What will Multi-variate Gaussian Distribution become when the quadratic form in the exponent is constant I'm reading the chapter 2 of PRML(Pattern Recognition and Machine Learning) and I don't understand the following statement in the first paragraph in page 80:

The Gaussian distribution will be constant on surfaces in $x$-space for which this quadratic form is constant.

My understanding is that the Gaussian will become an uniform distribution when the quadratic form is constant but I'm not sure. And when will the quadratic form be constant? 
The quadratic form is:
$$
    \Delta^2=(x-\mu)^T\Sigma^{-1}(x-\mu)
   $$
And the Multi-variate Gaussian is :
$$
N(x|\mu,\Sigma)=\frac{1}{(2\pi)^{(D/2)}}\frac{1}{|\Sigma|^{1/2}}\exp\lbrace -\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu) \rbrace
$$
 A: You are misunderstanding what the text is saying. A density (multivariate or
univariate) is a function that typically takes on different values for each
value of the argument: $f(x_1)$ and $f(x_2)$ are usually different numbers.
For uniform distributions, say $U(a,b)$, $f(x)$ has the same value $(b-a)^{-1}$
for all $x \in (a,b)$ and the same value $0$ for all $x \notin (a,b)$.
A (univariate) Gaussian density $f(x)$ has, in general, different values for
different $x$'s with minor exceptions: for each real number $t$,
$f(\mu+t)$ has the same value as $f(\mu-t)$ for the simple reason that
$$\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac 12 \left(\frac{(\mu+t)-\mu}{\sigma}\right)^2\right) 
=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac 12 \left(\frac{(\mu-t)-\mu}{\sigma}\right)^2\right)$$
both quantities being equal to $\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{t^2}{2\sigma^2}\right)$. Writing all of the above gobbledygook can be avoided
by noting that all we are looking for are the points where the quadratic
function $(x-\mu)^2$ has the same value, and it is easy to determine that
$x_1 = \mu+t$ and $x_2 = \mu-t$ give rise to the same value $t^2$ for
$(x-\mu)^2$. For a bivariate Gaussian distribution, the density
has constant value wherever a quadratic in two variables has the same value:
this quadratic corresponds to an ellipse with center $(\mu_X,\mu_Y)$ in the plane. More generally, the quadratic in $n \geq 2$ variables is the 
quadratic form
$(x-\mu)^T\Sigma^{-1}(x-\mu)$
that you quote, and the set of points where this function has the same value
is a surface in $n$-dimensional space.

The Gaussian distribution never becomes a uniform distribution in the
  usual sense of the word. A Gaussian density function can have the same
  numerical value for different values of the argument, but that is not the same
  as saying that the Gaussian random variable (or random vector)
  is uniformly distributed on those points. 

One could say that conditioned
on $X$ being in the set of points where the quadratic has fixed value,
$X$ is uniformly distributed in that set, but this conditional
distribution is not Gaussian in any sense of the word.  In the univariate
case, the conditional distribution is a discrete distribution: conditionally, $X$ takes
on values $\mu+t$ and $\mu-t$ with equal probability, in the bivariate
case, the conditional distribution of $(X,Y)$ is uniform on an ellipse
but it cannot be described by a joint density. While we can write something
like $$f_{X,Y\mid (X,Y)\in A}(x,y\mid A) = \begin{cases}c, & (x,y) \in A,\\0, &\text{otherwise}
\end{cases}$$
where $A$ is the ellipse in question, that $f$ is not a joint density: its
integral over the plane is $0$ instead of $1$, and Gaussianity cannot be
averred in any sense of the word. For example, the conditional
marginal distributions
of $X$ and $Y$ are not Gaussian distributions.
