Show that, in a normal coordinate chart, $\Gamma^{i}_{(x)(jk)}(p)=0$

Question

I am currently working through the following proof and have a question regarding the computations done by the author:

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Theorem: Let $(M, \mathcal{O}, \mathcal{A}, \nabla )$ be an arbitrary affine manifold and let $p \in M$. Then one can construct a chart $(U,x) \in \mathcal{A}$ with $p \in U$ such that $\Gamma^{i}_{(x)(jk)}(p)=0$, where $$\Gamma^{i}_{(x)(jk)}= \frac{1}{2}(\Gamma^{i}_{(x)jk}-\Gamma^{i}_{(x)kj})$$ with respect to the chart $(U,x)$ (denoted with a subscript in the gamma terms).

Proof: Let $(V,y) \in \mathcal{A}$ be any chart with $p \in V$. Thus, in general, $\Gamma^{i}_{(x)(jk)}(p) \neq 0$. Now consider a new chart $(U,x)$ where the chart transition map is given by

$$(x \circ y^{-1})(\alpha^{1},…, \alpha^{\text{dim}M}) := \alpha^{i} - \frac{1}{2}\alpha^{j}\alpha^{k}\Gamma^{i}_{(y)(jk)}(p)$$ where $y(p) = (\alpha^{1},…, \alpha^{\text{dim}M}) $. Now, this is where I am having trouble understanding: the author considers the change of vector components under change of chart

$$ \partial_{j}(x^{i} \circ y^{-1})(y(p)) = \delta^{i}_{j} - \alpha^{m}\Gamma^{i}_{(y)(jm)}(p). $$

I see how we obtain $\delta^{i}_{j}$, but how does one obtain the second term $-\alpha^{m}\Gamma^{i}_{(y)(jm)}(p)$ after differentiating? I don’t think I’m understanding how we differentiate the connection coefficients here, but these are all the steps the author gave so I’m not seeing how we arrive at this result; any help is appreciated.

Note: the rest of the proof is not pictured, but one can probably see how this leads to the symmetric part vanishing and the anti symmetric part surviving.

Answer 1

It should have said $y(p)=(0,\dots, 0)$, so because of this, the $\Gamma^{i}_{(y)\,jk}(p)$ are just numbers, so they come out of all the derivatives, i.e you do not differentiate them at all.

Anyway, let me outline how you can figure out the result ‘from scratch’.

• Fix a point $p\in M$
• Fix a chart $(V,y)$ around the given point $p$ such that $y(p)=0$.
• Construct a new chart $(U,x)$ around $p$ such that the transition map $x\circ y^{-1}$ is quadratic, i.e fix some numbers $A^{i}_{\,j}, B^{i}_{jk}\in\Bbb{R}$ and consider $(x^i\circ y^{-1})(\alpha^1,\dots,\alpha^n):=A^{i}_{\,j}\alpha^j+B^{i}_{jk}\alpha^j\alpha^k$.

Note that in order for this to actually be well-defined, we should assume that the matrix $[A^{i}_{\,j}]$ is invertible, because then the usual derivative of $x\circ y^{-1}$ at the origin will be the matrix $A$, which is invertible, so by the inverse function theorem, upon restricting to smaller neighborhoods, we get a $C^{\infty}$ diffeomorphism. Ok, so let’s suppose we shrink all the neighborhoods small enough. Now, calculate the following:

• what is $x(p)$?
• what is $\frac{\partial x^i}{\partial y^j}(p)$?
• what is $\frac{\partial^2x^i}{\partial y^j\partial y^k}(p)$?

This should be a trivial calculation, because we’re simply dealing with polynomials. With this information, and the knowledge of how the $\Gamma^{i}_{(x)\,jk}(p)$ is related to the $\Gamma$’s in the $y$ chart, you should be able to figure out how to choose the numbers $A^{i}_{\,j}$ and $B^{i}_{\,jk}$ in order to ensure such and such property of the $\Gamma$ in the new $x$ chart (at the single point $p$).

Finally you may wonder what prompted me to consider such quadratic changes of variables? The answer is simply that the transformation property of the $\Gamma$’s involves both the first and second derivatives of the chart-transition map. So, if I want to modify $\Gamma$’s at a point, then I need to ensure I am able to modify the first and second derivatives of the chart-transition as I wish; and finally quadratic polynomials are the simplest such functions.