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Set $f(x,y,z) = x + y + z + x^2 + y^2 + z^2$. Consider the surface

$$S = \{f(x,y,z) = 0\} \subset \mathbb{R}^3$$

near the origin $o = (0,0,0) \in S$. Write down the equation of the tangent plane $T_oS$. Prove that $(y,z)$ form a system of local co-ordinates on $S$ near $o$ and compute the Taylor series of the function

$$g = (x + xy + 3yz)|_S$$ at the point $o$ in the coordinates $(y,z)$ up to degree $2$.

I think for the first bit, to work out the tangent plane I need to partially differentiate $f$ at the origin, which would give me

$$T_oS = \frac{\partial f(o)}{\partial x}(x - o) + \frac{\partial f(o)}{\partial y}(y - o) + \frac{\partial f(o)}{\partial z}(z - o).$$

Is this correct for that bit?

I'm a little bit stuck on the second bit. Do I use this tangent space to get new coordinates for $x,y,z$ which I would then sub into $g$ and then try and expand to get the Taylor series?

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  1. About the equation of the tangent plane : just add "$=0$" and it will be correct. Indeed, recall that when a submanifold is defined as the zero set of a submersion (check that $f$ is one near $0$, i.e. that $Df(0) \neq 0$), then the tangent plane can be described as the kernel of $Df$.

  2. To prove that $(y,z)$ define local coordinates : use the implicit function theorem. (I'll go into more details if you want me to).

  3. Once you have proved that $(y,z)$ are local coordinates, write $x=\phi(y,z)$ where $\phi$ is the function defined by the implicit function theorem. Then note that for $(x,y,z) \in S$ near 0, $g(x,y,z)=\phi(y,z) + y \phi(y,z) + yz = \psi(y,z)$ and then compute the derivatives up to order 2 of $\psi$. You'll need the derivatives of $\phi$, which may be computed from the property $f(\phi(y,z),y,z)=0$ (as usual with the implicit function theorem).

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This implicit function theorem thing is confusing me a little. I tried writing down the Jacobian matrix of this at $o$ and got the $3 \times 3$ identity matrix, is this correct? Do I then say $x = -y - z - x^2 - y^2 - z^2$ and then sub $x$ into $g$ and then expand and re-sub it if I need to get it in terms of just $y$ and $z$? – Kaish Mar 6 '13 at 19:09
the function $f$ has domain $\mathbb{R}^3$ and range $\mathbb{R}$, so the same goes for its differential : it is a linear map from $\mathbb{R}^3$ to $\mathbb{R}$ so not a 3*3 matrix. the differential of $f$ at 0 is $Df(0)(x,y,z)=x+y+z$. in particular, $\frac{\partial f}{\partial x}=1 \neq 0$, so you may apply the implicit function theorem here. – Glougloubarbaki Mar 6 '13 at 20:11

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