2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

The formulas all look correct. With the definition of $\nabla_X\theta$, one gets \begin{align*} [\nabla_X\theta](Y,Z)&=X[\theta(Y,Z)]-\theta(\nabla_XY,Z)-\theta(Y,\nabla_XZ)\\ &=X[\omega(Y,Z,\xi)]-\omega(D_XY,Z,\xi)-\omega(Y,D_XZ,\xi)-\omega(Y,Z,D_X\xi)+\tau(X)\theta(Y,Z)\\ &=[D_X\omega](Y,Z,\xi)+\tau(X)\theta(Y,Z)\;. \end{align*} So the missing thing is the assumption that $D_X\omega=0$, that is, the covariant derivative is volume preserving.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank a lot for the answer. $D_X\omega=0$ is a reasonable assumption if $D$ is the standard covariant derivative on $\mathrm R^3$, and $\omega$ is the determinant. Thanks again. $\endgroup$ – Fly by Night Apr 11 '16 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.