Christoffel symbols coherence.

Although the CS (Christoffel symbols) aren't tensor component, they have a change of coordinate description, I tried to test the (transitivity ) coherence and I get a trouble, let me explain:

Given an affine connection $\nabla$ on a (smooth) variety, the CS associated to a chart $ (x_1, \dots, x_ m)$ are defined as:

$${}^x\Gamma^k_{i,j}\cdot \vec{x}_k=\nabla_{\vec{x}_i}\vec{x}_j$$

where $ \vec{x}_i:=\frac{\partial}{\partial x^i}$ and repeated indices are under summation.

given the chart's (on a same open) $(x_1, \ldots, x_m),\ (y_1, \ldots, y_m),\ (z_1, \ldots, z_m)$ we have (see Christoffel Symbols and the change of transformation law.):

1 ) $$ {}^y\Gamma^t_{r,s}= \dfrac{\partial^2 x^u}{\partial y^s\partial y^r} \cdot \dfrac{\partial y^t}{\partial x^u}+ {}^x\Gamma^k_{i,j}\cdot \dfrac{\partial y^t}{\partial x^k}\cdot \dfrac{\partial x^j}{\partial y^s} \cdot \dfrac{\partial x^i}{\partial y^r}.$$

2 ) $${}^z\Gamma^a_{b,c}= \dfrac{\partial^2 y^u}{\partial z^b\partial z^c} \cdot \dfrac{\partial z^a}{\partial y^u}+ {}^y\Gamma^t_{r,s}\cdot \dfrac{\partial z^a}{\partial y^t}\cdot \dfrac{\partial y^r}{\partial z^b} \cdot \dfrac{\partial y^s}{\partial z^c}$$

then for replacement (by smoothness we can change derivation order and apply the differentiation rule of composed function ):

3 ) $${}^z\Gamma^a_{b,c}= \dfrac{\partial^2 y^u}{\partial z^b\partial z^c} \cdot \dfrac{\partial z^a}{\partial y^u}+ \left[\dfrac{\partial^2 x^u}{\partial y^s\partial y^r} \cdot \dfrac{\partial y^t}{\partial x^u}+ {}^x\Gamma^k_{i,j}\cdot \dfrac{\partial y^t}{\partial x^k}\cdot \dfrac{\partial x^j}{\partial y^s} \cdot \dfrac{\partial x^i}{\partial y^r}\right]\cdot \dfrac{\partial z^a}{\partial y^t}\cdot \dfrac{\partial y^r}{\partial z^b} \cdot \dfrac{\partial y^s}{\partial z^c}=$$

3.1) $$\dfrac{\partial^2 y^u}{\partial z^b \partial z^c} \cdot \dfrac{\partial z^a}{\partial y^u}+ \dfrac{\partial^2 x^u}{\partial y^r\partial y^s} \cdot \dfrac{\partial y^t}{\partial x^u}\cdot \dfrac{\partial z^a}{\partial y^t}\cdot \dfrac{\partial y^r}{\partial z^b} \cdot \dfrac{\partial y^s}{\partial z^c} +$$ $${}^x\Gamma^k_{i,j}\cdot \dfrac{\partial y^t}{\partial x^k}\cdot \dfrac{\partial x^j}{\partial y^s} \cdot \dfrac{\partial x^i}{\partial y^r}\cdot \dfrac{\partial z^a}{\partial y^t}\cdot \dfrac{\partial y^r}{\partial z^b} \cdot \dfrac{\partial y^s}{\partial z^c}=$$

3.2) $$\dfrac{\partial^2 y^u}{\partial z^b\partial z^c} \cdot \dfrac{\partial z^a}{\partial y^u}+ \dfrac{\partial^2 x^u}{\partial z^b\partial z^c} \cdot \dfrac{\partial y^t}{\partial x^u}\cdot \dfrac{\partial z^a}{\partial y^t} +$$ $${}^x\Gamma^k_{i,j}\cdot \dfrac{\partial z^a}{\partial x^k}\cdot \dfrac{\partial x^j}{\partial z^c} \cdot \dfrac{\partial x^i}{\partial z^b}=$$

3.3) $$\dfrac{\partial^2 y^u}{\partial z^b\partial z^c} \cdot \dfrac{\partial z^a}{\partial y^u}+ \dfrac{\partial^2 x^u}{\partial z^b\partial z^c} \cdot \dfrac{\partial z^a}{\partial x^u} +{}^x\Gamma^k_{i,j}\cdot \dfrac{\partial z^a}{\partial x^k}\cdot \dfrac{\partial x^i}{\partial z^b}\cdot \dfrac{\partial x^j}{\partial z^c}.$$

but:

4 ) $${}^z\Gamma^a_{b,c}= \dfrac{\partial^2 x^u}{\partial z^b\partial z^c} \cdot \dfrac{\partial z^a}{\partial x^u}+ {}^x\Gamma^k_{i,j}\cdot \dfrac{\partial z^a}{\partial x^k}\cdot \dfrac{\partial x^i}{\partial z^b} \cdot \dfrac{\partial x^j}{\partial z^c}$$

the penultimate result $3\to 3.1\to 3.2\to 3.3$ is wrong, because form $(4)$ would follow: $$\dfrac{\partial^2 y^u}{\partial z^b \partial z^c} \cdot \dfrac{\partial z^a}{\partial y^u}=0$$

then form $(2)$ the CS's would represent a tensor. Where the passages in $(3)$ are wrong?

Solution 1:

Going from 3.1 to 3.2 is just plain wrong. The second-order partial derivatives are not tensorial.

In particular, \begin{align*} \frac{\partial^2 x^u}{\partial z^b\partial z^c} &= \frac{\partial}{\partial z^b}\left(\frac{\partial x^u}{\partial z^c}\right) = \frac{\partial}{\partial z^b}\left(\frac{\partial x^u}{\partial y^r}\frac{\partial y^r}{\partial z^c}\right) = \frac{\partial x^u}{\partial y^r}\frac{\partial^2 y^r}{\partial z^b\partial z^c} + \frac{\partial y^r}{\partial z^c}\frac{\partial}{\partial z^b}\left(\frac{\partial x^u}{\partial y^r}\right) \\ &=\frac{\partial x^u}{\partial y^r}\frac{\partial^2 y^r}{\partial z^b\partial z^c}+\frac{\partial y^r}{\partial z^c}\frac{\partial y^s}{\partial z^b}\frac{\partial^2 x^u}{\partial y^s\partial y^r}. \end{align*} I believe you totally omitted the first terms here in your calculation. I did not continue further.

REMARK: The presence of the first term explains why the Hessian of $f$ gives a well-defined quadratic form on the tangent space only at a critical point of $f$.