Let $a:\mathbb{R}^d\rightarrow\mathcal{S}_{d\times d}$ be a smooth map that takes its values in the space of $d\times d$ symmetric matrices and suppose there exists a $d\times d$ matrix valued $\sigma$ such that $a=\sigma \sigma^T$.
Let $\mathcal{L}^a:=\frac{1}{2}\nabla(a\nabla \cdot)=\frac{1}{2}\sum_{i,j=1}^d\partial_i(a^{ij}\partial_j\cdot)=\frac{1}{2}\sum_{i,j}[a^{ij}\partial_i\partial_j+(\partial_ia^{ij}\partial_j)]$.
It is not difficult to see, that the associated diffusion is a solution of the following SDE
$dX_t=\sigma(X_t)dB_t+b(X_t)dt$, where $b=(b^j)=(\sum_{i=1}^d \partial_i a^{ij})$.
Now, I have read that the associated Stratonovich SDE is given by
$ dX_t=\sigma(X_t)\circ dB_t. (*) $
I cannot verify this, as the usual formula for switching between Ito and Strato SDEs only gives
$ dX_t=\sigma(X_t)\circ dB_t+\tilde{b}(X_t)dt, $
where $\tilde{b}:=(\tilde{b}_j)=(\sum_{i=1}^d\partial_ia^{ij}-\frac{1}{2}\sum_{k,l=1}^d \sigma_{kl}\partial_k\sigma_{jl})$.
Using that
$ \sum_{i=1}^d \partial_i a^{ij}=\sum_{i,k=1}^d \partial_i (\sigma_{ik}\sigma_{kj}^T)=\sum_{i,k=1}^d \partial_i (\sigma_{ik}\sigma_{jk})=\sum_{i,k=1}^d (\sigma_{ik} \partial_i \sigma_{jk}+\sigma_{jk} \partial_i \sigma_{ik}) $
does not seem to make the expression for $\tilde{b}$ any easier, leave alone vanish.
Am I missing something or is (*) simply a mistake?
In the case, where $d=1$, we have the following situation:
$\mathcal{L}^af=\frac{1}{2}[\sigma^2f''+2\sigma'\sigma f']$ and the associated diffusion solves $dX_t=\sigma(X_t)dB_t+2\sigma'\sigma dt$. In this case, the formula to switch between Strato and Ito SDEs is given by
$ dX_t=b(X_t)dt+\sigma(X_t)dB_t \leftrightarrow dX_t=(b(X_t)-\frac{1}{2}(\sigma'\sigma)(X_t))dt+\sigma(X_t)\circ dB_t $
which in our case means that
$ dX_t=\frac{3}{2}\sigma'\sigma(X_t)dt+\sigma(X_t)\circ dB_t, $
which still does not result in the disappearance of the drift term.