# Differential geometry-geodesics

Let $\nabla_1$ and $\nabla_2$ be two affine connections of manifolds $M_1$ and $M_2$ and $\nabla$ induced connection on product $M_1\times M_2$. Prove the following:

If $\gamma_1$, $\gamma_2$ are curves on $M_1$, $M_2$ and $X_1$, $X_2$ are vector fields parallel along $\gamma_1$, $\gamma_2$, then the field $d_{j_1}(X_1)+d_{j_2}(X_2)$ is parallel along $\gamma_1 +\gamma_2$, and conversely also holds (every field parallel along $\gamma_1 +\gamma_2$ has this form).

Can we conclude that geodesics on $M_1\times M_2$ are products of geodesics on $M_1$ and $M_2$?

Detailed solutions and explanations are welcome. Thanks in advance.

• Out of curiosity, by "$d_{j_1}(X_1)$, do you mean $D_{\gamma_1}(X_1)$ (the covariant derivative of $X_1$ along $\gamma_1$)? In any case, a product of geodesics is a (flat) surface in $M_1 \times M_2$, so your proposed conclusion is not quite right. (However, a geodesic in the product projects to a geodesic in either factor, so there's a theorem lurking behind your intuition.) Dec 15, 2013 at 12:44
• I suspect that by $j_1$, you mean something like "injection on the first factor." So I suppose you have a chosen point $m_i \in M_i$, so that $j_1(t) = (t, m_2)$ and $j_2(t) = (m_1, t)$, the first map being defined on $M_1$ and the second being defined on $M_2$. Am I right? Then by $\gamma_1 + \gamma_2$, do you mean $$\gamma: [0, 1] \rightarrow M_1 \times M_2 : t \mapsto (\gamma_1(t), \gamma_2(t))$$ or something else? If so, it might be better to call it $(\gamma_1, gamma_2)$. Assuming I've got all this right, I don't immediately see why "conversely also holds". Can you explain? Dec 19, 2013 at 13:54

For $M=M_1\times M_2$ we have $TM=TM_1 \times TM_2$; at each point $p=(p_1,p_2)\in M$ the tangent space $T_pM$ naturally splits as the direct sum $T_{p_1}M_1\oplus T_{p_2}M_2$.
Now, if $\nabla^i$ are connections on $M_i, i=1,2$, then the standard connection on $M$ is the sum of two connections $\nabla^1 + \nabla^2$ meaning that for vector fields $X_i, Y_i, i=1,2$ on $M$ and $X=(X_1,X_2), Y=(Y_1,Y_2)$, we have $$\nabla_X(Y)= \nabla^1_{X_1}(Y_1) + \nabla^2_{X_2}Y_2.$$ Suppose that $c=c(t)=(c_1(t),c_2(t))$ is a smooth curve in $M$ with $c_i$ smooth curves in $M_i, i=1,2$. Then (in view of the formula for $\nabla$ above) $$\nabla_{c'(t)} c'(t)= \nabla^1_{c_1'(t)} c_1'(t) + \nabla^2_{c_2'(t)}c_2'(t).$$ The curve $c$ is a geodesic iff $\nabla_{c'(t)} c'(t)=0$. Note that a vector $v=(v_1, v_2)\in T_{(p_1,p_2)}(M)$ is zero if and only if both components of $v$ are zero. Therefore, $$\nabla_{c'(t)} c'(t)=0 \iff \nabla^1_{c_1'(t)} c_1'(t) =0, \nabla^2_{c_2'(t)}c_2'(t)=0.$$ Hence, $c$ is a geodesic if and only if both curves $c_1, c_2$ are geodesics.