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We know that the standard result derived in this post

$$\delta \Gamma_{bc}^{a} = \frac{1}{2}g^{da}(\nabla_{b}(\delta g_{dc}) + \nabla_{c}(\delta g_{db}) - \nabla_{d}(\delta g_{bc})) \, .$$

is a tensor. Now, I want to derive how it would look by lowering the index $a$. I am expecting the result to look similar to this

$$g_{ad}\delta \Gamma^{d}_{bc}=\delta \Gamma_{abc} = \frac{1}{2} \left( \nabla_b \delta g_{ac} + \nabla_c \delta g_{ab} - \nabla_a \delta g_{bc} \right)$$

based on the answer presented here. I was expecting it to be a tensor based on the following arguments.

\begin{align} \Gamma_{abc}' &= \frac{\partial x^d}{\partial x'^a}\frac{\partial x^e}{\partial x'^b}\frac{\partial x^f}{\partial x'^c}\Gamma_{def}+\frac{\partial^2 x^d}{\partial x'^b\partial x'^c}\frac{\partial x_d}{\partial x'^a} \end{align} Under variation \begin{align} \Gamma_{abc}'+\delta \Gamma_{abc}'&= \frac{\partial x^d}{\partial x'^a}\frac{\partial x^e}{\partial x'^b}\frac{\partial x^f}{\partial x'^c}(\Gamma_{def}+\delta \Gamma_{def})+\frac{\partial^2 x^d}{\partial x'^b\partial x'^c}\frac{\partial x_d}{\partial x'^a}\\ \delta \Gamma_{abc}' &=\frac{\partial x^d}{\partial x'^a}\frac{\partial x^e}{\partial x'^b}\frac{\partial x^f}{\partial x'^c}\delta \Gamma_{def} \end{align}

However, when I tried to derive $ \delta \Gamma_{abc} $ using the Leibniz rule from $ \delta (g_{ad} \Gamma^d_{bc})$, I got:

$$ \delta (g_{ad} \Gamma^d_{bc}) = (\delta g_{ad}) \Gamma^d_{bc} + g_{ad} (\delta \Gamma^d_{bc}).$$

Substituting $ \delta \Gamma^d_{bc} $,

$$ \delta \Gamma_{abc} = (\delta g_{ad}) \Gamma^d_{bc} + \frac{1}{2} \left( \nabla_b \delta g_{ac} + \nabla_c \delta g_{ab} - \nabla_a \delta g_{bc} \right). $$

There is an extra term $ (\delta g_{ad}) \Gamma^d_{bc} $ from what I was expecting. One possibility that cross my mind is that $\delta \Gamma_{abc}$ is not a tensor at all and the transformation property I used when considering variation is wrong. Could anyone provide clarification as to where I went wrong?

Chandra Prakash
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1 Answers1

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I realized the mistake I was making. If we look at the transformation law of variation we find:

\begin{align} \delta (g_{ad} \Gamma^d_{bc}) &= (\delta g_{ad}) \Gamma^d_{bc} + g_{ad} (\delta \Gamma^d_{bc})\\ \delta (g'^{ad} \Gamma'^d_{bc}) &= \frac{\partial x^e}{\partial x'^a} \frac{\partial x^g}{\partial x'^b} \frac{\partial x^h}{\partial x'^c} \delta g_{ef} \Gamma^f_{gh} + \delta g_{ef} \frac{\partial x^e}{\partial x'^a} \frac{\partial x^f}{\partial x'^b \partial x'^c}+ \frac{\partial x^e}{\partial x'^a} \frac{\partial x^g}{\partial x'^b} \frac{\partial x^h}{\partial x'^c} g_{ef} \delta \Gamma^f_{gh}\\ &= \frac{\partial x^e}{\partial x'^a} \frac{\partial x^g}{\partial x'^b} \frac{\partial x^h}{\partial x'^c} \delta (g_{ef} \Gamma^f_{gh}) + \delta g_{ef} \frac{\partial x^e}{\partial x'^a} \frac{\partial x^f}{\partial x'^b \partial x'^c} \end{align} So, if we are varying the metric then connection term of first kind is not a tensor. The mistake I made was not varying the metric in the original transformation: \begin{align} \Gamma_{abc}' &= \frac{\partial x^d}{\partial x'^a}\frac{\partial x^e}{\partial x'^b}\frac{\partial x^f}{\partial x'^c}\Gamma_{def}+g_{ef}\frac{\partial^2 x^e}{\partial x'^b\partial x'^c}\frac{\partial x^f}{\partial x'^a}\\ \Gamma_{abc}'+\delta \Gamma_{abc}'&= \frac{\partial x^d}{\partial x'^a}\frac{\partial x^e}{\partial x'^b}\frac{\partial x^f}{\partial x'^c}(\Gamma_{def}+\delta \Gamma_{def})+g_{ef}\frac{\partial^2 x^e}{\partial x'^b\partial x'^c}\frac{\partial x^f}{\partial x'^a}+\delta g_{ef}\frac{\partial^2 x^e}{\partial x'^b\partial x'^c}\frac{\partial x^f}{\partial x'^a}\\ \delta \Gamma_{abc}' &=\frac{\partial x^d}{\partial x'^a}\frac{\partial x^e}{\partial x'^b}\frac{\partial x^f}{\partial x'^c}\delta \Gamma_{def}+\delta g_{ef}\frac{\partial^2 x^e}{\partial x'^b\partial x'^c}\frac{\partial x^f}{\partial x'^a} \end{align}

Therefore, connection term of first kind is not a tensor under the variation of metric.

Chandra Prakash
  • 482