Inquiry about the vector field calculation of the SO3FM function

[mrzzmrzz]

Hey,
First of all, thank you very much for the excellent work by the authors. I am currently reading through the paper and the code, and I have encountered a small issue that has left me a bit puzzled. Specifically, the problem is as follows. Due to the differences in the setup of Flow Matching in FoldFlow (where $t = 0$ is the target distribution and $t = 1$ is the random distribution), I noticed a discrepancy when calculating the SO(3) vector field. In the paper, it is written:

However, after reviewing the code example, the implementation in the code looks like this:

def vectorfield(self, rot_0, rot_t, t):
    ...
    rot_t_minus_0 = rot_0.transpose(-1, -2) @ rot_t
    if self.inference_scaling < 0:
        u_t = rot_t @ (
            log(rot_t_minus_0)
            / torch.clamp(t[:, None, None], min=-self.inference_scaling)
        )
    else:
        u_t = rot_t @ (log(rot_t_minus_0) * self.inference_scaling)
    ...
    return None, u_t

The issue I don’t fully understand is that the paper describes the computation as $r_t^T r_0$, while the code implements it as $r_0^T r_t$. This seems to differ from what is described in the paper.

Another point I’m confused about is that the paper mentions this computation can be simplified to $\frac{\log_{r_t} (r_0)}{t}$. Would it be possible to directly compute it as log_not_from_identity(r_t, r_0})/ t?

Lastly, considering the difference in the notation of $t$ between the paper (where $t = 0$ is the random distribution and $t = 1$ is the target distribution) and the usual flow matching notation, if we were to change the notation to the usual one, could the interpolation on SO(3) be represented as:

$$ r_t = \exp_{r_1} \left( t \log_{r_1} (r_0) \right)? $$

Thank you for your time!