Let $\phi: (M,g)\hookrightarrow (N,\tilde{g})$ be an isometric embedding of a Riemannian manifold $M$ of dimension $m$ into a Riemannian manifold $N$ of dimension $n$. I am interested in trying to do a Taylor expansion of this mapping in Riemannian Normal coordinates.

I have shown previously that for any real-valued function $f: M\rightarrow \mathbb{R}$, the covariant $k$-th derivative $$ \nabla^{k+1} f [X_1,...,X_{k+1}]= X_{k+1}(\nabla^{k}[X_1,...,X_k])-\sum_{i=1}^k\nabla^k f[X_1,...,X_{i-1},\nabla_{X_{k+1}}X_i,...,X_k] $$

written Riemannian normal coordinates (or in in any coordinates where the Christoffel symbols vanish) is simply the expression $$ \sum_{I}X^I \frac{\partial^I f}{\partial X^I} $$ Where $I$ is a multiindex of size $k$ varying over all choices of $m$ integers from the set $\{1,...,m\}$ (This looks like the $k$-th derivative.)

This suggests that when I choose normal coordinates in $M$, and perform a Taylor expansion of $f$, I am actually computing the coordinate expression of $$ \nabla f[X] + \frac{1}{2!}\nabla^2 f[X,X] + \frac{1}{3!}\nabla^3f [X,X,X] + ... $$

If I wish to more generally understand the taylor expansion of a mapping between manifolds, $\phi: (M,g)\hookrightarrow (N,\tilde{g})$ as above, I can choose a coordinate mapping $\eta:U\rightarrow \mathbb{R}^n$ in $N$ and regard the $i$-th component function of the mapping $\eta\circ\phi$.

This suggests to me that there might be a way to Taylor expand $\phi$ in terms of it's covariant derivatives that is independent of choosing a particular chart $\eta$ in $N$. Is this the case? I suppose that the question more generally is is there a way to view a linear connection $\nabla_X Y$ as some combination of covariant derivatives on scalar functions?

  • $\begingroup$ Your second displayed equation doesn't look right to me. It holds if and only if the Christoffel symbols vanish to infinite order at the point (where the exponential coordinates are centered). But that doesn't hold unless the curvature vanishes to infinite order at the point. $\endgroup$ – Deane Yang Feb 15 at 16:19
  • $\begingroup$ In fact, any kind of Taylor expansion of a map like this is quite messy to calculate beyond the second or third order term. You can, however, try to adapt the approach in the paper of Heintze-Karcher (numdam.org/article/ASENS_1978_4_11_4_451_0.pdf) to do this using Jacobi fields (I mention them a lot. They're the answer to almost any question involving the exponential map). Would you be willing to say why you want to do this? $\endgroup$ – Deane Yang Feb 15 at 16:58
  • $\begingroup$ Thank you for the reply! I looked over my calculation for the second expression, and I can't see my mistake. I used an inductive argument, and relied on the fact that the Christoffel symbols vanish at the centered point. It's unfortunately a bit messy of a calculation, but I tried to clean it up as best as possible. rvaughnmath.files.wordpress.com/2019/02/… $\endgroup$ – Ryan Vaughn Feb 17 at 1:00
  • $\begingroup$ I often work with asymptotic comparisons of intrinsic and extrinsic distance of an embedded submanifold of $\mathbb{R}^n$ using Taylor expansions. I thought it would be interesting to try to generalize this to arbitrary isometric embeddings of one manifold into to another. I thought it would be interesting to express Taylor's Theorem in a coordinate-free manner using a linear connection. Thanks again! $\endgroup$ – Ryan Vaughn Feb 17 at 1:23
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    $\begingroup$ I think I see what you are saying. In my argument, I remove the term involving the Christoffel symbols since they are zero when evaluated at $p$. This is okay initially, but when I calculate the $k+1$th derivative, I need to keep in the term involving the Christoffel symbols from the $k$-th derivative because we have not yet evaluated at $p$. This will cause me to take derivatives of the Christoffel symbols. That is very interesting. I will see if I can prove this is true iff curvature vanishes to infinite order now. Thank you for pointing that out! $\endgroup$ – Ryan Vaughn Feb 17 at 14:33

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