Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm working on exercise 1(a) of chapter 6 in do Carmo's Riemannian Geometry:

Let $M_1$ and $M_2$ be Riemannian manifolds, and consider the product $M_1\times M_2$, with the product metric. Let $\nabla^1$ be the Riemannian connection of $M_1$ and let $\nabla^2$ be the Riemannian connection of $M_2$. Part (a): Show that the Riemannian connection $\nabla$ of $M_1\times M_2$ is given by $\nabla_{Y_1+Y_2}(X_1+X_2) = \nabla_{Y_1}^1 X_1 + \nabla_{Y_2}^2 X_2$, where $X_i,Y_i\in \Gamma(TM_i)$.

Of course the first thing is to show that $\nabla$ is a connection at all, and this is turning out to be more subtle than I had originally thought. First and foremost, it's not even immediately clear that the given formula uniquely determines $\nabla$, since $\Gamma(T(M_1\times M_2))\supsetneq \Gamma(TM_1)\oplus \Gamma(TM_2)$.

I'm having particular trouble showing that the Leibniz rule $\nabla_X(fZ)=X(f)\cdot Z+f\nabla_XZ$ holds. My original thought was to write $X=X_1+X_2$ and $Z=Z_1+Z_2$ and then calculate \begin{equation*} \nabla_X(fZ) = \nabla^1_{X_1}(fZ_1)+ \nabla^2_{X_2}(fZ_2) \end{equation*} \begin{equation*} = (X_1(f)\cdot Z_1 + f\nabla^1_{X_1}Z_1) + (X_2(f)\cdot Z_2 + f\nabla^2_{X_2}Z_2) = f\nabla_XZ + (X_1(f)Z_1+X_2(f)Z_2). \end{equation*} But this is definitely not looking like what I want. This is right iff $X(f)Z = X_1(f)Z_1+X_2(f)Z_2$, which is certainly not going to hold in general. Of course this shouldn't be right, because it's not like $Z=Z_1+Z_2 \in \Gamma(T(M_1\times M_2))$ is going have $Z_i$ be pulled back via the projections.

So my next guess was instead to integrate $X$ by a curve $\alpha:(-\epsilon,\epsilon)\rightarrow M_1\times M_2$, which I can even assume is a geodesic (meaning it projects to a geodesic in both factors). Then, along $\alpha$ I can hope to decompose $Z=Z_1+Z_2$, where $Z_i\in \Gamma(\alpha^* TM_i)$ (where I'm considering $TM_i \rightarrow M_1\times M_2$ as a subbundle of the tangent bundle $T(M_1\times M_2)$). In other words, I'm hoping to turn $Z|_\alpha$ into a sum of pullbacks. But whether or not I can even do this (which I can't in general if $\alpha$ is constant in one or the other factor), this gives me the same equations as above, which just as above is a problem.

In the above paragraph, I think I'm actually modifying $f$ to be a pullback too, but I think this should be alright since ultimately the only thing that matters is the value of $fZ$ along $\alpha$.

So, questions: (1) Is $\nabla$ uniquely determined by the given formula? (2) What am I doing wrong?

share|cite|improve this question
Probably a stupid question: My reflex would be to try to use Koszul's formula and check whether what you want to know holds true or not. Why doesn't this work? – t.b. Apr 6 '11 at 2:12
My reflex was to use the fact that a connection is determined by parallel transport. Given a path $\gamma = (\gamma_1, \gamma_2)$, then you can get a parallel transport on the tangent spaces of $M$ by using the parallel transport on the two sub-pieces. This should give a connection on the product manifold. (This parallel transport is an orthonormal transformation as the sum of two orthonormal transformations, so the connection is the Levi-Civita one.) – Akhil Mathew Apr 6 '11 at 2:31
@Theo: That sounds right, I hadn't thought of that. Thanks for the suggestion. – Aaron Mazel-Gee Apr 7 '11 at 8:14
@Akhil: Probably a stupid question, but is it then obvious that – Aaron Mazel-Gee Apr 7 '11 at 14:53
@Akhil: Sorry somehow I never finished that question! Are you implying that if a connection gives orthonormal parallel transports then it's the Levi-Civita connection? I've never heard that before. I can certainly imagine a nonstandard and yet orthonormal parallel transport on e.g. a vector bundle over the circle, but maybe there's something special here since we're talking about the tangent bundle? – Aaron Mazel-Gee Apr 9 '11 at 20:41
up vote 6 down vote accepted

The problem of verifying the Leibniz rule is closely related to your question as to what extent the formula given actually gives a well-defined map.

The point is that any vector field on $M_1\times M_2$ can locally be written as a linear combination of vector fields on $M_1$ and $M_2$, with the coeffients being functions on the product $M_1\times M_2$. The formula given should then be extended to all vector fields by assuming the Leibniz rule. (In short, the candidate connection on the product will satisfy the Leibniz rule by definition.)

You then have to check that what you have is well-defined, torsion-free, and compatible with the product metric. These should all be straightforward exercises, though.

share|cite|improve this answer
I'm still a little confused. What if as we move in the $M_2$ direction, the projection of $X$ to $\mathscr{TM}_1 \subseteq T(M_1\times M_2)$ changes directions? (I just made up the notation, of course.) I don't think this can be written as $f_1X_1+f_2X_2$. – Aaron Mazel-Gee Apr 6 '11 at 3:45
@Aaron: Dear Aaron, If $x_1,\ldots,x_m$ are local coordinates on $M_1$, and $y_1,\ldots,y_n$ are local coordinates on $M_2$, then any vector field is locally of the form $$f_1 \partial_{x_1} + \ldots + f_m \partial_{x_m} + g_1\partial_{y_1} + \ldots + g_n\partial_{y_n},$$ so it is a linear combination (with the coefficients being functions) of vector fields on $M_1$ and $M_2$ (namely the $\partial_{x_i}$ and $\partial_{y_j}$). However, writing this out, I see that I blundered in my answer: we may need more just one $f$ and one $g$. I will correct my answer. Regards, – Matt E Apr 6 '11 at 3:58
@Aaron: Note: "we may need more just one" should read "we may need more than just one". Also, the functions $f_i$ and $g_j$ are functions on the product $M_1\times M_2$. – Matt E Apr 6 '11 at 4:04
I see. This makes sense. I think the problem was poorly worded. Thanks a lot. – Aaron Mazel-Gee Apr 7 '11 at 8:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.