If $p$ is homogeneous of order $k$, then $p^{-1}(a)$ is diffeomorphic to $p^{-1}(b)$












1












$begingroup$


Given homogeneous function $p$ with order $m$, how can I show that



$$p^{-1}(a) = left(frac{a}{b}right)^{frac{1}{m}}p^{-1}(b)?$$



The original question is:




Let $p$ be any homogeneous polynomial in $k$-variables. Homogeneity means
$$p(tx_1, dots, tx_k) = t^mp(x_1, dots, x_k).$$
Prove that the set of points $x$, where $p(x) = a$, is a $k-1$ dimensional submanifold of $mathbb{R}^k$, provided that $a neq 0$. Show that the manifolds obtained with $a > 0$ are all diffeomorphic, as are those with $a <0$.




So I am wondering, is $m$ is odd, $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?
Thanks.






differential-topology smooth-manifolds diffeomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $p^{-1}(a):=frac 1 {p(a)}$?
    $endgroup$
    – user64494
    Jul 6 '13 at 18:51










  • $begingroup$
    I don't know really. But from my previous experience in the book, it means inverse function. Thanks @user64494!
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:23












  • $begingroup$
    Did you mean to type $(frac{a}{b})^{frac{1}{n}}$? You haven't told us what $m$ is.
    $endgroup$
    – Robert Lewis
    Jul 6 '13 at 19:34










  • $begingroup$
    Oh yes, let me correct them - $m$ is order. Thank you @RobertLewis
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:39






  • 1




    $begingroup$
    $p^{-1}(a)$ is a set. Since $p$ is a homogeneous polynomial of degree $m$: If $m = 0$, $p$ is constant, and so $p^{-1}(a)$ is empty for all $a$ except one, where it is all of $mathbb{R}^k$. If $m > 0$, for $a neq 0$, the level set is either empty, or a nonempty submanifold of dimension $k-1$. For $a = 0$, it may be a more complicated set. It's a relation between sets, the dilation $x mapsto lambdacdot x$ induces a bijection between the level set $p^{-1}(a)$ and the level set $p^{-1}(lambda^mcdot a)$, that can succinctly be written $lambdacdot p^{-1}(a) = p^{-1}(lambda^mcdot a)$.
    $endgroup$
    – Daniel Fischer♦
    Jul 6 '13 at 23:41


















1












$begingroup$


Given homogeneous function $p$ with order $m$, how can I show that



$$p^{-1}(a) = left(frac{a}{b}right)^{frac{1}{m}}p^{-1}(b)?$$



The original question is:




Let $p$ be any homogeneous polynomial in $k$-variables. Homogeneity means
$$p(tx_1, dots, tx_k) = t^mp(x_1, dots, x_k).$$
Prove that the set of points $x$, where $p(x) = a$, is a $k-1$ dimensional submanifold of $mathbb{R}^k$, provided that $a neq 0$. Show that the manifolds obtained with $a > 0$ are all diffeomorphic, as are those with $a <0$.




So I am wondering, is $m$ is odd, $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?
Thanks.






differential-topology smooth-manifolds diffeomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $p^{-1}(a):=frac 1 {p(a)}$?
    $endgroup$
    – user64494
    Jul 6 '13 at 18:51










  • $begingroup$
    I don't know really. But from my previous experience in the book, it means inverse function. Thanks @user64494!
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:23












  • $begingroup$
    Did you mean to type $(frac{a}{b})^{frac{1}{n}}$? You haven't told us what $m$ is.
    $endgroup$
    – Robert Lewis
    Jul 6 '13 at 19:34










  • $begingroup$
    Oh yes, let me correct them - $m$ is order. Thank you @RobertLewis
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:39






  • 1




    $begingroup$
    $p^{-1}(a)$ is a set. Since $p$ is a homogeneous polynomial of degree $m$: If $m = 0$, $p$ is constant, and so $p^{-1}(a)$ is empty for all $a$ except one, where it is all of $mathbb{R}^k$. If $m > 0$, for $a neq 0$, the level set is either empty, or a nonempty submanifold of dimension $k-1$. For $a = 0$, it may be a more complicated set. It's a relation between sets, the dilation $x mapsto lambdacdot x$ induces a bijection between the level set $p^{-1}(a)$ and the level set $p^{-1}(lambda^mcdot a)$, that can succinctly be written $lambdacdot p^{-1}(a) = p^{-1}(lambda^mcdot a)$.
    $endgroup$
    – Daniel Fischer♦
    Jul 6 '13 at 23:41
















1












1








1


1



$begingroup$


Given homogeneous function $p$ with order $m$, how can I show that



$$p^{-1}(a) = left(frac{a}{b}right)^{frac{1}{m}}p^{-1}(b)?$$



The original question is:




Let $p$ be any homogeneous polynomial in $k$-variables. Homogeneity means
$$p(tx_1, dots, tx_k) = t^mp(x_1, dots, x_k).$$
Prove that the set of points $x$, where $p(x) = a$, is a $k-1$ dimensional submanifold of $mathbb{R}^k$, provided that $a neq 0$. Show that the manifolds obtained with $a > 0$ are all diffeomorphic, as are those with $a <0$.




So I am wondering, is $m$ is odd, $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?
Thanks.






differential-topology smooth-manifolds diffeomorphism







share|cite|improve this question











$endgroup$




Given homogeneous function $p$ with order $m$, how can I show that



$$p^{-1}(a) = left(frac{a}{b}right)^{frac{1}{m}}p^{-1}(b)?$$



The original question is:




Let $p$ be any homogeneous polynomial in $k$-variables. Homogeneity means
$$p(tx_1, dots, tx_k) = t^mp(x_1, dots, x_k).$$
Prove that the set of points $x$, where $p(x) = a$, is a $k-1$ dimensional submanifold of $mathbb{R}^k$, provided that $a neq 0$. Show that the manifolds obtained with $a > 0$ are all diffeomorphic, as are those with $a <0$.




So I am wondering, is $m$ is odd, $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?
Thanks.






differential-topology smooth-manifolds diffeomorphism




differential-topology smooth-manifolds diffeomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 13:01









Brahadeesh

6,30742361




6,30742361










asked Jul 6 '13 at 18:21









WishingFishWishingFish

9951028




9951028












  • $begingroup$
    Do you mean $p^{-1}(a):=frac 1 {p(a)}$?
    $endgroup$
    – user64494
    Jul 6 '13 at 18:51










  • $begingroup$
    I don't know really. But from my previous experience in the book, it means inverse function. Thanks @user64494!
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:23












  • $begingroup$
    Did you mean to type $(frac{a}{b})^{frac{1}{n}}$? You haven't told us what $m$ is.
    $endgroup$
    – Robert Lewis
    Jul 6 '13 at 19:34










  • $begingroup$
    Oh yes, let me correct them - $m$ is order. Thank you @RobertLewis
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:39






  • 1




    $begingroup$
    $p^{-1}(a)$ is a set. Since $p$ is a homogeneous polynomial of degree $m$: If $m = 0$, $p$ is constant, and so $p^{-1}(a)$ is empty for all $a$ except one, where it is all of $mathbb{R}^k$. If $m > 0$, for $a neq 0$, the level set is either empty, or a nonempty submanifold of dimension $k-1$. For $a = 0$, it may be a more complicated set. It's a relation between sets, the dilation $x mapsto lambdacdot x$ induces a bijection between the level set $p^{-1}(a)$ and the level set $p^{-1}(lambda^mcdot a)$, that can succinctly be written $lambdacdot p^{-1}(a) = p^{-1}(lambda^mcdot a)$.
    $endgroup$
    – Daniel Fischer♦
    Jul 6 '13 at 23:41




















  • $begingroup$
    Do you mean $p^{-1}(a):=frac 1 {p(a)}$?
    $endgroup$
    – user64494
    Jul 6 '13 at 18:51










  • $begingroup$
    I don't know really. But from my previous experience in the book, it means inverse function. Thanks @user64494!
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:23












  • $begingroup$
    Did you mean to type $(frac{a}{b})^{frac{1}{n}}$? You haven't told us what $m$ is.
    $endgroup$
    – Robert Lewis
    Jul 6 '13 at 19:34










  • $begingroup$
    Oh yes, let me correct them - $m$ is order. Thank you @RobertLewis
    $endgroup$
    – WishingFish
    Jul 6 '13 at 19:39






  • 1




    $begingroup$
    $p^{-1}(a)$ is a set. Since $p$ is a homogeneous polynomial of degree $m$: If $m = 0$, $p$ is constant, and so $p^{-1}(a)$ is empty for all $a$ except one, where it is all of $mathbb{R}^k$. If $m > 0$, for $a neq 0$, the level set is either empty, or a nonempty submanifold of dimension $k-1$. For $a = 0$, it may be a more complicated set. It's a relation between sets, the dilation $x mapsto lambdacdot x$ induces a bijection between the level set $p^{-1}(a)$ and the level set $p^{-1}(lambda^mcdot a)$, that can succinctly be written $lambdacdot p^{-1}(a) = p^{-1}(lambda^mcdot a)$.
    $endgroup$
    – Daniel Fischer♦
    Jul 6 '13 at 23:41


















$begingroup$
Do you mean $p^{-1}(a):=frac 1 {p(a)}$?
$endgroup$
– user64494
Jul 6 '13 at 18:51




$begingroup$
Do you mean $p^{-1}(a):=frac 1 {p(a)}$?
$endgroup$
– user64494
Jul 6 '13 at 18:51












$begingroup$
I don't know really. But from my previous experience in the book, it means inverse function. Thanks @user64494!
$endgroup$
– WishingFish
Jul 6 '13 at 19:23






$begingroup$
I don't know really. But from my previous experience in the book, it means inverse function. Thanks @user64494!
$endgroup$
– WishingFish
Jul 6 '13 at 19:23














$begingroup$
Did you mean to type $(frac{a}{b})^{frac{1}{n}}$? You haven't told us what $m$ is.
$endgroup$
– Robert Lewis
Jul 6 '13 at 19:34




$begingroup$
Did you mean to type $(frac{a}{b})^{frac{1}{n}}$? You haven't told us what $m$ is.
$endgroup$
– Robert Lewis
Jul 6 '13 at 19:34












$begingroup$
Oh yes, let me correct them - $m$ is order. Thank you @RobertLewis
$endgroup$
– WishingFish
Jul 6 '13 at 19:39




$begingroup$
Oh yes, let me correct them - $m$ is order. Thank you @RobertLewis
$endgroup$
– WishingFish
Jul 6 '13 at 19:39




1




1




$begingroup$
$p^{-1}(a)$ is a set. Since $p$ is a homogeneous polynomial of degree $m$: If $m = 0$, $p$ is constant, and so $p^{-1}(a)$ is empty for all $a$ except one, where it is all of $mathbb{R}^k$. If $m > 0$, for $a neq 0$, the level set is either empty, or a nonempty submanifold of dimension $k-1$. For $a = 0$, it may be a more complicated set. It's a relation between sets, the dilation $x mapsto lambdacdot x$ induces a bijection between the level set $p^{-1}(a)$ and the level set $p^{-1}(lambda^mcdot a)$, that can succinctly be written $lambdacdot p^{-1}(a) = p^{-1}(lambda^mcdot a)$.
$endgroup$
– Daniel Fischer♦
Jul 6 '13 at 23:41






$begingroup$
$p^{-1}(a)$ is a set. Since $p$ is a homogeneous polynomial of degree $m$: If $m = 0$, $p$ is constant, and so $p^{-1}(a)$ is empty for all $a$ except one, where it is all of $mathbb{R}^k$. If $m > 0$, for $a neq 0$, the level set is either empty, or a nonempty submanifold of dimension $k-1$. For $a = 0$, it may be a more complicated set. It's a relation between sets, the dilation $x mapsto lambdacdot x$ induces a bijection between the level set $p^{-1}(a)$ and the level set $p^{-1}(lambda^mcdot a)$, that can succinctly be written $lambdacdot p^{-1}(a) = p^{-1}(lambda^mcdot a)$.
$endgroup$
– Daniel Fischer♦
Jul 6 '13 at 23:41












2 Answers
2






active

oldest

votes


















1












$begingroup$

We are given that $p : Bbb{R}^k to Bbb{R}$ is a homogeneous function of order $m$, and we have that $p^{-1}(a) = { x in Bbb{R}^k : p(x) = a }$ by the definition of $p^{-1}$. Now, if $a, b > 0$, then let $t = (b/a)^{1/m}$. Suppose $x in p^{-1}(a)$. Then $p(x) = a$, which implies that $$p(tx) = t^m p(x) = (b/a) cdot a = b.$$ Hence, $tx in p^{-1}(b)$. Conversely, suppose $x in Bbb{R}$ such that $tx in p^{-1}(b)$. Then $p(tx) = b$. But, $p(tx) = t^m p(x) = (b/a) cdot p(x)$. Hence, $$p(tx) = b implies (b/a) cdot p(x) = b implies p(x) = a.$$
Hence, $x in p^{-1}(a)$.



Thus, we have shown that $x in p^{-1}(a) iff tx in p^{-1}(b)$, where $t = (b/a)^{1/m}$. This can also be expressed as
$$
p^{-1}(a) = (a/b)^{1/m} p^{-1}(b),tag{$*$}
$$
where it is understood that
$$
(a/b)^{1/m} p^{-1}(b) := { (a/b)^{1/m} x : x in p^{-1}(b) }.
$$




Note that we have to assume that $m neq 0$ in the question, otherwise $p^{-1}(a)$ is either empty or all of $Bbb{R}^k$, neither of which are $(k-1)$-dimensional submanifolds of $Bbb{R}^k$.



Regarding the additional question about whether or not $m$ is odd, the answer is no, there is no such restriction on $m$. You are asked to show that $p^{-1}{a}$ and $p^{-1}(b)$ are diffeomorphic when $a$ and $b$ are both positive and when $a$ and $b$ are both negative. When they have opposite sign they may or may not be diffeomorphic, but the above analysis does not answer this question (and that is also not part of the original problem in your post). I am not entirely sure what you mean when you say




$a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?




My guess is as above, that you are wondering about whether $p^{-1}(a)$ and $p^{-1}(b)$ can be diffeomorphic when $a$ and $b$ have opposite sign. If you meant something else then I would be happy to hear any clarifications in the comments.




Lastly, there seems to be some confusion regarding the notation $p^{-1}$. It is not a function, so there is no meaning to the question of whether $p^{-1}(a)$ is unique. (I suppose you mean to ask whether it is well-defined?) Note that ($*$) is an equality of sets, not of real numbers; this might possibly be the source of your misconception.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    According to @Daniel Fischer's kind advice:



    $$vec{v} in p^{-1}(a)$$
    $$lambda vec{v} in p^{-1}(lambda^m a) Rightarrow vec{v} in frac{1}{lambda} p^{-1}(lambda^m a)$$
    So assuming the preimage is unique, we have
    $$p^{-1}(a) = frac{1}{lambda} p^{-1}(lambda^m a)$$
    Set $lambda = big(frac{b}{a}big)^{frac{1}{m}}$, we get
    $$p^{-1}(a) = Big(frac{b}{a}Big)^{-frac{1}{m}} p^{-1}Big(Big(frac{b}{a}Big)^{frac{1}{m} cdot m}aBig) Rightarrow p^{-1}(a) = Big(frac{a}{b}Big)^frac{1}{m} p^{-1}(a)$$



    Though, I'm not certain about assuming uniqueness.






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f437632%2fif-p-is-homogeneous-of-order-k-then-p-1a-is-diffeomorphic-to-p-1%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      We are given that $p : Bbb{R}^k to Bbb{R}$ is a homogeneous function of order $m$, and we have that $p^{-1}(a) = { x in Bbb{R}^k : p(x) = a }$ by the definition of $p^{-1}$. Now, if $a, b > 0$, then let $t = (b/a)^{1/m}$. Suppose $x in p^{-1}(a)$. Then $p(x) = a$, which implies that $$p(tx) = t^m p(x) = (b/a) cdot a = b.$$ Hence, $tx in p^{-1}(b)$. Conversely, suppose $x in Bbb{R}$ such that $tx in p^{-1}(b)$. Then $p(tx) = b$. But, $p(tx) = t^m p(x) = (b/a) cdot p(x)$. Hence, $$p(tx) = b implies (b/a) cdot p(x) = b implies p(x) = a.$$
      Hence, $x in p^{-1}(a)$.



      Thus, we have shown that $x in p^{-1}(a) iff tx in p^{-1}(b)$, where $t = (b/a)^{1/m}$. This can also be expressed as
      $$
      p^{-1}(a) = (a/b)^{1/m} p^{-1}(b),tag{$*$}
      $$
      where it is understood that
      $$
      (a/b)^{1/m} p^{-1}(b) := { (a/b)^{1/m} x : x in p^{-1}(b) }.
      $$




      Note that we have to assume that $m neq 0$ in the question, otherwise $p^{-1}(a)$ is either empty or all of $Bbb{R}^k$, neither of which are $(k-1)$-dimensional submanifolds of $Bbb{R}^k$.



      Regarding the additional question about whether or not $m$ is odd, the answer is no, there is no such restriction on $m$. You are asked to show that $p^{-1}{a}$ and $p^{-1}(b)$ are diffeomorphic when $a$ and $b$ are both positive and when $a$ and $b$ are both negative. When they have opposite sign they may or may not be diffeomorphic, but the above analysis does not answer this question (and that is also not part of the original problem in your post). I am not entirely sure what you mean when you say




      $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?




      My guess is as above, that you are wondering about whether $p^{-1}(a)$ and $p^{-1}(b)$ can be diffeomorphic when $a$ and $b$ have opposite sign. If you meant something else then I would be happy to hear any clarifications in the comments.




      Lastly, there seems to be some confusion regarding the notation $p^{-1}$. It is not a function, so there is no meaning to the question of whether $p^{-1}(a)$ is unique. (I suppose you mean to ask whether it is well-defined?) Note that ($*$) is an equality of sets, not of real numbers; this might possibly be the source of your misconception.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        We are given that $p : Bbb{R}^k to Bbb{R}$ is a homogeneous function of order $m$, and we have that $p^{-1}(a) = { x in Bbb{R}^k : p(x) = a }$ by the definition of $p^{-1}$. Now, if $a, b > 0$, then let $t = (b/a)^{1/m}$. Suppose $x in p^{-1}(a)$. Then $p(x) = a$, which implies that $$p(tx) = t^m p(x) = (b/a) cdot a = b.$$ Hence, $tx in p^{-1}(b)$. Conversely, suppose $x in Bbb{R}$ such that $tx in p^{-1}(b)$. Then $p(tx) = b$. But, $p(tx) = t^m p(x) = (b/a) cdot p(x)$. Hence, $$p(tx) = b implies (b/a) cdot p(x) = b implies p(x) = a.$$
        Hence, $x in p^{-1}(a)$.



        Thus, we have shown that $x in p^{-1}(a) iff tx in p^{-1}(b)$, where $t = (b/a)^{1/m}$. This can also be expressed as
        $$
        p^{-1}(a) = (a/b)^{1/m} p^{-1}(b),tag{$*$}
        $$
        where it is understood that
        $$
        (a/b)^{1/m} p^{-1}(b) := { (a/b)^{1/m} x : x in p^{-1}(b) }.
        $$




        Note that we have to assume that $m neq 0$ in the question, otherwise $p^{-1}(a)$ is either empty or all of $Bbb{R}^k$, neither of which are $(k-1)$-dimensional submanifolds of $Bbb{R}^k$.



        Regarding the additional question about whether or not $m$ is odd, the answer is no, there is no such restriction on $m$. You are asked to show that $p^{-1}{a}$ and $p^{-1}(b)$ are diffeomorphic when $a$ and $b$ are both positive and when $a$ and $b$ are both negative. When they have opposite sign they may or may not be diffeomorphic, but the above analysis does not answer this question (and that is also not part of the original problem in your post). I am not entirely sure what you mean when you say




        $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?




        My guess is as above, that you are wondering about whether $p^{-1}(a)$ and $p^{-1}(b)$ can be diffeomorphic when $a$ and $b$ have opposite sign. If you meant something else then I would be happy to hear any clarifications in the comments.




        Lastly, there seems to be some confusion regarding the notation $p^{-1}$. It is not a function, so there is no meaning to the question of whether $p^{-1}(a)$ is unique. (I suppose you mean to ask whether it is well-defined?) Note that ($*$) is an equality of sets, not of real numbers; this might possibly be the source of your misconception.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          We are given that $p : Bbb{R}^k to Bbb{R}$ is a homogeneous function of order $m$, and we have that $p^{-1}(a) = { x in Bbb{R}^k : p(x) = a }$ by the definition of $p^{-1}$. Now, if $a, b > 0$, then let $t = (b/a)^{1/m}$. Suppose $x in p^{-1}(a)$. Then $p(x) = a$, which implies that $$p(tx) = t^m p(x) = (b/a) cdot a = b.$$ Hence, $tx in p^{-1}(b)$. Conversely, suppose $x in Bbb{R}$ such that $tx in p^{-1}(b)$. Then $p(tx) = b$. But, $p(tx) = t^m p(x) = (b/a) cdot p(x)$. Hence, $$p(tx) = b implies (b/a) cdot p(x) = b implies p(x) = a.$$
          Hence, $x in p^{-1}(a)$.



          Thus, we have shown that $x in p^{-1}(a) iff tx in p^{-1}(b)$, where $t = (b/a)^{1/m}$. This can also be expressed as
          $$
          p^{-1}(a) = (a/b)^{1/m} p^{-1}(b),tag{$*$}
          $$
          where it is understood that
          $$
          (a/b)^{1/m} p^{-1}(b) := { (a/b)^{1/m} x : x in p^{-1}(b) }.
          $$




          Note that we have to assume that $m neq 0$ in the question, otherwise $p^{-1}(a)$ is either empty or all of $Bbb{R}^k$, neither of which are $(k-1)$-dimensional submanifolds of $Bbb{R}^k$.



          Regarding the additional question about whether or not $m$ is odd, the answer is no, there is no such restriction on $m$. You are asked to show that $p^{-1}{a}$ and $p^{-1}(b)$ are diffeomorphic when $a$ and $b$ are both positive and when $a$ and $b$ are both negative. When they have opposite sign they may or may not be diffeomorphic, but the above analysis does not answer this question (and that is also not part of the original problem in your post). I am not entirely sure what you mean when you say




          $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?




          My guess is as above, that you are wondering about whether $p^{-1}(a)$ and $p^{-1}(b)$ can be diffeomorphic when $a$ and $b$ have opposite sign. If you meant something else then I would be happy to hear any clarifications in the comments.




          Lastly, there seems to be some confusion regarding the notation $p^{-1}$. It is not a function, so there is no meaning to the question of whether $p^{-1}(a)$ is unique. (I suppose you mean to ask whether it is well-defined?) Note that ($*$) is an equality of sets, not of real numbers; this might possibly be the source of your misconception.






          share|cite|improve this answer











          $endgroup$



          We are given that $p : Bbb{R}^k to Bbb{R}$ is a homogeneous function of order $m$, and we have that $p^{-1}(a) = { x in Bbb{R}^k : p(x) = a }$ by the definition of $p^{-1}$. Now, if $a, b > 0$, then let $t = (b/a)^{1/m}$. Suppose $x in p^{-1}(a)$. Then $p(x) = a$, which implies that $$p(tx) = t^m p(x) = (b/a) cdot a = b.$$ Hence, $tx in p^{-1}(b)$. Conversely, suppose $x in Bbb{R}$ such that $tx in p^{-1}(b)$. Then $p(tx) = b$. But, $p(tx) = t^m p(x) = (b/a) cdot p(x)$. Hence, $$p(tx) = b implies (b/a) cdot p(x) = b implies p(x) = a.$$
          Hence, $x in p^{-1}(a)$.



          Thus, we have shown that $x in p^{-1}(a) iff tx in p^{-1}(b)$, where $t = (b/a)^{1/m}$. This can also be expressed as
          $$
          p^{-1}(a) = (a/b)^{1/m} p^{-1}(b),tag{$*$}
          $$
          where it is understood that
          $$
          (a/b)^{1/m} p^{-1}(b) := { (a/b)^{1/m} x : x in p^{-1}(b) }.
          $$




          Note that we have to assume that $m neq 0$ in the question, otherwise $p^{-1}(a)$ is either empty or all of $Bbb{R}^k$, neither of which are $(k-1)$-dimensional submanifolds of $Bbb{R}^k$.



          Regarding the additional question about whether or not $m$ is odd, the answer is no, there is no such restriction on $m$. You are asked to show that $p^{-1}{a}$ and $p^{-1}(b)$ are diffeomorphic when $a$ and $b$ are both positive and when $a$ and $b$ are both negative. When they have opposite sign they may or may not be diffeomorphic, but the above analysis does not answer this question (and that is also not part of the original problem in your post). I am not entirely sure what you mean when you say




          $a,b$ does not need to have the same sign to be homegeneous, since $lambda = left(frac{b}{a}right)^{1/m}$ still make sense?




          My guess is as above, that you are wondering about whether $p^{-1}(a)$ and $p^{-1}(b)$ can be diffeomorphic when $a$ and $b$ have opposite sign. If you meant something else then I would be happy to hear any clarifications in the comments.




          Lastly, there seems to be some confusion regarding the notation $p^{-1}$. It is not a function, so there is no meaning to the question of whether $p^{-1}(a)$ is unique. (I suppose you mean to ask whether it is well-defined?) Note that ($*$) is an equality of sets, not of real numbers; this might possibly be the source of your misconception.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 13:04

























          answered Dec 11 '18 at 12:55









          BrahadeeshBrahadeesh

          6,30742361




          6,30742361























              0












              $begingroup$

              According to @Daniel Fischer's kind advice:



              $$vec{v} in p^{-1}(a)$$
              $$lambda vec{v} in p^{-1}(lambda^m a) Rightarrow vec{v} in frac{1}{lambda} p^{-1}(lambda^m a)$$
              So assuming the preimage is unique, we have
              $$p^{-1}(a) = frac{1}{lambda} p^{-1}(lambda^m a)$$
              Set $lambda = big(frac{b}{a}big)^{frac{1}{m}}$, we get
              $$p^{-1}(a) = Big(frac{b}{a}Big)^{-frac{1}{m}} p^{-1}Big(Big(frac{b}{a}Big)^{frac{1}{m} cdot m}aBig) Rightarrow p^{-1}(a) = Big(frac{a}{b}Big)^frac{1}{m} p^{-1}(a)$$



              Though, I'm not certain about assuming uniqueness.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                According to @Daniel Fischer's kind advice:



                $$vec{v} in p^{-1}(a)$$
                $$lambda vec{v} in p^{-1}(lambda^m a) Rightarrow vec{v} in frac{1}{lambda} p^{-1}(lambda^m a)$$
                So assuming the preimage is unique, we have
                $$p^{-1}(a) = frac{1}{lambda} p^{-1}(lambda^m a)$$
                Set $lambda = big(frac{b}{a}big)^{frac{1}{m}}$, we get
                $$p^{-1}(a) = Big(frac{b}{a}Big)^{-frac{1}{m}} p^{-1}Big(Big(frac{b}{a}Big)^{frac{1}{m} cdot m}aBig) Rightarrow p^{-1}(a) = Big(frac{a}{b}Big)^frac{1}{m} p^{-1}(a)$$



                Though, I'm not certain about assuming uniqueness.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  According to @Daniel Fischer's kind advice:



                  $$vec{v} in p^{-1}(a)$$
                  $$lambda vec{v} in p^{-1}(lambda^m a) Rightarrow vec{v} in frac{1}{lambda} p^{-1}(lambda^m a)$$
                  So assuming the preimage is unique, we have
                  $$p^{-1}(a) = frac{1}{lambda} p^{-1}(lambda^m a)$$
                  Set $lambda = big(frac{b}{a}big)^{frac{1}{m}}$, we get
                  $$p^{-1}(a) = Big(frac{b}{a}Big)^{-frac{1}{m}} p^{-1}Big(Big(frac{b}{a}Big)^{frac{1}{m} cdot m}aBig) Rightarrow p^{-1}(a) = Big(frac{a}{b}Big)^frac{1}{m} p^{-1}(a)$$



                  Though, I'm not certain about assuming uniqueness.






                  share|cite|improve this answer











                  $endgroup$



                  According to @Daniel Fischer's kind advice:



                  $$vec{v} in p^{-1}(a)$$
                  $$lambda vec{v} in p^{-1}(lambda^m a) Rightarrow vec{v} in frac{1}{lambda} p^{-1}(lambda^m a)$$
                  So assuming the preimage is unique, we have
                  $$p^{-1}(a) = frac{1}{lambda} p^{-1}(lambda^m a)$$
                  Set $lambda = big(frac{b}{a}big)^{frac{1}{m}}$, we get
                  $$p^{-1}(a) = Big(frac{b}{a}Big)^{-frac{1}{m}} p^{-1}Big(Big(frac{b}{a}Big)^{frac{1}{m} cdot m}aBig) Rightarrow p^{-1}(a) = Big(frac{a}{b}Big)^frac{1}{m} p^{-1}(a)$$



                  Though, I'm not certain about assuming uniqueness.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 6 '13 at 23:19

























                  answered Jul 6 '13 at 23:12









                  WishingFishWishingFish

                  9951028




                  9951028






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f437632%2fif-p-is-homogeneous-of-order-k-then-p-1a-is-diffeomorphic-to-p-1%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more