Lie derivative of a metric determinant

[Frank Castle]

I’m hoping to clear up some confusion I have over what the Lie derivative of a metric determinant is.

Consider a 4-dimensional (pseudo-) Riemannian manifold, with metric $g_{\mu\nu}$. The determinant of this metric is given by $g:=\text{det}(g_{\mu\nu})$. Given this, now consider the quantity $\sqrt{-g}$. Is it correct that the Lie derivative of this, with respect to a vector field $X$, is given by $$\mathcal{L}_{X}(\sqrt{-g})=X^{\mu}\partial_{\mu}(\sqrt{-g})$$ The reason I’m unsure about this is that $\sqrt{-g}$ is technically a scalar density rather than just a simple scalar, and so I’m unsure whether or not it behaves like a scalar as far as the Lie (and covariant for that matter) derivative is concerned.

 

[andrewkirk]

I don't understand the distinction you are trying to make with the term 'scalar density'. $g=\mathrm{det}(g_{\mu\nu})$ is a function from the manifold to $\mathbb R$, and is hence a real scalar field. Therefore $\sqrt{-g}$ is also a scalar field, once we have chosen which branch of the square root function we wish to use.

I am rusty on Lie derivatives, but if the above formula is true for any scalar field $s$ when that is substituted for $\sqrt{-g}$ above, would it not automatically be true for $\sqrt{-g}$, since that too is a scalar field?

 

[Orodruin]

I don't understand the distinction you are trying to make with the term 'scalar density'. $g=\mathrm{det}(g_{\mu\nu})$ is a function from the manifold to $\mathbb R$, and is hence a real scalar field. Therefore $\sqrt{-g}$ is also a scalar field, once we have chosen which branch of the square root function we wish to use.

The metric determinant certainly is not a scalar function since it is coordinate system dependent. Instead, as Frank pointed out, it is a scalar density of weight 2. This is why you can multiply $dx^1 \wedge \ldots \wedge dx^N$ with $\sqrt{g}$ and obtain a coordinate independent volume element, since $dx^1\wedge \ldots \wedge dx^N$ is a tensor density of weight -1.

For the original question, I suggest having a look at section 1.2.9 of https://arxiv.org/pdf/0911.0334.pdf

 

[Frank Castle]

The metric determinant certainly is not a scalar function since it is coordinate system dependent. Instead, as Frank pointed out, it is a scalar density of weight 2. This is why you can multiply $dx^1 \wedge \ldots \wedge dx^N$ with $\sqrt{g}$ and obtain a coordinate independent volume element, since $dx^1\wedge \ldots \wedge dx^N$ is a tensor density of weight -1.

For the original question, I suggest having a look at section 1.2.9 of https://arxiv.org/pdf/0911.0334.pdf

Thanks for the link. I’m a bit unsure about the notation though, I can’t see anywhere where the author defines what $S_{i}$ is?!

Also, there is the identity, for a volume form: $$\mathcal{L}_{X}\left(\sqrt{-g}\,d^{4}x\right)=\nabla_{\mu}X^{\mu}\sqrt{-g}\,d^{4}x$$ and I can’t see how this agrees with how the notes (that you linked to) state how a scalar density behaves under a Lie derivative?

 

[Frank Castle]

Has anyone got any ideas on this as I’m still confused?!

 

[Orodruin]

Sorry not to get back to you. This is at the level where I have to sit down myself for some time with the appropriate references to make sure I do not say anything stupid. As I am on vacation I unfortunately do not have neither that time nor direct access to the references I would require to give you a well thought through answer.

 

[Frank Castle]

Sorry not to get back to you. This is at the level where I have to sit down myself for some time with the appropriate references to make sure I do not say anything stupid. As I am on vacation I unfortunately do not have neither that time nor direct access to the references I would require to give you a well thought through answer.

Ok, no problem. Appreciate your conscientiousness

 

[stevendaryl]

Thanks for the link. I’m a bit unsure about the notation though, I can’t see anywhere where the author defines what $S_{i}$ is?!

Also, there is the identity, for a volume form: $$\mathcal{L}_{X}\left(\sqrt{-g}\,d^{4}x\right)=\nabla_{\mu}X^{\mu}\sqrt{-g}\,d^{4}x$$ and I can’t see how this agrees with how the notes (that you linked to) state how a scalar density behaves under a Lie derivative?

I think I can derive that formula. For simplicity of explanation, let me simplify to two spacetime dimensions, and I'm going to use a coordinate basis $u, v$.

Since $\mathcal{L}_X (\sqrt{|g|} du \wedge dv)$ is a scalar, then we can evaluate it in any coordinate system. So I'm going to pick a locally Cartesian coordinate system, $x, y$. In a locally Cartesian coordinate system, $\sqrt{|g|} = 1$, so we're just evaluating:

$\mathcal{L}_X (dx \wedge dy)$

The rules for the Lie derivative of a form is this:

$\mathcal{L}_X (\omega) = \mathcal{i}_X d \omega + d \mathcal{i}_X \omega$

where $\mathcal{i}_X$ means contraction of the vector $X$ on the first "slot" of the tensor $\omega$, and $d$ means the exterior derivative.

In our case, $\omega = dx \wedge dy$. So $d \omega = 0$ (the exterior derivative of an exterior derivative is zero).
$\mathcal{i}_X (\omega) = X^x dy - X^y dx$. So we have:

$\mathcal{L}_X (\omega) = d (X^x dy - X^y dx) = \partial_\mu X^x dx^\mu \wedge dy - \partial_\mu X^y dx^\mu \wedge dx$

In 2-D spacetime, $dx^\mu \wedge dy = 0$ unless $dx^\mu = dx$, and similarly, $dx^\mu \wedge dx = 0$ unless $dx^\mu = dy$. So the above expression simplifies to:

$\mathcal{L}_X (\omega) = \partial_x X^x dx \wedge dy - \partial_y X^y dy \wedge dx = \partial_x X^x dx \wedge dy + \partial_y X^y dx \wedge dy$
(Because $dx \wedge dy = - dy \wedge dx$). So we have:

$\mathcal{L}_X (\omega) = (\partial_x X^x + \partial_y X^y) dx \wedge dy$

In a locally Cartesian coordinate system, $\partial_x X^x = \nabla_x X^x$ and $\partial_y X^y = \nabla_y X^y$. So we can write:

$\mathcal{L}_X (\omega) = (\nabla_\mu X^\mu) dx \wedge dy = (\nabla_\mu X^\mu) \omega$

Converting back to the coordinate system $u,v$, $\omega = \sqrt{|g|} du \wedge dv$. So we have:

$\mathcal{L}_X (\omega) = (\nabla_\mu X^\mu) \sqrt{|g|} du \wedge dv$

The generalization to any number of dimensions is straight-forward.

 

[stevendaryl]

$\mathcal{L}_X (\omega) = (\nabla_\mu X^\mu) \sqrt{|g|} du \wedge dv$

The generalization to any number of dimensions is straight-forward.

Okay, after having written that, I'm having doubts about whether the argument makes sense. The argument that (in the 2-D case):

$\mathcal{L}_X (dx \wedge dy) = \partial_\mu X^\mu (dx \wedge dy)$

works exactly the same, regardless of whether $x, y$ are locally Cartesian, or not. So we can just as well write, for locally non-Cartesian coordinates $u,v$

$\mathcal{L}_X (du \wedge dv) = \partial_\mu X^\mu (du \wedge dv)$

But the previous conclusion is that:

$\mathcal{L}_X (\sqrt{|g|} du \wedge dv) = \nabla_\mu X^\mu (\sqrt{|g|} du \wedge dv) = (\partial_\mu X^\mu + \Gamma^\mu_{\mu \lambda} X^\lambda) (\sqrt{|g|} du \wedge dv)$

Putting them together gives:

$\mathcal{L}_X (\sqrt{|g|} du \wedge dv) = \sqrt{|g|} \mathcal{L}_X (du \wedge dv) + \Gamma^\mu_{\mu \lambda} X^\lambda (\sqrt{|g|} du \wedge dv)$

If we assume that the Leibniz rule works for Lie derivatives, this means:

$(\mathcal{L}_X (\sqrt{|g|})) du \wedge dv + \sqrt{|g|} \mathcal{L}_X (du \wedge dv) = \sqrt{|g|} \mathcal{L}_X (du \wedge dv) + \Gamma^\mu_{\mu \lambda} X^\lambda (\sqrt{|g|} du \wedge dv)$

So

$(\mathcal{L}_X (\sqrt{|g|})) du \wedge dv = \Gamma^\mu_{\mu \lambda} X^\lambda \sqrt{|g|} du \wedge dv$

Which implies that $(\mathcal{L}_X (\sqrt{|g|})) = \Gamma^\mu_{\mu \lambda} X^\lambda \sqrt{|g|}$?

 

[Frank Castle]

Okay, after having written that, I'm having doubts about whether the argument makes sense. The argument that (in the 2-D case):

$\mathcal{L}_X (dx \wedge dy) = \partial_\mu X^\mu (dx \wedge dy)$

works exactly the same, regardless of whether $x, y$ are locally Cartesian, or not. So we can just as well write, for locally non-Cartesian coordinates $u,v$

$\mathcal{L}_X (du \wedge dv) = \partial_\mu X^\mu (du \wedge dv)$

But the previous conclusion is that:

$\mathcal{L}_X (\sqrt{|g|} du \wedge dv) = \nabla_\mu X^\mu (\sqrt{|g|} du \wedge dv) = (\partial_\mu X^\mu + \Gamma^\mu_{\mu \lambda} X^\lambda) (\sqrt{|g|} du \wedge dv)$

Putting them together gives:

$\mathcal{L}_X (\sqrt{|g|} du \wedge dv) = \sqrt{|g|} \mathcal{L}_X (du \wedge dv) + \Gamma^\mu_{\mu \lambda} X^\lambda (\sqrt{|g|} du \wedge dv)$

If we assume that the Leibniz rule works for Lie derivatives, this means:

$(\mathcal{L}_X (\sqrt{|g|})) du \wedge dv + \sqrt{|g|} \mathcal{L}_X (du \wedge dv) = \sqrt{|g|} \mathcal{L}_X (du \wedge dv) + \Gamma^\mu_{\mu \lambda} X^\lambda (\sqrt{|g|} du \wedge dv)$

So

$(\mathcal{L}_X (\sqrt{|g|})) du \wedge dv = \Gamma^\mu_{\mu \lambda} X^\lambda \sqrt{|g|} du \wedge dv$

Which implies that $(\mathcal{L}_X (\sqrt{|g|})) = \Gamma^\mu_{\mu \lambda} X^\lambda \sqrt{|g|}$?

Thanks for taking a look.

This is where I get stuck. If the Lie derivative of $\sqrt{\lvert g\rvert}$ is the same as for a scalar, i.e. $$\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right)=X^{\mu}\nabla_{\mu}\left(\sqrt{\lvert g\rvert}\right)=\frac{1}{2}X^{\mu}\sqrt{\lvert g\rvert}g^{\lambda\nu}\partial_{\mu}g_{\nu\lambda}=X^{\mu}\Gamma^{\nu}_{\mu\nu}\sqrt{\lvert g\rvert}$$ (where we’ve used that $\frac{1}{2}g^{\lambda\nu}\partial_{\mu}g_{\nu\lambda}=\Gamma^{\nu}_{\mu\nu}$.) then I can derive the correct expression, but I’m unsure whether this approach is correct or not?! Someone told me that it doesn’t matter that $\sqrt{\lvert g\rvert}$ is a scalar density, since we are evaluating it’s Lie derivative in a given coordinate system, and so, for all intents and purposes, it behaves as a scalar.

 

[stevendaryl]

Thanks for taking a look.

This is where I get stuck. If the Lie derivative of $\sqrt{\lvert g\rvert}$ is the same as for a scalar, i.e. $$\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right)=X^{\mu}\nabla_{\mu}\left(\sqrt{\lvert g\rvert}\right)=\frac{1}{2}X^{\mu}\sqrt{\lvert g\rvert}g^{\lambda\nu}\partial_{\mu}g_{\nu\lambda}=X^{\mu}\Gamma^{\nu}_{\mu\nu}\sqrt{\lvert g\rvert}$$ (where we’ve used that $\frac{1}{2}g^{\lambda\nu}\partial_{\mu}g_{\nu\lambda}=\Gamma^{\nu}_{\mu\nu}$.) then I can derive the correct expression, but I’m unsure whether this approach is correct or not?! Someone told me that it doesn’t matter that $\sqrt{\lvert g\rvert}$ is a scalar density, since we are evaluating it’s Lie derivative in a given coordinate system, and so, for all intents and purposes, it behaves as a scalar.

I did not know the expression for $\nabla_\mu det(g)$

$\nabla_\mu det(g) = det(g) g^{\lambda \nu} \partial_\mu g_{\nu \lambda}$

But it's a nice confirmation that my derivation got the same conclusion without assuming that.

 

[Frank Castle]

I did not know the expression for $\nabla_\mu det(g)$

$\nabla_\mu det(g) = det(g) g^{\lambda \nu} \partial_\mu g_{\nu \lambda}$

It is if you assume that $\text{det}(g_{\mu\nu})$ is a scalar, but I’m unsure if it’s correct to do so?!

 

[stevendaryl]

The role of $\sqrt{|g|} dx^4$ is as a 4-form; it takes 4 vectors and returns a scalar. So there is no other way to interpret $\sqrt{|g|}$ except as a scalar.

 

[Frank Castle]

The role of $\sqrt{|g|} dx^4$ is as a 4-form; it takes 4 vectors and returns a scalar. So there is no other way to interpret $\sqrt{|g|}$ except as a scalar.

But it transforms like a tensor density under coordinate transformations though, i.e. $$\sqrt{\lvert g’\rvert}=\frac{1}{J}\sqrt{\lvert g\rvert}$$ where $J=\lvert\text{det}\left(\frac{\partial x’^{\mu}}{\partial x^{\nu}}\right)\rvert$.

Isn’t this required to ensure that $\sqrt{|g|} dx^4$ is invariant under coordinate transformations (as $d^{4}x$ transforms as $d^{4}x’=\lvert J\rvert d^{4}x$).

 

[stevendaryl]

Look, for any tensor whatsoever, the components of the tensor transform in the opposite way as the basis tensor. For example, for a vector field $V = V^\mu e_\mu$, $V^\mu$ transforms in the opposite way as the basis vectors $e_\mu$.

Here's a paradox for you: If $e_0, e_1, e_2, e_3$ are the four basis vectors for some coordinate system, and $\phi^0, \phi^1, \phi^2, \phi^3$ are 4 scalars, then you can create a vector $V = \phi^0 e_0 + \phi^1 e_1 + ...$. Because a scalar times a vector is a vector, and a linear combination of vectors is still a vector. The components of $V$ are given by: $V^\mu = \phi^\mu$.

So here's the paradox: on the one hand, $\phi^\mu$ is a scalar, so it doesn't transform at all under a coordinate change. On the other hand, $V^\mu$ is a component of a vector, so it transforms. But they are equal! What gives?

Well, the answer is that $\phi^\mu$, being a scalar, has the same value in any coordinate system, but it's only equal to $V^\mu$ in one particular coordinate system.

In a similar way, there is a scalar field $Q$ which in a specific coordinate system is equal to $\sqrt{|g|}$. In that coordinate system, $dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ is a 4-form, and so is the combination:
$\omega = Q x^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$.

In a different coordinate system, you have different basis 1-forms, $dy^0, dy^1, dy^2, dy^3$. The same 4-form $\omega$ can be written in terms of either basis:

$\omega = Q x^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 = Q' dy^0 \wedge dy^1 \wedge dy^2 \wedge dy^3$

Both $Q$ and $Q'$ are scalar fields, but they are different scalar fields.

 

[Frank Castle]

Look, for any tensor whatsoever, the components of the tensor transform in the opposite way as the basis tensor. For example, for a vector field $V = V^\mu e_\mu$, $V^\mu$ transforms in the opposite way as the basis vectors $e_\mu$.

Here's a paradox for you: If $e_0, e_1, e_2, e_3$ are the four basis vectors for some coordinate system, and $\phi^0, \phi^1, \phi^2, \phi^3$ are 4 scalars, then you can create a vector $V = \phi^0 e_0 + \phi^1 e_1 + ...$. Because a scalar times a vector is a vector, and a linear combination of vectors is still a vector. The components of $V$ are given by: $V^\mu = \phi^\mu$.

So here's the paradox: on the one hand, $\phi^\mu$ is a scalar, so it doesn't transform at all under a coordinate change. On the other hand, $V^\mu$ is a component of a vector, so it transforms. But they are equal! What gives?

Well, the answer is that $\phi^\mu$, being a scalar, has the same value in any coordinate system, but it's only equal to $V^\mu$ in one particular coordinate system.

In a similar way, there is a scalar field $Q$ which in a specific coordinate system is equal to $\sqrt{|g|}$. In that coordinate system, $dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ is a 4-form, and so is the combination:
$\omega = Q x^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$.

In a different coordinate system, you have different basis 1-forms, $dy^0, dy^1, dy^2, dy^3$. The same 4-form $\omega$ can be written in terms of either basis:

$\omega = Q x^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 = Q' dy^0 \wedge dy^1 \wedge dy^2 \wedge dy^3$

Both $Q$ and $Q'$ are scalar fields, but they are different scalar fields.

Ah ok. So is the point that $\sqrt{|g|}$ corresponds to a scalar in each given coordinate system, but does not correspond to the same scalar in two different coordinate systems?

 

[stevendaryl]

Ah ok. So is the point that $\sqrt{|g|}$ corresponds to a scalar in each given coordinate system, but does not correspond to the same scalar in two different coordinate systems?

Yes, but more generally, when it comes to Lie derivatives and covariant derivatives, coefficients of vectors and tensors are treated like scalars.

For example: $\nabla_\mu V = \nabla_\mu (V^\nu e_\nu) = \nabla_\mu (V^\nu) e_\nu + V^\nu \nabla_\mu e_\nu$
$= \partial_\mu (V^\nu) e_\nu + V^\nu \Gamma^\lambda_{\mu \nu} e_\lambda$
which after renaming dummy indices becomes $= (\partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda) e_\nu$

This uses that for the scalar $V^\nu$, $\nabla_\mu V^\nu = \partial_\mu V^\nu$, and for basis vectors, $\nabla_\mu e_nu = \Gamma^\lambda_{\mu \nu} e_\lambda$

This might contradict what people often write: $\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda$. The reason for the difference is that some people use the notation $V^\mu$ to mean the vector, as a whole, and some people use it to mean the components of the vector. If you mean the components, then they are scalars, and the covariant derivative is just the partial derivative. If you mean the whole vector, then there is the additional connection term. But I think it's clearer to think of the connection coefficients as being due to derivatives of the basis vectors, instead of being an operation on components.

If you think of $V^\mu$ as a component of a vector, then there is a distinction between

1. $\nabla_\mu V^\nu$
2. $(\nabla_\mu V)^\nu$

The first means: take the covariant derivative of a component of vector $V$.
The second means: take the covariant derivative of vector $V$ and then consider the $\nu$ component of that.

 

[Orodruin]

The reason for the difference is that some people use the notation VμVμV^\mu to mean the vector, as a whole, and some people use it to mean the components of the vector.

Well, it is the choice to let $\nabla_\nu V^\mu$ mean $(\nabla_\nu V)^\mu$ rather than $\nabla_\nu(V^\mu)$. Should one wish to be very clear one can write the parentheses.

 

[stevendaryl]

Well, it is the choice to let $\nabla_\nu V^\mu$ mean $(\nabla_\nu V)^\mu$ rather than $\nabla_\nu(V^\mu)$. Should one wish to be very clear one can write the parentheses.

It usually causes no problems. Except when you are explicitly talking about basis vectors, then you would write $V = V^\mu e_\mu$. The covariant derivative of that expression has to treat $V^\mu$ as a scalar field, and the connection coefficients come from the covariant derivatives of $e_\mu$.

 

[Frank Castle]

Yes, but more generally, when it comes to Lie derivatives and covariant derivatives, coefficients of vectors and tensors are treated like scalars.

For example: $\nabla_\mu V = \nabla_\mu (V^\nu e_\nu) = \nabla_\mu (V^\nu) e_\nu + V^\nu \nabla_\mu e_\nu$
$= \partial_\mu (V^\nu) e_\nu + V^\nu \Gamma^\lambda_{\mu \nu} e_\lambda$
which after renaming dummy indices becomes $= (\partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda) e_\nu$

This uses that for the scalar $V^\nu$, $\nabla_\mu V^\nu = \partial_\mu V^\nu$, and for basis vectors, $\nabla_\mu e_nu = \Gamma^\lambda_{\mu \nu} e_\lambda$

This might contradict what people often write: $\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda$. The reason for the difference is that some people use the notation $V^\mu$ to mean the vector, as a whole, and some people use it to mean the components of the vector. If you mean the components, then they are scalars, and the covariant derivative is just the partial derivative. If you mean the whole vector, then there is the additional connection term. But I think it's clearer to think of the connection coefficients as being due to derivatives of the basis vectors, instead of being an operation on components.

If you think of $V^\mu$ as a component of a vector, then there is a distinction between

1. $\nabla_\mu V^\nu$
2. $(\nabla_\mu V)^\nu$

The first means: take the covariant derivative of a component of vector $V$.
The second means: take the covariant derivative of vector $V$ and then consider the $\nu$ component of that.

Ok, thanks for the info. So is the following a correct derivation of the Lie derivative of the volume form? $$\mathcal{L}_{X}\left(\sqrt{-g}d^{4}x\right)=\mathcal{L}_{X}\left(\sqrt{-g}\right)d^{4}x+\sqrt{-g}\mathcal{L}_{X}\left(d^{4}x\right)\\ \qquad\qquad\qquad\qquad=X^{\mu}\partial_{\mu}\left(\sqrt{-g}\right)d^{4}x+\sqrt{-g}\partial_{\mu}X^{\mu}d^{4}x\\ \qquad\qquad\qquad\qquad\qquad=\frac{1}{2}\sqrt{-g}X^{\mu}g^{\lambda\nu}\partial_{\mu}g_{\nu\lambda}d^{4}x+\sqrt{-g}\partial_{\mu}X^{\mu}d^{4}x\\ \qquad\qquad\qquad\quad=X^{\nu}\Gamma^{\mu}_{\;\mu\nu}\sqrt{-g}d^{4}x+\sqrt{-g}\partial_{\mu}X^{\mu}d^{4}x\\=\nabla_{\mu}X^{\mu}\sqrt{-g}d^{4}x$$

 

[MASK1729]

If X is a geodesic field (Jacobi field) then divergence of X i.e. ∇_μX^μ must be equal to some expression involving Ricci tensor. Similar to Jacobi equation giving deviation between nearby geodesics in terms of Riemannian curvature. Does anyone know this expression?