1
$\begingroup$

Let $(N,h)$ be a Riemannian manifold and $M$ be a submanifold of $N$ with the induced metric $g$. Then the operator $\hat{\nabla}:C^{\infty}(TM)\times\ C^{\infty}(TM) \rightarrow C^{\infty}(TM)$, given by

$\hat{\nabla}_{\hat{X}}\hat{Y}=(\nabla_XY)^T$

is the Levi-Civita Connection of $(M,g)$.

Here $X$ and $Y$ are the local extensions of $\hat{X}$, $\hat{Y}$ $\in C^{\infty}(TM)$.

My doubts

I was trying to show the described operator satisfies Koszul's formula. But I struggled with finding the tangential part $(\nabla_XY)^T$ of the operator $\nabla_XY$.

$\endgroup$
1
  • $\begingroup$ You can do it with Koszul formula, but it is much quicker to just go back to the definition and show that it is indeed a torsion-free connection compatible with the induced metric $\endgroup$
    – Didier
    Commented Apr 10 at 21:01

1 Answer 1

2
$\begingroup$

Repeating my comment: You can do it with Koszul formula, but it is much quicker to just go back to the definition and show that it is indeed a torsion-free connection compatible with the induced metric. See this already existing question for instance.

However, since the question specifically mentions Koszul formula, here is a proof using it. It is much longer than just going back to the definition.

Consider $\hat{X},\hat{Y},$ and $\hat{Z}$ three vector fields on $M$, and $X,Y$, and $Z$ three local extensions to $N$. Koszul formula yields \begin{align} 2h(\nabla_XY,Z) &= Xh(Y,Z) + Yh(Z,X) - Zh(X,Y)\\ &\quad + h([X,Y],Z) + h([Z,X],Y) - h([Y,Z],X). \end{align} Notice that $[X,Y]$, $[Y,Z]$ and $[Z,X]$ are local extensions of $[\hat{X},\hat{Y}]$, $[\hat{Y},\hat{Z}]$ and $[\hat{Z},\hat{X}]$. Hence, along $M$: \begin{align} 2h(\nabla_XY,Z) &= \hat{X}h(\hat Y,\hat Z) + \hat Yh(\hat Z,\hat X) - \hat Zh(\hat X,\hat Y)\\ &\quad + h([\hat X,\hat Y],\hat Z) + h([\hat Z,\hat X],\hat Y) - h([\hat Y,\hat Z],\hat X). \end{align} Since everything is tangent to $M$, by definition of the induced metric $g$, one has \begin{align} 2h(\nabla_XY,Z) &= \hat{X}g(\hat Y,\hat Z) + \hat Yg(\hat Z,\hat X) - \hat Zg(\hat X,\hat Y)\\ &\quad + g([\hat X,\hat Y],\hat Z) + g([\hat Z,\hat X],\hat Y) - g([\hat Y,\hat Z],\hat X). \end{align} Hence, by Koszul formula again, one has $$ g(\hat\nabla_{\hat X}\hat Y,\hat Z) = h(\nabla_XY,Z). $$ Along $M$, write $\nabla_XY = (\nabla_XY)^{\perp} + (\nabla_XY)^{\top}$ with $(\nabla_XY)^{\perp}$ (respectively $(\nabla_XY)^{\top}$) orthogonal (respectively tangent) to $M$. Then \begin{align} g(\hat \nabla_{\hat X} \hat Y,\hat Z) &= h(\nabla_XY,Z) \\ &= h((\nabla_XY)^{\perp}+(\nabla_XY)^{\top},Z) \\ &= h((\nabla_XY)^{\top},Z) & \text{since } Z\perp (\nabla_XY)^{\perp}\\ &= h((\nabla_XY)^{\top},\hat Z) & \text{since } Z|_M = \hat Z\\ &= g((\nabla_XY)^{\top},\hat Z) & \text{by definition of } g \end{align} and it follows that $\hat\nabla_{\hat X}\hat Y = (\nabla_XY)^{\top}$ for any local extensions $X$ and $Y$.

$\endgroup$
1
  • $\begingroup$ Thank you! This is very helpful! $\endgroup$
    – White Give
    Commented Apr 11 at 10:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .