Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the second fundamental form?

  1. Dec 22, 2005 #1
    OK, this is more of a spot for an elaboration on a question I just posted in another thread. Not quite duplicating threads, I hope, I just wanted to have this not buried in another spot...

    So, the question is this:

    Let's say that we have a smooth manifold [tex]\mathcal{M}[/tex]that may be viewed as a surface in [tex]\mathbb{R}^3[/tex] given by an embedding [tex]x[/tex]. Denote by [tex]n[/tex] the outward-pointing unit normal to [tex]\mathcal{M}[/tex] (yes, we are assuming that [tex]\mathcal{M}[/tex] is orientable), and by [tex]g[/tex] and [tex]h[/tex] the metric and second fundamental form, resp., defined by

    [tex]g_{ij}=\left<\dfrac{\partial x}{\partial u^i},\dfrac{\partial x}{\partial u^j}\right>[/tex]

    [tex]h_{ij}=-\left<\dfrac{\partial^2x}{\partial u^i\partial u^j},n\right>[/tex]

    with respect to some local coordinates [tex]\left\lbrace u^1,u^2\right\rbrace[/tex] for some (open) region of [tex]\mathcal{M}[/tex].

    What I need to understand, basically, is how this notation works. I can easily see how one would obtain

    [tex]g_{ij}=\dfrac{\partial y^k}{\partial u^i}\dfrac{\partial y^k}{\partial u^j};\hspace{0.75cm}k\text{ summed}[/tex]

    from the definiton that [tex]g=g_{ij}du^i\otimes du^j[/tex], [tex]y^k=y^k\left(u^1,u^2\right)[/tex], but I don't see how to go from here to the pairing above, and am even more confused about the pairing used to define [tex]h_{ij}[/tex] above.

    Any thoughts?
     
    Last edited: Dec 23, 2005
  2. jcsd
  3. Dec 27, 2005 #2

    gvk

    User Avatar

    I am only confused with the sign "-" in the definition for second fundamental form. From my prospective there should be "+",
    but it does not effect the results. By the way, what kind of book do you use?
    The most important thing is that the second fundamental form envolves the second derivitives of coordinate transformation!
    Let suppose, you have a smooth manifold [tex]\mathcal{M}[/tex] that may be viewed as a 2D (!) surface in 3D space with coordinates
    [tex]x=(x^1,x^2,x^3)[/tex] given by the equation [tex]x^3=f(x^1,x^2)[/tex]. Then two coordinates [tex](u^1,u^2)[/tex] on the surface can
    be [tex]u^1=x^1, u^2=x^2[/tex]. For simplicity I'll use the notion [tex]u=u^1, v=u^2[/tex] and parametric equations of the smooth
    manifold [tex]\mathcal{M}[/tex] are [tex]x^1=u, x^2=v, x^3=f(u,v)[/tex]
    Then
    [tex]x_{u}=(1,0, \dfrac{\partial f}{\partial u})[/tex],
    [tex]x_{v}=(0,1, \dfrac{\partial f}{\partial v})[/tex].
    The second derivitives:
    [tex]x_{u,u}=(0,0, \dfrac{\partial^2 f}{\partial u \partial u})[/tex],
    [tex]x_{u,v}=x_{v,u} =(0,0, \dfrac{\partial^2 f}{\partial u \partial v})[/tex],
    [tex]x_{v,v}=(0,0, \dfrac{\partial^2 f}{\partial v \partial v})[/tex].
    From here you can easily calculate [tex]g_{ij}[/tex] and [tex]h_{ij}[/tex].
     
  4. Dec 27, 2005 #3
    Thanks! I think...

    If I understand you correctly, the above notation, then, is somwhat abusive in that typically, a pairing is usually between an element of one space, and something that is dual to it, like

    [tex]\left<\dfrac{\partial}{\partial x},dx\right>[/tex]

    Not an iron fisted rule, mind you, but you see what I'm getting at--something like stating [tex]g_{ij}[/tex] as above is... not quite in line with the usual methods, in particular because it requires something like the "k summed" note above to fix the fact that summation convention is, technically, violated. Moreover, as per your example above, a note like this would also be required to calculate [tex]h_{ij}[/tex], right?

    Assuming then that I have this right, the next question would be how to extend this notation? That is, if [tex]\mathcal{M}[/tex] is an [tex]m[/tex]-dimensional manifold, and [tex]x:\mathcal{M}\rightarrow\mathbb{R}^{m+1}[/tex], then we may choose a Darboux frame [tex]\left(x;e_1,\ldots,e_m,e_{m+1}\right)[/tex] on [tex]\mathcal{M}[/tex]. But, in doing this, would this allow us to construct [tex]e_{m+1}[/tex] as a "normal" vector to the "surface", as would be the case if [tex]m=\left\lbrace 1,2\right\rbrace[/tex]?

    Looking forward to all of your input!

    Also, I'm not working with any one book in particular, rather quite a conglomarate, as well as a stack of papers. However, the notation quoted above is explicitly used in http://wwwmaths.anu.edu.au/research.reports/mrr/98/031/MRR98-031.pdf, which is pretty goovy read!
     
  5. Dec 29, 2005 #4

    gvk

    User Avatar

    No, it's not abusive. It's just simple scalar product. Of course you can do it if your space has a metric.
    Consider a smooth hypersurface in Euclidean [tex]\mathbb{R}^{m+1}[/tex] with Euclidean coordinates [tex]x^1, ....., x^{m+1}[/tex], which we shall assume to be given in the graphical form [tex]x^{m+1} = f(x^1, ..., x^m)[/tex]. Then the normal vector to the hypersurface is is given by
    [tex]n= \dfrac{(-f_{x^1},...,-f_{x^{m}},1)}{\sqrt{1+(f_{x^1})^2+...+(f_{x^m})^2}}[/tex]
    It's interested. Why are you doing this?
     
    Last edited: Dec 29, 2005
  6. Dec 31, 2005 #5
    Recall that the coefficients of the II fundamental form are

    [tex]L{i,j}= -\partial_i (r)\cdot\partial_j (n) = {\partial_{i,j}}^2(r)\cdot (n)[/tex]

    where I have used "[tex]\codt [/tex]" instead of pairing.
     
  7. Feb 1, 2006 #6
    So, how could we generalize and make definitions in higher dimension?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: What is the second fundamental form?
Loading...