I am reading a paper, in which the author considers a geodesic $\gamma(t)$ on a surface with $\gamma(0)=p,\gamma'(0)=v, \|v\|=1$ and wants to express the second derivative.
He gets
$$\gamma''(t) = - \gamma(t) - h(\gamma',\gamma')\nu(\gamma(t)),$$
where $\nu$ is the gauss map and $h$ is the Weingarten map/Second fundamental form.
I see how he gets the first part, but i can't see how he gets the normal part. Can anyone help?
So we have
$$\gamma \subset \Sigma \subset \mathbb S^3 \subset \mathbb R^4.$$
and $\gamma$ is a geodesic in $\Sigma$. Let $\nabla$ be the connection on $\Sigma$. Then $\nabla_{\gamma'} \gamma'=0$. By definition of $\nabla$, it's
$$\nabla_{ \gamma'} \gamma' = \gamma'' - (\gamma'')^\perp,$$
where $(v)^\perp$ denote the part $v$ which is orthogonal to $T\Sigma$. Since $(T\Sigma)^\perp$ is spanned by $\nu, \gamma$, we have
$$ (\gamma'')^\perp = \langle \gamma'' , \gamma\rangle \gamma + \langle \gamma'', \nu\rangle \nu. $$
Firstly, since $\langle \gamma, \gamma\rangle=1$, differentiate twice gives
$$\langle \gamma '', \gamma\rangle = - \langle \gamma', \gamma'\rangle = -\|v\| =-1.$$
Second, $\langle\gamma'', \nu\rangle = h(\gamma',\gamma')$. Thus
$$\gamma''= (\gamma'')^\perp = - \gamma + h(\gamma', \gamma') \nu$$
I am somehow missing a sign. Probably because Brendle is using $h(\gamma',\gamma') := \langle \gamma', \nabla_{\gamma'} \nu\rangle$?