If $p$ is homogeneous of order $k$, then $p^{-1}(a)$ is diffeomorphic to $p^{-1}(b)$
$begingroup$
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
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
asked Jul 6 '13 at 18:21
WishingFishWishingFish
9951028
$endgroup$
|
show 8 more comments
$begingroup$
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
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
asked Jul 6 '13 at 18:21
WishingFishWishingFish
9951028
$endgroup$
$begingroup$
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
|
show 8 more comments
$begingroup$
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
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
asked Jul 6 '13 at 18:21
WishingFishWishingFish
9951028
$endgroup$
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
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
asked Jul 6 '13 at 18:21
WishingFishWishingFish
9951028
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
asked Jul 6 '13 at 18:21
WishingFishWishingFish
9951028
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
edited Dec 11 '18 at 13:01
Brahadeesh
6,30742361
6,30742361
asked Jul 6 '13 at 18:21
WishingFishWishingFish
9951028
asked Jul 6 '13 at 18:21
WishingFishWishingFish
9951028
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
|
show 8 more comments
$begingroup$
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
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
$endgroup$
add a comment |
$begingroup$
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.
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
$endgroup$
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.
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
edited Dec 11 '18 at 13:04
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
answered Dec 11 '18 at 12:55
BrahadeeshBrahadeesh
6,30742361
6,30742361
add a comment |
add a comment |
$begingroup$
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.
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
$endgroup$
add a comment |
$begingroup$
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.
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
$endgroup$
add a comment |
$begingroup$
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.
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
$endgroup$
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.
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
edited Jul 6 '13 at 23:19
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
answered Jul 6 '13 at 23:12
WishingFishWishingFish
9951028
9951028
add a comment |
add a comment |
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$begingroup$
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