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Guillemin-Pollack 1.2.8:

What is the tangent space to the hyperboloid defined by $x^2+y^2-z^2=a$ at $(\sqrt a,0,0)$, where $a>0$?

Is there a way to compute it using Characterization of the tangent space in terms of velocity vectors? I cannot apply the usual definition of Guillemin and Pollack because I don't know how exactly to parametrize a point with $z=0$ on the hyperboloid (see Hyperboloid is a manifold)

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  • $\begingroup$ Do you have questions to my answer below? $\endgroup$ – exchange Jan 31 '18 at 14:23
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    $\begingroup$ Thanks for your answer! Yes, I have some questions but haven't been able to formulate them yet since some of them are lengthy. Probably I'll do that some time in the future. I've accepted your answer though. $\endgroup$ – user500094 Jan 31 '18 at 18:14
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You can calculate the tangent space via the inverse of a function that defines your hyperboloid.

Let $f:\mathbb{R}^3\rightarrow \mathbb{R}$ be defined by $f(x,y,z):=x^2+y^2-z^2-a$. Then \begin{equation*} \begin{split} & df = (2x,2y,-2z), \end{split} \end{equation*} which is isomorphic to a basis vector of $T_{f(x)}(\mathbb{R})\simeq \mathbb{R}$ if any of the $x,y,z$ are non-zero. This must be the case for $f(x,y,z)=0$, so by the preimage theorem $H:=f^{-1}(0)$ defines the manifold of the hyperboloid.

Now because $df_x^{-1}(0)=T_x(f^{-1}(0))$, we can obtain the tangent space $T_x(H)$ by looking for all vectors $v$ for which $df_x(v)=0$. On $f^{-1}(0)$, we have the condition $z=\sqrt{x^2+y^2-a}$, so \begin{equation*} df_x(v)= \begin{pmatrix} 2x\\ 2y\\ - 2\sqrt{x^2+y^2-a} \end{pmatrix} \begin{pmatrix} v_1\\ v_2\\ v_3 \end{pmatrix} = 2(xv_1+yv_2-\sqrt{x^2+y^2-a}v_3) \end{equation*} This is zero if $v_1 = (\sqrt{x^2+y^2-a}~v_3-yv_2)/x$ (and if one wants the tangent space at $x=0$, then one must choose other combinations like $v_3=(xv_1+yv_2)/\sqrt{x^2+y^2-a}$ etc), so the tangent space is two-dimensional, depends on $x,y,z$ and is given (except at $x=0$) by \begin{equation} T_{x,y,z}(H)=\left\{(v_1,v_2,v_3)~|~v_2,v_3\in\mathbb{R},~v_1=(\sqrt{x^2+y^2-a}~v_3-yv_2)/x\right\}. \end{equation} In particular, like the whole space, it is only defined for $x^2+y^2\ge a$.

At the point $x=(\sqrt{a},0,0)$, the tangent space is thus \begin{equation} T_{\sqrt{a},0,0}(H)=\left\{(v_1,v_2,v_3)~|~v_2,v_3\in\mathbb{R},~v_1=0\right\}\simeq \mathbb{R}^2. \end{equation}

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