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Let $V_1,\ldots,V_k$ be a transversal set of smooth compact orientable sub-manifolds of a compact orientable manifold $M$, and set $V=\bigcap V_i$.

Is it always possible to equip a neighborhood $U$ of $V$ in $M$ with a metric $g$ such that every $V_i\cap U$ is totally geodesic?

or that the exponential map $exp: N_V^M=TV^{\perp g} \to U$ maps very $(N_V^M \cap TV_i)|_V$ into $V_i$?

If not, what are the obstructions?

Edit (Misha): Just to clarify things: Smooth submanifolds $M_1,...,M_k$ of a smooth manifold $M$ are said to intersect transversally at a point $x$ if $$ codim_{T_x M} \bigcap_{i=1}^k T_x M_i= \sum_{i=1}^k codim_M M_i. $$ In other words, codimension of intersection (at the tangent space level) is the sum of codimensions.

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  • $\begingroup$ You can always make a closed submanifold totally geodesic for some metric on the ambient manifold. I don't see why you're bringing in intersections. $\endgroup$ – Ryan Budney Feb 14 '14 at 3:14
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    $\begingroup$ all the $V_i$ at the same time Sir! not just $V$. Read the question carefully before you vote to close it. $\endgroup$ – Mohammad F. Tehrani Feb 14 '14 at 3:52
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    $\begingroup$ This is an OK question for MO. Even if $k=2$ and $V$ has trivial normal bundle the answer is unclear since the problem is to extend the metric from a neighborhood of the intersection while keeping each $V_i$ totally geodesic. (Partition of unity will not work directly, of course.) $\endgroup$ – Misha Feb 14 '14 at 6:14
  • $\begingroup$ @MohammadF.Tehrani I think we must be careful to use "transversal". What of the following two statements you are considering for transversality: $\endgroup$ – Ali Taghavi Feb 14 '14 at 13:40
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    $\begingroup$ Just for the record, here is the precise definition of transversality: Smooth submanifolds $M_1,...,M_k$ in a smooth manifold $M$ intersect transversally at a point $x$ if $codim_{T_xM} \cap_{i=1}^k T_x M_i= \sum_{i=1}^k codim_M(M_i)$. $\endgroup$ – Misha Feb 15 '14 at 14:45
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It is not possible in general. Here is a counterexample. Let $V_1$ be the $xy$ plane in $\mathbb{R}^3$, $V_2$ be the $xz$ plane in $\mathbb{R}^3$, and let $W_3$ be the surface $z=\exp(-1/x^2)$ for $x>0 $ and $z=0$ for $x\leq 0$. Then let $V_3$ be $W_3$ rotated by $45$ degrees in the $yz$-coordinates. It is easy to see that $V_1$, $V_2$, and $V_3$ are pairwise transverse. Their intersection $V$ is the interval $(-\infty,0]$ on the $x$-axis. At the origin, it is clear that not all three surfaces can be made totally geodesic with any smooth metric, because then two geodesics could be constructed so that they are different yet have the same initial velocity. I realize that $V$ is a manifold with boundary in this case. Perhaps this example could be tweaked.

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  • $\begingroup$ A transverse set of 3 hypersurfaces in $\mathbb{R}^3$ has only points in their triple intersection. Your example is not a transverse intersection (although every two of them are transverse). $\endgroup$ – Mohammad F. Tehrani Feb 15 '14 at 14:23
  • $\begingroup$ Being transverse is a property which involves all the manifolds at the same time, transverse manifolds have minimal possible intersection on every subset. $\endgroup$ – Mohammad F. Tehrani Feb 15 '14 at 14:24

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