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Show orthogonal matrices are manifolds (Munkres Analysis on Manifolds)

  1. Jun 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Let ##O(3)## denote the set of all orthogonal 3 by 3 matrices, considered as a subspace of ##\mathbb{R}^9##.
    (a) Define a ##C^\infty## ##f:\mathbb{R}^9 \rightarrow \mathbb{R}^6## such that ##O(3)## is the solution set of the equation ##f(x) = 0##.
    (b) Show that ##O(3)## is a compact 3-manifold in ##\mathbb{R}^9## without boundary. [Hint: Show the rows of ##Df(x)## are independent if ##X \in O(3)##.

    2. Relevant equations
    I think the following theorem:

    Let ##O## be open in ##\mathbb{R}^n##; let ##f:O \rightarrow \mathbb{R}## be of class ##C^r##. Let ##M## be the set of points ##x## for which ##f(x) = 0##; let ##N## be the set of points for which ##f(x) \geq 0##. Suppose ##M## is non-empty and ##Df(x)## has rank 1 at each point of ##M##. Then ##N## is an n-manifold in ##\mathbb{R}^n## and the boundary of ##N## is ##M##.

    might be useful, but I don't see an opening for how I can apply it to the problem.


    3. The attempt at a solution
    I'm stuck on part (a) of the question, I think because the construction of the function seems kind of open-ended to me, as well as unmotivated in relation to part (b) (e.g. Why ##f:\mathbb{R}^9 \rightarrow \mathbb{R}^6##?). I feel like the function satisfying the required conditions may not be unique, like maybe something like
    ##f(\textbf{x},\textbf{y},\textbf{z}) = (\textbf{x} \cdot \textbf{y}, \textbf{x} \cdot \textbf{z}, \textbf{y} \cdot \textbf{z}, \textbf{x} \cdot \textbf{y}, \textbf{x} \cdot \textbf{z}, \textbf{y} \cdot \textbf{z} ) ##
    or some permutation of it will satisfy the conditions for ##f## but is not unique (where ##\textbf{x}, \textbf{y} , \textbf{z}## are the orthogonal vectors making up the orthogonal matrix). Am I completely off the mark? I'm quite lost on this problem, any help would be greatly appreciated!
     
    Last edited: Jun 24, 2012
  2. jcsd
  3. Jun 24, 2012 #2
    I had to stare at it a while, because at first I was pulled towards the whole det(X)=1 idea. But I think your idea is good, the columns should be orthogonal. What else should they be?
     
  4. Jun 24, 2012 #3
    Hi algebrat,

    Thanks again for helping!

    I'm not sure what other properties our function should have. I suppose we can make each component function independent as well, so that our jacobian will have full rank?

    So maybe the function could be like:

    ##f(\textbf{x}, \textbf{y}, \textbf{z}) = (\textbf{x} \cdot \textbf{y}, \textbf{x} \cdot \textbf{z}, \textbf{y} \cdot \textbf{z}, \textbf{x} \cdot (\textbf{y} + \textbf{z}), \textbf{y} \cdot (\textbf{x} + \textbf{z}), \textbf{z} \cdot (\textbf{x} + \textbf{y})) ##?

    But my bigger problem is I still don't understand why we're trying to construct this function? How does it help with part (b) of the problem? Especially since all the dimensions are different. We're trying to show ##O(3)## is a 3-manifold but we're constructing a function whose range is ##\mathbb{R}^6##?
     
  5. Jun 24, 2012 #4
    What additional condition must the columns satisfy? Find some examples of O(n) for various spaces.

    There's one more conceptually/geometrically understandable fact about them you can know, besides just being orthogonal to each other. What is it?
     
  6. Jun 24, 2012 #5

    micromass

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    A tiny bit more help: what is [itex]\mathbf{x}\cdot \mathbf{x}[/itex] for columns [itex]\mathbf{x}[/itex]??
     
  7. Jun 24, 2012 #6
    Hi,

    Hmm, I see! So since the vectors making up the matrix are orthonormal, we should also have ##||\textbf{x}||^2 = ||\textbf{y}||^2 = ||\textbf{z}||^2 = 1##. So, here's another stab at it:

    ##f(\textbf{x},\textbf{y},\textbf{z}) = (||\textbf{x}||^2 - 1 , ||\textbf{y}||^2 - 1, ||\textbf{z}||^2 - 1, \textbf{x} \cdot \textbf{ y}, \textbf{x} \cdot \textbf{ z}, \textbf{y} \cdot \textbf{ z}) ##

    So the advantage of this function now is ##f(A)=0## if and only if ##A \in O(3)##.
    Before in my previous example, the matrix ##\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}## would have satisfied the conditions required, but ##A## is clearly not an orthogonal matrix. Was this along the lines of what you two were hinting at? I guess I should have realized that sooner.

    Now, I believe the function satisfies the requirements for part (a) since ##f## is also ##C ^ \infty##. However, the question remains of how do I relate this result to part (b)? I'm thinking that I want to somehow transform this function so that I can apply the theorem in the original post, but the problem is the theorem requires a real-valued function, while I have one whose range is ##\mathbb{R}^6## (Maybe I can just define a new function ##g## as the sum of the component functions of ##f##?), along with other problems. Any more suggestions or hints for how I should proceed?

    Thanks again for all the help!
     
  8. Jun 24, 2012 #7

    micromass

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    Use exercise 2 of the same chapter.
     
  9. Jun 24, 2012 #8
    Which book are y'all using.
     
  10. Jun 24, 2012 #9

    micromass

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    Analysis on Manifolds by Munkres. This is exercise 5 of section 24.
     
  11. Jun 24, 2012 #10
    Aha! I hadn't looked at that problem, since I was only trying to work through the odds. Well, I guess that clears things up! I knew there had to be a reason why we were doing what we're doing in part (a). Guess I should've looked one page back. Onto proving exercise 2 then!

    Yep, the book is Analysis on Manifolds by James Munkres. Here's the theorem Micromass is referring to, in case you're wondering (which needs to be proven):

    Let ##f: \mathbb{R}^{n+k} \rightarrow \mathbb{R}^n## be of class ##C^r##. Let ##M## be the set of all ##x## such that ##f(x) = 0##. Assume that ##M## is non-empty and that ##Df(x)## has rank ##n## for ##x \in M##. Then ##M## is a k-manifold without boundary in ##\mathbb{R}^{n+k}##. Furthermore, if ##N## is the set of all ##x## for which

    ##f_1(x) = ... = f_{n-1}(x) = 0 \text{ and } f_n(x) \geq 0,##
    and if the matrix
    ##\partial(f_1,...,f_{n-1})/\partial x##

    has rank ##n-1## at each point of ##N##, then ##N## is a ##k+1## manifold, and ##M## is the boundary of ##N##
     
  12. Jun 25, 2012 #11
    So I did some thinking, and I'm still stuck on the proof of the theorem in post 10. Particularly, I'm not sure how to show that ##N## is a ##k+1## manifold. But I did try to attempt the first part of the proof. Please help me out if you can!

    Here is my proof of the first part of the theorem:

    We know that ##Df(x)## has rank ##n## for ##x \in M##. Then, if we had a square matrix of size ##n+k \times n+k## with ##Df(x)## embedded inside, we just need ##k## more linearly independent rows so that it has full rank. For simplicity's sake, assume the function ##g: \mathbb{R}^{n+k} \rightarrow \mathbb{R}^{n+k}## defined by

    ##g(x_1,...,x_{n+k}) = (x_1,...,x_k,f_1(x),...,f_n(x))##

    has rank ##n+r## (If it doesn't we can shift around the ##x_k's## in our function until it does). Then, ##Det(Dg(x)) \not = 0## for ##x \in M##. Hence, by the inverse function theorem, ##g## is a diffeomorphism between a neighborhood ##V## containing ##x## with an open neighborhood ##U## in ##\mathbb{R}^{n+k}##. Moreover, ##g## will carry ##V \cap M## in a one-to-one fashion onto the set ##U_0 = U \cap \mathbb{R}^k \times 0^n##. Now, we define the projection function ##\pi : U_0 \rightarrow W## as the projection of ##U_0## onto its first ##k## coordinates. This is a bijective function, since the last ##n## coordinates of elements in ##U_0## are 0. Then, the composite function ##\alpha = g^{-1} \circ \pi ^{-1}## represents a coordinate patch from ##W## to ##V \cap M##. Hence, ##M## is a k-manifold.
    (Check: ##\alpha## is of class ##C^r## since ##\pi ^{-1}## is of ##C^\infty## and ##g^{-1}## is of ##C^r## since it is a diffeomorphism. ##\alpha ^{-1}## is continuous since ##g## and ##\pi## are continuous. ##D\alpha (x)## has rank ##k## for ##x \in M## since the first ##k## rows of ##D\alpha## are the standard bases vectors.)

    The key part of my proof was that ##DetDg(x) \not = 0## so that I could use the inverse function theorem to create a diffeomorphism. However, that doesn't seem to work for the second part of the proof, where ##Df(x)## can have possibly rank only ##n-1## at each point of ##N##. Thus, I can't define a function that I can apply the inverse function theorem on if I can't be sure it has full rank.

    Sorry if I'm a little slow on this subject. Your help is much appreciated! Any critique on my proof of the first part of the theorem would be great as well!
     
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