The explicit expression for integral of forms Could anyone please help me with the following three questions? They are simple questions, but I am confused.


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*With a $2$-form $F=\frac{1}{2}F_{\mu\nu}dx^{\mu}\wedge dx^{\nu}$ in 4 dimension, what is the explicit expression of $\int F\wedge F$ in coordinates? 
I got something like $$ \int F\wedge F= \frac{1}{4}\int F_{\mu\nu}F_{\rho\sigma}\epsilon^{\mu\nu\rho\sigma}d^4x.$$
My question is: How is that $\epsilon$ defined? I found several definitions and they differ from each other by a factor of $\sqrt{|g|}$ infront. If I am going to do it in 4-D spherical coordinate, how does $\epsilon$ look like ? 

*What is the explicit expression of this one in a 3-D coordinate? 
$$\int_{R^3} d(\phi df\wedge dg)$$ 
$\phi$, $f$ and $g$ are $0$-form functions. What if I use the Stoke's theorem,
$$\int_{\partial R^3} \phi df\wedge dg$$ 
how does it look like explicitly in some coordinate? 
 A: 1. question 
The totally antisymmetric symbol $\varepsilon_{i_1,i_2,\dots,i_n}$ ($=1$ if indexes not permuted) is not a tensor. Usually, it is just a symbol. The total antisymmetry means that it is antisymmetric in all indexes, meaning that it changes sign to $-1$ if the permutation of indexes is odd, and remains $+1$ if the permutation is even, and it vanishes if any index repeats.
To see how it transforms, we will take a slight detour, and consider how it can be used to define determinants. So, let $\mathbf M^{n\times n}$ denote $n\times n$ matrices over lets say $\mathbb R$, and define $\det:\mathbf M^{n\times n}\to\mathbb R$ such that $$A=(a_{ij})\mapsto \det A := \sum_{i_1,i_2,\dots,i_n} \varepsilon_{i_1,i_2,\dots,i_n}a_{1i_1}a_{2i_2}\cdots a_{ni_n}.$$
To get the transformation law, we will write down this equation differently, and then set $A$ to be the Jacobian of a change of coordinates. This will give us what we need. 
Let's use the summation convention from now on. First note that $\varepsilon_{i_1,i_2,\dots,i_n}\varepsilon_{i_1,i_2,\dots,i_n}=n!$, because the number of permutations of a set of $n$ elements is $n!$ - this product of Levi-Civitas will now give 1 whenever there is any permutation, and 0 when it's not, so it just counts the permutations.
Secondly, convince yourself that we could write
$$\det A = \frac{1}{n!}\varepsilon_{j_1,j_2,\dots,j_n}\varepsilon_{i_1,i_2,\dots,i_n}a_{j_1i_1}a_{j_2i_2}\cdots a_{j_ni_n}.$$
We are doing the same sum as before, but overcounting by $n!$, hence the factor $1/n!$. Now multiply by $\varepsilon_{j_1,j_2,\dots,j_n}$ and sum:
$$\varepsilon_{j_1,j_2,\dots,j_n}\det A = \varepsilon_{i_1,i_2,\dots,i_n}a_{j_1i_1}a_{j_2i_2}\cdots a_{j_ni_n}.$$
Good, now we can see about the $\sqrt{g}$ factors. Put $A=\big(\frac{\partial x_m}{\partial y_n}\big):$ 
$$\varepsilon_{i_1,i_2,\dots,i_n}\frac{\partial x_{i_1}}{\partial y_{j_1}}\frac{\partial x_{i_2}}{\partial y_{j_2}}\cdots \frac{\partial x_{i_n}}{\partial y_{j_n}}=\det\big(\frac{\partial x_m}{\partial y_n}\big)\varepsilon_{j_1,j_2,\dots,j_n}.$$
Therefore, we can't think of $\varepsilon_{i_1,i_2,\dots,i_n}$ as components of some $(0,n)$-tensor $$\varepsilon_{abc\dots} = \sum_{i_1,i_2,\dots,i_n}\varepsilon_{i_1,i_2,\dots,i_n} \mathrm d x^{i_1}\otimes \mathrm d x^{i_2}\otimes\dots\otimes\mathrm d x^{i_n}, $$
since it's components transform as $$\varepsilon_{j_1,j_2,\dots,j_n}=J^{-1}\varepsilon_{i_1,i_2,\dots,i_n}\frac{\partial x_{i_1}}{\partial y_{j_1}}\frac{\partial x_{i_2}}{\partial y_{j_2}}\cdots \frac{\partial x_{i_n}}{\partial y_{j_n}}$$ where $J$ is the Jacobian determinant, as before. The Jacobian shouldn't be there for this to be a tensor - the situation we have is that $\varepsilon_{abc\dots}$ is a tensor density, which is common enough to have a name! If our space is admits a metric tensor, however, we know that in a new basis $$\tilde g\equiv\det\left(  \tilde g_{ab}\right)=\det\left(g_{\mu\nu}\frac{\partial x_{\mu}}{\partial y_{a}} \frac{\partial x_{\nu}}{\partial y_{b}}\right) \\=\det\left(\frac{\partial x_{\mu}}{\partial y_{a}}\right)\det\left(\frac{\partial x_{\nu}}{\partial y_{b}}\right)\det\left( g_{\mu\nu}\right)= J^2 g.$$
This means that while $\varepsilon$ gets an extra factor of $J^{-1}$, $g$ gets an extra factor of $J^2$. Therefore, the following transforms correctly:
$$e_{abc\dots}:=\sqrt{|g|} \varepsilon_{abc\dots}.$$
Now we raise the indices using the (inverse) metric: 
$$e^{j_1\dots j_n}=\sqrt{|g|} g^{i_1 j_1}\cdots g^{i_n j_n}\varepsilon_{i_1\dots i_n}$$
where we use our second definitions for the determinant (but with all indices up!): 
$$e^{j_1\dots j_n}=\sqrt{|g|}\:\frac{1}{g}\:\varepsilon^{j_1\dots j_n}=\frac{1}{\sqrt{|g|}}\varepsilon^{j_1\dots j_n}.$$
Therefore, in your question, $\epsilon^{\mu\nu\rho\sigma}d^4x$ lacks a factor of $1/\sqrt{|g|}$... however, $d^4 x$ lacks a factor of $\sqrt{|g|}$! So the two precisely cancel each other out!
If you are't familiar with the reason the volume element is $\sqrt{|g|} \mathrm d^n x$, it's similar to the above. Because the exerior (wedge) product is totally antisymmetric, exactly like the Levi-Civita, we can write
$$\mathrm d^n x = \mathrm d x^{i_1}\wedge\dots\wedge\mathrm d x^{i_n}= \frac{1}{n!}\sum \varepsilon_{i_1,\dots,i_n}\mathrm d x^{i_1}\otimes\dots\otimes\mathrm d x^{i_n} $$
and exactly the same discussion as before implies that the extra factor is needed. 
In spherical coordinates, we see that the symbol $\varepsilon$ picks up $J^{-1}$. On wikipedia we can find the Jacobian for n-spherical cooridnates. Moreover, $\varepsilon$ picks up all those derivatives. 
2. question
I urge you to try this one yourself, it's a really simple exercise, but I'll write it out. Try it before reading! We can write $\mathrm d f$ as $$\mathrm d f=\frac{\partial f}{\partial x}\mathrm d x+\frac{\partial f}{\partial y}\mathrm d y+\frac{\partial f}{\partial z}\mathrm d z$$
and similarly with $g$, so that 
$$\mathrm df\wedge\mathrm d g = \left(f_x g_y - g_x f_y\right) \mathrm d x \wedge \mathrm dy+\left(f_x g_z - g_x f_z\right) \mathrm d x \wedge \mathrm dz+\left(f_y g_z - g_y f_z\right) \mathrm d y \wedge \mathrm dz.$$
$\phi\,\mathrm df\wedge\mathrm d g$ is just this multiplied by $\phi$. Note that your second integral is zero, since $\mathbb R^3$ has no boundary! I don't have time to explicitly calculate the exterior derivative of this, but the result will definitely be very long, and you'll have $(\text{lots of factors involving partial derivatives})\mathrm d x \wedge\mathrm d y\wedge\mathrm d z$.
