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Let $(B_t)$ a Brownian motion on $\mathbb R$. The following puzzled. So, I read for example here that $(e^{iB_t})_{t\geq 0}$ is a Brownian motion on the circle. But using the stereographic projection, $\left(\frac{2B_t}{1+B_t^2},\frac{B_t^2-1}{1+B_t^2}\right)$ should also be a Brownian motion, right ? But it looks that both seem to be very different in the sens that the first one will reach the north pole in infinite time a.s. (and looks more faithful of the Brownian motion on $\mathbb R$), whereas the other one will reach all point of the circle in finitely many time. So, at the end, I'm not so sure why these two brownian motion are so different, and I'm suspicious if $\left(\frac{2B_t}{1+B_t^2},\frac{B_t^2-1}{1+B_t^2}\right)$ is really a Brownian motion of the circle. Can some one explain a bit better how things behave ?

tiko
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1 Answers1

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Let $$W_t:=\left(\frac{2B_t}{1+B_t^2},\frac{B_t^2-1}{1+B_t^2}\right),\quad t\geq 0.$$ Then $(W_t)_{t\geq 0}$ is not a Brownian motion on the circle. Nevertheless, it's a Brownian motion on $\mathbb S^{1}\setminus \{(0,1)\}$.

Surb
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  • You didn't explain why it is not a brownian motion or it is a brownian motion on $S^1 \ {0,1}$ – NN2 Mar 01 '21 at 11:48
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    @NN2 : It's not a Brownian motion on the sphere because it doesn't leave on the sphere but on $\mathbb S^1\setminus {(0,1)}$. It is a Brownian motion on $\mathbb S^1\setminus {(0,1)}$ because its generator is $\frac{1}{2}\Delta _{LB}$ where $\Delta _{LB}$ is the Laplace-Beltrami operator on $\mathbb S^1\setminus {(0,1)}$. – Surb Mar 01 '21 at 11:52