TITLE: Tangent Space of 3-Sphere at a point P
QUESTION [0 upvotes]: Say I have a four variable function $S^3$ that is
$x^2+y^2+z^2+w^2=1$
and I want to find the tangent space at a given point P = (0,0,0,1). So far I have approached the problem by allowing for a function $\gamma(t)=(x(t),y(t),z(t),w(t))$ which belongs to $S^3$ such that we have
$x(t)^2+y(t)^2+z(t)^2+w(t)^2=1$
Now I am confused as to how the derivative of such a function, $\gamma(t)$, can be evaluated to be
$\gamma'(t)=2xx'+2yy'+2zz'+2ww'$
and how we know that $x(0)=y(0)=z(0)=0$ and that $w(0)=1$ so therefore $w'(0)=1$. From this it is said that apparently the set of vectors of the form $(a,b,c,0)$ make up the tangent plane at the point $P$.

REPLY [1 votes]: If $M$ is an hypersurface of $\mathbb{R}^n$ defined by the submerion $f:U\subset\mathbb{R}^n\rightarrow \mathbb{R}$ where $U$ is an open subset of $\mathbb{R}$, such that $M=\{x:f(x)=0\}$, for every $x\in M, T_xM=\{x:df_x(v)=0\}$, apply this to $f(x)=x^2+y^2+z^2+w^2-1$.
{"set_name": "stack_exchange", "score": 0, "question_id": 3578622}