# I Computing the Weingarten Map L by raising an index

1. Nov 27, 2016

### Leo Mar

Hello,

I have to prove that the Weingarten Map L for the unit sphere is + or - the identity "by computing the Lik in a coordinate patch and raising an index".

S^2 : x(Φ,θ)=(sinΦcosθ, sinΦsinθ, cosΦ)

I have computed the first (g) and the second (Λ) fundamental forms and I have found :

L=g-1Λ= ( -1 0 ) = -Identity
_________( 0 -1 )

The plus identity is obtained similarly by choosing a parametrization with inward pointing normal.

But what does "raising an index" mean?

Thank you.

2. Nov 28, 2016

### Ben Niehoff

Well, we have basically no context to go off of, since you haven't told us what a Weingarten map is, or where you pulled this notation from.

But if the thing you call $g$ is the metric tensor (which I think is the same thing as the first fundamental form), then it certainly looks to me as though you have already "raised an index", which just means contracting a slot of a lower-index tensor with $g^{-1}$.

3. Nov 30, 2016

### lavinia

Here is the general context. Although it seems to want a lowered not a raised index.

The second fundamental form is classically written as $edx^{2} + 2fdxdy + gdy^{2}$ in coordinates $(x,y)$
The metric tensor is $Edx^{2} + 2Fdxdy + Gdy^{2}$

You are being asked to write out the Wiengarten map in terms of these these two. If $N$ is the unit normal then for a tangent vector $X$ the Wiengarten map $W(X)$ is equal to $X⋅N$ , the derivative of $N$ with respect to $X$.

Hint: Think of the second fundamental form as the tensor $edx⊗dx +fdx⊗dy+fdy⊗dx+gdy⊗dy$
Think of the Weingarten map as a tensor.

Last edited: Nov 30, 2016