Parallel transport of general tensors

Question

Everything is happening on a Riemannian manifold $(N,g)$ with the Levi-Civita connection in the following discussion. In a paper I'm reading, the author defines on a codimension 2 submanifold, $M\subset N$, a metric $\omega$ where the then claims that $M$ is $\omega$ is defined on the boundary of a tubular nbhd of $M$, namely onto the hypersurface $M'_r$ where $$M'_r:=\{x\in N: d_g(x,M)=r \}.$$ My question is that am I correct to assume that he parallelly transported $\omega$ from $M$ to $M_r'$ along the radial directions? I was aware that you can parallel transport vectors, but I wasn't aware that you could for general tensors. Can you also parallelly transport functions?

Answer 1

As far as I understand, there is no parallel transport involved but an identification in some natural coordinates. Let me summarize what I think I have understand from a quick reading of the paper (available at arXiv:1708.08211). I would be happy to receive any comment regarding any misunderstanding from anyone. If a specialist has a really important remark, may they feel free to edit this answer to fix any fundamental error.

Let $N$ be a codimension $2$ submanifold of the Riemannian manifold $(M,g)$. Consider $\nu N$ the normal bundle of $N$ in $M$, that is $$ \nu N = \{(x,v) \in N\times T_xM \mid v \in v \perp T_xN \}. $$ It is a rank $2$ vector bundle over $N$. The normal exponential map is the map $$ \begin{array}{r|ccc} E\colon & \nu N & \longrightarrow & M \\ & (x,v) & \longmapsto & \exp_x(v) \end{array}. $$ Fix $x\in N$. It can be shown that $dE(x,0) \colon T_xN \oplus \nu_xN \to T_xM$ is the identity map, and from the inverse function Theorem there exists an open neighbourhood $V_x\subset N$ of $x$ in $N$, and a real number $r_0>0$, such that $E$ induces a diffeomorphism from $\{(x,v) \in \nu N \mid x\in V_x, |v|_g < r_0\}$ onto its image $U_x$. By shrinking $V_x$ into a smaller neighbourhood, one can find a local trivialization $$ \nu N|_{V_x} \simeq V_x \times \Bbb R^2, $$ and therefore we have a diffeomorphism (denoting $B(0,r_0)$ the open disk of radius $r_0$ in $\Bbb R^2$): $$ V_x \times B(0,r_0) \longrightarrow U_x. $$ If $\{x^1,\ldots,x^{n-2}\}$ are coordinates for $N$ on $V_x$ and $(r,\theta)$ are polar coordinates on $B(0,r_0)$, then $\{x^1,\ldots,x^{n-2},r,\theta\}$ are now coordinates on $U_x$. It seems that the hypothesis of the authors (see equation (3.4), page 9 of the arXiv version) are that in this coordinates system, the metric reads $$ g|_{\{x^1,\ldots,x^{n-2},r,\theta\}} = f^2 dr^2 + r^2 (d\theta + \sigma)^2 + \omega|_{\{x^1,\ldots,x^{n-2}\}}, $$ where $f$ is a function on $U_x$, $\sigma$ is a $1$-form on $\Bbb S^1$, and $\omega$ is a metric on $N$ (the restriction of $g$).

In fact, it seems that the authors assume that this can be done globally: the normal bundle of $N$ seems to be assumed trivial (we say that $N$ is co-orientable), and the whole tubular neighbourhood $U$ of $N$ seems to be endowed with a global trivialization $$ N \times B(0,r_0) \longrightarrow U, $$ so that equation (3.4) defines a metric on all of $U$. (To be perfectly rigorous, I should have consider polar coordinates with $0<r<r_0$ instead of $0\leqslant r<r_0$, and the diffeomorphism is in fact bewteen $N \times (B(0,r_0)\setminus \{0\})$ and $U\setminus N$.)