I'm reading Nomizu & Sasaki's "*Affine Differential Geometry: Geometry of Affine Immersions*" and I'm having some trouble with Proposition 1.4.

I have an immersed surface in $M \hookrightarrow \mathbb R^{3}$. Assume that we have a volume form, say $\omega$ on $\mathbb R^{3}$. This could, for example, be the determinant. We also have a covariant derivative on $\mathbb R^{3}$, say $D$.

Let $Y$ and $Z$ be smooth, linearly independent vector fields on $M$, and let $\xi$ be a smooth transverse vector field on $M$, i.e. $T_pM \oplus \mathrm{span}(\xi) = T_p\mathbb R^3$ for all $p \in M$. We have the two equations \begin{eqnarray*} D_YZ &=& \nabla_YZ + h(Y,Z)\xi \\ \\ D_Y\xi &=& -SY + \tau(Y)\xi \end{eqnarray*} We can define a volume element on $M$, say $\theta$, as $\theta(Y,Z) := \omega(Y,Z,\xi)$.

**Proposition 1.4:** $\nabla_X\theta = \tau(X)\theta$ for all $X \in T_pM$.

**My attempt to follow the proof:**

$$X[\omega(Y,Z,\xi)] = [D_X\omega](Y,Z,\xi)+\omega(D_XY,Z,\xi)+\omega(Y,D_XZ,\xi)+\omega(Y,Z,D_X\xi)$$ where $X[\omega(Y,Z,\xi)]$ is the directional derivative of the function $\omega(Y,Z,\xi)$.

First, notice that $X[\omega(Y,Z,\xi)]=X[\theta(Y,Z)]$. Second \begin{eqnarray*} \omega(D_XY,Z,\xi) &=& \omega(\nabla_XY+h(X,Y)\xi,Z,\xi) \\ \\ &=& \omega(\nabla_XY,Z,\xi) \\ \\ &=& \theta(\nabla_XY,Z) \end{eqnarray*} Similarly $\omega(Y,D_XZ,\xi) = \theta(Y,\nabla_XZ)$. Finally we have \begin{eqnarray*} \omega(Y,Z,D_X\xi) &=& \omega(Y,Z,-SX+\tau(X)\xi) \\ \\ &=& \omega(Y,Z,\tau(X)\xi) \\ \\ &=& \tau(X)\omega(Y,Z,\xi) \\ \\ &=& \tau(X)\theta \end{eqnarray*} Substituting all of these simplifications into the original equations gives $$X[\theta(Y,Z)] = [D_X\omega](Y,Z,\xi)+\theta(\nabla_XY,Z)+\theta(Y,\nabla_XZ)+\tau(X)\theta(Y,Z)$$

**Where I get lost:**

The authors then suddenly jump to this: \begin{eqnarray*} [\nabla_X\theta](Y,Z) &=& X[\theta(Y,Z)] - \theta(\nabla_XY,Z)-\theta(Y,\nabla_XZ) \\ \\ &=& \tau(X)\theta(Y,Z) \end{eqnarray*}

I think they have done something with $[D_X\omega](Y,Z,\xi)$ but I don't know what.