Covariance, inner product and independence Let $X$ and $Y$ be two normally distributed random variables.
I understand that if $\operatorname{cov}(X,Y)=0$ then they are stochastically independent (right?).
Now, is it true that if  $X \cdot Y = 0$ they are independent?, are those two results equivalent? 
 A: Note that there are two ways in which $X$ and $Y$ can be normally distributed:


*

*Way A. When each is normally distributed, i.e. $X \sim \mathcal N(\mu_X, \sigma^2_X)$ and $Y \sim \mathcal N(\mu_Y, \sigma^2_Y)$.

*Way B. When $X$ and $Y$ are jointly normally distributed. In this case, one could think of a new random variable $(X,Y) \sim  \mathcal N(\mu, \Sigma)$. For definition see here.
As is immediately apparent from the definition, Way A is strictly weaker than Way B. So when someone tells you

Let $X$ and $Y$ be two normally distributed random variables.

you should assume (since the keywords "jointly" or "multivariate normal" are missing) that they only mean Way A.
And assuming Way A, the following 

if $\operatorname{cov}(X,Y)=0$ then they are stochastically independent

does not hold. For counterexamples and some pictures see this. For more pictures and intuition see this.
By the way, assuming Way B we do have 

if $\operatorname{cov}(X,Y)=0$ then they are stochastically independent.

To prove it, show that for any real $x$ and $y$ holds $f_{(X,Y)}(x,y) = f_X(x)f_Y(y)$ (where $f_{(X,Y)}$ is the density of $(X,Y)$ a.k.a. the joint density of $X$ and $Y$).

Now to the second part of the question: 

Now, is it true that if  $X \cdot Y = 0$ they are independent?

First of all, what is $X\cdot Y$? Well, I see a dot there, is this a dot product? Since the things you plug into it are random variables, I think (see dot vs inner product) that you meant it to be an inner product.
And an inner product (on a vector space $V$ over a field $F$) is just any map
\begin{align}
 \langle \cdot , \cdot \rangle: V\times V \to F
\end{align}
that satisfies certain properties. 
Nothing stops us from defining the following map:
\begin{align}
\operatorname{cov} : L^2 \times L^2 & \to \mathbb R\\
(X,Y) & \mapsto \mathbb E \Big( \big[(X - \mathbb E[X])(Y - \mathbb E[Y])\big] \Big)
\end{align}
where you can roughly think of $L^2$ as the vector space of all random variables (on some underlying probability space) that have finite variance.
As it turns out, the map $\operatorname{cov}$ does indeed satisfy the properties to call itself an inner product.
So in the context of random variables, if someone writes $X\cdot Y$ it is (unless clearly noted otherwise) by definition the same as $\operatorname{cov}(X,Y)$.
