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Consider the set

$X:=\{(x,y,z)\in\mathbb{R}^3 | x^2+3y^2=1+z^2\}$

I have to show that $X$ is a submanifold of $\mathbb{R}^3$ (and this is trivial); then, using on $T\mathbb{R}^3$ the standard coordinates $x,y,z,\partial_x,\partial_y,\partial_z$ I have to find the equations that describe $TX\subseteq T\mathbb{R}^3$ and $NX\subseteq T\mathbb{R}^3$.

Any suggestion? Thanks...

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+1 for this great question: all boooks on differential geometry (or even advanced calculus) construct the tangent bundle of a submanifold of $\mathbb R^n$ in abstract terms, but I have never seen a concrete example like the one mentioned here described explicitly. –  Georges Elencwajg Feb 12 '13 at 22:17
    
@Frankenstein The symbols $\partial_x$, $\partial_y$ and $\partial_z$ are not coordinates, they are vectors - more specifically: coordinate vector fields. Just like when you write $(x,y,z)$ you mean $x$ lots of the first basis vector, $y$ lots of the second basis vector and $z$ lots of the third basis vector: $(x,y,z) \sim x \, \vec{v}_1 + y \, \vec{v}_2 + z \, \vec{v}_3$. The tangent space has coordinates $(a,b,c)$ where $(a,b,c) \sim a \, \partial_x + b \, \partial_y + c \, \partial_z$. If this is a homework question, then be careful. Not all answer on this site are always correct. –  Fly by Night Feb 12 '13 at 23:10

2 Answers 2

up vote 4 down vote accepted

The tangent bundle $TX$ is the set of $6$-tuples $(x,y,z;\partial_ x,\partial_ y,\partial_ z)\in T\mathbb R^3=\mathbb R^3\times \mathbb R^3$ satisfying the equations $x^2+3y^2-z^2-1=0$ and $2x\partial_ x+6y\partial_y-2z\partial_z=0$.

The normal bundle $NX$ is the set of $6$-tuples $(x,y,z;2xr,6yr,-2zr)\in T\mathbb R^3=\mathbb R^3\times \mathbb R^3$ with $x^2+3y^2-z^2-1=0$ and $r\in \mathbb R$ .

Edit
At Frankenstein's request I'll add a few words of explanation.
The equation $x^2+3y^2-z^2-1=0$ just says that we are studying a point $P\in X$: this concerns the first three coordinates of our $6$-tuple $(x,y,z;\partial_ x,\partial_ y,\partial_ z)\in T\mathbb R^3=\mathbb R^3\times \mathbb R^3$.
At $P$ the gradient of $f(x,y,z)=x^2+3y^2-z^2-1$ is the vector $\nabla f(P)=(2x,6y,-2z)$.
A vector $v=(\partial_ x,\partial_ y,\partial_ z)\in T_P\mathbb R^3$ is tangent to $X$ if $v$ is orthogonal to $\nabla f(P)$, which translates into $\nabla f(P)\cdot (\partial_ x,\partial_ y,\partial_ z)=2x\partial_ x+6y\partial_y-2z\partial_z=0$.
A vector $v=(\partial_ x,\partial_ y,\partial_ z)\in T_P\mathbb R^3$ is normal to $X$ if $v$ is proportional to $\nabla f(P)$, which translates into $ (\partial_ x,\partial_ y,\partial_ z)=r \nabla f(P)=r(2x, 6y,-2z)=(2xr,6yr,-2zr)$.

A remark
Notice that the description of $TX$ is purely by equations whereas that of $NX$ is a mixture of an equation and a parametrization.
This is a linear algebra phenomenon:
$\bullet$ A plane in $\mathbb R^3$ is best described by the equation $l(x)=0$ where $l$ is a linear form, unique up to a non-zero constant.
$\bullet \bullet$ However a line must be described by two linear forms and there is absolutely no canonical choice for these.
So it is better to describe a line by choosing a non zero vector $v$ on it , which will be unique up to a constant, and take multiples $rv$ of that vector.
In other words, a line is better described parametrically than equationally.

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Can you explain me why? I can't visualize the situation... –  Frankenstein Feb 12 '13 at 21:01
    
Dear Frankenstein: I have added a little edit which might help you. –  Georges Elencwajg Feb 12 '13 at 21:31
    
@GeorgesElencwajg Your description of the tangent bundle is incorrect. You describe $TX$ as 6-tuplies $(x,y,z,\partial_x,\partial_y,\partial_z)$. However, $\partial_x$, $\partial_y$ and $\partial_z$ are the basis vectors themselves. The equation $2x\,\partial_x + 6y \, \partial_y - 2z\,\partial_z = 0$ holds if, and only if, $x=y=z=0$ because $\partial_x$, $\partial_y$ and $\partial_z$ are linearly independent vectors (strictly speaking they are coordinate vector fields). –  Fly by Night Feb 12 '13 at 22:46
    
Dear @Fly by Night: in my notation, $\partial_ x,\partial_ y,\partial_ z$ are real numbers, not tangent vectors . This is perfectly clear from the context. As Humpty-Dumpty said: "When I use a word, it means just what I choose it to mean—neither more nor less." Also I find your explaining to me such very elementary (and irrelevant) facts on tangent fields and your down-voting of a rival answer (which I wrote before yours) quite amusing. –  Georges Elencwajg Feb 12 '13 at 23:12
    
@GeorgesElencwajg Your notation goes against everything standard in differential geometry. The down-vote will remain until you correct your mistake. (Just do a little search for coordinate vector fields, you'll soon see that $\partial_{x_i}$ and $\partial_i$ are universally used for vector fields and not numbers.) If the facts I explained were so elementary then I would have expected you to use the notation correctly and to not be so amazed by such a trivial example (as your comments to the OP seem to suggest). –  Fly by Night Feb 12 '13 at 23:29

The gradient vector $\nabla f = f_x\,\partial_x + f_y\,\partial_y + f_z \, \partial_z$ is at right angles to the zero-level set $f=0$. In your case $f(x,y,z) = x^2+3y^2-z^2-1$ and so:

$$\nabla f = 2x \, \partial_x + 6y \, \partial_y - 2z \, \partial_z \, .$$

A vector belongs to $X$ if it is based at a point of $X$ and is tangent to $X$ at that point. Equivalently, a vector must be based at a point of $X$ and be perpendicular to $\nabla f$. Such a vector must be based at a point $(x,y,z)$ for which $x^2+3y^2-z^2-1=0$ and be of the form $a \, \partial_x + b \, \partial_y + c \, \partial_z$ for whch $2xa + 6yb -2zc = 0$. This last condition means that $$a \, \partial_x + b \, \partial_y + c \, \partial_z \perp 2x \, \partial_x + 6y \, \partial_y - 2z \, \partial_z \, .$$

Putting this all together, it follows that:

$$TX = \{ (x,y,z;a,b,c) : x^2+3y^2-z^2 = 1 \ \text{ and } \ xa + 3yb - zc = 0 \} \, . $$

For the normal bundle, you want vectors based on $X$ and parallel to $\nabla f$ at that point. Hence:

$$NX = \left\{ (x,y,z;2kx,6ky,-2kz) : x^2+3y^2-z^2 = 1 \ \text{ and } \ k \in \mathbb{R} \right\} \, . $$

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