What does it mean by multiplying two differentials in stochastic calculus?

Question

I'm learning stochastic calculus right now at school and I'm coming across multiplication rules such as $$dW_t \cdot dW_t = dt$$ and I don't really quite understand what they meant. I tried looking up but I couldn't find any resources that explains this properly either. Originally, I thought it's supposed to mean that we can "multiply" two stochastic integrals in the sense that $$ \int_0^t f(s, \omega) dW_s \cdot \int_0^t g(s, \omega) \, dW_s = \int_0^t f(s, \omega) \cdot g(s, \omega) ds$$ where $f$ and $g$ are two functions on $[0,\infty) \times \Omega$ but this is not correct since if $f, g \equiv 1$ then the left hand side is random while the right hand side is deterministic. Can anyone explain this to me properly?

Answer 1

As mentioned in the comments, they are simply referring to the quadratic variation

$$[X]_{t}=\lim_{|P|\to 0}\sum_{k=1}^{n}(X_{t_{k}}-X_{t_{k-1}})^{2},$$

where $P$ ranges over partitions of the interval $[0,t]$. For Brownian motion this converges to $t$ eg. Is the fact that $dW_t\sim (dt)^{1/2}$ come from the $1/2-$holder property of Brownian motion?