# Verify an operator is a levi-civita connection of the Riemannian submanifold $(M,g)$

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 $$\nabla_XY$$.

• 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 Commented Apr 10 at 21:01

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 $$Y$$.