3
$\begingroup$

I'm following Carrol's book on general relativity and in Chap. 2 he talks about tensor densities, differential forms and integration.

He says we can identify the integration measure $dx^{n}$ as $dx^{0} \wedge...\wedge dx^{n-1}$ where $\wedge$ is the wedge product between differential forms. Then he points out that the RHS is misleading because it looks like a tensor (an n form). In fact if you take two functions $f$ and $g$ and apply the exterior derivative you get the one-forms $df$ and $dg$, and then you can do the wedge product to get $df\wedge dg$ which is a 2-form. But in the case of $dx^{n}$ as $dx^{0} \wedge...\wedge dx^{n-1}$ he argues it's different because the functions appearing here are the coordinate functions themselves.

He then proceeds to prove that $dx^{n}$ as $dx^{0} \wedge...\wedge dx^{n-1}$ is actually a tensor density of weight 1 writing:

$dx^{0}\wedge...\wedge dx^{n-1}=\frac{1}{n!}\tilde \epsilon_{\mu1..\mu n} dx^{\mu1} \wedge...\wedge dx^{\mu n}$ (here $\tilde \epsilon $ is the Levi Civita symbol) Then under a transformation of coordinate $\tilde \epsilon$ stays the same by definition while $dx^{\mu 1} \wedge ... \wedge dx^{\mu n}$ transforms as $\frac{dx^{\mu 1'}} {dx^{\mu1 }}... \frac{dx^{\mu n'}}{dx^{\mu n}} dx'^{\mu 1} \wedge ... \wedge dx'^{\mu n}$. So he concludes $\tilde \epsilon_{\mu 1... \mu n} dx^{mu 1}\wedge ... \wedge dx^{\mu n}=|\frac{\partial x^{\mu}}{\partial x^{\mu' }}| \tilde \epsilon_{\mu 1'... \mu n'} dx^{mu 1'}\wedge ... \wedge dx^{\mu n'} $ where $|\frac{\partial x^{\mu}}{\partial x^{\mu' }}|$ is the determinant of the Jacobian of the coordinate transformation.

what I don't understand is :

1)Why can he use in the proof that $dx^{\mu 1} \wedge ... \wedge dx^{\mu n}$ transforms as $\frac{dx^{\mu 1'}} {dx^{\mu1 }}... \frac{dx^{\mu n'}}{dx^{\mu n}} dx'^{\mu 1} \wedge ... \wedge dx'^{\mu n}$ which is a TENSOR transformation law when he stated at first that $dx^{\mu 1} \wedge ... \wedge dx^{\mu n}$ was not a tensor?

2)why $\frac{dx^{\mu 1'}} {dx^{\mu1 }}... \frac{dx^{\mu n'}}{dx^{\mu n}}= |\frac{\partial x^{\mu}}{\partial x^{\mu '}}|$ ? That would be true if the matrix of coordinate transformation was diagonal, so the determinant is just the product of the diagonal elements. How can it be true in general?

$\endgroup$

3 Answers 3

2
$\begingroup$

I'll restrict my attention to the 2D case at the moment; the general case is exactly the same, and just requires more writing.

  1. Under a coordinate transformation $x\mapsto y$, a 1-form $\mathrm dx^\mu$ transforms as $\mathrm dx^\mu \mapsto \frac{\partial x^\mu}{\partial y^\nu} \mathrm dy^\nu$. As a result, the wedge product of two 1-forms $\mathrm dx^\mu \wedge \mathrm dx^\nu$ transforms as $$\mathrm dx^\mu \wedge \mathrm dx^\nu \mapsto \left(\frac{\partial x^\mu}{\partial y^\alpha} \mathrm dy^\alpha\right)\wedge\left(\frac{\partial x^\nu}{\partial y^\beta} \mathrm dy^\beta\right) = \frac{\partial x^\mu}{\partial y^\alpha} \frac{\partial x^\nu}{\partial y^\beta} \mathrm dy^\alpha \wedge \mathrm dy^\beta$$ The extension to a generic $n$-form is straightforward.

  2. Consider the volume form $\mathrm dx^0 \wedge \mathrm dx^1 = \frac{1}{2!} \epsilon_{\mu\nu} \mathrm dx^\mu \wedge \mathrm dx^\nu$. Applying the transformation law above,

$$\epsilon_{\mu\nu} \mathrm dx^\mu \wedge \mathrm dx^\nu \mapsto \epsilon_{\mu\nu} \frac{\partial x^\mu}{\partial y^\alpha} \frac{\partial x^\nu}{\partial y^\beta} \mathrm dy^\alpha \wedge \mathrm dy^\beta$$ Note that for a $2\times 2$ matrix $A$ with entries $A^i_{\ \ j}$, we have

$$\mathrm{det}(A) = A^0_{\ \ 0} A^1_{\ \ 1}- A^0_{\ \ 1} A^1_{\ \ 0}= \epsilon_{i j} A^i_{\ \ 0} A^j_{\ \ 1}$$ $$\implies \epsilon_{nm} \mathrm{det}(A) = \epsilon_{ij} A^i_{\ \ n} A^j_{\ \ m}$$

Comparing this with what we see above, $$\epsilon_{\mu\nu} \frac{\partial x^\mu}{\partial y^\alpha} \frac{\partial x^\nu}{\partial y^\beta}= \mathrm{det}(J) \epsilon_{\alpha\beta}$$ where $J$ is the Jacobian of the transformation. Therefore, we obtain $$\mathrm dx^0 \wedge \mathrm dx^1 = \frac{1}{2!}\epsilon_{\mu\nu}\mathrm dx^\mu \wedge \mathrm dx^\nu \mapsto \frac{\mathrm{det}(J)}{2!} \epsilon_{\alpha\beta} \mathrm dy^\alpha\wedge \mathrm dy^\beta = \mathrm{det}(J) \mathrm dy^0 \wedge \mathrm dy^1$$

$\endgroup$
3
  • 1
    $\begingroup$ Thank you this is very clear. The only thing I'm still unsure about is : you wrote d$x^{\mu}$ which looks like an exterior derivative acting on a vector, while I know that the exterior derivative is defined to act only on forms. Also you called it a one form, as if there was only one lower index (the one of the exterior derivative), but you also have the upper $\mu$ index. Could you clarify that? $\endgroup$
    – Mathew
    Nov 7, 2021 at 8:42
  • 1
    $\begingroup$ @Mathew It sounds like you're being tripped up by index notation. The set of objects $\{x^0,x^1\}$ are the coordinate functions (which are scalars/0-forms) in whatever chart you're using. The objects $\big\{\frac{\partial}{\partial x^0}, \frac{\partial}{\partial x^1}\big\}$ constitute the corresponding basis for the tangent space, and are vectors. On the other hand, $\{\mathrm dx^0,\mathrm dx^1\}$ are the exterior derivatives of the coordinate functions. They are 1-forms/covectors, and constitute a basis for the cotangent space. $\endgroup$
    – J. Murray
    Nov 7, 2021 at 13:05
  • 1
    $\begingroup$ @Mathew A generic vector can be written $\mathbf V = V^1 \frac{\partial}{\partial x^1} + V^2 \frac{\partial}{\partial x^2}$, which becomes $V^\mu \frac{\partial}{\partial x^\mu}$ in index notation. Thus we say that the components of a vector have an index upstairs (a vector itself has no indices) while the basis vectors have downstairs indices. Similarly, the components of a 1-form have an index downstairs, while the basis 1-forms have upstairs indices. With that in mind, $\mathrm dx^\mu$ is just the $\mu^{th}$ basis 1-form, i.e. either $\mathrm dx^0$ or $\mathrm dx^1$ for $\mu=0,1$. $\endgroup$
    – J. Murray
    Nov 7, 2021 at 13:09
2
$\begingroup$
  1. I don't have a copy of Carroll's book next to me, so I can't comment on what precisely he says. However, saying the wedge of $dx^i$'s is not a tensor (field) is just wrong. On a manifold we can take a chart $(U,\varphi)$, so $\varphi:U\to\varphi[U]\subset \Bbb{R}^n$. The $n$ component functions of $\varphi$ are traditionally written $(x^1,\dots, x^n)$ (or if you want to start your indices as $(x^0,\dots, x^{n-1})$, be my guest). Each $x^i$ is a function $x^i:U\to \Bbb{R}$. As such we are perfectly allowed to take its exterior derivative $dx^i$. This is now a differential $1$-form on the open set $U\subset M$. The wedge product of $n$ differential 1-forms on $U$ is again (by definition) a differential $n$-form on $U$, so $dx^1\wedge \cdots \wedge dx^n$ is a differential $n$-form on $U$.

Also, differential forms (on $U$) ARE tensor fields (on $U$), so it is also perfectly correct to say that $dx^1\wedge \cdots \wedge dx^n$ is a tensor field on $U$.

  1. It seems you should look at one of the many equivalent ways of defining a determinant. Let's say $A$ is an $n\times n$ matrix; we denote the entry in row $i$, column $j$ as $a_{ij}$. Then, one can define \begin{align} \det A&=\sum_{\sigma \in S_n}\text{sgn}(\sigma)\cdot a_{1,\sigma(1)}\cdots a_{n,\sigma(n)} \end{align} This is known as Leibniz's formula for determinants. The physics way of writing this is that rather than writing a permutation $\sigma:\{1,\dots, n\}\to\{1,\dots, n\}$ as a function, we just display its outputs as $(\mu_1,\dots, \mu_n)$. The Levi-Civita symbol $\epsilon$ is literally the same thing as the sign of a permutation. So, one can write the above as \begin{align} \det A&= \sum \epsilon_{\mu_1\cdots \mu_n}\cdot a_{1,\mu_1}\cdots a_{n,\mu_n}. \end{align} If you're instead more familiar with the Laplace formula i.e. "expansion along minors", then let us, temporarily, assume the equivalence between the Leibniz permutation and Laplace minor formulas. Then, by considering two charts $(U,\varphi=(x^1,\dots, x^n))$ and $(V,\psi=(y^1,\dots, y^n))$, with non-empty overlap $U\cap V\neq \emptyset$, we have \begin{align} dx^1\wedge \cdots \wedge dx^n&=\left(\frac{\partial x^1}{\partial y^{\nu_1}}\,dy^{\nu_1}\right)\wedge \cdots \wedge \left(\frac{\partial x^n}{\partial y^{\nu_n}}\,dy^{\nu_n}\right)\\ &=\left(\frac{\partial x^1}{\partial y^{\nu_1}}\cdots \frac{\partial x^n}{\partial y^{\nu_n}}\right)\cdot \left(dy^{\nu_1}\wedge \cdots \wedge dy^{\nu_n}\right)\\ &=\left(\frac{\partial x^1}{\partial y^{\nu_1}}\cdots \frac{\partial x^n}{\partial y^{\nu_n}}\right)\cdot \left(\epsilon^{\nu_1,\dots, \nu_n}\,dy^1\wedge \cdots \wedge dy^n\right)\\ &=\left(\epsilon^{\nu_1,\dots, \nu_n}\frac{\partial x^1}{\partial y^{\nu_1}}\cdots \frac{\partial x^n}{\partial y^{\nu_n}}\right)\,dy^1\wedge \cdots dy^n\\ &=\det \left(\frac{\partial x}{\partial y}\right)\, dy^1\wedge \cdots \wedge dy^n. \end{align} Throughout, $\epsilon^{\nu_1,\cdots, \nu_n}$ is just a symbol denoting the sign of the permutation which sends $i\in\{1,\dots, n\}$ to $\nu_i\in\{1,\dots, n\}$.

About Tensor Densities.

Tensor densities are completely different beasts from tensor fields and differential forms (which are also tensor fields, as I mentioned above). For now I shall only focus on scalar densities. A scalar density of weight 1 can be obtained by "taking absolute values" of differential forms. Explicitly, recall that a differential $n$-form $\omega$ on an $n$-dimensional manifold $M$ is by definition an object (a mapping) which assigns to each point $p\in M$ an alternating multilinear functional $\omega_p:\underbrace{T_pM\times \cdots \times T_pM}_{\text{$n$ times}} \to \Bbb{R}$, i.e it eats $n$ vectors $v_1,\dots, v_n\in T_pM$ and spits out a real number $\omega_p(v_1,\dots, v_n)\in\Bbb{R}$. By taking absolute values, I just mean the object $|\omega|$ which assigns to each $p\in M$ and vectors $v_1,\dots, v_n\in T_pM$, the real number $|\omega_p(v_1,\dots, v_n)|\in\Bbb{R}$.

So, any differential $n$-form $\omega$ on $M$ has the property that once we specify a coordinate chart $(U,\varphi=(x^1,\dots, x^n))$, we can write (as an equality of differential forms on $U$) \begin{align} \omega&=f_{(x)}\cdot dx^1\wedge \cdots \wedge dx^n, \end{align} for some unique function $f_{(x)}:U\to\Bbb{R}$ (the subscript $x$ is just to emphasize that everything is with respect to the chart $\varphi=(x^1,\dots, x^n)$). Given a scalar density $\rho$ on $M$ of weight $1$, we can write it as \begin{align} \rho&= \widetilde{f}_{(x)}\cdot |dx^1\wedge \cdots dx^n|, \end{align} for some other function $\widetilde{f_{(x)}}:U\to\Bbb{R}$.\

The natural objects to integrate on manifolds are precisely scalar densities of weight $1$, because the transformation law for the coefficients is \begin{align} \rho&=\widetilde{f_{(x)}}\cdot |dx^1\wedge \cdots dx^n| = \tilde{f_{(x)}}\cdot \left|\det\left(\frac{\partial x}{\partial y}\right)\,dy^1\wedge \cdots dy^n\right|\\ &=\widetilde{f_{(x)}}\cdot \left|\det \frac{\partial x}{\partial y}\right|\, \left|dy^1\wedge \cdots dy^n\right| \end{align}

This factor of $\left|\det \frac{\partial x}{\partial y}\right|$ is exactly the reciprocal of the factor we pick up from the change-of-variables theorem for integral calculus in $\Bbb{R}^n$. i.e this cancellation is what allows us to define in a chart-independent manner the integral of a density over a chart domain as \begin{align} \int_U\rho&:= \int_{\varphi[U]}\widetilde{f_{(x)}}\circ \varphi^{-1}\,\, d\lambda\\ &\equiv \int_{\varphi[U]}\widetilde{f_{(x)}}\left(\varphi^{-1}(a)\right)\,\,d\lambda(a)\\ &\equiv \int_{\varphi[U]}\widetilde{f_{(x)}}\left(\varphi^{-1}(a)\right)\,\,d^na, \end{align} here the $\equiv$ means "same thing different notation". Here, on the RHS, we're taking the scalar density $\rho$ and integrating its chart-representative function $\widetilde{f_{(x)}}\circ \varphi^{-1}:\varphi[U]\subset \Bbb{R}^n\to\Bbb{R}$ with respect to $n$-dimensional Lebesgue measure. i.e it's a standard integral that you've always seen in multivariable calculus clsses, prior to learning anything manifolds. (ok if you don't know about Lebesgue measure and integrals, then just assume all the functions are smooth, and that all the sets are "nice", so that the RHS can be interpreted as a Riemann-integral in $n$-dimensions, so it really is the basic vanilla object).

By a technical partition of unity argument, one can then define $\int_M\rho$, i.e an integral over the entire manifold.

Differential forms come into the picture for integration because if we have an orientation for the whole manifold, then one can construct an isomorphism between the set of scalar densities and the set of differential $n$-forms. This allows us to define the integral of $n$-forms over $n$-dimensional oriented manifolds. We like working with differential forms because they're very easy to calculate with and work with (primarily because of the simplicity of the Cartan calculus, i.e the nice manner in which all the operations like exterior derivative, Lie derivative, interior product all behave).


Equivalence of Determinant Definitions.

Regarding the equivalence between the Leibniz and Laplace formulae, I think the most efficient way to do it is by induction on the size, $n$, of the matrix. When $n=1$, they're clearly equal. Let us suppose as the induction hypothesis they are equal for matrices of size $n-1$. Then, for a matrix $A$ of size $n$, we have \begin{align} \sum_{\sigma\in S_n}\text{sgn}(\sigma)\cdot \prod_{i=1}^na_{i,\sigma(i)}&=\sum_{j=1}^n\sum_{\substack{\sigma\in S_n\\\sigma(1)=j}}\text{sgn}(\sigma)\cdot a_{1,j}\cdot \prod_{i=2}^na_{i,\sigma(i)} \end{align} All I've done here is note that we can split up the sum $\sum_{\sigma\in S_n}$ into $n$ parts, by specifying what $\sigma(1)$ can be. i.e adding up over all permutations is the same as adding up all permutations with $\sigma(1)=1$ plus all permutations with $\sigma(1)=2$ and so on. Note that in the product $\prod_{i=2}^na_{i,\sigma(i)}$, clearly none of the elements of $A$ from row $1$, nor column $j$ appear. In other words, these entries are precisely the matrix entries of the $(1,j)$ minor matrix, $C_{1,j}$ (i.e the matrix $C_{1,j}$ is obtained by deleting row $1$ and column $j$ from $A$).

Now, this is where one has to be slightly careful with indices and stuff. I feel like the more one tries to explain it, the more confusing it becomes, but I'll try my best (though at the end of the day, once you have this "outline", you should verify the details for yourself). Specifying a permutation $\sigma\in S_n$ with $\sigma(1)=j$ is the same thing as specifying a certain permutation $\tau\in S_{n-1}$ (because specifying a permutation in $S_n$ means we need to define where each of the $n$ elements get sent to, but we're already fixing $\sigma(1)=j$, so we only need to define where the remaining $n-1$ terms get mapped to). The relationship between these two is that $\text{sgn}(\sigma)=(-1)^j\text{sgn}(\tau)$. So we have \begin{align} \sum_{j=1}^n\sum_{\substack{\sigma\in S_n\\\sigma(1)=j}}\text{sgn}(\sigma)\cdot a_{1,j}\cdot \prod_{i=2}^na_{i,\sigma(i)}&= \sum_{j=1}^n\sum_{\tau\in S_{n-1}}(-1)^j\text{sgn}(\tau)a_{1,j}\cdot \prod_{i=1}^{n-1}(C_{1,j})_{i,\tau(i)}\\ &=\sum_{j=1}^n(-1)^j\cdot a_{i,j}\cdot \left(\sum_{\tau \in S_{n-1}}\text{sgn}(\tau)\cdot \prod_{i=1}^{n-1}(C_{1,j})_{i,\tau(i)}\right)\\ &=\sum_{j=1}^n(-1)^j\cdot a_{1,j}\cdot \det(C_{1,j}). \end{align} Note that the product $\prod_{i=2}^na_{i,\sigma(i)}$ became a product $\prod_{i=1}^{n-1}(C_{1,j})_{i,\tau(i)}$, because the exact same elements appear in both terms; we're just writing it with different notation. The last equality was by our induction hypothesis (which we can use since $C_{1,j}$ is an $(n-1)\times (n-1)$ matrix). This shows that the Leibniz and Laplace formulas agree for the $n\times n$ matrix $A$, thereby completing the induction.

$\endgroup$
2
  • $\begingroup$ This is the only way I can make sense of what Carroll says. On the other hand, it seems that physicists have a (supposedly) a different notion of tensor density. I asked this question about it. $\endgroup$ Mar 26, 2023 at 12:09
  • $\begingroup$ The argument you gave is what I myself would say as well, but may I ask you what do you think of how some other people justify that part in Carroll (which is also the point 2. in the above link). That seems to be the dominant idea here, checking other posts or the other answer. $\endgroup$ Mar 26, 2023 at 12:13
0
$\begingroup$

I think I know where I was wrong for the fist question: in $dx^{0}\wedge\ldots\wedge dx^{n-1}$ $0$...$n-1$ are labels while in $dx^{\mu_1}\wedge\ldots\wedge dx^{\mu_n}$ $\mu_1 \mu_n$ are real indices. So actually $dx^{\mu_1}\wedge\ldots\wedge dx^{\mu_n}$ is a form. But still the second question remains.

$\endgroup$
1
  • $\begingroup$ You can edit your question instead of posting the edits as an answer. $\endgroup$
    – Lost
    Nov 6, 2021 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.