Kernel of a Normal Distribution From  Wikipedia ,

The kernel of a probability density function (pdf) or probability mass function (pmf) is the form of the pdf or pmf in which any factors that are not functions of any of the variables in the domain are omitted. Note that such factors may well be functions of the parameters of the pdf or pmf.
An example is the normal distribution. Its probability density function is :
$$p(x| \mu,\sigma^2)=\frac{1}{\sqrt {2\pi\sigma^2}}\exp[-\frac{1}{2}(\frac{x-\mu}{\sigma})^2],\quad -\infty<x<\infty$$
and the associated kernel is
$$p(x| \mu,\sigma^2)\propto \exp[-\frac{1}{2}(\frac{x-\mu}{\sigma})^2]$$
Note that the factor in front of the exponential has been omitted, even though it contains the parameter \sigma^2 , because it is not a function of the domain variable x .

But my question is formula of $\sigma$ contains $x$'s , that is ,
$\sigma^2=\frac{1}{N}\sum_{i=1}^{N}(x_i-\mu)^2$
Then why $\frac{1}{\sigma}$ part is omitted from being a kernel part ? What do they want to mean by : function of the domain variable x ?
 A: You seem to be confusing the population variance $\sigma^2$ with a version of the sample variance $S^2$. Recall that
$$
S^2 = {1 \over n-1} \sum_i (X_i - \bar{X})^2.
$$
This is a statistic and thus can completely be calculated from the data. This is not the population variance.
For a continuous random variable the population variance is defined as
$$
\sigma^2 = E\left[(X - E[X])^2\right] = \int \limits_{-\infty}^\infty (x-\mu)^2 f_X(x)dx
$$
for $E(X) = \mu$. Note how we integrate over all $X$ in order to get $\sigma^2$ (we "integrate $X$ out"). That's why it is not a function of $X$.
Update:
No matter what the population looks like we have defined $\sigma^2 = E\left[(X - E[X])^2\right]$. The outermost expectation is over all $X$ so it doesn't matter whether a sum or an integral is used, we are averaging over the entire population and so $\sigma^2$ is not a function of $X$. Suppose we had a population where $X \in \{1, 2, 3\}$ and $P(X=1) = .5$, $P(X = 2) = .3$, and $P(X = 3) = .2$. Then we'll have that
$$
\mu = E[X] = \sum_{x=1}^3 x P(X = x) = 1*.5 + 2*.3 + 3*.2 = 1.7
$$
and therefore
$$
\sigma^2 = E\left[(X - E[X])^2\right] = \sum \limits_{x = 1}^3 (x - \mu)^2 P(X = x)
$$
$$
= (1 - 1.7)^2 * .5 + (2 - 1.7)^2 * .3 + (3 - 1.7)^2 * .2 = 0.61.
$$
So even though we computed $\sigma^2$ using a formula like what you have, it is not defined to be this. That formula is purely a result of taking an expectation of a discrete random variable. We are summing out $X$ just as if we instead integrated it out. 
A: The formula you provide is an estimator. There are many ways to get estimators: least squares, maximum likelihood, Bayesian, etc. 
The normal distribution has two parameters. Typically (but not always), the parameters are the mean $\mu$ and variance $\sigma^2$. These do NOT depend on any $x_i$, but instead they describe the distribution of those $x_i$.
The maximum likelihood estimators for $\mu$ and $\sigma^2$ are 
$$ \hat{\mu} = \frac{1}{N} \sum_{i=1}^N x_i, \quad \hat{\sigma}^2 = \frac{1}{N} (x_i-\hat{\mu})^2. $$
Notice that these estimators have the caret symbol to emphasize that they are estimates of the parameters $\mu$ and $\sigma^2$ (but not their actual values).
So if $X\sim N(\mu,\sigma^2)$, i.e. the random variable $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$, then the pdf is 
$$f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{1}{2\sigma^2} (x-\mu)^2}.$$
The domain variables are variables representing the random variables. For this normal distribution $x$ is the domain variable. 
The kernel, dropping all terms from the pdf that do not include the domain variable x, is 
$$f_X(x) \propto e^{-\frac{1}{2\sigma^2} (x-\mu)^2}.$$
