Consider the following SDE on $\mathbf R^d$:
\begin{equation}\tag{*}
dX_t^i = -\frac{d-1}{2}X_t^i dt + \sum_{j=1}^d(\delta^{ij}-X_t^iX_t^j)dW_t^j, \quad i=1,2,...,d,
\end{equation}
where $W = (W^1,W^2,...,W^d)$ is a $d$-dimensional standard Brownian motion. The paper ['Kinetic Brownian motion on Riemannian manifolds', Subsection 2.2](https://projecteuclid.org/download/pdf_1/euclid.ejp/1465067216) has the following claim,
> **Claim:** The $\mathbf R^d$-valued process $X=(X^1,X^2,...,X^d)$ lies on the sphere $\mathbf S^{d-1}$.

But why?

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I try to solve this problem in two options. But neither works.

**OPTION 1.** I rewrite the SDEs (*) to the Stratonovich form. To this purpose, let $\sigma_j^i(x) = \delta^{ij} - x^ix^j$. Then $\partial_k\sigma_j^i = -(\delta^{ik}x^j+\delta^{jk}x^i)$. Thus,
\begin{equation}
\begin{split}
dX_t^i &= -\frac{d-1}{2}X_t^i dt + \sum_{j=1}^d \sigma^i_j(X_t)dW_t^j \\ 
&= -\frac{d-1}{2}X_t^i dt + \sum_{j=1}^d \sigma^i_j(X_t)\circ dW_t^j - \frac{1}{2} \sum_{j,k=1}^d \sigma^k_j\partial_k\sigma_j^i(X_t) dt\\
&= \left(1-\sum_{j=1}^d(X_t^j)^2\right) X_t^i dt + \sum_{j=1}^d(\delta^{ij}-X_t^iX_t^j)\circ dW_t^j.
\end{split}
\end{equation}

Let $Z_t=(t,W_t)$ and 
\begin{align}
V_0(x) &= \left(1-\sum_{j=1}^d(x^j)^2\right) x^i \frac{\partial}{\partial x^i}, \\
V_j(x) &= (\delta^{ij}-x^i x^j)\frac{\partial}{\partial x^i}.
\end{align}
Then we can write
$$dX_t = \sum_{j=0}^d V_j(X_t) \circ dZ_t^j.$$
If the vector fields $V_j, j=0,1,...,d$ are all tangent to $\mathbf S^{d-1}$, then the claim follows by [Proposition 1.2.8 of Hsu's book 'Stochastic Analysis On Manifolds'](https://books.google.com/books?id=GDEPCgAAQBAJ&pg=PA22&lpg=PA22&dq=Stochastic+Analysis+On+Manifolds+Proposition+1.2.8&source=bl&ots=mrP-5cJU8J&sig=ACfU3U2TtOh3dVPkKkLMnCynbhVbGpRV9A&hl=zh-CN&sa=X&ved=2ahUKEwiK2LPd0v7hAhVIbbwKHSf2D_QQ6AEwAnoECAkQAQ#v=onepage&q=Stochastic%20Analysis%20On%20Manifolds%20Proposition%201.2.8&f=false). But
\begin{align}
V_0 \cdot x &= \sum_i V_0^i x^i = \left(1-\sum_j(x^j)^2\right) \left( \sum_i (x^i)^2 \right), \\
V_j \cdot x &= \sum_i V_j^i x^i = \left(1-\sum_i(x^i)^2\right) x^j,
\end{align}
they are not zero. Hence we cannot assert $V_j, j=0,1,...,d$ to be tangent to $\mathbf S^{d-1}$.

**OPTION 2.**  For the function $r(x) = \|x\| = \sqrt{\sum_{i=1}^d (x^i)^2}$, we have $\partial_i r = \frac{x^i}{r}$, $\partial_i \partial_j r = \frac{\delta^{ij}}{r} - \frac{x^ix^j}{r^3}$. Let $R_t = \|X_t\|$. Using Ito's formula,
\begin{equation}
\begin{split}
dR_t &= \sum_i \partial_i r(X_t) dX_t^i + \frac{1}{2}\sum_{i,j} \partial_i \partial_j r(X_t) d\langle X^i, X^j\rangle_t \\
&= -\frac{d-1}{2}R_t dt + \sum_{j=1}^d\left(\frac{1}{R_t}-R_t\right)X^j_tdW_t^j + \frac{d-1}{2R_t} dt \\
&= \frac{d-1}{2}\left(\frac{1}{R_t}-R_t\right) dt + \sum_{j=1}^d\left(\frac{1}{R_t}-R_t\right)X^j_tdW_t^j.
\end{split}
\end{equation}
We cannot get a zero for $dR_t$, hence we cannot assert $R_t$ to be constant.

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Can anyone figure out how to prove the claim? Or point out the mistake in my arguments? TIA...

**PS:** This is a crosspost from [math.stackexchange](https://math.stackexchange.com/questions/3211869/how-to-judge-the-solution-process-of-an-sde-to-lie-on-the-sphere).