# What is the Levi-Civita connection of product of semi-Riemann manifolds

Let $$(M_i,g_i)$$ be semi-Riemann manifold for $$i=1,2$$, and $$M=(M_1 \times M_2, g_1 +g_2)$$ is the product semi-Riemann manifold, with $$\pi_i: M \rightarrow M_i$$ be the canonical projection.

If $$X_i$$ is a vector field on $$M_i$$, then define $$\overline{X_i}$$ as a vector field on $$M$$ such that $$\pi_i.\overline{X_i}=X_i$$ and $$\pi_j.\overline{X_i}=0$$ for $$j \ne i$$; in other words, $$\overline{X_1}=(X_1,0)$$ and $$\overline{X_2}=(0,X_i)$$. Then, $$[\overline{X_1},\overline{X_2}]=0$$.

Now, let $$\nabla^i$$ be the Levi-Civita connection on $$M_i$$ and $$\nabla$$ the Levi-Civita conection on $$M$$. Then we need to show that $$\nabla_{\overline{X_i}}\overline{Y_i}=\overline{\nabla^i_{X_i}Y_i}$$ and $$\nabla_{\overline{X_1}} \overline{Y_2}=0$$.

I am not sure how can I use work on the Levi-Civita connection operator on ordered pairs..

Any hint is appreciated.

With each projection $$\pi_i\colon M_1\times M_2 \to M_i$$, pull back the Levi-Civita connection $$\nabla^i$$ of $$(M_i,g_i)$$ to a connection $$\pi_i^*(\nabla^i)$$ in the pull-back bundle $$\pi_i^*(TM_i) \to M_1\times M_2$$. Then form the direct sum bundle $$\pi_1^*(TM_1) \oplus \pi_2^*(TM_2) \to M_1\times M_2,$$and equip it with the direct sum connection $$\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2)$$. This is the Levi-Civita connection of $$(M_1\times M_2 , \pi_1^*g_1 + \pi_2^*g_2)$$, since $$\pi_1^*(TM_1) \oplus \pi_2^*(TM_2) \cong T(M_1\times M_2).$$Let $$(x^i)$$ and $$(y^\mu)$$ be local coordinates for $$M_1$$ and $$M_2$$, respectively, giving local coordinates $$(x^i \circ \pi_1, y^\mu \circ \pi_2)$$ for $$M_1\times M_2$$. Any vector field $$X$$ on $$M_1 \times M_2$$ may be locally written as $$X_{(x,y)} = \underbrace{\sum_i X^i(x,y) \frac{\partial}{\partial (x^i\circ \pi_1)}\bigg|_{(x,y)}}_{\mbox{\doteq X_1(x,y)\in T_xM_1}} + \underbrace{\sum_\mu X^\mu(x,y) \frac{\partial}{\partial (y^\mu \circ \pi_2)}\bigg|_{(x,y)}}_{\mbox{\doteq X_2(x,y)\in T_yM_2}},$$and similarly for a second vector field $$Y$$.

The problem here is that $$X_1$$ is not a vector field on $$M_1$$ because it depends also on the point $$y \in M_2$$ --- likewise for $$X_2$$ not being a vector field on $$M_2$$. They are, in fact, sections of $$\pi_1^*(TM_1)$$ and $$\pi_2^*(TM_2)$$ (this implies that they are particular vector fields on the full manifold $$M_1\times M_2$$).

For simplicity, let's omit the $$\pi$$'s and denote the coordinate vector fields by $$\partial_i$$ and $$\partial_\mu$$ only (either on each $$M_i$$ or on the full product $$M_1\times M_2$$ --- there will be a slight abuse of notation in what follows, but note the use of letters in different alphabets to indicate these objects live, a priori, in different spaces).

The business with pull-back bundles is the way to work around this "bug".

Using the characteristic properties of direct sums and pull-backs, i.e.,

1. $$(\nabla' \oplus \nabla'')_X(\psi'\oplus \psi'') = \nabla'_X\psi' + \nabla_X''\psi''$$

2. $$(F^*\nabla)_X(Y \circ F) = \nabla_{F_\ast X}Y$$

one has $$(\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2))_XY = (\pi_1^*(\nabla^1))_{X_1}Y_1 + (\pi_2^*(\nabla^2))_{X_2}Y_2,$$and each term in the right side above is computed in the same way --- let's do the first: \begin{align*} (\pi_1^*(\nabla^1))_{X_1}Y_1 &= (\pi_1^*(\nabla^1))_{X_1}\left( \sum_i Y^i \partial_i\right) \\ &= \sum_i X_1(Y^i) \partial_i + \sum_i Y^i (\pi_1^*(\nabla_1))_{X_1}\partial_i \\ &= \sum_i X_1(Y^i)\partial_i + \sum_{i,j} Y^iX^j (\pi_1^*(\nabla^1))_{\partial_j}\partial_i \\ &= \sum_k X_1(Y^k)\partial_k + {\color{red}{\sum_{i,j} X^iY^j \nabla^1_{\partial_i}\partial_j}} \\ &= \sum_k X_1(Y^k)\partial_k + \sum_{i,j,k} X^iY^j (\Gamma^1)_{ij}^k\partial_k \end{align*}In the term in red, the abuse of notation $$x^i \equiv x^i \circ \pi_1$$ kicked in. Similarly, one gets $$(\pi_2^*(\nabla^2))_{X_2}Y_2 = \sum_{\lambda} X_2(Y^\lambda)\partial_\lambda + \sum_{\mu,\nu,\lambda} X^\mu Y^\nu (\Gamma^2)_{\mu\nu}^\lambda \partial_\lambda.$$So, how does one find the Christoffel symbols $$\Gamma_{ab}^c$$ for this connection? If all $$a,b,c$$ are not all in the "same alphabet" (i.e., all mid-alphabet latin letters, or all greek letters), we get zero. Else, we get $$(\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2))_{\partial_i}\partial_j = (\Gamma^1)_{ij}^k \partial_k \qquad \mbox{and} \quad (\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2))_{\partial_\mu}\partial_\nu = (\Gamma^2)_{\mu\nu}^\lambda \partial_\lambda.$$

Those results coincide with what you get using the coordinate-version of the Koszul formula $$\Gamma_{ab}^c = \frac{1}{2} \sum_d g^{cd}(\partial_ag_{bd} + \partial_b g_{ad} - \partial_dg_{ab}),$$since we have (in block form) that $$\begin{pmatrix} ((g_1)_{ij})_{i,j} & 0 \\ 0 & ((g_2)_{\mu\nu})_{\mu,\nu} \end{pmatrix}^{-1} = \begin{pmatrix} ((g_1)^{ij})_{i,j} & 0 \\ 0 & ((g_2)^{\mu\nu})_{\mu,\nu} \end{pmatrix}.$$

• Also, for a related and detailed computation using pull-back connections, without abusing notation at all, see this post. Dec 3, 2020 at 6:23