- #1

yucheng

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- Homework Statement
- We are to show that $\frac{\partial Z}{\partial Z_{ij}} = Z Z^{ij}$ See https://math.stackexchange.com/questions/4141664/derivative-of-metric-tensor-with-respect-to-entries

P.S. this is actually part of the proof of the Voss-Weyl formula given in Pavel Grinfeld's book but the author does not care to derive it...

- Relevant Equations
- N/A

We know that the cofactor of determinant ##A##, is

$$\frac{\partial A}{\partial a^{r}_{i}} = A^{i}_{r} = \frac{1}{2 !}\delta^{ijk}_{rst} a^{s}_{j} a^{t}_{k} = \frac{1}{2 !}e^{ijk} e_{rst} a^{s}_{j} a^{t}_{k}$$

By analogy,

$$\frac{\partial Z}{\partial Z_{ij}} = \frac{1}{2 !}e^{ikl} e^{jmn} Z_{km} Z_{ln} = \frac{Z}{2 !} \epsilon^{ikl} \epsilon^{jmn} Z_{km} Z_{ln}$$?

As $$\epsilon^{ijk} = \frac{e^{ijk}}{\sqrt{Z}}$$. However, we should note that Z is **not** a tensor, since we only have relative levi-civita symbols, but if we substitute absolute levi-civita symbols, then it is indeed one (second equality)? Wait, I'm not sure whether one should call the former, ##\epsilon## 'absolute' and the latter, ##e## 'relative'...

Hmmm, this looks interesting. Would you give any hints as to what's wrong? Thanks in advance!

$$\frac{\partial A}{\partial a^{r}_{i}} = A^{i}_{r} = \frac{1}{2 !}\delta^{ijk}_{rst} a^{s}_{j} a^{t}_{k} = \frac{1}{2 !}e^{ijk} e_{rst} a^{s}_{j} a^{t}_{k}$$

By analogy,

$$\frac{\partial Z}{\partial Z_{ij}} = \frac{1}{2 !}e^{ikl} e^{jmn} Z_{km} Z_{ln} = \frac{Z}{2 !} \epsilon^{ikl} \epsilon^{jmn} Z_{km} Z_{ln}$$?

As $$\epsilon^{ijk} = \frac{e^{ijk}}{\sqrt{Z}}$$. However, we should note that Z is **not** a tensor, since we only have relative levi-civita symbols, but if we substitute absolute levi-civita symbols, then it is indeed one (second equality)? Wait, I'm not sure whether one should call the former, ##\epsilon## 'absolute' and the latter, ##e## 'relative'...

Hmmm, this looks interesting. Would you give any hints as to what's wrong? Thanks in advance!

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