1
$\begingroup$

I am working on a (Lorentzian) manifold $(M,g)$, and I know that in local coordinates $g$ takes the form \begin{equation} \label{1} g = h_{ij}\mathrm{d}x^i \mathrm{d}x^j + k_{\mu \nu} \mathrm{d} x^\mu \mathrm{d} x^\nu, \end{equation} where, say, the indices $i,j$ run from $0$ to $1$ and $\mu, \nu$ run from $2$ to $3$. I want to say that, in a suitable sense, the Levi-Civita connection of $g$ decomposes into a direct sum of Levi-Civita connections of $h$ and $k$, $$\nabla^{(g)} = \nabla^{(h)} \oplus \nabla^{(k)}.$$ I have a feeling that this is trivial, but got into a slight mess trying to prove it. My intuition is that the above form of the metric $g$ means that the tangent bundle $TM$ must be a sum of the form $TM = TH \oplus TK$. Any vector field $X \in \Gamma(TM)$ then decomposes into a sum $X = X^{(h)} \oplus X^{(k)}$, where $X^{(h)} \in \Gamma(TH)$, $X^{(k)} \in \Gamma(TK)$, and the only sensible definition of a direct of sum of connections I could come up with was $$ (\nabla^{(h)}\oplus \nabla^{(k)} )(X, Y) := (\nabla^{(h)}(X^{(h)}, Y^{(h)})) \oplus (\nabla^{(k)}(X^{(k)}, Y^{(k)})).$$ However, this doesn't seem to (although I feel like I am wrong) define a connection on $TM$ (I couldn't convince myself that the Leibniz rule holds).

I have a strong feeling that such a decomposition of $\nabla^{(g)}$ must hold, however. For one, I have calculated the Christoffel symbols of $g$ and it turns out that all the ones that mix Roman and Greek indices vanish. I feel like I am missing something obvious.

P.S. I do not a priori know that $M= H \times K$. Does the form of the metric imply this?

  • 0
    Possible duplicate of [the Levi-Civita connection on a product of Riemannian manifolds](http://math.stackexchange.com/questions/31231/the-levi-civita-connection-on-a-product-of-riemannian-manifolds) – 2017-01-29
  • 0
    I did wonder this - but I don't a priori know that my manifold is a product manifold. Does that follow from the form of the metric? If so, could you explain? – 2017-01-29
  • 0
    Your question is purely local and your assumption is that locally the manifold and the metric split. There is a stronger result (de Rham decomposition) which says if a complete connected Riemannian manifold splits locally then its universal cover splits globally, but this is irrelevant for your purposes. – 2017-01-29
  • 0
    Indeed - I definitely do not have completeness. In any case, are you saying that the local assumptions are enough to get the splitting of the connection? – 2017-01-29
  • 1
    Yes, the linked question implies that the connection splits locally. The statement about a global splitting does not make sense without global assumptions: You cannot even make sense of the formula $\nabla^h \oplus \nabla^k$ if you do not assume that the tangent bundle globally splits. – 2017-01-29
  • 0
    Okay, thanks for the input! – 2017-01-29

0 Answers 0