# Variation of determinant of a metric

1. Nov 8, 2015

### S. Leger

1. The problem statement, all variables and given/known data
I'm trying to calculate the variation of the following term for the determinant of the metric in the polyakov action:
$$h = det(h_{ab}) = \frac{1}{3!}\epsilon^{abc}\epsilon^{xyz}h_{ax}h_{by}h_{cz}$$
I know that there are some other ways to derive the variation of a metric, e.g. with the help of Jacobi's formula. But what I mean is the "trick" to derive it from exactly this term above.
Can someone show me?
How can I translate the result into factors of $h_{ab}$ and $h$?

2. Relevant equations

3. The attempt at a solution

Last edited: Nov 8, 2015
2. Nov 8, 2015

### fzero

You can multiply your result by $\delta^a_b = h_{bc}h^{ca}$. The resulting expression will have a factor of $\epsilon^{acd}\epsilon^{efg} h_{cf} h_{dg} h_{eb}$. You should be able to argue that this is, up to possible combinatorial factors equivalent to $h \delta^a_b$ by symmetry considerations.

3. Nov 8, 2015

### S. Leger

Thanks a lot fzero!
I'm working on a more general solution, which I will post as soon as I can express it.
It's based on the fact that the derivation of the above determinant is a subdeterminant of the determinant itself.
Multiplicated with the matrix will result in the determinant again - and so the derivation of the determinant is a multiple of the inverse of the matrix, i.e. $$\delta h=h^{ab}h$$
But it's not yet obvious for me why this imply the claim $$\delta h=h^{ab}\delta h_{ab}h$$ I'm looking for....?

Last edited: Nov 8, 2015
4. Nov 9, 2015

### S. Leger

This is my attempt for a general solution, for n dimensions:
$$h=det(h_{ab})=\frac{1}{n!}\epsilon^{\alpha_1...\alpha_n}\epsilon^{\beta_1...\beta_n} h_{\alpha_1\beta_1}...h_{\alpha_n\beta_n}$$
and so $$\delta h=\frac{1}{(n-1)!}\epsilon^{\alpha_2...\alpha_n}\epsilon^{\beta_2...\beta_n} h_{\alpha_2\beta_2}...h_{\alpha_n\beta_n}=h^{\alpha_1\beta_1}h$$ is that correct so far?
But now I stuck: what is the decisive step from $$\delta h = h^{\alpha_1 \beta_1}h$$ to $$\delta h = h^{\alpha \beta} \delta h_{\alpha \beta}h ?$$

5. Nov 9, 2015

### fzero

You have to compute the variation with respect to something, so what you're computing here is
$$\delta h=\frac{1}{(n-1)!}\epsilon^{\alpha_1...\alpha_n}\epsilon^{\beta_1...\beta_n} \delta h_{\alpha_1\beta_1} h_{\alpha_2\beta_2}...h_{\alpha_n\beta_n}.$$
There's some additional argument needed to argue that this is related to $h h_{\alpha\beta}$.

6. Nov 9, 2015

### S. Leger

the variation of the determinant should be w.r.t. the metric, i.e. $$h_{ab}$$
I thought that the second term is right because there is only a variation in "part" $$h_{\alpha_1 \beta_1}$$ of the field.
But you're right, there is no relation to the right term - and that's my problem.

Any hint how to variate the first equation correctly? I was sure, that there must be a kind of trick which leads to the last eq.

7. Nov 9, 2015

### samalkhaiat

Use the matrix identity $$\ln ( \det M ) = \mbox{Tr} ( \ln M )$$ Then apply any derivation operator (such as the $\delta$) to that identity: $$\frac{\delta(\det M)}{\det M} = \mbox{Tr}(M^{-1} \delta M) .$$ You get the result you want if you take $M = h_{\mu\nu}$, and $M^{-1}=h^{\rho \sigma}$.

8. Nov 9, 2015

### S. Leger

Thank you samalkhaiat,
Yes this is one of those shorthand and easy ways to variate the metric determinant.
But I am searching the derivation of the above equation ( with the levi cevita) behause I heard that this was an easy "trick" to get the result

9. Nov 9, 2015

### S. Leger

10. Nov 9, 2015

### fzero

11. Nov 9, 2015

### S. Leger

fzero,
I'm sorry if I have not yet completely understood your hint but does this mean that $$\delta^a_b$$ (Kronecker) is the same as $$\delta$$ (variation)?
And isn't the result I get from this not $$\delta h =h^{bc}h_{ca}h$$ rather than $$\delta h=h^{ab}\delta h_{ab}h$$?

Last edited: Nov 9, 2015
12. Nov 9, 2015

### fzero

No, unfortunately we are using $\delta$ for two different things. The Kronecker delta $\delta^a_b$ is not directly related to the variations $\delta h$ or $\delta h_{ab}$. What I meant is that you had an expression that we can write
$$\frac{\delta h}{\delta h_{\alpha_1\beta_1}} =\frac{1}{(n-1)!}\epsilon^{\alpha_1...\alpha_n}\epsilon^{\beta_1...\beta_n} h_{\alpha_2\beta_2}...h_{\alpha_n\beta_n}.$$
Now, we sum both sides against the identity $\delta^\gamma_{\beta_1} = h^{\gamma \epsilon} h_{\epsilon \beta_1}$, so on the RHS we have
$$\frac{1}{(n-1)!}\left( \epsilon^{\alpha_1...\alpha_n}\epsilon^{\beta_1...\beta_n} h_{\epsilon \beta_1} h_{\alpha_2\beta_2}...h_{\alpha_n\beta_n}\right) h^{\gamma \epsilon}.$$
You should be able to argue that
$$\epsilon^{\alpha_1...\alpha_n}\epsilon^{\beta_1...\beta_n} h_{\epsilon \beta_1} h_{\alpha_2\beta_2}...h_{\alpha_n\beta_n} \propto \delta^{\alpha_1}_\epsilon$$
by using the total antisymmetry of the factors of $\epsilon^{\alpha_1...\alpha_n}$ and the symmetry of $h_{\alpha\beta}$.

13. Nov 9, 2015

### samalkhaiat

Start with $$\delta h = \frac{\partial h}{\partial h_{c_{1}c_{2}}} \ \delta h_{c_{1}c_{2}} .$$ Show that
$$\frac{\partial h}{\partial h_{c_{1}c_{2}}} = \frac{1}{(n-1)!} \epsilon^{c_{1}a_{2} \cdots a_{n}} \epsilon^{c_{2}b_{2} \cdots b_{n}} h_{a_{2}b_{2}} \cdots h_{a_{n}b_{n}} . \ \ \ \ (1)$$ Since $h^{c_{1}c_{2}}$ is the only available tensor for the RHS of (1), you can set $$\epsilon^{c_{1}a_{2} \cdots a_{n}} \epsilon^{c_{2}b_{2} \cdots b_{n}} h_{a_{2}b_{2}} \cdots h_{a_{n}b_{n}} = h^{c_{1}c_{2}} A .$$ Contract this equation with $h_{c_{1}c_{2}}$ and get $$A = (n-1)! \ h .$$ This gives you $$\frac{\partial h}{\partial h_{c_{1}c_{2}}} = h \ h^{c_{1}c_{2}} .$$ And the starting equation leads to $$\delta h = h \ h^{\alpha \beta} \delta h_{\alpha \beta} .$$

14. Nov 10, 2015

### S. Leger

Yeah, that's it!
Sorry but it took some time to check all your steps in the calculations and to prove it handwritten by myself, especially the indices..
But everything turns out right now und it's pretty clear. With all this information I understand what the trick (in andriens post) is.

fzero and samalkhaiat, thank you very much!