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I'm stuck on the following exercise:

"Verify that $M:=\{(x,y,z)\in\mathbb{R}^3:x^2+3y^2+2z^2=3, x+y+z=0\}$ is a submanifold. Also, say what its dimension is and compute the Jacobian matrix of the function."

The definition of $C^k (k\geq 1)$ submanifold of dimension $p$ that I have in my notes is:

"$V\subset\mathbb{R}^n$ is a $C^k (k\geq 1)$ submanifold of dimension $1\leq p\leq n-1$ iff $\forall x_0\in V \exists U$ neighbourhood of $x_0$ in $\mathbb{R}^n$ and a function $f\colon U\to\mathbb{R}^{n-p}$, $f\in C^k(U)$ such that: (1) $x_0\in U$, (2) $V\cap U=\{x\in U:f(x)=0\}$, (3)rank $Jf(x)=n-p\ \forall x\in U$."

It's the first exercise I do on (sub)manifolds and I don't know how to get started, so I would appreciate if someone explained to me how to do this kind of exercise.

Best regards,

lorenzo.

2 Answers 2

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You have a set in $\mathbb{R}^3$ defined by 2 equations, so, as in linear algebra, you can start guessing that it will have dimension 1. If you are familiar with the equations, you will know that your are cutting a cone with a plane, which results in a curve.

Following your definition, I would define the function

$f(x,y,z)= ( x^2+3y^2+2z^2-3, x+y+z )$.

It is obvious that $f(p)=0$ if and only if $p\in M$. Now you have to check the other conditions.

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    Tell me if you have trouble prooving what is left. – 2017-01-16
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    Thank you for your answer; I was thinking about defining the function $f:B_1(x,y,z)\to\mathbb{R}^3, f(x,y,z)=(x^2+3y^2+2z^2-3, x+y+z)$ for each point $(x,y,z)\in\mathbb{R}^3$ (where $U$ would be $B_1(x,y,z)$) but how can I be sure that $U\cap V=\{x\in U:f(x)=0\}$ in this case? Also I think the Jacobian should be $Jf(x,y,z)=((2x,6y,4z),(1,1,1))$; to compute its rank should I evaluate it at some point in particular? – 2017-01-16
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    Well, in this kind of definition it is not obvious that you can take the "same" neighbourhood $U$ for every $x_0$. – 2017-01-16
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    For **every** set $U$, you have $\{x\in U:f(x)=0\}\subseteq V$ since you have defined $f$ for this to happen. ( If you still have trouble here, try to prove both inclusions ) – 2017-01-16
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    Finally, have in mind that your definitions says "$Jf(x)=n-p\ \forall x\in U$". This means that you should compute it´s rank for **every** $(x,y,z)$. Some points will have problems, but they won´t be in $V$, so you can cut them off by specifying the neighbourhoods $U$. – 2017-01-16
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You can use the fact that the pre-image of a regular value for a differentiable mapping $f: \mathbb{R}^n \to \mathbb{R}^m$ is a submanifold of dimension $n-m$. Here, our condition can be reduced to

$$3x^2 + 5y^2 + 4xy - 3 = 0, $$ so our $f: \mathbb{R^2} \to \mathbb{R}$ is given by $f(x,y) = 3x^2 + 5y^2 + 4xy - 3$. This reduces to showing that the Jacobian is full rank at your point (let me know if you're having problems with this).

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    It's a homework problem. Try to lead him without solving it. That way there is some value to what we are doing. As opposed to helping folks plagiarize their way to a worthless advanced degree in mathematics. – 2017-01-16
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    Generally granted, but he/she specifically phrased: "I would appreciate if someone explained to me how to do *this kind* of exercise," which led me to believe he/she was tackling a problem set with similar questions where a general principle would be helpful. – 2017-01-16
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    But I suppose I could have applied the principle to a similar problem to get the same point across in a less hand-holding way. – 2017-01-16
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    @Andrew Whelan Thank you for your answer; shouldn't the condition be $3x^2+5y^2+4xy-3=0$? Also by what you say I understand that $U$ should be the subset of $\mathbb{R}^3$ where the Jacobian $Jf(x,y,z)=((2x,6y,4z),(1,1,1))$ has rank $2$, am I right? – 2017-01-16
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    Correct - I've edited my post above. – 2017-01-16