*}$$ where I've switched to subscripts $X_p = X(p)$ for a slight ease of notation. (Also, $\perp$ denotes the normal projection.)
(3) $\nabla$ is symmetric.
Finally, symmetry follows from noting that
$$\begin{align*} (\nabla_XY)(p) - (\nabla_YX)(p) & = (\overline{\nabla}_{X^*}Y^*)(p)^\top - (\overline{\nabla}_{Y^*}X^*)(p)^\top \\ & = \pi^\top\!\left( (\overline{\nabla}_{X^*}Y^*)(p) - (\overline{\nabla}_{Y^*}X^*)(p) \right) \\ & = \pi^\top([X^*, Y^*]_p) \\ & = \pi^\top([X, Y]_p) \\ & = [X, Y]_p \end{align*}$$ since $[X,Y]$ is tangent to $M$ whenever $X$ and $Y$ are.
Best Answer
This is somewhat delicate to do with complete detail and involves exactly the kind of issues on which Do Carmo doesn't bother to expand.
Let $\gamma : [0,1] \rightarrow M$ be a smooth curve and let $P,P'$ be two smooth parallel vector fields along $\gamma$. Define $f(t) = g(P(\gamma(t)), P'(\gamma(t)))$. The function $f : [0,1] \rightarrow M$ is smooth. We will show that the derivative $f'(t)$ is identically zero.
Let $t_0 \in [0,1]$. Choose some vector field $Z \in \Gamma(M)$ such that $Z(\gamma(t_0)) = \dot{\gamma}(t_0)$. Consider three cases:
(1) $\dot{\gamma}(t_0) \neq 0$. Then, for small enough $\epsilon > 0$, the image $\gamma([t_0 - \epsilon, t_0 + \epsilon])$ is an embedded compact submanifold of $M$ with boundary. This implies that $P,P'$ (restricted to the image) are defined on a compact submanifold of $M$ and so can be extended to vector fields $X,X' \in \Gamma(M)$, defined $\textbf{globally}$ on $M$, such that $X(\gamma(t)) = P(\gamma(t)), X'(\gamma(t)) = P'(\gamma(t))$ for $t \in [t_0 - \epsilon, t_0 + \epsilon]$. Thus, $$ f'(t_0) = \dot{\gamma}(t_0)g(X,X') = \left. Zg(X,X')\right|_{p=\gamma(t_0)} = \left. g(\nabla_Z X, X')\right|_{p=\gamma(t_0)} + \left. g(X, \nabla_Z X')\right|_{p=\gamma(t_0)} = g(\left. \frac{DP}{dt} \right|_{t=t_0}, X'(\gamma(t_0))) + g(X(\gamma(t_0)), \left. \frac{DP'}{dt} \right|_{t=t_0}) = 0. $$
(2) $\dot{\gamma}(t_0) = 0$, but there is a sequence $t_n \rightarrow t_0$ such that $\dot{\gamma}(t_n) \neq 0$. For example, you can think about a curve $\gamma$ that traces a line segment from $t = 0$ to $t_0 = \frac{1}{2}$, slows down smoothly as $t$ approaches $\frac{1}{2}$, and then turns back on its trace. In this case, you can't necessarily extend the vector fields $P, P'$ to $\textbf{global}$ vector fields, but you get from continuity of $f'$ and from the previous analysis that $f'(t_0) = 0$.
(3) $\dot{\gamma}(t) \equiv 0$ in a neighborhood of $t_0$. This implies that $\gamma(t)$ is constant around $t_0$. Since parallel transport along a constant curve is constant, the function $f(t)$ itself is constant around $t_0$ and in particular $f'(t_0) = 0$.