Prove independence by using copula Suppose that I have multivariate random variable $(x_1,...,x_n)$ whose pdf $f(x_1,...,x_n)$ exist and does not depend on $x_i, 1 \leq i \leq n$. An example is random vector uniformly distributed on a sphere.
Next I find that the product of marginal pdf $\Pi_i f_{X_i}(x_i)$ does not depend on $x_i$ either.


*

*By using Sklar theorem $f(x_1,...,x_n) = c(x_1,...,x_n) \Pi_i f_{X_i}(x_i)$, is it safe to say that the copula density $c(x_1,...,x_n)$ does exist?

*If the copula density exist, it does not depend on $x_i$. Is that the sufficient condition to say that random variables $x_i$ are independent?

 A: It appears you are dealing with absolutely continuous random variables, in which case the copula density certainly exists. 
The Copula  and the copula density for continuous rv's have as their arguments the distribution functions of the rv's (viewed as uniform $U(0,1)$ rv's).
You are examining the case where the joint pdf does not depend on the rv's. Since the joint pdf results from differentiating the joint distribution function, $F_J$, it follows that the variables appear in the joint distribution function in an affine way. In most cases this necessitates that the marginal distribution functions are themselves an affine function of the variable: 
$$F_i(x_i) = a_i + b_ix_i$$
The Copula is 
$$C[(F_1(x_1),...,F_n(x_n)]$$
and the copula density is 
$$c[(F_1(x_1),...,F_n(x_n)] = \frac {\partial^n C[(F_1(x_1),...,F_n(x_n)]}{\partial F_1...\partial F_n}$$
Note that we differentiate with respect to the distribution functions, not with respect to the variables. It follows that the copula density is not free of the $x_i$'s.
The necessary and sufficient condition for independence is to have the copula density equal to unity. This will happen if the Copula is equal to the product of the marginal distribution functions (viewed as uniform $U(0,1)$ rv's). In the literature this is called the Independence Copula.
A: For your two questions:


*

*Due to Sklar's theorem, for any multivariate function, there is a copula function. If the margins distributions are continuous then copula function is unique.  

*No. Copula density depends on the random variables. However, due to the complexity of copula computation, in the simplified version of the copula, copula depends on the random variables only through its arguments. If the random variables are independent, then copula density is equal to the product of the marginal distribution of the random variables. That is:
$c(u, v) = u.v = 1$ (since the variables are uniform on [0,1]). Hence, copula here is the product or independent copula. 
As I mentioned in the comment. If copula function shows independence between the random variables, then these variables are independent.  Here is a theorem:
THEOREM 2.2 (from http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/tutorials/xfghtmlnode13.html). Equation (2.4) is Sklar's theorem:
Let $ R_1$ and $ R_2$ be random variables with continuous distribution 
functions $ F_1$ and $ F_2$ and joint distribution function $ H$. Then 
$ R_1$ and $ R_2$ are independent if and only if  $ C_{R_1 R_2} = \Pi$.
From Sklar's Theorem we know that there exists a unique copula $ C$ with
$\displaystyle \textrm{P}(R_1 \le r_1, R_2 \le r_2) = H(r_1,r_2) = C(F_1(r_1),F_2(r_2)) \, .$ (2.6)
Independence can be seen using Equation (2.4) for the joint distribution function $ H$ and the definition of 
$ \Pi$,
 $\displaystyle H(r_1,r_2) = C(F_1(r_1),F_2(r_2)) = F_1(r_1) \cdot F_2(r_2) \; .$
