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I Computing the Weingarten Map L by raising an index

  1. Nov 27, 2016 #1
    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. jcsd
  3. Nov 28, 2016 #2

    Ben Niehoff

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    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}##.
     
  4. Nov 30, 2016 #3

    lavinia

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    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
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