Results 1 to 5 of 5

Math Help - Questions on a Theorem and its proof in baby Rudin

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    120

    Questions on a Theorem and its proof in baby Rudin

    This is Theorem 10.27 in Page 267-268 of Rudin's "Principles of Mathematical Analysis", as the picture below shows.


    I have three questions:
    First, what is "oriented rectilinear k-simplex"? Is it the same as oriented affine k-simplex defined in the previous page?
    Second, if \sigma is in open set E, that is, its range \subseteq E, how to guarantee that \bar\sigma is also in E? If not, integral \int\nolimits_{\bar\sigma} \omega may not be meaningful at all.
    Three, the proof deals only with the relation of Jacobians of \sigma and \bar\sigma, what about the coefficient function? In page 254, the integral is defined as \int\nolimits_D \sum a_{i_1...i_k}(\sigma or \bar\sigma(\mathbf u))J_{i_1...i_k}d\bf u, where D is the parameter domain, but in general \sigma(\mathbf u)\ne\bar\sigma(\mathbf u), how can we say that the two integrals differ only in the sign?
    If any other information is needed, please reply too. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,224
    Thanks
    1791
    An "oriented rectilinear k-simplex" is a special case of an "oriented affine k-simplex" in that it requires that the boundaries be straight lines, planes, etc. in a given coordinate system which an oriented affine k-simplex does not.

    Are you clear on the distinction between \sigma and \overline{\sigma}? Unfortunately you didn't include the "(75)" where \sigma was given but I think it is simply a simplex. \overline{\sigma}, we are told, "is obtained from \sigma by interchanging p_0 and p_j". That is, \overline{\sigma} is exactly the same point set as \sigma with possibly a different orientation. Being "in E" is a property of the point set, regardless of orientation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2009
    Posts
    120
    Thanks for the reply. The following pictures are other pages of the book containing definitions needed.
    the following is the definition of "oriented affine k-simplex"(Page 266):


    This page contains the meaning of \bar\sigma(Page 267):


    This page defines the action of a k-form on k-surfaces(page 254), where \frac{\partial(x_{i_1},...,x_{i_k})}{\partial(u_1,  ...,u_k)} in (35) is just the J_{i_1...i_k} I use in my first post.


    Now, what is the formulation of "oriented rectilinear k-simplex"? Is it all points \mathbf x=(x_1,...,x_k) in \mathbb R^k satisfying \sum\limits_{i = 1}^k x_i=1 and x_i\geq 0 for each i? Maybe its definition can explain why \sigma(\mathbf u)=\bar\sigma(\mathbf u).
    Sorry for the missing parts of the book. If any other related information is needed, please reply, Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    120
    The second question can be proved as follows:
    It's enough to prove the situation where only one pair of vertices are interchanged, since general case is a series of compositions of this special case. The problem can be divided into two cases: a) \mathbf p_0 is not interchanged, b) \mathbf p_0 is interchanged. Suppose \forall\mathbf u=(\alpha_1, ..., \alpha_k)\in Q^k. For the first case, let us suppose \mathbf p_1 and \mathbf p_2 are interchanged. Then \bar\sigma(\mathbf u)=\sigma((\alpha_2,\alpha_1,\alpha_3,...,\alpha_k  )), where (\alpha_2,\alpha_1,\alpha_3,...,\alpha_k) obviously belongs to Q^k, so \sigma((\alpha_2,\alpha_1,\alpha_3,...,\alpha_k))\  in E, that is, \bar\sigma(\mathbf u)\in E. For the second case, without loss of generality, I suppose the pair \mathbf p_0 and \mathbf p_1 are exchanged. Then by (76) \bar\sigma(\mathbf u)=\mathbf p_1+\alpha_1(\mathbf p_0-\mathbf p_1)+\sum\limits_{i = 2}^k\alpha_i(\mathbf p_i-\mathbf p_1). Adding and subtracting \mathbf p_0, we get \mathbf p_0+(1-\alpha_1-\sum\limits_{i = 2}^k\alpha_i)(\mathbf p_1-\mathbf p_0)+\sum\limits_{i = 2}^k\alpha_i(\mathbf p_i-\mathbf p_0). This is just the value of vector (1-\alpha_1-\sum\limits_{i = 2}^k\alpha_i,\alpha_2,...,\alpha_k) under \sigma. Noting that 0\leq 1-\alpha_1-\sum\limits_{i = 2}^k\alpha_i\leq 1 since \sum\limits_{i = 1}^k\alpha_i\leq 1, and that the sum of all components is 1-\alpha_1\leq 1, this vector is in Q^k. So the above value \in E, as desired. The same argument can be applied to prove that if \sigma is in E, then \partial\sigma is also in E.
    I hope someone could help me with the other two questions: the formulation of "oriented rectilinear k-simplex" and why the two integrals are equal when considering the coefficient function. Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2009
    Posts
    120
    OK, I finally found the answer to the 3rd question:
    the invariance of the value of the integral in spite of the fact that \sigma(\mathbf u)\ne\bar\sigma(\mathbf u) and therefore a_{i_1...i_k}(\sigma(\mathbf u))\ne a_{i_1...i_k}(\bar\sigma(\mathbf u)) results from the theorem of change of variable. We have \int_\sigma \omega=\int_{Q^k}\sum a_{i_1...i_k}(\sigma(\mathbf u))J_{i_1...i_k}(\mathbf u)d\bf u and \int_{\bar\sigma}\omega=\int_{Q^k}\sum a_{i_1...i_k}(\bar\sigma(\mathbf u)){\bar J}_{i_1...i_k}(\mathbf u)d\bf u, where J is the Jacobian generated from \sigma and \bar J is the Jacobian generated from \bar\sigma. According to the proof given by the book, \bar J(\mathbf u)=-J(\mathbf u). As my last post indicated, \bar\sigma(\mathbf u)=\sigma(B(\mathbf u)) if \mathbf p_0 is not involved in the interchange, where B is a flip defined in P249, or \bar\sigma(\mathbf u)=\sigma(\bf G(u)) otherwise, where \bf G is a primitive defined in P248 with m=1 and g(\mathbf u)=1-\alpha_1-\sum\limits_{i = 2}^k\alpha_i. For the former case, the differential of oriented affine simplex is just the linear transformation A in (78) which is fixed for any \mathbf u\in\mathbb R^k, so -J(\mathbf u)=-J(B(\bf u)). Thus, \int_{\bar\sigma}\omega=-\int_{Q^k}\sum a_{i_1...i_k}(\sigma(B(\mathbf u)))J_{i_1...i_k}(B(\mathbf u))d\bf u. Since B merely interchanges two coordinates, it differs from \int_\sigma \omega=\int_{Q^k}\sum a_{i_1...i_k}(\sigma(\mathbf u))J_{i_1...i_k}(\mathbf u)d\bf u only in the order in which k integrations are carried out, but integral \int_{Q^k} is independent of the integration order (See Example 10.4 in P247), then we get the equality. For the latter case, \int_{\bar\sigma}\omega=-\int_{Q^k}\sum a_{i_1...i_k}(\sigma(\mathbf G(\mathbf u)))J_{i_1...i_k}(\mathbf G(\mathbf u))d\bf u. We try to prove {\bar f}_{k-1} (\alpha_2,...,\alpha_k)=f_{k-1} (\alpha_2,...,\alpha_k) where {\bar f}_{k-1} (\alpha_2,...,\alpha_k)=\int_0^1 \sum a_{i_1...i_k}(\sigma(\mathbf G(\mathbf u)))J_{i_1...i_k}(\mathbf G(\mathbf u))d\alpha_1 and f_{k-1} (\alpha_2,...,\alpha_k)=\int_0^1 \sum a_{i_1...i_k}(\sigma(\mathbf u))J_{i_1...i_k}(\mathbf u)d\alpha_1, 0\leq\alpha_i\leq 1, 1\leq i\leq k-1, because if this is the case, subsequent integral will lead to the same consequence \int_{Q^k} as desired. If \sum\limits_{i = 2}^k\alpha_i\geq 1, both integrand are 0 by the definition of \int_{Q^k} and {\bar f}_{k-1}=f_{k-1} trivially. If 0\leq\sum\limits_{i = 2}^k\alpha_i\leq 1, denoting \sum\limits_{i = 2}^k\alpha_i as S, f_{k-1}=\int_0^{1-S}\sum a_{i_1...i_k}(\sigma(\alpha_1,\alpha_2,...,\alpha_  k))J_{i_1...i_k}(\alpha_1,\alpha_2,...,\alpha_k)d\  alpha_1 because \alpha_1\geq 1-S would get out of the standard simplex Q^k and make the integrand zero. Since \alpha_2,...,\alpha_k is fixed, g(\mathbf u)=g(\alpha_1)=1-\alpha_1-S, g'(\alpha_1)=-1, this means g is strictly decreasing, so g(0)=1-S and g(1-S)=0. In addition, g is continuous on [0,1-S] due to its differentiability. Applying the Theorem of change of variable (not the one in page 132-133 of this book because g is not increasing, but the one in page 144-145 of Apostol's "Mathematical analysis") and Th6.17, we get \int_0^{1-S}\sum a_{i_1...i_k}(\sigma(g(x_1),\alpha_2,...,\alpha_k)  )J_{i_1...i_k}(g(x_1),\alpha_2,...,\alpha_k)d\alph  a_1 (note that \int_{1-S}^0=-\int_0^{1-S}). But for the same consideration of integral limits, {\bar f}_{k-1}=\int_0^{1-S}\sum a_{i_1...i_k}(\sigma(\mathbf G(\mathbf u)))J_{i_1...i_k}(\mathbf G(\mathbf u))d\alpha_1 which is just the above integral.
    It is worth noting that I was not using the Theorem 10.9 of Change of variables in P252 because some conditions does not meet, but the idea in its proof, in particular its first paragraph, is almost the same as the above argument.
    Since I have proved the general case of oriented affine k-simplex, my first question on the definition of the so-called "oriented rectilinear k-simplex" is no longer important (actually the exposition of the book that follows all uses the general oriented affine simplex case of the theorem).
    Last edited by zzzhhh; December 5th 2009 at 05:09 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Theorem from Baby Rudin (Thm. 2.36)
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: January 13th 2011, 09:45 PM
  2. Baby Rudin problem 1.5
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: December 29th 2010, 07:11 AM
  3. Baby Rudin - ch 2, ex 27
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 21st 2010, 01:24 PM
  4. A Baby Rudin Related Question
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: September 23rd 2010, 04:18 PM
  5. Baby Rudin - Existence of logarithm
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: August 31st 2010, 04:59 AM

Search Tags


/mathhelpforum @mathhelpforum