# Some manipulations on a Riemannian manifold

Let $\phi$ be a point on a Riemannian manifold $(M,g)$ and $\xi \in T_\phi M$. Then I want to understand the proof/meaning of the following three identities I ran into,

• $\nabla _\nu \xi ^i = \partial _\nu \xi ^i + \Gamma ^{i}_{jk} \partial _\nu \phi ^j \xi ^k$

(...to my mind the above looks very different from how the connection/covariant derivative is supposed to act on a vector field!..)

• $(e^\xi \phi)^i = \phi^i + \xi ^i - \frac{1}{2}\Gamma ^{i}_{jk}(\phi)\xi ^j \xi ^k + O(\xi^3)$

(..apart from the mysterious RHS I don't understand what is this quantity "$e^\xi \phi$"!?..one clearly couldn't have been multiplying two points on a general Riemannian manifold!..)

• $\frac{1}{2}\nabla _\phi (\sum _{j=1}^N d^2_g(\phi, e^{\xi_j} \phi) ) = \sum _{j=1}^N \xi_j$

(..the above apparently has an interpretation of defining the "barycenter" of the set of points $\{ e^\xi_j \phi \}$..)

It would be great if someone can help understand the above three identities.

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Could you perhaps provide the context of these identities? Where did you run into them? – Jesse Madnick Oct 7 '12 at 0:28
These occur at top of page 53, bottom of page 58 and top of page 59 of this math.ias.edu/QFT/fall/NewGaw.ps – user6818 Oct 17 '12 at 16:51

From the first your identity it is clear that a coordinate system $(x^i)$ around point $\phi$ is assumed, so we have $\phi^i = x^i (\phi)$ seen as functions of point $\phi$. The coordinate frame is formed by vectors $\partial_i := \tfrac{\partial}{\partial x^i}$. Vector $\nu$ in direction of which vector $\xi$ is being differentiated can be written in coordinates as $\nu = \nu^j \partial_j$.

We can consider directional derivatives $\partial_{\nu} \phi^i$ in the direction of vector $\nu$ that is given by $$\partial_{\nu} \phi^j = \nu^l \partial_l \phi^j$$ where the Einstein summation convention is used. Observe that these derivative are nothing else than the directional derivatives of the coordinate functions $$\partial_{\nu} \phi^j = \partial_{\nu} x^j = \nu^l \partial_l x^j = \nu^l \delta_{l}{}^{j} = \nu^j$$

In other words, $\partial_{\nu} \phi^j$ is a fancy way of saying $\nu^j$.

With this in mind eliminating the components of vector $\nu$ form the first identity $$\nu^j \nabla _j \xi ^i = \nu^j \partial _j \xi ^i + \Gamma ^{i}_{jk} \nu^j \xi ^k$$ we recover the usual formula for the covariant derivative: $$\nabla _j \xi ^i = \partial _j \xi ^i + \Gamma ^{i}_{jk} \xi ^k$$

With regards to your second identity, $(e^\xi \phi)^i$ is the $i$-th coordinate of the exponential map of point $\phi$ in the direction of vector $\xi$, and this identity is the Taylor expansion of the exponential map in the coordinate neighborhood. It looks to me that here that normal coordinates are needed to have this expansion in this form.

The last identity looks too bizarre to me. In particular, I am struggling to find an interpretation for $\nabla_{\phi}$ (because $\phi$ is a point!), so more context is required to understand the meaning of it.

Edit. Since we are given the context now (see the OP's comment to the question), it becomes a little easier to understand the two other identities in the consideartion. I have to admit that I dont' know complete answers because my background is significantly different from what people are doing in that text, but nevertheless I am able to comment on certain things there.

The exponential map in Riemannian geometry is usually denoted by $$\operatorname{exp}_{p} \colon T_{p} M \to M$$ which in the text that we are looking at is abbreviated to $e^{\xi} p$ for the reasons that roughly speaking $e^{\xi}e^{\zeta} \approx e^{\xi + \zeta}$. The precise definition of the exponential map uses the geodesic $\gamma_{\xi}(t)$ such that $\gamma_{\xi}(0) = p$ and $\dot{\gamma_{\xi}}(0) = \xi$. Then $$\operatorname{exp}_{p}(\xi) = \gamma_{\xi}(1)$$ for $\xi \in T_{p}M$. A very useful note for this stuff is here.

In the third identity of the question, $\nabla_{\phi} f$ seems to be the gradient of function $f \colon M \to \Bbb R$ at point $\phi$. To prove the identity one needs to differentiate the function there, properly unwielding all the definitions. Some people (not me!) even may say that intuitively this identity is obvious. Perhaps, if we can understand this identity for the flat case (in $\Bbb R^{n}$ with the Euclidean metric), then it is more or less clear in general. But in the flat case it is indeed obvious (even to me!)

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Thanks for the help. You can see my comment above for a reference to these. Can you explain what you mean by exponential map of point $\phi$ in the diretion if $\xi$" - can you kindly elaborate on that? Like I would have thought that $e^\xi$ is already a point on the manifold then what is $e^{\xi}\phi$? - what is this implicit "multiplication" ? – user6818 Oct 17 '12 at 16:53
$e^\xi \phi$ is the notation in your reference for the exponential map based at point $\phi$. "The direction" is the tangent vector $\xi$ at point $\phi$. See my edit for further details. – Yuri Vyatkin Oct 17 '12 at 23:46