# Embedding curvature

I want to compute the embedding curvature of a curve embedded in some manifold, as defined (eg., by Amari in Methods of Information Geometry) by : $$H(X,Y) = \nabla_X Y - \nabla_X^{(\pi)} Y$$ where $\pi : T_p(S) \rightarrow T_p(M)$ is the mapping from the tangent space of the manifold $S$ to the tangent space of my curve. For the case of my curve, $H$ is thus scalar and we only have the case $H(X,X)$.

However, there is something I am missing when trying to compute the term $\nabla_X X$. This term is computed by : $$\nabla_X X = X^i \{\partial_i X^k + X^j\Gamma^k_{ij}\}\partial_k$$ My problem lies in the computation of $\partial_i X^k$. In particular, $X$ is my tangent vector on the curve, and now, I need to differentiate it in the direction $\partial_i$ which is not tangent to the curve (it's $\frac{\partial}{\partial u^i}$ for some parameterization $u$ of the manifold) : hence, I need to know how much my tangent vector is varying in a direction that is not along my curve.... How is it possible to do that ? Did I misunderstand something ? (I guess!!)

Thanks !

Nicolas

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So $M$ is the curve, which is embedded in an ambient manifold $S$... is that right? Also, what do you mean by "$H$ is thus scalar"? – Jesse Madnick Sep 15 '12 at 9:41
$M$ is the curve embedded is $S$, yes. I mean $H$ is scalar because a curve is 1D, so there is a single tangent per point, so the single entry for H(X,Y) is H(X,X) (by H is scalar, I mean, H is a 1x1 tensor). – WhitAngl Sep 15 '12 at 16:56
Unless I misunderstood something. In particular, when computing the Christoffel symbol of $\nabla^{(\pi)}$, I only have a single entry ($\nabla^{(\pi)}$ is only defined along the curve, right?) – WhitAngl Sep 15 '12 at 17:03
@WhitAngl $H$ is not a $1 \times 1$-tensor on $TM$ because it takes values in $TS$ – Yuri Vyatkin Sep 16 '12 at 2:41
mmm.. then I don't understand anything :s Do we agree at least that $\Gamma^{(\pi)}$, the Christoffel symbol of the projected connection $\nabla^{(\pi)}$ is a scalar in my case ? That's what I understand from "Methods of Information Geometry" by Amari, page 22 (preview available on google book : books.google.com/… ). – WhitAngl Sep 16 '12 at 5:57

This is not a complete answer, but rather a series of remarks, because I don't see the precise question. I deliberately leave some of the points open (marked with question marks), because (1) I just want to sketch the direction; (2) I am not an expert in the affine differential geometry, and my understanding is based on my experience in the Riemannian situation. Perhaps, someone who knows the affine geometry would address this question more specifically.

The setting that we deal with in this question is an immersion (or, locally, an embedding) $f\colon M \to S$ where $S$ is a space with a connection $\nabla$, and $M$ is equipped with a linear (?) map (more precisely, a family of maps) $\pi_{p} : T_p(S) \rightarrow T_p(M)$ that may need to satisfy certain properties for the constructions to be valid.

With these data we may attempt to define a connection $\nabla^{(\pi)}$ on $M$ by $$(\nabla^{(\pi)}_{X}{Y})_p = \pi_{p}((\nabla_{X}{Y})_{p})$$ at each point $p \in M$. Notice, that $\nabla$ is not a connection on $M$, in particular, it "does not know" how to act on vector fields $X,Y \in T(M)$. To make the above definition correct, we need to consider extensions $\tilde{X}, \tilde{Y}$ of fields $X, Y$ to a (tubular?) neighborhood of $M$ in $S$, and then set $$(\nabla^{(\pi)}_{X}{Y})_p = \pi_{p}((\nabla_{\tilde{X}}{\tilde{Y}})_{p}) = \pi_{p}((\nabla_{\tilde{X}_{p}}{\tilde{Y}})_{p}) = \pi_{p}((\nabla_{X_{p}}{\tilde{Y}})_{p})$$ where the last two equalities come from the linearity in the slot $X$. Somehow we need to understand that this definition does not depend on the extensions. Or it does?

Assuming that we understand why the definition of $\nabla^{(\pi)}_{X}{Y}$ works, we may want to compare the covariant derivatives introducing the quantity $H(X,Y) = \nabla_X Y - \nabla_X^{(\pi)} Y$. Again, one needs to explain why (or why not) this quantity is independent of extensions. When it is done it is clear that $H(X,Y)$ is bilinear, however it may not be symmetric for an arbitrary connection $\nabla$ on $M$.

Now, when we are trying to apply this construction to a curve $M$ in $S$ it is not clear why $H(X,Y)$ should be a scalar, because $\nabla_{X}{Y}$ is a vector in $S$ (or, one may reduce the target space somehow?). In the Riemannian case, for example, $H(X,Y)$ has a many components as the co-dimension of $M$ in $S$ is.

When we attempt to compute $\nabla_{X}{X}$ in coordinates, we immediately stumble upon the necessity to have an extension of the field $X$ to be able to "differentiate in the direction ... which is not tangent to the curve". This could be circumvented with the observation that with a choice of coordinates one obtains a standard way to get extensions (constant along the coordinate lines).

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It does not really answers my question but that helped me understand (see chat), so I validate the answer. In the meantime, I found out an alternative way to formulate the problem in a pure Riemannian setting (writing the connection in a frame with $\dot\gamma(t)$ as one of the elemnts of the basis), with much help from a professor here. Thanks all ! – WhitAngl Sep 22 '12 at 18:16