Derivative of Determinant of Metric Tensor With Respect to Entries

In summary: Just a quick question, what is $$\frac{\partial {Z}}{\partial Z_{ij}}$$? Do we take the partial derivative of each entry of the metric tensor and get back the matrix with the same dimensions?Yes, that is correct. You can also write it as \begin{align*}\frac{\partial {Z}}{\partial Z_{ij}} = \frac{\partial {Z_{lm}}}{\partial Z_{ij}} = \delta^l_i \delta^m_j.\end{align*}
  • #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!
 
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  • #3
Use that ln(det(X)) = tr(ln(X)) and differentiate.
 
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  • #4
Orodruin said:
Use that ln(det(X)) = tr(ln(X)) and differentiate.
I see that you've given this hint somewhere else as well. Reminds me of exponentiating an operator... Oh well, I have got to learn it... Thanks!
 
  • #5
Did I not format the homework statement properly?... $$\frac{\partial Z}{\partial Z_{ij}} = Z Z^{ij}$$

Where ##Z## is the determinant.
 
  • #6
It's okay. Orodruin already gave you a hint:\begin{align*}
\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}} &= \dfrac{\partial \log |Z|}{\partial Z_{ij}} = \mathrm{Tr} \left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}} \right)
\end{align*}Do you know how to take the trace?
 
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  • #7
ergospherical said:
It's okay. Orodruin already gave you a hint:\begin{align*}
\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}} &= \dfrac{\partial \log |Z|}{\partial Z_{ij}} = \mathrm{Tr} \left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}} \right)
\end{align*}Do you know how to take the trace?
I am confused with traces and logs of matrices actually, but let me have a look. (I know the definition, trace is the sum of the diagonal terms, log: expand log in terms of the power series of the matrix)
 
  • #8
ergospherical said:
It's okay. Orodruin already gave you a hint:\begin{align*}
\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}} &= \dfrac{\partial \log |Z|}{\partial Z_{ij}} = \mathrm{Tr} \left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}} \right)
\end{align*}Do you know how to take the trace?
wait

\begin{align*}
\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}} = \mathrm{Tr} \left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}} \right)
\end{align*}
and isn't

\begin{align*}
\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}} = \left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}} \right)
\end{align*}

?

Because they are scalars!

Wait this is confusing did you intend ##|Z|## to be the determinant while ##Z## to be the metric tensor?
 
  • #9
No. Remember that the trace of a matrix is ##\mathrm{Tr}(A) = {A^i}_i## with summation implied over repeated indices. Matrix multiplication is ##A = BC \iff {A^i}_j = {B^i}_k {C^k}_j## and the trace of this product is then ##{A^i}_i = {B^i}_k {C^k}_i##.

In the context of this step, \begin{align*}
{\left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}}\right)^k}_l = Z^{km} {\frac{\partial {Z_{ml}}}{\partial Z_{ij}}}
\end{align*}What is the trace?
 
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  • #10
Why am I switching between matrices and tensors o:)
 
  • #11
yucheng said:
Wait this is confusing did you intend ##|Z|## to be the determinant while ##Z## to be the metric tensor?
Yes, I am denoting by ##Z## a matrix.
 
  • #12
Finally, found my pen and paper.

\begin{align*}
\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}} &= \mathrm{Tr} \left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}} \right)
\end{align*}

since $$ Z = [Z_{ij}]$$
, $$ Z^{-1} = [Z^{ij}]$$
, $$\mathrm{Tr} \left(AB\right) = A^{i}_{j} B^{j}_{i}$$

We see that
$$RHS = Z^{lm} \frac{\partial Z_{lm}}{\partial Z_{ij}}$$

and combining LHS and RHS

$$\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}}= Z^{lm} \frac{\partial Z_{lm}}{\partial Z_{ij}} = Z^{lm} \delta^{ij}_{lm} = Z^{ij}$$

In conclusion,

$$\dfrac{1}{|Z|} \dfrac{\partial |Z|}{\partial Z_{ij}}= Z^{ij}$$

or

$$\dfrac{\partial |Z|}{\partial Z_{ij}}= |Z| Z^{ij}$$

Thanks @Orodruin and @ergospherical !
 
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  • #13
Nice. :oldeyes:
 
  • #14
ergospherical said:
No. Remember that the trace of a matrix is ##\mathrm{Tr}(A) = {A^i}_i## with summation implied over repeated indices. Matrix multiplication is ##A = BC \iff {A^i}_j = {B^i}_k {C^k}_j## and the trace of this product is then ##{A^i}_i = {B^i}_k {C^k}_i##.

In the context of this step, \begin{align*}
{\left( Z^{-1} \frac{\partial Z}{\partial Z_{ij}}\right)^k}_l = Z^{km} {\frac{\partial {Z_{ml}}}{\partial Z_{ij}}}
\end{align*}What is the trace?
Just a quick question, what is $$\frac{\partial {Z}}{\partial Z_{ij}}$$? Do we take the partial derivative of each entry of the metric tensor and get back the matrix with the same dimensions?
 
  • #15
Yeah, it's a matrix with components ##\dfrac{\partial Z_{kl}}{\partial Z_{ij}}## in the ##(k,l)^{\mathrm{th}}## position.
 
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1. What is the derivative of the determinant of a metric tensor with respect to its entries?

The derivative of the determinant of a metric tensor with respect to its entries is a mathematical expression that represents the rate of change of the determinant when any of its entries are varied. It is commonly used in differential geometry and general relativity to calculate the curvature of a space.

2. Why is the derivative of the determinant of a metric tensor important?

The derivative of the determinant of a metric tensor is important because it provides a way to measure the curvature of a space. It also allows for the calculation of other important quantities, such as the Christoffel symbols and the Riemann curvature tensor, which are essential in understanding the geometry of a space.

3. How is the derivative of the determinant of a metric tensor calculated?

The derivative of the determinant of a metric tensor is calculated using the Leibniz formula, which states that the derivative of a determinant is equal to the sum of the products of the determinant of the cofactor matrix and its corresponding entry in the original matrix. This process can be quite complex and is often done using computer software or specialized calculators.

4. Can the derivative of the determinant of a metric tensor be negative?

Yes, the derivative of the determinant of a metric tensor can be negative. This indicates that the curvature of the space is decreasing in a particular direction. However, in general relativity, the curvature of a space is often described as positive or negative, rather than increasing or decreasing, so the sign of the derivative may not have a physical interpretation.

5. Are there any applications of the derivative of the determinant of a metric tensor?

Yes, there are many applications of the derivative of the determinant of a metric tensor. It is used extensively in differential geometry and general relativity to study the curvature of space and to make predictions about the behavior of matter and light in the presence of massive objects. It is also used in other fields, such as computer graphics and computer vision, to measure the distortion of images and surfaces.

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